<?xml version='1.0' encoding='UTF-8'?><?xml-stylesheet href="http://www.blogger.com/styles/atom.css" type="text/css"?><feed xmlns='http://www.w3.org/2005/Atom' xmlns:openSearch='http://a9.com/-/spec/opensearchrss/1.0/' xmlns:georss='http://www.georss.org/georss' xmlns:gd='http://schemas.google.com/g/2005' xmlns:thr='http://purl.org/syndication/thread/1.0'><id>tag:blogger.com,1999:blog-3478091816824629985</id><updated>2011-09-10T06:35:30.977-05:00</updated><category term='exam'/><category term='l&apos;hospital&apos;s rule'/><category term='implicit'/><category term='Volume'/><category term='derivative'/><category term='Washers'/><category term='ninth'/><category term='integral'/><category term='Disks'/><category term='area'/><category term='Limits'/><title type='text'>B-Rob's AP Calculus Blog 09 - 10</title><subtitle type='html'></subtitle><link rel='http://schemas.google.com/g/2005#feed' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/posts/default'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default?max-results=100'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/'/><link rel='hub' href='http://pubsubhubbub.appspot.com/'/><link rel='next' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default?start-index=101&amp;max-results=100'/><author><name>Archimedes</name><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='23' height='32' src='http://1.bp.blogspot.com/_M51pVgQmAi4/TD5ngKLMfFI/AAAAAAAAABQ/kvu77fl4cIg/S220/DSC_0210.jpg'/></author><generator version='7.00' uri='http://www.blogger.com'>Blogger</generator><openSearch:totalResults>857</openSearch:totalResults><openSearch:startIndex>1</openSearch:startIndex><openSearch:itemsPerPage>100</openSearch:itemsPerPage><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-308519559362184219</id><published>2010-12-13T08:49:00.002-06:00</published><updated>2010-12-13T09:25:24.875-06:00</updated><title type='text'>Review Blog</title><content type='html'>So this week, like sarah stated, we've gone over AB stuff (AKA stuff that cam back to haunt us).  Although I thought it was the hardest thing last year, it is relatively easy this time around.  Here are some tips that you may need to remember when practicing AB for the AP exam. :P&lt;br /&gt;&lt;br /&gt;RELATED RATES:&lt;br /&gt;your steps are:&lt;br /&gt;1. Identify what you have (dr/dt etc.)&lt;br /&gt;2. Determine which formula you are dealing with (a lot of times its volume or area...)&lt;br /&gt;3. You usually take the derivative somewhere around this step (like of the equation...)&lt;br /&gt;4. Plug in&lt;br /&gt;5. Solve&lt;br /&gt;&lt;br /&gt;EXAMPLE Air is being pumped into a spherical balloon at a rate of 5 cm3/min.  Determine the rate at which the radius of the balloon is increasing when the diameter of the balloon is 20 cm.&lt;br /&gt;&lt;br /&gt; &lt;br /&gt;&lt;br /&gt;Solution&lt;br /&gt;&lt;br /&gt;The first thing that we’ll need to do here is to identify what information that we’ve been given and what we want to find.  Before we do that let’s notice that both the volume of the balloon and the radius of the balloon will vary with time and so are really functions of time.&lt;br /&gt;&lt;br /&gt;We know that air is being pumped into the balloon at a rate of 5 cm3/min.  This is the rate at which the volume is increasing.  Recall that rates of change are nothing more than derivatives and so we know that, dv/dt=5.&lt;br /&gt;&lt;br /&gt;We want to determine the rate at which the radius is changing.  Again, rates are derivatives and so it looks like we want to determine, &lt;br /&gt;dr/dt= ?&lt;br /&gt;r= diameter/2=10 cm&lt;br /&gt;&lt;br /&gt;Note that we needed to convert the diameter to a radius.&lt;br /&gt;&lt;br /&gt; &lt;br /&gt;&lt;br /&gt;Now that we’ve identified what we have been given and what we want to find we need to relate these two quantities to each other.  In this case we can relate the volume and the radius with the formula for the volume of a sphere.&lt;br /&gt;&lt;br /&gt;V= 4/3 pi r^3&lt;br /&gt;&lt;br /&gt;Now we don’t really want a relationship between the volume and the radius.  What we really want is a relationship between their derivatives.  We can do this by differentiating both sides with respect to t.  In other words, we will need to do implicit differentiation on the above formula.  Doing this gives,&lt;br /&gt;&lt;br /&gt;dV/dt=4 pi r^2 dr/dt&lt;br /&gt;&lt;br /&gt;Now I would just plug in giving me:&lt;br /&gt;&lt;br /&gt;dr/dt = 1/80pi cm/min.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-308519559362184219?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/308519559362184219/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/12/review-blog.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/308519559362184219'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/308519559362184219'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/12/review-blog.html' title='Review Blog'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-8303835992668657165</id><published>2010-12-12T14:32:00.002-06:00</published><updated>2010-12-12T14:39:09.012-06:00</updated><title type='text'>12/12 post</title><content type='html'>okayyy, so this past week we have been going over calculus AB stuff, reviewing for the ap test at the end of the year since it is half ab stuff and half bc stuff.&lt;br /&gt;we are done learning calc bc stuff for the year! :D yayyyyyy. hehe&lt;br /&gt;&lt;br /&gt;soooo, i shall go over AP CALC AB information for you fellas.&lt;br /&gt;&lt;br /&gt;limit rules as x approaches infinty.&lt;br /&gt;1. if top degreee &gt; bottom degree = +/- infinity&lt;br /&gt;2. if top degree &lt; bottom degree = 0&lt;br /&gt;3. if top degree = bottom degree = divide leading coefficients.&lt;br /&gt;&lt;br /&gt;when to use l'hopitals rule.&lt;br /&gt;only if the limit is in indeterminate form, then this can be applied! you take the derivative of the top &amp; bottom of the fraction, (separately) then take the limit again &amp; see if it gives you a number/indeterminate form. repeat as many times as necessary.&lt;br /&gt;&lt;br /&gt;rate of change problems.&lt;br /&gt;write down all information given to you. figure out what is being asked for. memorize any formulas you may need, (volume of cube, cylinder, area of things, etc.), &lt;-- use any of those formulas needed, plug in all information given, &amp; solve. pretty simple, just make sure you utilize your time wisely and figure out what is being asked of you to find in the problem!&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;HELP:&lt;br /&gt;please, someone go over any sequences &amp; series stuff from calc bc. it drives me NUTS.&lt;br /&gt;*truth is, I really just need to memorize all the formulas*&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-8303835992668657165?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/8303835992668657165/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/12/1212-post.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8303835992668657165'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8303835992668657165'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/12/1212-post.html' title='12/12 post'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6682859954219279938</id><published>2010-12-02T09:18:00.003-06:00</published><updated>2010-12-02T09:33:43.050-06:00</updated><title type='text'>Blog instead of Comments</title><content type='html'>So, since we really didn't do blogs over the holidays there isn't anything to comment on...so I will just review in a blog.&lt;br /&gt;&lt;br /&gt;Taylor: they will give you c&lt;br /&gt;f(c) + f^first deriv(c)(x) + f^second deriv(c)/2! (x-c)^2 + f^third dirv(c)/3! (x-c)^3&lt;br /&gt;&lt;br /&gt;Maclauirn: is centered at 0&lt;br /&gt;f(0) + f^first deriv(0)(x) + f^second deriv(0)/2! (x^2) + f^third dirv(0)/3! (x^3)&lt;br /&gt;&lt;br /&gt;EXAMPLE:&lt;br /&gt;Maclaurin up to the third degree&lt;br /&gt;&lt;br /&gt;f=e^x&lt;br /&gt;f(0)=e^0 = 1&lt;br /&gt;f^1(0)= e^x(1) = 1&lt;br /&gt;f^11(0) = e^x(1)(1) = 1&lt;br /&gt;&lt;br /&gt;1+x+1/2! x^2&lt;br /&gt;1+x+1/2(x^2)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Now don't forget about sequences and series! my fav :)&lt;br /&gt;HERE ARE JUST A FEW THINGS TO REMEMBER ABOUT THE RULES:&lt;br /&gt;&lt;br /&gt;p-series&lt;br /&gt;1/n^p &lt;br /&gt;p &gt;1 CONVERGES&lt;br /&gt;p &lt;1 or =1 DIVERGES&lt;br /&gt;&lt;br /&gt;geometric&lt;br /&gt;(5/4)^n&lt;br /&gt;n &lt;1 CONVERGES&lt;br /&gt;n &gt;1 DIVERGES&lt;br /&gt;&lt;br /&gt;limit comparison and direct comparison&lt;br /&gt;*you must compare it to something easier&lt;br /&gt;*use a different test&lt;br /&gt;*then use this test to confirm first one&lt;br /&gt;&lt;br /&gt;root and ratio&lt;br /&gt;&lt;1 CONVERGES&lt;br /&gt;&gt;1 or infinity DIVERGES&lt;br /&gt;=1 INCONCLUSIVE&lt;br /&gt;&lt;br /&gt;alternating series&lt;br /&gt;*must take out the (-1)^n thing&lt;br /&gt;*take limit MUST =0 or CANNOT be used&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;and something else we covered---&lt;br /&gt;&lt;br /&gt;radius and interval of convergence:&lt;br /&gt;*use the ratio test&lt;br /&gt;*take the limit&lt;br /&gt;*set up -1&lt; x &gt;1&lt;br /&gt;*may have to solve&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6682859954219279938?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6682859954219279938/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/12/blog-instead-of-comments.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6682859954219279938'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6682859954219279938'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/12/blog-instead-of-comments.html' title='Blog instead of Comments'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5540879795354930807</id><published>2010-11-30T12:22:00.001-06:00</published><updated>2010-11-30T12:23:57.178-06:00</updated><title type='text'>Post.</title><content type='html'>So we have a test. So here's a little review.&lt;br /&gt;&lt;br /&gt;1. A sequence converges if it's limit is a number. It diverges if there's an infinity anywhere in it. For instance:&lt;br /&gt;Given the sequence represented by the equation (n+1)/(n^2), say whether the sequence converges or diverges...at this point you would take the limit as n approaches infinity. In this case if would approach 0 because your limit rules say that if the degree of the top is less than the degree of the bottom, the limit approaches 0. Got it? So the entire sequence converges to 0 (a number)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2. Now for the difference between a sequence and a series. Indeed, I believe we learned this back in Advanced Math, but BRob stressed to us that Tir had issues with it, so might as well knock it in there a couple of times.&lt;br /&gt;&lt;br /&gt;A sequence is just a list of numbers...aka...1, 3, 5, 7,...&lt;br /&gt;**Note for this one it would be all odd numbers&lt;br /&gt;&lt;br /&gt;A series is basically the same thing as a sequence, except that you have like addition signs in it...for example...3+4+5+6+7..&lt;br /&gt;**Se those addition signs?? yeah, they're the ones you look out for..&lt;br /&gt;&lt;br /&gt;3. Okay, so where I got a little tripped up was when we were saying: "If___, then___" But now, I think I've got it right..&lt;br /&gt;&lt;br /&gt;Comment if you agree with this, "If the sequence of the series converges, then the series converges"&lt;br /&gt;&lt;br /&gt;"If the sequence of the series diverges, then the series diverges"&lt;br /&gt;&lt;br /&gt;My question is what are the conditions for ratio and root tests?  I forgot those on the last quiz. :D&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5540879795354930807?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5540879795354930807/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post_30.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5540879795354930807'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5540879795354930807'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post_30.html' title='Post.'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-458417109165363793</id><published>2010-11-30T12:22:00.000-06:00</published><updated>2010-11-30T12:23:56.304-06:00</updated><title type='text'>Post.</title><content type='html'>So we have a test. So here's a little review.&lt;br /&gt;&lt;br /&gt;1. A sequence converges if it's limit is a number. It diverges if there's an infinity anywhere in it. For instance:&lt;br /&gt;Given the sequence represented by the equation (n+1)/(n^2), say whether the sequence converges or diverges...at this point you would take the limit as n approaches infinity. In this case if would approach 0 because your limit rules say that if the degree of the top is less than the degree of the bottom, the limit approaches 0. Got it? So the entire sequence converges to 0 (a number)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2. Now for the difference between a sequence and a series. Indeed, I believe we learned this back in Advanced Math, but BRob stressed to us that Tir had issues with it, so might as well knock it in there a couple of times.&lt;br /&gt;&lt;br /&gt;A sequence is just a list of numbers...aka...1, 3, 5, 7,...&lt;br /&gt;**Note for this one it would be all odd numbers&lt;br /&gt;&lt;br /&gt;A series is basically the same thing as a sequence, except that you have like addition signs in it...for example...3+4+5+6+7..&lt;br /&gt;**Se those addition signs?? yeah, they're the ones you look out for..&lt;br /&gt;&lt;br /&gt;3. Okay, so where I got a little tripped up was when we were saying: "If___, then___" But now, I think I've got it right..&lt;br /&gt;&lt;br /&gt;Comment if you agree with this, "If the sequence of the series converges, then the series converges"&lt;br /&gt;&lt;br /&gt;"If the sequence of the series diverges, then the series diverges"&lt;br /&gt;&lt;br /&gt;My question is what are the conditions for ratio and root tests?  I forgot those on the last quiz. :D&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-458417109165363793?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/458417109165363793/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/458417109165363793'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/458417109165363793'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post.html' title='Post.'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1904001717652511769</id><published>2010-11-29T18:57:00.002-06:00</published><updated>2010-11-29T19:03:05.290-06:00</updated><title type='text'>birfday blog :)</title><content type='html'>okkkk, so we had thanksgiving week off. and we had a takehome test. i shall do some problems from this test.&lt;br /&gt;&lt;br /&gt;1. a sub n = (-2/3) ^ n&lt;br /&gt;&lt;br /&gt;you plug in values starting at one, then two, three, etc...&lt;br /&gt;&amp; find your answer.&lt;br /&gt;&lt;br /&gt;2. determine the convergence/divergence of the sequence with given nth term. if the sequence converges, find its limit.&lt;br /&gt;a sub n = 3^n/5^n&lt;br /&gt;geometric test. rewrite as (3/5) ^ n. &lt;br /&gt;abs. value of r &lt; 1 then it converges. 3/5 = r. after that, take the limit of the original problem. &amp; it converges to that #.&lt;br /&gt;&lt;br /&gt;these are just some examples. can someone explain integral test to me?&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1904001717652511769?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1904001717652511769/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/birfday-blog.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1904001717652511769'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1904001717652511769'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/birfday-blog.html' title='birfday blog :)'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1949392008405987632</id><published>2010-11-28T18:36:00.002-06:00</published><updated>2010-11-28T18:55:00.735-06:00</updated><title type='text'>Combined late and now bloggg..</title><content type='html'>Alright..so i'll start off by saying these holidays went by entirely too fast..and obviously, you can't post blogs on iphones..because i actually did try in the airport going to new york..and it doesn't work...your keyboard does not pull up when you press in the blog box..odd. i know.&lt;br /&gt;&lt;br /&gt;but anyway, lets get this thing started.&lt;br /&gt;&lt;br /&gt;So, lets go over some throw back stuff..&lt;br /&gt;&lt;br /&gt;1.  Substitution: when the derivative is in the equation&lt;br /&gt;  ex:  Scosxsinx&lt;br /&gt;           u = sinx&lt;br /&gt;           du = cos x&lt;br /&gt;     S u du&lt;br /&gt;    =1/2u^2   = 1/2sin^2(x) + c&lt;br /&gt;&lt;br /&gt;2. By-Parts:  when the derivative is not in the equation but you can sort of manipulate to figure it out...&lt;br /&gt;    ex:  Sarctan(x)&lt;br /&gt;           u = arctan(x)        dv = dx&lt;br /&gt;           du = 1/ 1+x^2       v = x&lt;br /&gt;     = arctan(x) (x) - S x/1+x^2&lt;br /&gt;     = arctan(x) (x) - 1/2ln(2)&lt;br /&gt;&lt;br /&gt;3.  Partial Fractions:  when you have a fraction where the bottom can be broken up or factored&lt;br /&gt;    ex:  umm..i can't think of one bc im not so good at these&lt;br /&gt;  But basically you have to break up the bottom and separate it to different fractions with differetn letters "naming" the fractions then you solve different systems to find your letters and then plug in to find the answer.&lt;br /&gt;&lt;br /&gt;So, i've been working on my packet and i'm having alot of trouble remembering this stuff...and i'm getting kinda confused on some things..so come prepared on monday-wednesday..cause i'll have PLENTY questions for the test thursday :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1949392008405987632?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1949392008405987632/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/combined-late-and-now-bloggg.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1949392008405987632'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1949392008405987632'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/combined-late-and-now-bloggg.html' title='Combined late and now bloggg..'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6319218666409821845</id><published>2010-11-17T20:50:00.002-06:00</published><updated>2010-11-17T21:14:23.033-06:00</updated><title type='text'>11/17 (late blog)</title><content type='html'>Sorry I am late, I just hate computers and avoid them at all cost :).&lt;br /&gt;&lt;br /&gt;Taylor Polynomials and Approximations:&lt;br /&gt;&lt;br /&gt;The form of a convergent power series:&lt;br /&gt;&lt;br /&gt;"In this section you will study a general procedure for deriving the power series for a function that has derivatives of all orders.  The following theorem gives the form that every convergent power series must take."&lt;br /&gt;&lt;br /&gt;If f is represent by a power series f(x) = E an(x-c)^n for all x in an open interval l containing c, then an = f^(n)(c)/n! and&lt;br /&gt;f(x) = f(c) + f'(c)(x-c) + f''(c)/2! * (x-c)^2 +...+ f^n(c)/n! * (x-c)^n +... .&lt;br /&gt;&lt;br /&gt;Definition of Taylor and Maclaurin Series:&lt;br /&gt;&lt;br /&gt;If a function f has derivatives of all orders at x = c, then the series&lt;br /&gt;E(from n=0 to infinity) f^n(c)/n! * (x-c)^n = f(c) + f'(c)(x-c) + f''(c)/2! * (x-c)^2 +...+ f^n(c)/n! * (x-c)^n +...&lt;br /&gt;is called the Taylor series for f(x) at c.  Moreover, if c = 0, then the series is the Maclaurin series for f. &lt;br /&gt;&lt;br /&gt;"If you know the pattern for the coefficients of the Taylor polynomials for a function, you can extend the pattern easily to form the corresponding Taylor series."&lt;br /&gt;&lt;br /&gt;The convergence of a Tyalor series will always equal f^n(c)/n! * (x-c)^n if lim(as n -&gt; infinity) Rn = 0.&lt;br /&gt;&lt;br /&gt;Guidelines for Finding A Taylor Series:&lt;br /&gt;&lt;br /&gt;1.)  Differentiate f(x) several times and evaluate each derivative at c. &lt;br /&gt;      f(c), f'(c), f''(c), f'''(c), ... , f^n(c), ...&lt;br /&gt;&lt;br /&gt;2.)  Use the sequence developed in the first step to form the Taylor coefficients an = f^n(c)/n!, and determine the interval of convergence for the resulting power series&lt;br /&gt;       f(c) + f'(c)(x-c) + f''(c)/2! * (x-c)^2 +...+ f^n(c)/n! * (x-c)^n +... .&lt;br /&gt;&lt;br /&gt;3.)  Within the interval of convergence, determine whether the series converges to f(x).&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Everyone should look at and put to memory the chart of page 684 about power series for elementary functions.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6319218666409821845?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6319218666409821845/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/1117-late-blog.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6319218666409821845'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6319218666409821845'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/1117-late-blog.html' title='11/17 (late blog)'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-7973719647831174508</id><published>2010-11-15T06:49:00.002-06:00</published><updated>2010-11-15T06:55:37.868-06:00</updated><title type='text'>11/14 post</title><content type='html'>what we did this week in calc class, we basically went over some more stuff with power series and taylor polynomials and maclaurin serires and what not. same old chapter nine stuff that we been doing for the past 2 weeks or whatever.&lt;br /&gt;&lt;br /&gt;ok so POWER SERIES:&lt;br /&gt;what you do for this is &lt;br /&gt;1. do ratio test&lt;br /&gt;2. set lim of abs value less than 1&lt;br /&gt;3. solve&lt;br /&gt;&lt;br /&gt;i went over taylor polynomials and all that last week. so let's go over derivative rules, since we been using that lately&lt;br /&gt;sin = cos&lt;br /&gt;cox = -sin&lt;br /&gt;tan = sec^2&lt;br /&gt;sec = sectan&lt;br /&gt;1/x^2 = -2/x^3&lt;br /&gt;xsinx = product rule&lt;br /&gt;x/cosx = quotient rule&lt;br /&gt;&lt;br /&gt;that is just a few examples of some derivative formulas for ya.&lt;br /&gt;&lt;br /&gt;lim rules as n approaches infinity&lt;br /&gt;1. if top degree &gt; bottom degree = +/- infinity&lt;br /&gt;2. if top degree &lt; bottom degree = 0&lt;br /&gt;3. if top degree = bottom degree = divide leading coefficients&lt;br /&gt;&lt;br /&gt;yayyyyyyyyyyyyy :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-7973719647831174508?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/7973719647831174508/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/1114-post.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7973719647831174508'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7973719647831174508'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/1114-post.html' title='11/14 post'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-4280272858669362118</id><published>2010-11-14T21:07:00.001-06:00</published><updated>2010-11-14T21:07:44.903-06:00</updated><title type='text'>Post # 11</title><content type='html'>&lt;span xmlns=''&gt;&lt;p&gt;Okayy, so I'm going to go over some derivative rules and identities because while I was doing the homework I quickly realized that I do NOT remember how to do derivatives because I'm so in integral mode..&lt;br /&gt;&lt;/p&gt;&lt;p&gt;So, let's get it.&lt;br /&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;**The formula for quotient rule is vu^1-uv^1/v^2 or the derivative of the top times the bottom – the derivative of the bottom times the top over the bottom squared. &lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;An example is sin x / x-1. Take the derivative of the top which is (cos x) times the bottom (x-1) – the derivative of the bottom (1) times the top (sin x) over the bottom squared (x-1) ^2. &lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;From there, it is just simple algebra. &lt;br/&gt;The answer comes out to (cos x) (x-1)-sin x/(x-1) ^2. &lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;*You do not use quotient rule when there is only an x on the bottom. &lt;br/&gt;*You just bring the x to the top and make the exponent negative then use the formula U^n.&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;**Another thing is the product rule..if everyone remembers it.&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;(first)(derivative of second) + (second)(derivative of first)&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;**Third, you need to remember when to use chain rule..&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;*when you have something inside something or something raised to something with a variable..&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;Also, remember all the trig functions are THE OTHER way around..&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;Like &lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;Sin = cos&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;Cos =-sin&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;Tan = sec^2&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;Sec = sectan&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;And so on…&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063'&gt;&lt;span style='font-family:Helvetica'&gt;The thing I need most help with is where to go after root test when doing power series…like I get an answer then take the limit then what? And also, what are the quiz orders and when are we taking the HUGE test on everything? THANKSSSS &lt;/span&gt;&lt;span style='font-family:Wingdings'&gt;J&lt;/span&gt;&lt;/span&gt;&lt;/p&gt;&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-4280272858669362118?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/4280272858669362118/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-11_14.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4280272858669362118'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4280272858669362118'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-11_14.html' title='Post # 11'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1770262611953799975</id><published>2010-11-14T17:15:00.002-06:00</published><updated>2010-11-14T17:24:07.051-06:00</updated><title type='text'>Blog</title><content type='html'>Power Series is something relatively easy, and I found the homework for both this and thee Maclaurin/Taylor Polynomials to be extremely redundant.  However, there's like only two basic rules for Power Series.&lt;br /&gt;&lt;br /&gt;1.Do the ratio test.&lt;br /&gt;2 set the limit of the absolut value less than 1&lt;br /&gt;3. solve.&lt;br /&gt;&lt;br /&gt;Basically this is just a review on the ratio test.  HOWEVER.  Say I have after the ratio test &lt;br /&gt;&lt;br /&gt;limit as n-inf. of abs(x^2/2!)&lt;br /&gt;&lt;br /&gt;Now this is the thing.  the limit of the abs value is set to less than one right?  well, if I plug in infinity, it'll give me inf over a number. which is just infinity.  Therefore it diverges.  HOWEVER,once again, if I plug in say 1 for x, I'll be left with 1/2 which is less than 1 but not greater than -1 (coming from the absolute value thing where you put -ve &lt; inside of abs&lt; +ve) value .  if I plug in 0, I'll be left we something less than 1.  therefore at both x=1 and 0, the polynomial converges (aka no infinity)&lt;br /&gt;&lt;br /&gt;For what I do not get, and perhaps a question for Brob is exactly how this will be phrased on the AP.  Also, I would like to know what to do with the graph ones??  Thanks oh so much.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1770262611953799975?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1770262611953799975/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/blog.html#comment-form' title='4 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1770262611953799975'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1770262611953799975'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/blog.html' title='Blog'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>4</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-8466503020967095933</id><published>2010-11-08T22:17:00.002-06:00</published><updated>2010-11-08T22:26:35.990-06:00</updated><title type='text'>Post #11?</title><content type='html'>Well we worked on the Taylor and Maclaurin Polynomial/Series something things.&lt;br /&gt;&lt;br /&gt;Taylor:&lt;br /&gt;Pn(x) = f(c) + F'(c)(x-c) + (f''(c)(x-c)^2)/2! + ... (f^n(c)(x-c)^n)/n!&lt;br /&gt;*C is used b/c it is not centered a zero.&lt;br /&gt;&lt;br /&gt;Maclaurin:&lt;br /&gt;It is simply the same formula, but 0 replaces the c.&lt;br /&gt;*This is one is centered at zero.&lt;br /&gt;&lt;br /&gt;This will be a very short blog because....I don't understand this really.&lt;br /&gt;I don't have my notebook..so I can't really think of problem. But I do know you follow the formula like you plug in c for the equation so that's the first term, then do + take the derivative and also put (x-c)&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;Can someone show me example pleassssse? I just really need an easy step by step for this.&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-8466503020967095933?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/8466503020967095933/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-11.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8466503020967095933'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8466503020967095933'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-11.html' title='Post #11?'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1772376906290309122</id><published>2010-11-08T06:29:00.002-06:00</published><updated>2010-11-08T06:38:47.214-06:00</updated><title type='text'>11/7 post.</title><content type='html'>so this week in calc bc we finally quit learning the hard stuff in chapter nine, (like sequences and series) and moved on to something that i'm starting to understand better! thank you! haha. &lt;br /&gt;we learned taylor polynomial and some other stuff. so this is the formula thing for it...&lt;br /&gt;&lt;br /&gt;Pn(x) = f(c) + F'(c)(x-c) + (f''(c)(x-c)^2)/2! + ... (f^n(c)(x-c)^n)/n!&lt;br /&gt;&lt;br /&gt;also we learned somehting called Macclaurin series. this uses the same formula, except everywhere you see c, you put 0. because Macclaurin's formula is centered at 0.&lt;br /&gt;&lt;br /&gt;it was all-in-all a pretty simple week. you just follow that formula. i don't think i had any questions. let me just throw in a little something though to make this blog a little longer.&lt;br /&gt;&lt;br /&gt;limit rules as n approaches infinity.&lt;br /&gt;1. if top degree &gt; bottom degree = +/- infinity.&lt;br /&gt;2. if top degree &lt; bottom degree = 0.&lt;br /&gt;3. if top degree = bottom degree = divide leading coefficients.&lt;br /&gt;&lt;br /&gt;yayyyyyyy :D&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1772376906290309122?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1772376906290309122/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/117-post.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1772376906290309122'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1772376906290309122'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/117-post.html' title='11/7 post.'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-3666546229206124701</id><published>2010-11-07T20:42:00.002-06:00</published><updated>2010-11-07T20:56:54.875-06:00</updated><title type='text'>Post for 11/7</title><content type='html'>So this week in the wonderful world of Calculus BC:  Taylor Polynomial and Approximations.&lt;br /&gt;&lt;br /&gt;I would just like to say that I have absolutley no clue at all what these are or how you do them. So that is pretty much my question for the week.&lt;br /&gt;&lt;br /&gt;After teaching Mu A practice I realized that I am kind of rusty with my derivatives.&lt;br /&gt;&lt;br /&gt;So today's post will be about derivatives.&lt;br /&gt;&lt;br /&gt;d/dx [uv] = uv' + vu'&lt;br /&gt;&lt;br /&gt;d/dx [u/v] = (vu' - uv') / v^2&lt;br /&gt;&lt;br /&gt;d/dx [sinu] = (cosu)(u')&lt;br /&gt;&lt;br /&gt;d/dx [cosu] = -(sinu)(u')&lt;br /&gt;&lt;br /&gt;d/dx [tanu] = (secu)^2(u')&lt;br /&gt;&lt;br /&gt;d/dx [cotu] = -(cscu)^2(u')&lt;br /&gt;&lt;br /&gt;d/dx [secu] = (secu)(tanu)(u')&lt;br /&gt;&lt;br /&gt;d/dx [cscu] = -(cscu)(cotu)(u')&lt;br /&gt;&lt;br /&gt;d/dx [ln(u)] = u' / u&lt;br /&gt;&lt;br /&gt;d/dx [u] = (u)(u') / (u)&lt;br /&gt;&lt;br /&gt;d/dx [e^u] = e^u * u'&lt;br /&gt;&lt;br /&gt;One rule for derivative that the Mu A's weren't really getting was chain rule.  I told that the way I remember it was to work from the outside in.&lt;br /&gt;&lt;br /&gt;Peace Out,&lt;br /&gt;Ryan&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-3666546229206124701?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/3666546229206124701/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-for-117.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3666546229206124701'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3666546229206124701'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-for-117.html' title='Post for 11/7'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-8339248017389469235</id><published>2010-11-07T18:45:00.002-06:00</published><updated>2010-11-07T18:53:10.996-06:00</updated><title type='text'>Yet another blog.</title><content type='html'>Let me just say that today I sat down and did all my homework.  Rather attempted all my homework.  I found myself terribly confused once you get to remainders and such.  I know for a fact that I need help on that.  However I do know how to find whatever degree polynomial functions.&lt;br /&gt;&lt;br /&gt;So Taylor is the series your generating.  it's given as:&lt;br /&gt;&lt;br /&gt;Pn(x)=f(c) + F'(c)(x-c) + (f''(c)(x-c)^2)/2!...&lt;br /&gt;&lt;br /&gt;C is going to be any number really, and you're just going to plug in the number to the original (they give you it) and then take the derivative and plug into the formula.  &lt;br /&gt;&lt;br /&gt;Now the difference between Maclaurin is that Maclaurin is centered at 0, meaning that your c is zero.  so you're just going to start off by plugging in zero to the original the just taking the derivative and repeatedly plugging in the zero.  And after approximating a value to the degree the problem it tells you to, you're get your answer (use this when you have like cos(1.1), or something you know you cant to without a calculator easily)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;For stuff I DO NOT GET.  I do not get how to find the remainder.  Also, what do you do if you're approximating something and you find a pattern?  like after the 4th term you get what you started with?  Just a couple of questions. :DD&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-8339248017389469235?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/8339248017389469235/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/yet-another-blog.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8339248017389469235'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8339248017389469235'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/yet-another-blog.html' title='Yet another blog.'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2242976444030023941</id><published>2010-11-01T20:23:00.002-05:00</published><updated>2010-11-01T20:25:59.241-05:00</updated><title type='text'>Post # 10</title><content type='html'>So, since i hardly ever ask questions..and always in need of knowledge.  I'll ask many questions to hopefully be answered.&lt;br /&gt;&lt;br /&gt;1.  When do you know if a test fails you and you need to do something else.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2. What tests are "inconclusive"?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3.  What does inconclusive mean?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;4.  What came first:  the chicken or the egg?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;5.  How do you do series with trig functions..like cos(pi/x)..?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;6.  How do you do the word problem things?&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;For the most part, i understood this chapter..these six questions always confused me though.. and i think on the test today i began to become more worried about that than what i  needed to be worried about!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2242976444030023941?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2242976444030023941/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-10_01.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2242976444030023941'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2242976444030023941'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-10_01.html' title='Post # 10'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-7272644112382257232</id><published>2010-11-01T09:00:00.002-05:00</published><updated>2010-11-01T09:38:16.203-05:00</updated><title type='text'>Post #10</title><content type='html'>Sequence:&lt;br /&gt;take the limit to see if it diverges or converges&lt;br /&gt;+if it has a limit it converges&lt;br /&gt;+if you have a type of infinity it diverges&lt;br /&gt;&lt;br /&gt;Finding terms: plug into equation&lt;br /&gt;Partial Sums: add the term before it&lt;br /&gt;&lt;br /&gt;Series:&lt;br /&gt;&lt;br /&gt;Arithmetic sum: n(t+tn)/2&lt;br /&gt;Geometric sum: t1/1-r&lt;br /&gt;&lt;br /&gt;nth term test:&lt;br /&gt;take the limit, may have to use L"H rule&lt;br /&gt;+if you get zero must use different test&lt;br /&gt;+if you get a number it diverges&lt;br /&gt;&lt;br /&gt;Integral Test: &lt;br /&gt;if you can integrate it easily&lt;br /&gt;+if x is greater than or equal to one it converges&lt;br /&gt;+anything else diverges&lt;br /&gt;&lt;br /&gt;p-series test:&lt;br /&gt;1/n^p&lt;br /&gt;+if p is greater than 1 it converges&lt;br /&gt;+if it is less than or equal to 1 it diverges&lt;br /&gt;&lt;br /&gt;Geometric:&lt;br /&gt;+if absolute value of r is less than 1 it converges&lt;br /&gt;+greater than or equal to one it diverges&lt;br /&gt;&lt;br /&gt;direct comparison test:&lt;br /&gt;compare it to something easier&lt;br /&gt;you then might get a geometric or p-series&lt;br /&gt;use the test that works&lt;br /&gt;+if you get it converges, stop&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;+if it diverges you must know ?&lt;/span&gt; &lt;br /&gt;&lt;br /&gt;Limit Comparison Test:&lt;br /&gt;compare to something easier&lt;br /&gt;use another test&lt;br /&gt;+divide original by compared&lt;br /&gt;+take limit, if +ve number it converges&lt;br /&gt;+if infinity, it diverges&lt;br /&gt;&lt;br /&gt;alternating series:&lt;br /&gt;will have like (-1)^n+1&lt;br /&gt;+take the alternating part out for new&lt;br /&gt;+take limit, must get zero&lt;br /&gt;*if you get a number it diverges&lt;br /&gt;+now add +1 to each n, less than or equal to, new one&lt;br /&gt;+if true converges&lt;br /&gt;+if not true diverges&lt;br /&gt;&lt;br /&gt;Ratio Test:&lt;br /&gt;usually used with ! or variable exponents&lt;br /&gt;+add +1 to each n divided by the original&lt;br /&gt;+things should cancel&lt;br /&gt;* remember 2^n+1 can be written has 2^n 2^1 and (n+1)! is like n!(n+1)&lt;br /&gt;+take limit&lt;br /&gt;+less than 1 it converges, greater than 1 or infinity it diverges&lt;br /&gt;+if you get one it is inconclusive&lt;br /&gt;&lt;br /&gt;root test:&lt;br /&gt;used when something is raised to the n&lt;br /&gt;+take the nth root of the absolute value of the original&lt;br /&gt;+take limit&lt;br /&gt;+if you get less than 1 it converges, if you et greater than one or infinity it diverges&lt;br /&gt;+if you get one it is inconclusive &lt;br /&gt;&lt;br /&gt;Absolute Convergence: if absolute value of an converges then an converges&lt;br /&gt;Conditional Convergence: if an converges but absolute vale does not converge&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-7272644112382257232?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/7272644112382257232/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-10.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7272644112382257232'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7272644112382257232'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/post-10.html' title='Post #10'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6795152178361929794</id><published>2010-11-01T00:10:00.002-05:00</published><updated>2010-11-01T00:20:11.863-05:00</updated><title type='text'>Maleries post</title><content type='html'>Okay.  So I, for one, would like to know WHEN DOES THIS CHAPTER END???  I am BEYOND tired of sequences and series....&lt;br /&gt;&lt;br /&gt;AHH.. okay. Math right?  well perhaps this will help us in future endeavors.&lt;br /&gt;&lt;br /&gt;Okay, so first, to see if a SERIES converges or diverges, you follow that chart.  &lt;br /&gt;&lt;br /&gt;First things first, nth term.  you just take the limit as x goes to infinity.  If you get anything other than zero, the series diverges right of the bat.  If you get zero, the nth term test is inconclusive.&lt;br /&gt;&lt;br /&gt;Next, you can determine whether or not it's a pseries, geometric, or something you can integrate.&lt;br /&gt;&lt;br /&gt;PSERIES is when its n raised to some exponent&lt;br /&gt;GEOMETRIC is when it's some fraction raised to the n&lt;br /&gt;Integral is when its something easy..so say ln integration would be easy to do (i.e. 1/n)...or if you feel like messing with by parts, go for it.&lt;br /&gt;&lt;br /&gt;Pseries-if p(exponent) is greater than one it converges, if its less than or equal to it diverges&lt;br /&gt;&lt;br /&gt;Geometric-if abs(r) (your thing being raised) is less than one, it converges.  Not it diverges.&lt;br /&gt;&lt;br /&gt;Integral-if you get after integrating infinity anywhere, it diverges.# it converges&lt;br /&gt; &lt;br /&gt;If the above do not apply you have a couple of options.&lt;br /&gt;&lt;br /&gt;There's the ROOT TEST (where you just force a root...I'm a little shady on this one).  RATIO TEST (where you add one to every n and put that over your original and take the limit) ALTERNATING SERIES TEST (which is a little tricky..check your book.)  DIRECT COMPARISON TEST (compare it to something bigger and try to use pseries, geo, integral, etc.)  and LIMIT COMPARISON TEST where you just compare it to something and put your original over what you're comparing it to and take the limit.)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6795152178361929794?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6795152178361929794/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/maleries-post.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6795152178361929794'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6795152178361929794'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/11/maleries-post.html' title='Maleries post'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-8269665160110150259</id><published>2010-10-31T21:47:00.002-05:00</published><updated>2010-10-31T22:09:53.945-05:00</updated><title type='text'>post 10.</title><content type='html'>soooo, we got a big test tomorrow in calc. so let's go over what is gonna be on that!&lt;br /&gt;&lt;br /&gt;p-series:&lt;br /&gt;used whenever you have n^#.&lt;br /&gt;if # is &lt;/= 1, it diverges. if # is &gt; 1 it converges.&lt;br /&gt;&lt;br /&gt;geometric series:&lt;br /&gt;used whenever you have #^n&lt;br /&gt;if abs.value of # &lt;&gt; 1, continue on to testing..&lt;br /&gt;you would divide original by what you compared it to. then take the limit as n goes to infinty of that. &amp;amp; if you get # &gt; 1, it diverges. if not, it is inconclusive.&lt;br /&gt;&lt;br /&gt;nth term test:&lt;br /&gt;you use this when you don't exactly know what else to do &amp;amp; you are just testing.&lt;br /&gt;take the lim as x goes to infinity.&lt;br /&gt;if you get anything besides 0, it diverges. if you get 0, it's inconclusive.&lt;br /&gt;&lt;br /&gt;integral test:&lt;br /&gt;this is used whenever you have absolutely nothing else to do. you take the integral from n to infinity&lt;br /&gt;if you get infinity, it diverges. if you get a number, it converges.&lt;br /&gt;&lt;br /&gt;alternating series test:&lt;br /&gt;used whenever you have -1 or -2 raised to n.&lt;br /&gt;you take out that portion of the problem, then continue on. but if you have -2, you just take out the negative.&lt;br /&gt;after taking it out, you take the limit as n goes to infinity.&lt;br /&gt;if you get anything besides 0, it diverges. if you get 0, continue on &amp;amp; add one to every n.&lt;br /&gt;then compare to the original after you took out the -# ^ n by doing (n+1 equation) &lt;/= (original equation)&lt;br /&gt;^ if above is true, then converges. if false, then diverges.&lt;br /&gt;&lt;br /&gt;sum of geometric series:&lt;br /&gt;first term/1-r&lt;br /&gt;&lt;br /&gt;sum of nongeometric series:&lt;br /&gt;n(t1 + tn)/2&lt;br /&gt;&lt;br /&gt;ratio test:&lt;br /&gt;use if it tells you to.. pretty much.&lt;br /&gt;add one to each n. then divide &lt;-- that by the original. then take lim as n goes to inifinty.&lt;br /&gt;if you get # &lt;&gt; 1 or infinity, it diverges.&lt;br /&gt;(also you use this if you have a factorial!)&lt;br /&gt;&lt;br /&gt;NOW for what I don't understand...&lt;br /&gt;-direct comparison test&lt;br /&gt;-limit comparison test&lt;br /&gt;-absolute convergence&lt;br /&gt;-conditional convergence&lt;br /&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-8269665160110150259?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/8269665160110150259/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-10.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8269665160110150259'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8269665160110150259'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-10.html' title='post 10.'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2859238221724788363</id><published>2010-10-25T08:27:00.002-05:00</published><updated>2010-10-25T08:39:52.516-05:00</updated><title type='text'>Post #9</title><content type='html'>Well does it converge or diverge?&lt;br /&gt;&lt;br /&gt;You should probably try the nth term test before trying another method.&lt;br /&gt;&lt;br /&gt;P-Series: n^p&lt;br /&gt;if p is greater than 1--&gt;converge&lt;br /&gt;if less than or eqaul to 1--&gt;diverge&lt;br /&gt;&lt;br /&gt;Geometric: (1/2)^n&lt;br /&gt;if the absoultue value of r(which is the n) is less than 1--&gt;converge&lt;br /&gt;if the absoultue value of r greater than 1--&gt;diverge&lt;br /&gt;&lt;br /&gt;Intergral Test: Probably use when everything fails.&lt;br /&gt;First, take the limit from 0 to infinity.&lt;br /&gt;Second, integrate like normal.&lt;br /&gt;Third, plug in 0 and infinity.&lt;br /&gt;Finally, take the limit.&lt;br /&gt;&lt;br /&gt;If you still get any form of infinity--&gt;diverge&lt;br /&gt;if you get a number--&gt;converge.&lt;br /&gt;&lt;br /&gt;Comparing: You can compare the problem to an eaiser one.&lt;br /&gt;1. direct comparison&lt;br /&gt;2. limit comparison&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;My Question: I need someone to explain the difference between direct and limit (the steps and everything).  I'm afraid that I'm mixing them up or using a combination of both.  Please help.&lt;/strong&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2859238221724788363?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2859238221724788363/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-9_25.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2859238221724788363'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2859238221724788363'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-9_25.html' title='Post #9'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1655474434370406887</id><published>2010-10-24T19:33:00.003-05:00</published><updated>2010-10-24T19:43:15.650-05:00</updated><title type='text'>A blog</title><content type='html'>So steph came to me and asked me how to break down the steps for finding whether or not a series converges or diverges.  So...I believe they're the following...&lt;br /&gt;&lt;br /&gt;1. nth term test-so basically you're going to just take the limit as n goes to infinity right off the bat...if it equals something other than zero, you know for sure that it diverges.  However, If you get 0, you have to proceed to next step...&lt;br /&gt;&lt;br /&gt;2.  Okay.  for step two you can choose between a, b, and c.  &lt;br /&gt;         a.  p series&lt;br /&gt;         b.  geometric&lt;br /&gt;         c.  integral test&lt;br /&gt;&lt;br /&gt;**basically, you have to recognize that if it's 1/n^(exp), it's a pseries.  If not you see if it's geometric (i.e. something raised to the n).  And if it's something that looks easy to integrate, integrate it.&lt;br /&gt;&lt;br /&gt;If step 2 fails, go onto step 3.&lt;br /&gt;&lt;br /&gt;3.  Unless the problem tells you otherwise, you can use either direct comparison OR limit comparison.  For direct, you just compare it to something bigger (so if i have a fraction, the number on the bottom will be smaller...I know, confusing...).  So once I find something bigger and similar, I just go through step 2.  Same goes for limit comparison test, although, all you need is something similar, doesn't have to be either bigger or smaller than the original.  Then all you do is take the limit as n goes to infinity.  &lt;br /&gt;&lt;br /&gt;simple enough??&lt;br /&gt;&lt;br /&gt;I know it's confusing, you just have to think through all the steps.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;QUESTIONS:&lt;br /&gt;&lt;br /&gt;I have NO CLUE WHATSOEVER how to determine if a test is "inconclusive".  So if you have that practice packet, problem 16...I got the integral test done, but I got infinity minus something.  Does that mean it's inconclusive???  kbye.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1655474434370406887?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1655474434370406887/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/blog.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1655474434370406887'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1655474434370406887'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/blog.html' title='A blog'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-663917082167306057</id><published>2010-10-24T18:52:00.002-05:00</published><updated>2010-10-24T19:00:20.115-05:00</updated><title type='text'>post 9</title><content type='html'>so, still confused with chapter 9 stuff..&lt;br /&gt;i'll just go over what i know.&lt;br /&gt;so if you are given something, (idk if it matters if it is a sequence or a series).. you automaticaly take the nth term test.&lt;br /&gt;so take lim as n goes to infinty. if you get 0, you have to move on to take another test. if not, it automatically diverges. the next tests are:&lt;br /&gt;geometric, p-series, or integral teset.&lt;br /&gt;nowwww, i know geometric is when you have an exponent, and p-series is when you haev a fraction. that's about all i know. therefore idk how to choose one, what to do after i choose one, or anything like that. i'm just so lost, idk why. i've been having a bad past few weeks.&lt;br /&gt;&lt;br /&gt;anything else that i didn't mention in here, please explain to me. i think i need a tutor for this chapter or something.. idk.&lt;br /&gt;&lt;br /&gt;I DO NOT UNDERSTAND:&lt;br /&gt;direct comparison test&lt;br /&gt;limit comparison test&lt;br /&gt;^^i completely failed that quiz. ha&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-663917082167306057?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/663917082167306057/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-9_24.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/663917082167306057'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/663917082167306057'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-9_24.html' title='post 9'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1399836724988305443</id><published>2010-10-24T12:40:00.002-05:00</published><updated>2010-10-24T12:58:59.673-05:00</updated><title type='text'>Ryan - October 24 Post</title><content type='html'>To be honest, I really have been lost since we started this chapter.  It's probably because I didn't do my homework the first night, and I've never been able to catch up.  So I'll start where I started to get lost (9.1).&lt;br /&gt;&lt;br /&gt;*Sequence - a list of numbers.&lt;br /&gt;&lt;br /&gt;*To find the terms in a sequence, simply plug in n (the term you are on) for x and solve.&lt;br /&gt;&lt;br /&gt;*Sequences converge if they have a limit.&lt;br /&gt;*Sequences diverge if they don't have a limit.&lt;br /&gt;*To determine if a sequnce has a limit, take the limit as n-&gt;infinity of a(sub)n.&lt;br /&gt;&lt;br /&gt;******Something we need to know!!!!!!!!&lt;br /&gt;Limit(as n goes to infinity) of (1+(1/n))^n  =  e&lt;br /&gt;&lt;br /&gt;*Sequence properties follow limit properties.&lt;br /&gt;&lt;br /&gt;**Sqeeze Theorem.&lt;br /&gt;- &lt;= an &lt;= +&lt;br /&gt;*For sequences - sqeeze with a convergent sequence related to a(sub)n.&lt;br /&gt;&lt;br /&gt;*Monotonic - if terms are always increasing or always decreasing.&lt;br /&gt;&lt;br /&gt;*If a sequence is:&lt;br /&gt;*bounded and monotonic - it converges&lt;br /&gt;*bounded and not monotonic - it diverges&lt;br /&gt;*not bounded and monotonic - it diverges&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1399836724988305443?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1399836724988305443/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/ryan-october-24-post.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1399836724988305443'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1399836724988305443'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/ryan-october-24-post.html' title='Ryan - October 24 Post'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6373757594927382527</id><published>2010-10-24T11:10:00.002-05:00</published><updated>2010-10-24T11:18:27.062-05:00</updated><title type='text'>Post #9</title><content type='html'>So, blogger is finally accepting my password and I can now do my blog the right way!  Yayy!&lt;br /&gt;&lt;br /&gt;So, this week, on the days I was here, we reviewed and took some quizzes.&lt;br /&gt;&lt;br /&gt;I will explain, to the best of my knowledge of how to tell if something converges or diverges.&lt;br /&gt;&lt;br /&gt;So, first things first you shoud check if it's geometric.  Geometric is when something is being multiplied to every term in the series.  If the thing that is getting multiplied to is less than one, the series CONVERGES.&lt;br /&gt;&lt;br /&gt;Next, is alot like geometric..P-Series.  Its p-series if you have a fraction and the bottom is n raised to the number exponent.  If your number is less than or equal to one it DIVERGES, if its greater than one it CONVERGES.&lt;br /&gt;&lt;br /&gt;Now, lets go over the limit comparison test, you take an equaiton and simplify it by taking the greatest exponent terms and taking the limit of that...&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;HERE COMES MY QUESTION:  WHAT DOES THIS TELL YOUUUUUU?&lt;/strong&gt;&lt;br /&gt;&lt;strong&gt;&lt;/strong&gt;&lt;br /&gt;&lt;strong&gt;I have the same question for Direct Comparison Test along with the integral test.&lt;/strong&gt;&lt;br /&gt;&lt;strong&gt;&lt;/strong&gt;&lt;br /&gt;&lt;strong&gt;Can anyone tell me how these three tests give you an answer..and how do you know if it doesn't help you?&lt;/strong&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6373757594927382527?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6373757594927382527/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-9.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6373757594927382527'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6373757594927382527'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-9.html' title='Post #9'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-7939954641575031128</id><published>2010-10-17T22:23:00.002-05:00</published><updated>2010-10-17T22:30:06.664-05:00</updated><title type='text'>post 8</title><content type='html'>this week we learned more about convergence, divergence, p-series, nth term, integral test, and we learned something new called direct comparison test.&lt;br /&gt;&lt;br /&gt;i'll be completely honest, and say that i am really confused with a lot of this stuff. idk why it just isn't sticking in my head.&lt;br /&gt;&lt;br /&gt;i need someone to go over pretty much all of it... especially direct comparison test.&lt;br /&gt;i also don't know the difference. whenever brob says what kind is this. idk how to tell if it's p-series, or something else. i'm really lost. idk, this just wasn't my best week.&lt;br /&gt;&lt;br /&gt;i know they all have to do with sigmas. and i know what to do if the steps are in front of me. but idk, i guess i'm just lost.&lt;br /&gt;&lt;br /&gt;sorry if this is a short post, i'm just in need of help mostly.&lt;br /&gt;i'll go over at least one thing.&lt;br /&gt;&lt;br /&gt;INFINITY RULES:&lt;br /&gt;top degree &gt; bottom degree = +/- infinity.&lt;br /&gt;top degree &lt; bottom degree = 0&lt;br /&gt;top degree = bottom degree = divide coefficients&lt;br /&gt;&lt;br /&gt;these are actually very helpful and never go away. a lot of the time we just automatically do l'hospitals rule whenever we could just be doing this!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-7939954641575031128?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/7939954641575031128/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-8_17.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7939954641575031128'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7939954641575031128'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-8_17.html' title='post 8'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5872647456872460728</id><published>2010-10-17T10:14:00.002-05:00</published><updated>2010-10-17T10:35:34.087-05:00</updated><title type='text'>Post #8?</title><content type='html'>Well lets just get started..&lt;br /&gt;&lt;br /&gt;Direct Comparison Test:&lt;br /&gt;-This is dealing with a sigma.&lt;br /&gt;-You have to find an easier one to compare it to.&lt;br /&gt;-You will us either the nth term test, p-series thing, integral test, geometric thing.&lt;br /&gt;&lt;br /&gt;EXAMPLE:&lt;br /&gt;Say you have (sigma thing): 4^n/(5^n +3)&lt;br /&gt;4^n/(5^n +3)--&gt;compare to 4^n/5^n --&gt;same thing as (4/5)^n&lt;br /&gt;*this is geometric because it would be multiply by 4/5 &lt;br /&gt;*so by the rule for geometric 4/5 &lt; 1 --&gt;converges&lt;br /&gt;&lt;br /&gt;P-Series:&lt;br /&gt;-These are so easy.&lt;br /&gt;-it is n^p&lt;br /&gt;-if p &gt; 1 --&gt;converges&lt;br /&gt;-if p &lt; or = --&gt;diverges&lt;br /&gt;&lt;br /&gt;EXAMPLE:&lt;br /&gt;1/n^2&lt;br /&gt;*check to make sure it is n^p (which yes it is)&lt;br /&gt;*p=2&lt;br /&gt;*by the rule p &gt; 1 --&gt;converges&lt;br /&gt;&lt;br /&gt;FEW THINGS TO SET STRAIGHT:&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;sequence: list of numbers&lt;/span&gt;&lt;br /&gt;-converges if it has a limit&lt;br /&gt;-diverges if it doesn't have a limit&lt;br /&gt;-monotonic-terms always increasing/decreasing&lt;br /&gt;-if bounded &amp; monotonic--&gt;converges&lt;br /&gt;-if monotonic &amp; not bounded--&gt;diverges&lt;br /&gt;-if bounded &amp; not monotonic--&gt;can be divergent&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;series: add/sub terms in a seq&lt;/span&gt;&lt;br /&gt;-if sequence of partial sums converges--&gt;series converges&lt;br /&gt;-if sequence of partial sums diverges--&gt;series diverges&lt;br /&gt;-arithmetic series never converges&lt;br /&gt;-geometric converges if absolute vale of r &lt;1&lt;br /&gt;&lt;br /&gt;* 1/infinity = 0&lt;br /&gt;* 1/0 = infinity&lt;br /&gt;* lim x-&gt;infinity of arctanx = pi/2&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;QUESTIONS FOR YOU TO COMMENT:&lt;br /&gt;&lt;br /&gt;I have questions on the homework form this weekend about direct comparison test.  What would you compare these to? &lt;br /&gt;&lt;br /&gt;ln n/n+1&lt;br /&gt;1/n!&lt;br /&gt;e^-n^2 --&gt; would you do something like 1/n^2?&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5872647456872460728?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5872647456872460728/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-8.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5872647456872460728'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5872647456872460728'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-8.html' title='Post #8?'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-4747110553721979070</id><published>2010-10-11T06:55:00.002-05:00</published><updated>2010-10-11T07:00:59.071-05:00</updated><title type='text'>post..</title><content type='html'>okay, so i'll start with things i don't understand.&lt;br /&gt;&lt;br /&gt;1. i still don't understand how to tell if it converges/diverges. i skip all those problems&lt;br /&gt;&lt;br /&gt;2. i don't understand improper integrals... they just don't click in my head.&lt;br /&gt;&lt;br /&gt;3. i need help with some of those random trig function problems that are like S sec^6(2x). can someone briefly go over some rules about those.&lt;br /&gt;&lt;br /&gt;4. trig sub... i can only do those on a good day. i'll get lost after finding x = ... (sqrt)x =..... and then i plug it in and i get stuck..&lt;br /&gt;&lt;br /&gt;5. i'm also a litttleeeee confused on whether on not something is bounded. i didn't really get that. i understand monotonic though. which is what i'll go over&lt;br /&gt;&lt;br /&gt;when it asks you if a sequence/series is monotonic... that means that it wants to know if it is always increasing/always decreasing.&lt;br /&gt;so you plug in about 4 or 5 numbers (0-5), and see if it constantly goes up, constantly goes down, or isn't constant. if it is constant, it's monotonic. if it is NOT constant, it's NOT monotonic.&lt;br /&gt;and i understand that usually your function is bounded by your first term... but i'm confused about the limit and stuff. you have to take the limit or something and sometimes it's bounded below? idk i'm kinda lost.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-4747110553721979070?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/4747110553721979070/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post_11.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4747110553721979070'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4747110553721979070'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post_11.html' title='post..'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-820088382407039026</id><published>2010-10-10T21:45:00.002-05:00</published><updated>2010-10-10T22:03:55.200-05:00</updated><title type='text'>Post #7</title><content type='html'>A couple example problems that when I see I know exactly what to do.&lt;br /&gt;&lt;br /&gt;1. S 3/(x-13)^6 dx&lt;br /&gt;&lt;br /&gt;*basic substitution&lt;br /&gt;u=x-13   du= 1&lt;br /&gt;substitute: 8 S 1/u^6 = 8 S u^-6&lt;br /&gt;integrate: (1/5)(8) S u^-5&lt;br /&gt;plug in: -8/5(x-13)^-5 +C&lt;br /&gt;which can be written as -8/5(x-13)^5 +C&lt;br /&gt;&lt;br /&gt;2. S sinxcos^4x dx&lt;br /&gt;&lt;br /&gt;*basic substitution b/c cos and sin are direct derivatives of each other&lt;br /&gt;u=cosx   du= -sinx&lt;br /&gt;substitute: -S u^4&lt;br /&gt;integrate: -1/5u^5&lt;br /&gt;plug in: (-1/5)cos^5x +C&lt;br /&gt;rewritten: -cos^5x/5 +C&lt;br /&gt;&lt;br /&gt;3. S cos^3 6x dx&lt;br /&gt;&lt;br /&gt;*break it up&lt;br /&gt;break: cos^26x(cos6x)&lt;br /&gt;identity: (1-sin^26x)cos6x&lt;br /&gt;multiply in: S cos 6x - cos6xsin^26x&lt;br /&gt;*substitute for each&lt;br /&gt;u=6x      u=sin 6x&lt;br /&gt;du=6      du=6cos6x&lt;br /&gt;1/6 S cosu - 1/6 S u^2&lt;br /&gt;3(sin6x/6) - sin^36x/18&lt;br /&gt;3sin6x/18 - sin^36x/18&lt;br /&gt;sin6x(3-sin^26x)/18&lt;br /&gt;&lt;br /&gt;4. lim         28-7x+4x^2/5x^2 -7&lt;br /&gt;x--&gt;infinity&lt;br /&gt;&lt;br /&gt;*you may think L"Hopital's Rule BUT it is as x--&gt;infinity so use your limit rules!&lt;br /&gt;exponents equal each other so divide coefficients&lt;br /&gt;=4/5&lt;br /&gt;&lt;br /&gt;5. infinity S 2   3/x^5   (diverge or converge?)&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;Let me know if this is right!&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;a S 2   3x^-5 = -3(1/4)x^-4&lt;br /&gt;3/4 x^-4 &lt;br /&gt;lim x--&gt; infinty  3/4x^4&lt;br /&gt;= 0 +3/64&lt;br /&gt;so it = 3/64?&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;NOW MY BIG QUESTIONS:&lt;br /&gt;How do you do improper integrals if its bounds are like 9 to 11, like no infinity?&lt;br /&gt;How do you know you can use synthetic division on an integral?&lt;br /&gt;Anyone know some tricks about trig sub?&lt;br /&gt;How you do chasing the rabbit again?&lt;br /&gt;&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-820088382407039026?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/820088382407039026/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-7.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/820088382407039026'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/820088382407039026'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-7.html' title='Post #7'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-958487265842511037</id><published>2010-10-10T12:56:00.002-05:00</published><updated>2010-10-10T13:12:08.932-05:00</updated><title type='text'>Post...</title><content type='html'>Okay.  So for the past couple of days we've been doing sequences and series.  So, basically there are a few cardinal rules you should follow..  For example:&lt;br /&gt;&lt;br /&gt;1.  A sequence converges if it's limit is a number.  It diverges if there's an infinity anywhere in it.  For instance:&lt;br /&gt; Given the sequence represented by the equation (n+1)/(n^2), say whether the sequence converges or diverges...at this point you would take the limit as n approaches infinity.  In this case if would approach 0 because your limit rules say that if the degree of the top is less than the degree of the bottom, the limit approaches 0.  Got it?  So the entire sequence converges to 0 (a number)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;2.  Now for the difference between a sequence and a series.  Indeed, I believe we learned this back in Advanced Math, but BRob stressed to us that Tir had issues with it, so might as well knock it in there a couple of times.  &lt;br /&gt;&lt;br /&gt;   A sequence is just a list of numbers...aka...1, 3, 5, 7,...&lt;br /&gt;**Note for this one it would be all odd numbers&lt;br /&gt;&lt;br /&gt;   A series is basically the same thing as a sequence, except that you have like addition signs in it...for example...3+4+5+6+7..&lt;br /&gt;**Se those addition signs?? yeah, they're the ones you look out for..&lt;br /&gt;&lt;br /&gt;3. Okay, so where I got a little tripped up was when we were saying:  "If___, then___"  But now, I think I've got it right..&lt;br /&gt;&lt;br /&gt;Comment if you agree with this, "If the sequence of the series converges, then the series converges"&lt;br /&gt;&lt;br /&gt;"If the sequence of the series diverges, then the series diverges"&lt;br /&gt;&lt;br /&gt;I have a feeling that when the AP comes around, we'll most likely have to put this SOMEWHERE....anywho.&lt;br /&gt;&lt;br /&gt;FOR STUFF YOU CAN COMMENT ON!&lt;br /&gt;&lt;br /&gt;I truly was shaky on the whole find the sequence of the series...is that where you just find the terms plugged into the formula?  &lt;br /&gt;&lt;br /&gt;Also, I need help with sigmas...I'm used to having Step by Step Steps, and I don't.  Could someone sum them up for me??? Thank you oh so much!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-958487265842511037?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/958487265842511037/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post_10.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/958487265842511037'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/958487265842511037'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post_10.html' title='Post...'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1194858881768055718</id><published>2010-10-10T10:40:00.002-05:00</published><updated>2010-10-10T10:54:32.964-05:00</updated><title type='text'>10/10/10 post</title><content type='html'>So since my last post, we have started to learn about sequences and series.&lt;br /&gt;&lt;br /&gt;A sequence is a list of numbers defined by some equation, and a series is the addition or substraction of this list of numbers.&lt;br /&gt;&lt;br /&gt;Converge vs. Diverge.&lt;br /&gt;Sequences converge if they have a limit.&lt;br /&gt;Sequence diverge if they don't have a limit.&lt;br /&gt;If the limit of a sequence at infinity is infinity, then the sequence diverges.&lt;br /&gt;If a sequence is bounded and monotonic* then it is converges.&lt;br /&gt;If a sequence is bounded and not monotonic then the sequence diverges.&lt;br /&gt;If a sequence is not bounded and monotonic then the sequence diverges.&lt;br /&gt;*Monotonic - if terms are always increasing or always decreasing.&lt;br /&gt;If a sequence of partial sums converge then the series converges.&lt;br /&gt;If a sequence of partial sums diverge then the series diverges.&lt;br /&gt;&lt;br /&gt;Series:&lt;br /&gt;If something asks you to find the nth partial sum this means to find the sum at the nth term.&lt;br /&gt;An arithmetic series will never converge. It will always diverge (as it approaches infinity).&lt;br /&gt;&lt;br /&gt;Need to know:&lt;br /&gt;Lim as n -&gt; infinity (1 + (1/n))^n = e&lt;br /&gt;&lt;br /&gt;Sequence properties follow limit properties (infinity limits at least).&lt;br /&gt;&lt;br /&gt;Questions:&lt;br /&gt;&lt;br /&gt;How do I use the squeeze theorem? I'm completely unsure of what to do.&lt;br /&gt;&lt;br /&gt;How to find equations of series.&lt;br /&gt;&lt;br /&gt;I'm somewhat unconfident of what to do anytime I see a sigma.&lt;br /&gt;&lt;br /&gt;Ryan B.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1194858881768055718?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1194858881768055718/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/101010-post.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1194858881768055718'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1194858881768055718'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/101010-post.html' title='10/10/10 post'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1532741924345973744</id><published>2010-10-07T06:36:00.002-05:00</published><updated>2010-10-07T06:40:10.089-05:00</updated><title type='text'>No questions...</title><content type='html'>Yeah I realized I didn't post questions too!  But I really do have questions.  Maybe someone can answer these if they haven't done their comments..since no really did questions.&lt;br /&gt;&lt;br /&gt;1. Like how you tell when it is divergent or convergent?&lt;br /&gt;&lt;br /&gt;2. For improper integrals, sometimes they don't have an infinity so how do you break it up?&lt;br /&gt;&lt;br /&gt;3. I need a reminder for how to do chasing the rabbit!&lt;br /&gt;&lt;br /&gt;4.  And how do you integrate 14x^27 cosx^14 dx?&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1532741924345973744?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1532741924345973744/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/no-questions_07.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1532741924345973744'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1532741924345973744'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/no-questions_07.html' title='No questions...'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2246178309375277124</id><published>2010-10-06T21:20:00.002-05:00</published><updated>2010-10-06T21:27:34.198-05:00</updated><title type='text'>No Questions.</title><content type='html'>Since there are no questions on anyone's blogs.  I'll just do a mini-blog.&lt;br /&gt;&lt;br /&gt;INTEGRATION:&lt;br /&gt;Anytime there is an x term and an e term in a problem, use by-parts.  The x term will always be your u and the e term will always be the dv.&lt;br /&gt;&lt;br /&gt;Anytime there is an e term and a trig term, use by-parts until you see chasing the rabbit.&lt;br /&gt;&lt;br /&gt;Anytime there is an x term and a trig term, use by-parts with your u as the x.&lt;br /&gt;&lt;br /&gt;Questions:&lt;br /&gt;&lt;br /&gt;S x^2 / x+3  I have no clue how to even start this.&lt;br /&gt;&lt;br /&gt;This is from Calc. 1, but can anyone tell me the difference between a washer and (I don't know the other thing). THANKS&lt;br /&gt;&lt;br /&gt;ryan&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2246178309375277124?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2246178309375277124/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/no-questions.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2246178309375277124'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2246178309375277124'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/no-questions.html' title='No Questions.'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2399209957551606240</id><published>2010-10-06T20:38:00.001-05:00</published><updated>2010-10-06T20:39:21.774-05:00</updated><title type='text'>sorry</title><content type='html'>no one posted questions on their blogs. that's why i don't have comments. NEXT TIME, please post questions :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2399209957551606240?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2399209957551606240/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/sorry.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2399209957551606240'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2399209957551606240'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/sorry.html' title='sorry'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2820531591951415690</id><published>2010-10-04T09:31:00.001-05:00</published><updated>2010-10-04T09:32:29.707-05:00</updated><title type='text'>Post...</title><content type='html'>Just some throwback steps straight from my old notebook.  thought they MIGHT be useful to some...&lt;br /&gt;&lt;br /&gt;First Derivative Test:&lt;br /&gt;1. Take the derivative of the original problem.&lt;br /&gt;2. Set the first derivative equal to Zero.&lt;br /&gt;3. Solve for x.&lt;br /&gt;4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)&lt;br /&gt;5. Pick a number in the intervals then plug that number in the first derivative for x.&lt;br /&gt;6. Solve.&lt;br /&gt;&lt;br /&gt;Second Derivative Test:&lt;br /&gt;1. Take the derivative of the first derivative.&lt;br /&gt;2. Set the second derivative equal to Zero.&lt;br /&gt;3. Solve for x.&lt;br /&gt;4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)&lt;br /&gt;5. Pick a number in the intervals then plug that number in the second derivative for x.&lt;br /&gt;6. Solve.&lt;br /&gt;&lt;br /&gt;limits:&lt;br /&gt;&lt;br /&gt;Rule #1 - When the degree (exponent) of the bottom is GREATER than the degree of the top, the limit is Zero.&lt;br /&gt;Rule #2 - When the degree (exponent) of the bottom is SMALLER than the degree of the top, the limit is infinity. (positive or negative)&lt;br /&gt;Rule #3 - When the degrees are equal, the limit is the coeffecients.&lt;br /&gt;&lt;br /&gt;linierazation:&lt;br /&gt;&lt;br /&gt;The steps for solving linearization problems are:&lt;br /&gt;1. Pick out the equation&lt;br /&gt;2. f(x)+f`(x)dx&lt;br /&gt;3. Figure out your dx&lt;br /&gt;4. Figure out your x&lt;br /&gt;5. Plug in everything you get&lt;br /&gt;&lt;br /&gt;implicit derivatives:&lt;br /&gt;&lt;br /&gt;First Derivative:&lt;br /&gt;1. take the derivative of both sides&lt;br /&gt;2. everytime you take the derivative of y note it with dy/dx or y^1&lt;br /&gt;3. solve for dy/dx&lt;br /&gt;&lt;br /&gt;Second Derivative:&lt;br /&gt;&lt;br /&gt;first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.&lt;br /&gt;you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.&lt;br /&gt;once you have everything plugged in and ready to go you then solve for d^2y/d^2x&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;HOW TO FIND THE EQUATION OF A TANGENT LINE:&lt;br /&gt;&lt;br /&gt;1. take f'(x)&lt;br /&gt;2. plug x in to find your slope m&lt;br /&gt;3. plug x into f(x)to get y&lt;br /&gt;4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1).&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2820531591951415690?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2820531591951415690/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2820531591951415690'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2820531591951415690'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post.html' title='Post...'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-8040514728675801627</id><published>2010-10-04T07:09:00.002-05:00</published><updated>2010-10-04T07:16:29.738-05:00</updated><title type='text'>Post #6</title><content type='html'>Rules for Limits if x---&gt;infinity&lt;br /&gt;1. degree of top equals degree of bottom--&gt;ANSWER: top coefficient over bottom coefficient&lt;br /&gt;2. degree is bigger than bottom degree--&gt;ANSWER: positive or negative infinity&lt;br /&gt;3. top degree is less than bottom degree--&gt;ANSWER: 0&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Also with limits---&gt;if indeterminate form is created&lt;br /&gt;L'Rule!  All you do is take derivative of top over derivative of bottome.  If indeterminate form is still created then simply repeat.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;NOTE: I have been getting a lot better at substitution.  With all the practice I'm starting to recognize things that need to be substituted.  Like if you have sin and cos, those are derivatives of each other so you would use substitution. &lt;br /&gt; &lt;br /&gt; &lt;br /&gt;NOTE: Let's not forget that the integrals of some things are on our chart.  Most of them are like ln of something.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;NOTE: Remember AREA means to INTEGRATE!  I saw that one popped up on the packet.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-8040514728675801627?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/8040514728675801627/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-6_04.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8040514728675801627'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8040514728675801627'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-6_04.html' title='Post #6'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-4926129988033064571</id><published>2010-10-03T22:03:00.002-05:00</published><updated>2010-10-03T22:16:58.934-05:00</updated><title type='text'>post 6</title><content type='html'>l'hospitals rule:&lt;br /&gt;if the limit is in indeterminate form, you use this.&lt;br /&gt;you take the derivative of the top and the bottom of the fraction (separately, do not use quotient rule) and then take the limit of it again.&lt;br /&gt;if you get indeterminate form, repeat as many times as necessary.&lt;br /&gt;&lt;br /&gt;this week in calc, we just went over all the ways to integrate pretty much. partial fractions, trig sub, all that good stuff. and we had a test on thursday. and now we have a take home test due this week sometime (i think either thursday/friday) if someone knows when it is due please tell me.&lt;br /&gt;&lt;br /&gt;i'm just gonna kinda review randomness today.&lt;br /&gt;sorry if it's short. i'm not really focused too well right now...&lt;br /&gt;&lt;br /&gt;partial fractions:&lt;br /&gt;whenever you simplify the bottom of the fraction as much as you can... if you have something like this n/n(n-1)&lt;br /&gt;you'd use the normal rules and just do A/n + B/n-1.&lt;br /&gt;but if you ever have something left in the bottom that is still squared, that's when the rules get tricky.&lt;br /&gt;if your exponent is on the outside of the equation...(example: x/(x-1)^2) then you would do A/(x-1) + B/(x-1)^2&lt;br /&gt;and if your exponent was 7, then you would go all the way up to 7.&lt;br /&gt;no if your exponent is on the inside of the equation....(x/(x^2-1) then you would do like this... A + Bx/(x^2-1)&lt;br /&gt;&lt;br /&gt;get it ? hopefully you do.&lt;br /&gt;now i am confused on what convergent and divergent is and how you can tell the difference.&lt;br /&gt;help?&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-4926129988033064571?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/4926129988033064571/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-6_03.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4926129988033064571'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4926129988033064571'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-6_03.html' title='post 6'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1413245127534844377</id><published>2010-10-03T20:11:00.001-05:00</published><updated>2010-10-03T20:11:31.199-05:00</updated><title type='text'>Post # 6</title><content type='html'>&lt;span xmlns=''&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;So, lets go through a few throwback/ things that are necessary to be known.  This week we prepared for the test and hopefully we all did good on it! Cross your fingers being that exams are coming up shortly!! Ahh, deep breath.  So, anyway, I'm going to go over a few things that need to be remembered because they are in all of the parts of what we've been up to..so here it goes…&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;br /&gt; &lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;SUBSTITUTION&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;Substitution takes the place of the derivative rules for problems such as product rule and quotient rule and should be used PLENTY right now…&lt;br/&gt;&lt;br /&gt;				&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;The steps to substitution are:&lt;br/&gt;1. Find a derivative inside the interval&lt;br/&gt;2. set u = the non-derivative&lt;br/&gt;3. take the derivative of u&lt;br/&gt;4. substitute back in&lt;br/&gt;&lt;br/&gt;e INTEGRATION&lt;br/&gt; - whatever is raised to the e power will be your u, and du will be the derivative of u.&lt;br/&gt;&lt;br/&gt;For example:&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;e^2x-1dxu=2x-1 du=2&lt;br/&gt;rewrite the function as:&lt;br/&gt;1/2{ e^u du, therefore&lt;br/&gt;= 1/2e^2x-1+C is your final answer.&lt;br/&gt;&lt;br/&gt;LIMITS:&lt;br /&gt;&lt;/span&gt;&lt;/p&gt;&lt;p&gt;&lt;span style='color:#204063; font-family:Helvetica'&gt;Rules for Limits:…&lt;br/&gt;1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient&lt;br/&gt;2. if top degree is bigger than bottom degree, the answer is positive or negative infinity&lt;br/&gt;3. if top degree is less than bottom degree, the answer is 0&lt;/span&gt;&lt;/p&gt;&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1413245127534844377?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1413245127534844377/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-6.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1413245127534844377'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1413245127534844377'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/10/post-6.html' title='Post # 6'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5927222965939924038</id><published>2010-09-27T13:17:00.002-05:00</published><updated>2010-09-27T13:23:57.842-05:00</updated><title type='text'>Post # 5</title><content type='html'>Okay, so i'm just going to kind of explain the things i don't understand..&lt;br /&gt;&lt;br /&gt;1.  So, let me start with the improper integrals. I dont understand how to plug in certain things to it? or what you do.&lt;br /&gt;&lt;br /&gt;2.  Will lo'hopitals rule work for all limits?&lt;br /&gt;&lt;br /&gt;3. How do you know if a problem with a squareroot should be solved by substitution, bi-parts..etc.&lt;br /&gt;&lt;br /&gt;4.  What is the best way to solve a limit.&lt;br /&gt;&lt;br /&gt;Okay, so lets get to explaining a little.&lt;br /&gt;&lt;br /&gt;When you do the thing with the chart on page A21, you should be aware to make sure all of the important things are being substituted for; as well as integrating if it still has an integral symbol in the equation!&lt;br /&gt;&lt;br /&gt;Sorry this blog isn't  very informative..but it helps me more when poeple answer my questions then saying what i know.  Hopefully someone can help!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5927222965939924038?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5927222965939924038/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-5_27.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5927222965939924038'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5927222965939924038'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-5_27.html' title='Post # 5'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1071276872595065259</id><published>2010-09-26T23:28:00.002-05:00</published><updated>2010-09-26T23:36:51.615-05:00</updated><title type='text'>Post #5</title><content type='html'>Well I didn't bring home my notebook...so lets see what I can remember.&lt;br /&gt;&lt;br /&gt;I know we talked about how if you have a definite integral with bounds that have an infinity in it. &lt;span style="font-weight:bold;"&gt; Also does this apply where you have a discontinuity or something?&lt;/span&gt; Well, anyways you replace it with an A or some other letter.  Then I think you integrate normally..&lt;span style="font-weight:bold;"&gt;and then take the limit?&lt;/span&gt;  I actually was confused on this last week.  &lt;span style="font-weight:bold;"&gt;Can anyone give me an example or explain it better?&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Well I will reexplain Lo'Hopital's Rule:&lt;br /&gt;&lt;br /&gt;This is when you have a limit.  You then plug in the  number the limit is approaching into the equation.  If you get anything like infinity/infinity, zero/zero, infinity/zero, zero/infinity you need Lo'Hopital's rule.  So, you take derivative of the top and derivative of the bottom.  Plug in the number from the limit.  If you still get anything like infinity/infinity..etc. then you repeat this process.  If not then you are done!&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;One last thing:&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;How do you know when something is to integrate with substitution?&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1071276872595065259?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1071276872595065259/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-5.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1071276872595065259'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1071276872595065259'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-5.html' title='Post #5'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5586100762208618645</id><published>2010-09-26T22:38:00.002-05:00</published><updated>2010-09-26T23:11:26.122-05:00</updated><title type='text'>Week 5 maybe?</title><content type='html'>Wellllllll, I forgot my notebook at school on friday so I guess I'll just explain some stuff by memory.&lt;br /&gt;&lt;br /&gt;Throwbacks from Calculus 1!&lt;br /&gt;&lt;br /&gt;Throwback from Calc. 1 that can help us in physics!&lt;br /&gt;Position, Velocity, Acceleration.&lt;br /&gt;Velocity is the derivative of Position.&lt;br /&gt;Acceleration is the derivative of Velocity.&lt;br /&gt;Physics explains it as:&lt;br /&gt;Velocity is how fast an x (position) is moving.&lt;br /&gt;Acceleration is how fast you are going faster.&lt;br /&gt;Calculus is basically an easier way to do this.&lt;br /&gt;&lt;br /&gt;Antiderivative = Integration.&lt;br /&gt;&lt;br /&gt;Anytime you see instantaneous rate of change, just take a derivative and plug in for x.&lt;br /&gt;&lt;br /&gt;Trig Problems:&lt;br /&gt;&lt;br /&gt;tancot = 1&lt;br /&gt;cot/csc = cos&lt;br /&gt;sin(2x) = 2sin(x)cos(x)&lt;br /&gt;sin(-x) = -sin(x)&lt;br /&gt;&lt;br /&gt;QUESTIONS, I'm looking at a MAO test right now that I have.&lt;br /&gt;&lt;br /&gt;Oblique asymptote?&lt;br /&gt;&lt;br /&gt;When you have an integral from 0 to x^2 and give you a value and say take the derivative. How you do that?&lt;br /&gt;&lt;br /&gt;How to do f^(-1)'(x).&lt;br /&gt;&lt;br /&gt;Any Volume problems.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5586100762208618645?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5586100762208618645/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/week-5-maybe.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5586100762208618645'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5586100762208618645'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/week-5-maybe.html' title='Week 5 maybe?'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2001166537551148698</id><published>2010-09-26T19:24:00.002-05:00</published><updated>2010-09-26T19:38:06.964-05:00</updated><title type='text'>Mal's Post</title><content type='html'>Whatever will I explain on this fine evening????&lt;br /&gt;&lt;br /&gt;Hmm...good question.&lt;br /&gt;&lt;br /&gt;How about some throwback?&lt;br /&gt;&lt;br /&gt;Okay, so derivatives...because I'm a little rusty on that.&lt;br /&gt;&lt;br /&gt;Okay. So, my thing is that lately I've been noticing that every time I'm trying to take the derivative of my u in py parts, I'm actually integrating.  Which isn't what your supposed to be doing.   So here's a little review.&lt;br /&gt;&lt;br /&gt;a derivative is a what?  &lt;br /&gt;&lt;br /&gt;SLOPE.  Good!  I'm glad we're off to a good start here...&lt;br /&gt;&lt;br /&gt;Okay.  How do you find a slope/derivative?&lt;br /&gt;Well, you multiply the the variable by your exponent and then subtract one from your exponent.  VOILA!!!  a derivative.  &lt;br /&gt;&lt;br /&gt;So if I'm given the position equation x(t)=2t^2+5t and I want to find the velocity at t=4, what do I do?  Well, I take the derivative of the equation first then plug in 4 to see just what the slope(velocity) is at that particular point.  So:&lt;br /&gt;&lt;br /&gt;my derivative equals 4t+5.  Now you plug in 4.&lt;br /&gt;&lt;br /&gt;16+5=21!!!!!  Congrats!!! okay.&lt;br /&gt;&lt;br /&gt;Now, if I want to find the acceleration, what do I do?  take the derivative twice.  The second derivative (hence the word twice)..&lt;br /&gt;&lt;br /&gt;so we found the first time that it was 4t+5 was the velocity.&lt;br /&gt;&lt;br /&gt;so the second derivative would equal what?  well the derivative of 5 is 0...constant derivative=0....and the derivative of 4t is just 4...so you acceleration at any point (even t=4, 5, or 6) is 4.&lt;br /&gt;&lt;br /&gt;Another quick review would be the following:&lt;br /&gt;&lt;br /&gt;If I want to get back from say acceleration to velocity. what would I do?  integrate.&lt;br /&gt;&lt;br /&gt;Now I realize that I'm being redundant and reviewing probably what's the easiest things in the big mighty calculus book, but I needed SOMETHING to talk about.  Love you guys!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2001166537551148698?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2001166537551148698/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/mals-post_26.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2001166537551148698'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2001166537551148698'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/mals-post_26.html' title='Mal&apos;s Post'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5288606783067327163</id><published>2010-09-19T22:46:00.002-05:00</published><updated>2010-09-19T23:04:29.160-05:00</updated><title type='text'>Mal's Post</title><content type='html'>So, partial fractions were the main focus this week, along with sharpening our integration skills.  Some things you need to remember when it comes to partial fractions, is that there are basically 4 cases, including synthetic division, that you can use when regular substitution fails, and by parts I think too.  So, basically, when you see a fraction...&lt;br /&gt;&lt;br /&gt;1. See if it can be solved using synthetic division (i.e. degree of top greater than degree of bottom)&lt;br /&gt;&lt;br /&gt;2. If synthetic won't work, then Factor the Bottom!&lt;br /&gt;&lt;br /&gt;Once you're dong factoring, you're gonna want to split it up with different letters over all the factors...so A B C, and so on.&lt;br /&gt;&lt;br /&gt;However.  If you have say a (x+1)^2, you have to put that particular factor over Cx+D...don't ask why, it's just the way of the gods..&lt;br /&gt;&lt;br /&gt;So once you decide what to do after factoring, you go through and plug in numbers for x (after finding common denominators) that cancel out the other letters, till you have values for all letters (variables) used. &lt;br /&gt;&lt;br /&gt;Once finding those values you go BACK, again, and plug those in in the appropriate places.  Once you get that, then you FINALLY integrate.  Yayy!!  finally!  (Most of the times you should get a lot of natural logs...just saying.   Got it?  good.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Okay...a question...what the heck is this integration table thing??  Unfortunately I was absent that day, so I was unable to get a good explanation....so if someone could do a problem or two and explain each step, it'd be much appreciated....:DD   Thanks!!!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5288606783067327163?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5288606783067327163/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/mals-post.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5288606783067327163'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5288606783067327163'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/mals-post.html' title='Mal&apos;s Post'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-420096675972881195</id><published>2010-09-19T21:41:00.002-05:00</published><updated>2010-09-19T21:50:10.833-05:00</updated><title type='text'>post 4</title><content type='html'>sorrryyyy i missed post 3. i fell asleep before i did it, haha. anyways...&lt;br /&gt;&lt;br /&gt;this week in calc we learned partial fractions and went over more trig sub! yayyyy. and we learned about integration tables on friday. which is pretty cool that all you have to do is plug it into a formula.&lt;br /&gt;&lt;br /&gt;partial fractions is used when you pretty much have no other choice. and it has to be fraction. and your top degree is smaller then your bottom degree.&lt;br /&gt;becuase when your top degree is larger than your bottom degree, you would use synthetic division, which may i say makes life so easy. haha&lt;br /&gt;&lt;br /&gt;anyways, here's how you do partial fractions...&lt;br /&gt;1. you factor the bottom.&lt;br /&gt;2. you do a/first term, b/second term, c/third term..etc &amp;amp; set that equal to your equation****&lt;br /&gt;3. you then do common denominator and get rid of all the fractions.&lt;br /&gt;4. you plug in CONVENIENT x values.&lt;br /&gt;5. solve for a, b, c, d and what not.&lt;br /&gt;6. then you go back to step 2 and plug in all the numbers you got.&lt;br /&gt;7. integrate &amp;amp; solve&lt;br /&gt;&lt;br /&gt;****now there is a tricky part for step two. if you have something squared when you factor. like x^5/(x+1)^2&lt;br /&gt;&lt;br /&gt;you would do this&lt;br /&gt;a/(x+1) + b/(x+1)^2&lt;br /&gt;... and if it was cubed, or to the fourth or whatever, you would go all the way up to that degree.&lt;br /&gt;&lt;br /&gt;ALSO, if you have something inside of it squared after factoring.. like this&lt;br /&gt;x^5/x(x^2+1)&lt;br /&gt;then you would put the term that has a squared in it like this...&lt;br /&gt;a/x + (bx+c)/(x^2+1)&lt;br /&gt;and you would do that for whatever letter you are at... for example if you already had an a &amp;amp; b, you would do cx + d/ whatever.&lt;br /&gt;&lt;br /&gt;get it? k good :)&lt;br /&gt;can someone go over synthetic division again. for some reason i keep forgetting it.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-420096675972881195?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/420096675972881195/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-4_8077.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/420096675972881195'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/420096675972881195'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-4_8077.html' title='post 4'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2823123479572557480</id><published>2010-09-19T20:42:00.002-05:00</published><updated>2010-09-19T23:13:40.218-05:00</updated><title type='text'>Week Number Four</title><content type='html'>This week we reviewed trig. substitution integration and we also learned how to solve integration using partial fractions.&lt;br /&gt;&lt;br /&gt;You can actually use partial fractions to break up fractions even if you aren't using it for integration.&lt;br /&gt;&lt;br /&gt;Some simple steps for partial fraction integration:&lt;br /&gt;1) First of all, make sure it isn't any other type of integration.&lt;br /&gt;2) If you can factor the top or bottom then do so (if you can cancel anything then do so).&lt;br /&gt;3) Once you've ruled out everything else, you then split the bottom factors into A/(factor1) + B/(factor2) +... = original.&lt;br /&gt;4) Get common denominators and add.&lt;br /&gt;5) Pick a convenient value for x (one that would give you zero once plugged into a factor) and plug in.&lt;br /&gt;6) Solve for A, B, ... .&lt;br /&gt;7) You then take the values of your variables and plug back in to when you first broke up the fraction.&lt;br /&gt;8)Integrate! It will almost always be natural log integration.  Remember that any number in front of a natural log is also it's exponent.&lt;br /&gt;&lt;br /&gt;Question:&lt;br /&gt;For integration using charts like we did in class on friday, how exactly do I know which formula to plug in to if the problem I'm facing doesn't have all of the necessary components. I.e. if the formula has an x and my problem doesn't.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2823123479572557480?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2823123479572557480/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/week-number-four.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2823123479572557480'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2823123479572557480'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/week-number-four.html' title='Week Number Four'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-3269376474280311934</id><published>2010-09-19T10:38:00.002-05:00</published><updated>2010-09-19T10:53:44.498-05:00</updated><title type='text'>Post #4</title><content type='html'>This week in Calculus, we learned partial fractions and how to use a table to integrate easier.  I think partial fractions are pretty easy, but that's being said with fingers crossed! So, let me try to explain what i don't understand.  &lt;strong&gt;How do you know what part of the equation to use to find the right equation in the table?&lt;/strong&gt;&lt;br /&gt;&lt;br /&gt;So, let me explain partial fractions.&lt;br /&gt;&lt;br /&gt;1. How do you know its a partial fraction?&lt;br /&gt;    Well duhhh sillies, its going to have a fraction with some quadratics&lt;br /&gt;&lt;br /&gt;2.  So, then what do you do?  You need to break the fraction up.&lt;br /&gt;&lt;br /&gt;3.  You break up the fraction, by factoring the bottom and rewriting it as multiple new fractions.&lt;br /&gt;&lt;br /&gt;4.  Take the separate denominators with a numerator of A, B, C, or D.&lt;br /&gt;&lt;br /&gt;5. Next, you create a common denominator and set that equal to the numerator of the initial fraction.&lt;br /&gt;&lt;br /&gt;6. You then pick convient values to solve for the variables of the equation.&lt;br /&gt;&lt;br /&gt;7.  After you find the values, you plug everything back in to the fractions you created&lt;br /&gt;&lt;br /&gt;8. Integrate&lt;br /&gt;&lt;br /&gt;9. And hopefully box off the correct answerrrr!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-3269376474280311934?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/3269376474280311934/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-4_19.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3269376474280311934'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3269376474280311934'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-4_19.html' title='Post #4'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-3182534587547590870</id><published>2010-09-18T15:09:00.002-05:00</published><updated>2010-09-18T15:24:08.370-05:00</updated><title type='text'>Post #4</title><content type='html'>Well this week we talked about partial fractions and B integration tables.  So I guess that is what I will be covering in this blog.&lt;br /&gt;&lt;br /&gt;Partial Fractions:&lt;br /&gt;problem: S 1/x^2+5x+6&lt;br /&gt;&lt;br /&gt;first: factor the bottom&lt;br /&gt;1/(x+3)(x+2)&lt;br /&gt;&lt;br /&gt;second: set up the fractions, start off with a over the first, set equal to problem&lt;br /&gt;A/x+3 + B/x+2 = 1/x^2+5x+6&lt;br /&gt;&lt;br /&gt;third: multiply to get denominates equal, which gives you&lt;br /&gt;A(x+2) + B(x+3) = 1&lt;br /&gt;&lt;br /&gt;fourth: find convenient values for x to solve for A and B&lt;br /&gt;A(x+2) + B(x+3)&lt;br /&gt;&lt;br /&gt;*choose -2 to plug into A b/c it will give you zero&lt;br /&gt;A(-2+2) + B(-2+3)=1&lt;br /&gt;A(0) + B(1) =1&lt;br /&gt;B=1&lt;br /&gt;*choose -3 for B /c it will give you zero&lt;br /&gt;A(-3+2) + B(-3+ -3)=1&lt;br /&gt;A(-1) + B(0)=1&lt;br /&gt;A=-1&lt;br /&gt;&lt;br /&gt;fifth: plug A and B back into the fractions A/x+3 + B/x+2&lt;br /&gt;S -1/x+3 + 1/x+2&lt;br /&gt;&lt;br /&gt;*1/x gives you a ln&lt;br /&gt;=-ln(x+3) + ln(x+2) +C&lt;br /&gt;*simplify&lt;br /&gt;ln(x+2/x+3) +C&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Note: if you were given S fx^2+20x+6/x(x+1)^2, your fractions would be, it is a rule&lt;br /&gt;A/x + B/x+1 + C/(x+1)^2&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;B integration tables:&lt;br /&gt;This is from A21 in the book.  It has a listing of all kinds of integrals and what they equal.  You basically have to figure out which integral applies to the problem.  If you need a u, a, or something you have to figure that out from the problem.  All there is left is to plug in and simplify.  The only tricky part is make sure you pick the right integral to use. &lt;br /&gt;&lt;br /&gt;My Question:&lt;br /&gt;Can anyone tell me which integral I would use for these two:&lt;br /&gt;S 1/squareroot x(1-cos squareroot x)&lt;br /&gt;or&lt;br /&gt;S x^7lnx&lt;br /&gt;*for this one there is a u^n formula nd a 1/u formula? but what do I use?&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-3182534587547590870?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/3182534587547590870/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-4.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3182534587547590870'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3182534587547590870'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-4.html' title='Post #4'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6478440500535834002</id><published>2010-09-12T21:41:00.002-05:00</published><updated>2010-09-12T22:36:39.736-05:00</updated><title type='text'>week THREE</title><content type='html'>This week was ALL about Trig. Sub.&lt;br /&gt;&lt;br /&gt;Trig. Sub. is used anytime there is a square root in the form of sqrt(a^2 - x^2), sqrt (x^2 - a^2), or sqrt(x^2 + a^2) in an integration problem.&lt;br /&gt;&lt;br /&gt;If you see sqrt(a^2 - x^2) then x=asin(theta) and sqrt(a^2 - x^2)=acos(theta).&lt;br /&gt;&lt;br /&gt;If you see sqrt (x^2 - a^2) then x=asec(theta) and sqrt (x^2 - a^2)= atan(theta).&lt;br /&gt;&lt;br /&gt;If you see sqrt(x^2 + a^2) then x=atan(theta) and sqrt(x^2 + a^2)=asec(theta).&lt;br /&gt;&lt;br /&gt;After figuring out which one you have to use, you then find dx by taking the derivative of x and if you have an (x) variable in the problem, then solve for that using the x equation.&lt;br /&gt;&lt;br /&gt;After integrating, you have to switch the theta versions of the problem back to x.&lt;br /&gt;&lt;br /&gt;Some issues I'm having with this is when you actually have to integrate.&lt;br /&gt;So Ryan, here's some integration formulas to remember:&lt;br /&gt;Ssec = lnsec + tan&lt;br /&gt;Ssec^2x = tan&lt;br /&gt;Ssec^3x = (1/2)sectan + (1/2)lnsec + tan&lt;br /&gt;Scsc*cot = -csc&lt;br /&gt;Stan = -lncos&lt;br /&gt;Scot = lnsin&lt;br /&gt;&lt;br /&gt;And some trigonometry formulas to remember:&lt;br /&gt;sin^2x + cos^2x = 1&lt;br /&gt;1 + tan^2x = sec^2x&lt;br /&gt;1 + cot^2x = csc^2x&lt;br /&gt;sin2x = 2sinxcos&lt;br /&gt;&lt;br /&gt;Ryan&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6478440500535834002?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6478440500535834002/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/week-three.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6478440500535834002'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6478440500535834002'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/week-three.html' title='week THREE'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-347845022760963099</id><published>2010-09-12T21:37:00.003-05:00</published><updated>2010-09-12T22:01:25.374-05:00</updated><title type='text'>Post 3</title><content type='html'>Okay.  So I would like to state a couple of things that Abbey and I learned while doing that partner thing:&lt;br /&gt;&lt;br /&gt;1. When, after doing trig sub, you get either SIN^2 or COS^2, that would be when you use the following Reduction Formulas:&lt;br /&gt;&lt;br /&gt;cos^2 = (1+cos2x)/2&lt;br /&gt;sin^2 = (1-cos2x)/2&lt;br /&gt;&lt;br /&gt;The way I remember that is, sin is negative because it's negative that its not sine. and cosine is cosine, so it's positive.&lt;br /&gt;&lt;br /&gt;2. So instead of doing by parts like the textbook says to do when you get S of sec to an odd power, you should just memorize s=S sec^3(x) because it shows up quite often.  I think it's something like 1/3secxtanx + 1/3ln(secx + tanx) + C&lt;br /&gt;&lt;br /&gt;3. When doing trig sub, you have to be able to see basic trig concepts, so to speak.  Say I have S 1/sec^2(x).  One of the problems Abbey had was realizing that that was the same as cos^2(x).  That's just something you have to watch out for.&lt;br /&gt;&lt;br /&gt;4. Also, I'm pretty sure everyone could brush up on some things...ie trig formulas.&lt;br /&gt;&lt;br /&gt;5. Memorize your triangles...I'm thinking.&lt;br /&gt;&lt;br /&gt;6. In normal trig integration.  You have to realize that sometimes...say you have tan and sec in an integral...if you can some how get a du (ie sec^2 or sectan) out of it, thats the way you want to go because by doing so, you just have to do normal substitution integration again.  Got it?  good.&lt;br /&gt;&lt;br /&gt;Okay, for stuff you can comment on...umm...&lt;br /&gt;&lt;br /&gt;Could someone explain the process of doing definite integrals for me?  I'm kinda sketchy on the part where you plug the original values in?? it's all kind of a blur.  Anyways..thanks!!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-347845022760963099?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/347845022760963099/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-3_2937.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/347845022760963099'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/347845022760963099'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-3_2937.html' title='Post 3'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-4261832689674905231</id><published>2010-09-12T19:23:00.002-05:00</published><updated>2010-09-12T19:54:55.331-05:00</updated><title type='text'>Post #3</title><content type='html'>Alrighty here we go...&lt;br /&gt;&lt;br /&gt;First, I would like to list some identities that I need to learn:&lt;br /&gt;&lt;br /&gt;sinx= 1/2 - 1/2cos2x&lt;br /&gt;cosx= 1/2 + 1/2cos2x&lt;br /&gt;&lt;br /&gt;cos^2x + sin^2x = 1&lt;br /&gt;1 + tan^2x = sec^2x&lt;br /&gt;1 + cot^2x = csc^2x&lt;br /&gt;&lt;br /&gt;tanx = sinx/cosx&lt;br /&gt;cotx = cosx/sinx&lt;br /&gt;&lt;br /&gt;sin(2x) = 2sinxcosx&lt;br /&gt;&lt;br /&gt;Well, we have been doing trig sub like all week.  So let me try to explain it the best way I can since I haven't gotten the hang of it just yet.&lt;br /&gt;&lt;br /&gt;*Say you have the square root of x^2/squarerootof 25-x^2 dx&lt;br /&gt;&lt;br /&gt;1. You see what box you need to use (which I need to memorize).  There are three different cases: (a is the number, u is the x)&lt;br /&gt;square root of a^2 - u^2 ---&gt;asin(t)&lt;br /&gt;square root of a^2 + u^2 ---&gt;atan(t)&lt;br /&gt;square root of u^2 - a^2 ---&gt;asec(t)&lt;br /&gt;This example would follow a^2 - u^2&lt;br /&gt;&lt;br /&gt;2. Then find your x and dx:&lt;br /&gt;x is from the three different cases so x= 5sin(t)&lt;br /&gt;dx is the derivative so dx= 5cos(t)&lt;br /&gt;&lt;br /&gt;3. Next you have the square root:&lt;br /&gt;squarerootof 25-x^2 ---&gt;5cos(t) &lt;br /&gt;*those are from your chart thing too, but Mal Pal said that it is usually like the opposite of the x&lt;br /&gt;&lt;br /&gt;4. and don't forget to account for the x^2:&lt;br /&gt;so take your x, x=5sin(t) and square it which gives you x^2=25sin^2(t)&lt;br /&gt;&lt;br /&gt;5. Now plug everything in!&lt;br /&gt;so you should get&lt;br /&gt;25sin^2(t)5cos(t)/5cos(t)&lt;br /&gt;*simplify: the 5cost cancel leaving 25sin^2(t) &lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight:bold;"&gt;Now I'm pretty sure you use the power reduction formulas here right?&lt;br /&gt;But this is where I get stuck..how do I use 1/2 - 1/2cos2x for 25sin^2(t)?&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;6. I know you integrate after that.&lt;br /&gt;&lt;br /&gt;7. form the triangle and use that to plug in to the trig functions&lt;br /&gt;*don't forget SOHCAHTOA&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-4261832689674905231?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/4261832689674905231/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-3_12.html#comment-form' title='4 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4261832689674905231'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4261832689674905231'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-3_12.html' title='Post #3'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>4</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5002141337395445125</id><published>2010-09-12T09:33:00.002-05:00</published><updated>2010-09-12T10:08:49.809-05:00</updated><title type='text'>Post # 3</title><content type='html'>Hello Blogmates.&lt;br /&gt;&lt;br /&gt;This week in Calculus we learned trig sub and then we reviewed integration.  Integration is something i really need help with becasue&lt;strong&gt; I can't seem to understand when to integrate regularly, bi part, substitute, or trig sub.  Hints?&lt;/strong&gt;&lt;br /&gt;&lt;strong&gt;&lt;/strong&gt;&lt;br /&gt;So, in order to help myself, i'll list a few formulas we should not forget.&lt;br /&gt;&lt;br /&gt;S sinx = -cos x + c&lt;br /&gt;S -sinx = cos x + c&lt;br /&gt;S cosx = sin x + c&lt;br /&gt;S tanx = ln /cosx/ + c&lt;br /&gt;S secx = ln /secx + tanx/ + c&lt;br /&gt;S cscx = - ln / secx + cotx/ + c&lt;br /&gt;S cotx = ln/sinx/ + c&lt;br /&gt;&lt;br /&gt;Next, it is very important that you learn these POWER REDUCTION FORMULAS:&lt;br /&gt;&lt;br /&gt;cosx = 1/2 + 1/2cos2x&lt;br /&gt;sinx = 1/2 - 1/2cox2x&lt;br /&gt;&lt;br /&gt;And this PYTHAGOREAN IDENTITY:&lt;br /&gt;&lt;br /&gt;cos^2x + sin^2x = 1&lt;br /&gt;&lt;br /&gt;Now, i think the things i need help with the most is with choosing which integration method to do..hopefully that atleast comes with time?&lt;br /&gt;&lt;br /&gt;The last thing i want to explain is a nice summary of trig sub.&lt;br /&gt;&lt;br /&gt;So, &lt;strong&gt;first&lt;/strong&gt;, you choose what box your problem is and find the information necessary; including: x, dx, sqrt, and whatever else is in the problem.&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Second&lt;/strong&gt;, you plug all of that back into the problem and hopefully cancel some things.&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Third&lt;/strong&gt;, integrate.&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Fourth&lt;/strong&gt;, form the triangle by the information given in the selected box.&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Fifth&lt;/strong&gt;, find the trig functions in the problem's answer by using SOHCAHTOA and the triangle.&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Lastly&lt;/strong&gt;, pray you got it right.&lt;br /&gt;&lt;br /&gt;Now, a quote from my favorite youtube video, glozell,&lt;br /&gt;&lt;br /&gt;PEACE AND BLESSINGS. PEACE AND BLESSINGS.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5002141337395445125?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5002141337395445125/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-3.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5002141337395445125'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5002141337395445125'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-3.html' title='Post # 3'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2406250635671194973</id><published>2010-09-06T20:57:00.002-05:00</published><updated>2010-09-06T21:37:22.027-05:00</updated><title type='text'>Post #2</title><content type='html'>Kay. Since I've yet to touch on L'Hopital's Rule, I shall do so now.&lt;br /&gt;&lt;br /&gt;The most important thing to learn about l'Hôpital's rule is when it should not be used:&lt;br /&gt;&lt;br /&gt;Definitely do NOT use it when the limits of the two parts are not both 0, or both infinity. In this case the rule is likely to give a wrong answer! &lt;br /&gt;&lt;br /&gt;Example:&lt;br /&gt;&lt;br /&gt;limx-&gt;0+ (cos x)/x&lt;br /&gt;&lt;br /&gt;is positive infinity, because the numerator approaches 1 while the denominator approaches 0. If we incorrectly apply l'Hôpital's rule, we get&lt;br /&gt;&lt;br /&gt;limx-&gt;0+ (- sin x)/1 = 0.&lt;br /&gt;&lt;br /&gt;So you DO use L'Hopital's Rule when you get an indeterminate in the first place...this is inf/inf, 0/0, etc.&lt;br /&gt;&lt;br /&gt;Okay, for Trig SUB!!!!!  I'm getting pretty good at this, so bear with me....&lt;br /&gt;&lt;br /&gt;My trick is:  Everytime I see a trig function to an odd power, I take out an even...After this I use an appropriate identity.  It's really not all that hard...I have my notecards somewhere...just ask me for them..&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;OkAY!!!! for things you can comment on....&lt;br /&gt;&lt;br /&gt;Does anyone know how to:&lt;br /&gt;&lt;br /&gt;1. Divide stuff?  like x^2 + x+ 7 all over x-8. the other day BRob tried to do a problem like that, and I failedddd miserably.  Easier way??&lt;br /&gt;&lt;br /&gt;2. Chasing the Rabbit.  One time I ended up with chasing the rabbit, but the answer was something super easy.  any hints as to when you should use by parts i.e. chasing the rabbit?&lt;br /&gt;&lt;br /&gt;alright. night.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2406250635671194973?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2406250635671194973/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-2_06.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2406250635671194973'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2406250635671194973'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-2_06.html' title='Post #2'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-3641617098604620990</id><published>2010-09-06T11:19:00.003-05:00</published><updated>2010-09-06T11:28:29.612-05:00</updated><title type='text'>blog 2</title><content type='html'>alright alright, so this week we had our first real test. it was kindaaaa hard, but i think i did well :) we are still working on integration .. by parts, trig sub, wallis formula, all that good stuff.&lt;br /&gt;&lt;br /&gt;so, first i'll tell you what wallice's formula is (by the way idk how to spell it so ignore that)...&lt;br /&gt;if you have an integration problem of sin or cos raised to a power.. this is when you use this&lt;br /&gt;S cos^5(x)&lt;br /&gt;alright, so if your degree is ODD, you do (2/3)(4/5)..(n-1/n) and simply multiply them together. so your answer would be 8/15.&lt;br /&gt;S sin^8(x)&lt;br /&gt;if your degree is EVEN, you do (1/2)(3/4)...(n-1/n) (pi/2). then multiply&lt;br /&gt;so your answer would be (1/2)(3/4)(5/6)(7/8)(pi/2). i don't feel like multiplying it out haha.&lt;br /&gt;&lt;br /&gt;HELP:&lt;br /&gt;alright, trig sub. it's pretty much a bunch of formulas telling you what to substitute in and when to do it when you are integrating trig functions.&lt;br /&gt;i know how to do these.. i just tend to mess up cuz i don't memorize when i have to do what. &amp;amp; i also didn't bring my book home to remember to put anything about it on here.. :x so.. could someone maybe go over a few formulas for me?&lt;br /&gt;&lt;br /&gt;when to do synthetic division:&lt;br /&gt;when your top function degree is larger than the top.&lt;br /&gt;&lt;br /&gt;so say you had S (x^2 + 2x +5)/(x-6)&lt;br /&gt;6 would go in your box, then 1, 2, 5...&lt;br /&gt;i think. if i'm wrong someone please let me know! also, i get kinda lost after that.. i stop and don't remember what to do next, the only thing i remember is that i have to put my remainder over the bottom of the fraction at the end... help please&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-3641617098604620990?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/3641617098604620990/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/blog-2.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3641617098604620990'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3641617098604620990'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/blog-2.html' title='blog 2'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-3093483444240841986</id><published>2010-09-05T22:14:00.003-05:00</published><updated>2010-09-06T19:23:45.782-05:00</updated><title type='text'>Post #2</title><content type='html'>Hello my Calculus BC friends,&lt;br /&gt;&lt;br /&gt;TRIG SUBSTITUTION!&lt;br /&gt;&lt;br /&gt;Some basic integrals:&lt;br /&gt;S sinu du = -cos u + C&lt;br /&gt;S cosu du = sin u + C&lt;br /&gt;S tan u du = -ln|cos u| + C&lt;br /&gt;S cot u du = ln|sin u| + C&lt;br /&gt;S secu du = ln|sec u + tan u| + C&lt;br /&gt;S cscu du = -ln|csc u + cot u| + C&lt;br /&gt;S sec^2 u du = tan u + C&lt;br /&gt;S csc^2 u du = -cot u + C&lt;br /&gt;&lt;br /&gt;Some identities:&lt;br /&gt;sin^2x + cos^2x = 1 .&lt;br /&gt;sin^2x = (1 - cos 2x)/2 &lt;br /&gt;cos^2x = (1 + cos 2x)/2 &lt;br /&gt;&lt;br /&gt;*What I try to do:  usually try to take out some kind of squared, then change the to an identity, distribute in, and substitute.  &lt;br /&gt;&lt;br /&gt;*ALL the Rules:&lt;br /&gt;    SIN &amp; COS guidelines:&lt;br /&gt;    1. If the power of the sine is odd and positive, save one sine&lt;br /&gt;    factor and convert the remaining factors to cosines. Then, expand&lt;br /&gt;    and integrate.&lt;br /&gt;    2. If the power of the cosine is odd and positive, save one cosine&lt;br /&gt;    factor and convert the remaining factors to sines. Then, expand&lt;br /&gt;    and integrate.&lt;br /&gt;    3. If the powers of both sine and cosine are even and&lt;br /&gt;    non negative, make repeated use of the half-angle identities for&lt;br /&gt;    sin^2x and cos^2x to convert the integrand to odd powers of the&lt;br /&gt;    cosine. Then proceed as in guideline 2.&lt;br /&gt;&lt;br /&gt;    SEC &amp; TAN guidelines:&lt;br /&gt;    1. If the power of the secant is even and positive, save a secantsquared&lt;br /&gt;    factor and convert the remaining factors to tangents.&lt;br /&gt;    Then expand and integrate.&lt;br /&gt;    2. If the power of the tangent is odd and positive, save a secanttangent&lt;br /&gt;    factor and convert the remaining factors to secants. Then&lt;br /&gt;    expand and integrate.&lt;br /&gt;    3. If there are no secant factors and the power of the tangent is&lt;br /&gt;    even and positive, convert a tangent-squared factor to a secantsquared&lt;br /&gt;    factor, then expand and repeat if necessary.&lt;br /&gt;    4. If the integral is of the form S secmx dx, where m is odd and&lt;br /&gt;    positive, use integration by parts.&lt;br /&gt;    5. If none of the first four guidelines applies, try converting to&lt;br /&gt;&lt;br /&gt;Wallis formula:&lt;br /&gt;Only works with sin and cos when going from 0 to pi/2.  n is the exponent  &lt;br /&gt;when n is ODD: (2/3)(4/5)(6/7)...(n -1)/n&lt;br /&gt;EVEN: (1/2)(3/4)(5/6)...((n-1)/n)(pi/2)&lt;br /&gt;&lt;br /&gt;HERE IS WHAT YOU CAN COMMENT ON:&lt;br /&gt;Now I understand everything, but I somehow cannot always work the problems.  Does anyone have some kind of trick on how to know when you look at a problem and know you have to either substitute, by part it, or trig sub?  Also, do you know something that can help me remember how to do trig sub? (like the steps explained easier or a trick to remember or the steps you follow EVERY time?)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-3093483444240841986?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/3093483444240841986/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-2_05.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3093483444240841986'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3093483444240841986'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-2_05.html' title='Post #2'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6581610426457921717</id><published>2010-09-05T21:44:00.002-05:00</published><updated>2010-09-05T21:53:17.616-05:00</updated><title type='text'>Post #2</title><content type='html'>Okay, so after what felt like the longest week ever, it's now time to do the blog.  This week in Calculus BC I was kinda discouraged by trig sub because its something i really don't understand.  I only do the problems right when i have the formulas in front of me...and after studying them for a week straight, and still getting mixed up on them, i'm finding it almost hopeless.&lt;br /&gt;&lt;br /&gt;Hopefully someone can show me their study techniques?&lt;br /&gt;&lt;br /&gt;But lets go over a few things...&lt;br /&gt;&lt;br /&gt;For trig sub, something i always get wrong is WHEN to actually do the method..so, i believe it is when you can't basically bi-part something? correct?&lt;br /&gt;&lt;br /&gt;Also, you should never bi-part or trig sub things when you only have the trig function and its derivative/ integral..just saying.   I do it all the time and it is definately the hard way.&lt;br /&gt;&lt;br /&gt;I really wish there is something i can explain that i know how to do, but there really isn't..&lt;br /&gt;&lt;br /&gt;I guess i'll explain Wallis Formula.&lt;br /&gt;&lt;br /&gt;So, you do this when you have sin or cos and the degree is EVEN:&lt;br /&gt;&lt;br /&gt;1. Start with 1/2 and multiply the chronological numbers until you get to the exponent. &lt;br /&gt;2. Then multiply by pi/two&lt;br /&gt;3. Add +c&lt;br /&gt;4. Box or circle your answer&lt;br /&gt;&lt;br /&gt;When the degree is ODD:&lt;br /&gt;&lt;br /&gt;1.  Start with 1/2 and multiply the chronological numbers unitl you get to the exponent.&lt;br /&gt;2. Put a + c&lt;br /&gt;3. Box or circle your answer&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;So, i really feel like a baby and hopefully someone can help me..&lt;br /&gt;i really just need all the helpful hints and basic problems explained to me.&lt;br /&gt;i'm not quite sure why my brain hasn't kicked into school mode yet..&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6581610426457921717?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6581610426457921717/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-2.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6581610426457921717'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6581610426457921717'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/09/post-2.html' title='Post #2'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2676929528053371013</id><published>2010-08-29T22:03:00.002-05:00</published><updated>2010-08-29T22:45:54.122-05:00</updated><title type='text'>Ryan's First Calculus BC Post!</title><content type='html'>This week in Calculus BC was pretty good. &lt;br /&gt;Tuesday we re-learned Integration by-parts. &lt;br /&gt;&lt;br /&gt;Integration by-parts:&lt;br /&gt;Sudv = uv - Svdu&lt;br /&gt;&lt;br /&gt;Example:&lt;br /&gt;Sxe^(x)dx&lt;br /&gt;Pick a u and dv and derive/integrate:&lt;br /&gt;u = x  du = 1 dx&lt;br /&gt;dv = e^x(dx)  v = e^x&lt;br /&gt;You then plug these into your equation:&lt;br /&gt;xe^(x) - Se^(x)dx&lt;br /&gt;And solve:&lt;br /&gt;xe^(x) - e^(x) + C&lt;br /&gt;&lt;br /&gt;We also reviewed "Chasing the Rabbit"&lt;br /&gt;This method of integration happens when you have an&lt;em&gt; e&lt;/em&gt; term and a trig term in an integration problem and you basically keep integrating by-parts until you arrive at an integration term the same as you started off with, you them set everything you have equal to the original problem.&lt;br /&gt;&lt;br /&gt;We learned something brand new this week that I'm having a little trouble with: Trig Integration.&lt;br /&gt;I get the basic problems, but I was having issues with the homework over the weekend.&lt;br /&gt;Two examples would be:&lt;br /&gt;Ssec^3(pix)dx&lt;br /&gt;and&lt;br /&gt;Stan^5(x/2)dx&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2676929528053371013?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2676929528053371013/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/ryans-first-calculus-bc-post.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2676929528053371013'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2676929528053371013'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/ryans-first-calculus-bc-post.html' title='Ryan&apos;s First Calculus BC Post!'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-4576628273425456764</id><published>2010-08-29T19:50:00.002-05:00</published><updated>2010-08-29T20:01:21.914-05:00</updated><title type='text'>Steph's First Calc BC Post</title><content type='html'>So, lets get this thing going again.  First i'll give a little summary of things I did this past week.  We went over lo'hospital's rule, integration, substitution, bi-parts, trig integration, and i think there's one more thing that just isn't clicking this moment.&lt;br /&gt;&lt;br /&gt;So since this week was basically a review, let me touch on everything we did.  Kinda like a re-review.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;So, lets do some reviewing...&lt;br /&gt;&lt;br /&gt;LO'HOSPITAL'S RULE:&lt;br /&gt;&lt;br /&gt;1.  This deals only with limits..so if you don't have a limit, dont do this!&lt;br /&gt;2.  Plug in the number the limit is approaching into the equation and make sure you get an infinitave..such as infinity/infinity, zero/zero, infinity/zero, zero/infinity.&lt;br /&gt;3. Next, you need to take the derivative of the top, and the derivitave of the bottom.&lt;br /&gt;&lt;br /&gt;******NOT QUOTIENT RULE, JUST TWO DERIVATIVES.&lt;br /&gt;&lt;br /&gt;4. Plug in the number the limit is approaching&lt;br /&gt;5. If you get another infinitive, repeat.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt; INTEGRATION:&lt;br /&gt;&lt;br /&gt;*So, i know this is a little review, but don't laugh...you mess up on these sometimes too!&lt;br /&gt;&lt;br /&gt;1.  You know it's an integral when you see the "s" looking thinggyy.&lt;br /&gt;2.  Add one to the exponent&lt;br /&gt;3.  Multiply the coefficient by the recriprocal of the new exponent.&lt;br /&gt;&lt;br /&gt;If indefinite, be sure to include +c, if not, solve.&lt;br /&gt;&lt;br /&gt;So, these are the two topics that i was most confortable with my quizzes. &lt;br /&gt;&lt;br /&gt;Now, the things i'm pretty sure i didn't do good on, bi-parts and trig substitution..&lt;br /&gt;These two topics just don't click in my head.  Any suggestions on how i should study thesee things?&lt;br /&gt;&lt;br /&gt;Also, how do you tell if its biparts or trig sub..?&lt;br /&gt;&lt;br /&gt;What would you do for   S xarctanx?&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-4576628273425456764?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/4576628273425456764/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/stephs-first-calc-bc-post.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4576628273425456764'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4576628273425456764'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/stephs-first-calc-bc-post.html' title='Steph&apos;s First Calc BC Post'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5610722802002092617</id><published>2010-08-29T19:16:00.002-05:00</published><updated>2010-08-29T19:35:10.751-05:00</updated><title type='text'>first calc BC post..</title><content type='html'>alright, so back to another year of blogs... YAY :D&lt;br /&gt;haha, anyways. this week we took 3 tests. oh my lord, yes i know.&lt;br /&gt;one on l'hopital's rule, one on basic integration, and one on integration by parts. i did great on the first one, hopefully i did good on the other two, too. :)&lt;br /&gt;&lt;br /&gt;ok, so let's go over some integration by parts.&lt;br /&gt;first you need to find your u and your dv.&lt;br /&gt;usually your u is whatever can be reduced.&lt;br /&gt;then after that, you find your du and your v.&lt;br /&gt;*remember sometimes your dv can be your dx and your v would then become x.&lt;br /&gt; then you simply plug into the formula and integrate.&lt;br /&gt;&lt;br /&gt;oh .. &amp;amp; the formula is&lt;br /&gt;uv - S vdu&lt;br /&gt;*p.s. - S is the integration symbol&lt;br /&gt;&lt;br /&gt;EXAMPLE:&lt;br /&gt;S xsin2x dx&lt;br /&gt;u = x           v = -1/2cos2x&lt;br /&gt;du = 1 dx    dv=sin2x&lt;br /&gt;&lt;br /&gt;x-1/2cos2x - S -1/2cos2x&lt;br /&gt;=&gt; your answer would be...&lt;br /&gt;-(x)1/2cos(2x) + 1/4sin(2x) + c&lt;br /&gt;&lt;br /&gt;easy right? yeah. just wait til you get to the hard ones. lol OH &amp;amp; there is chasing the rabbit. this happens whenever you integrate something using by parts twice, and you end up with the same thing you started with in the original problem. then you simply set it equal to the original and solve it like that.&lt;br /&gt;&lt;br /&gt;Another aspect of calculus bc that we reviewed this week, which i'll review briefly, is l'hopitals rule.&lt;br /&gt;this is whenever you try to find a limit of something, but it's in indeterminate form... then you have to use l'hopitals rule.&lt;br /&gt;which means you take the derivative of the top of the fraction and the derivative of the bottom of the fraction... SEPARATELY! no quotient rule. then plug in &amp;amp; solve.&lt;br /&gt;but, you must make sure that it is in fraction form.. because you cannot use l'hopitals rule if it is not.&lt;br /&gt;&lt;br /&gt;alright, well that's all for now.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5610722802002092617?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5610722802002092617/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/first-calc-bc-post.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5610722802002092617'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5610722802002092617'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/first-calc-bc-post.html' title='first calc BC post..'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2170825231523001004</id><published>2010-08-29T19:14:00.002-05:00</published><updated>2010-08-29T19:35:17.340-05:00</updated><title type='text'>Abbey's first BC blog!</title><content type='html'>Well first 2 week of school done!&lt;br /&gt;&lt;br /&gt;One thing I'm comfortable with is L'Hopital's Rule.  It is used when an indeterminate form occurs with a limit.  It could be in any form of: infinity-infinity, 0/0, infinity/infinity, 0(infinity), 0^0, 1^infinity, and infinity^0.  &lt;br /&gt;&lt;br /&gt;When you get an indeterminate form, you take the derivative top then derivative of bottom.  If you still get an indeterminate form just repeat the second step.  &lt;br /&gt;&lt;br /&gt;EXAMPLES:&lt;br /&gt;lim    e^2x - 1/x = e^2(0) - 1/1 = 1-1/0 = o/o  &lt;---indeterminate form&lt;br /&gt;x--&gt;0   &lt;br /&gt;&lt;br /&gt;So, take derivative of top then bottom.&lt;br /&gt;e^2x - 1/x = 2e^2x/1 = 2(1)/1 = 2&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;lim             lnx/x = infinity/infinity  &lt;---indeterminate form&lt;br /&gt;x--&gt;infinity &lt;br /&gt;&lt;br /&gt;1/x/1 = 1/x = 0&lt;br /&gt;With this example you would use your limit rules because it is as x approaches infinity.  Since the degree of bottom is larger than the degree of top it equal zero.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Another thing we have covered is substitution.  I can usually pick out my u and du, but sometimes I have trouble completing the problem.  I guess I just need more practice.&lt;br /&gt;&lt;br /&gt;EXAMPLE:&lt;br /&gt;S tsint^2dt&lt;br /&gt;&lt;br /&gt;u=t^2  du=2t&lt;br /&gt;&lt;br /&gt;1/2 S sinudu&lt;br /&gt;-1/2cost^2 +C&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;By parts is also something I can usually get my u, du, dv, and v.  I just sometimes have a hard time finishing the problem.  The formula is: uv - Svdu.  Remember that the u is something that you want to reduce and dv can be the dx.&lt;br /&gt;&lt;br /&gt;EXAMPLE:&lt;br /&gt;S xe^x&lt;br /&gt;&lt;br /&gt;u=x    v=e^x&lt;br /&gt;du=1   dv=e^xdx&lt;br /&gt;&lt;br /&gt;xe^x- S e^xdx&lt;br /&gt;xe^x-e^x +C&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;What I need help on!  The whole sin, cos, secant, tan stuff with the formulas.  I really didn't get the homework...&lt;br /&gt;&lt;br /&gt;How would you work this?&lt;br /&gt;S sec^4 5xdx&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2170825231523001004?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2170825231523001004/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/abbeys-first-bc-blog.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2170825231523001004'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2170825231523001004'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/abbeys-first-bc-blog.html' title='Abbey&apos;s first BC blog!'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-7614820887300285275</id><published>2010-08-29T17:07:00.003-05:00</published><updated>2010-08-29T18:35:56.415-05:00</updated><title type='text'>Mal's First BC Post...</title><content type='html'>So, I've got to get back into the routine of doing these things.  Luckily my dad's got my back and remembered for me.  &lt;br /&gt;Calculus. Right.  Here I go...&lt;br /&gt;A brief overview of basic integration (a few tips):&lt;br /&gt;&lt;br /&gt;1. Remember that normally your u is whatever is being raised to a power or under the square root or the bottom of a fraction, etc.  For example:&lt;br /&gt;&lt;br /&gt;∫x/√(x^2+1)&lt;br /&gt;u=x^2+1&lt;br /&gt;du=2x&lt;br /&gt;&lt;br /&gt;Now balance the 2 by putting a ½ in front and you have:&lt;br /&gt;&lt;br /&gt;1/2∫du/(√u)&lt;br /&gt;=√(x+1) + C&lt;br /&gt;&lt;br /&gt;2. You have to be able to recognize basic derivatives (i.e. trig ones)  So:&lt;br /&gt;∫secx tanx  &lt;br /&gt;&lt;br /&gt;We know that the derivative of sec x is sec tan, so obviously the integral equals:&lt;br /&gt;&lt;br /&gt;secx+C&lt;br /&gt;&lt;br /&gt;3. Also remember that if you see 1/x and your integrating..that’s natural log integration.&lt;br /&gt;&lt;br /&gt;4. Okay.  By parts.  You must remember that the formula is as follows:&lt;br /&gt; uv- ∫(v)du&lt;br /&gt;&lt;br /&gt;That, my friends is a crucial part.  Also remember that your u is usually that which will decrease more.  So if you have x and x^3, your u would be x and dv x^3.  Got it? Good.  HOWEVER, if you have a ln in the equation, that needs to be your u because you cannot integrate a ln…It’s just not conducive. &lt;br /&gt;&lt;br /&gt;5. Next subject: Trig Substitution.   I know we had some homework over the weekend, and for a lot of it, I was unsure.  I did find out, I believe, that the integral of sec x is always going to be:&lt;br /&gt;&lt;br /&gt;∫secx = ln(secx  +  tanx) + C&lt;br /&gt;&lt;br /&gt;Don’t ask me why, that’s just what I found.  You have to memorize it I believe?  Now my main problem is trying to figure out when to substitute in certain trig identities.  Does anyone know a way to remember it ?  In desperate need of help if I’m going to pass that quiz on Wednesday.  Muchas Gracias!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-7614820887300285275?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/7614820887300285275/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/mals-first-bc-post.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7614820887300285275'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7614820887300285275'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/08/mals-first-bc-post.html' title='Mal&apos;s First BC Post...'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6168167997774494560</id><published>2010-05-13T21:14:00.000-05:00</published><updated>2010-05-13T21:14:17.279-05:00</updated><title type='text'>Final Reflections by Ryne Trosclair</title><content type='html'>Finally, a whole year is gone.  There have been many things that we learned in Calculus, like derivatives, integrals, limits, related rates, optimization, angle of elevation, and other mathematical processes that could help us in our near future.  &lt;br /&gt;&lt;br /&gt;Calculus, for the most part, came easy to me, whether it was memorizing the long list of differentials to applying our knowledge of graphs to find how many cars passed by within 3 hours, it was like a fun, challenging puzzle.  One particular subject we were learning that through me off for a bit was integrating where the area was a base and the cross sections were squares or semi-circles.  After a bit of playing around with some problems, I finally figured out how to work them correctly and moved on to something new.&lt;br /&gt;&lt;br /&gt;One thing that Alex mentioned was by-parts integration.  With things winding down in the last couple of months of school, Brandi decided to teach some of us how to integrate by parts one day during lunch.  That was definitely a challenge, but John managed to create a funny saying that helped me learn it, 'uv minus vdu'.  Saying that over and over again, anyone can memorize the by-parts integration formula with ease, and eventually be able to integrate fairly tricky problems with ease.&lt;br /&gt;&lt;br /&gt;One last thing; I feel sorry for the juniors.  Spending the entire year with Alex during seventh hour showed me a bit of Calculus BC, which was very annoying.  There is going to be a lot to memorize and everything has little tricks that will annoy the hell out of you.  He's not joking about sequences and series!!!  One day, he showed me some work he was doing with Taylor and MacClaurin Series and it blew my mind.  Good luck and spend your last year of high school wisely Juniors.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6168167997774494560?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6168167997774494560/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/final-reflections-by-ryne-trosclair.html#comment-form' title='26 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6168167997774494560'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6168167997774494560'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/final-reflections-by-ryne-trosclair.html' title='Final Reflections by Ryne Trosclair'/><author><name>Ryne</name><uri>http://www.blogger.com/profile/10479244229585901625</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='33' height='26' src='http://4.bp.blogspot.com/_0cEWx_zn6bE/Soy-GMCRNQI/AAAAAAAAAAM/wTrnyNanspQ/S220/no+hax+4+u.JPG'/></author><thr:total>26</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5024678777490621618</id><published>2010-05-13T13:48:00.002-05:00</published><updated>2010-05-13T14:06:44.802-05:00</updated><title type='text'>Alex Tir's Final Reflection</title><content type='html'>Hello to all and farewell to all.&lt;br /&gt;This is my first post on this blog and my last post on this blog.&lt;br /&gt;I will say what I have to say for the end of this year and the start of another.&lt;br /&gt;&lt;br /&gt;It was a pleasure being with all of you this year in Claculus AB as an ASO student.  I usually never had many classes with my graduating class, but this gave me another class to be with you all this year.  For this final reflection, I am going to reflect on my entire two years of Calculus experience.&lt;br /&gt;&lt;br /&gt;Starting off with Calculus my first year, I had a very easy time learning limits and derivatives.  I would like to say that taking derivatives are probably the easiest thing to do, but when this concept is put into a word problem-format, sometimes I still make mistakes.  There are always key words to derivatives:  slope, rate of change, rate, maximum, minimum, etc.  For limits, one thing that I always forget is that a limit can exist at a removable.&lt;br /&gt;&lt;br /&gt;When I got to integration, things became a bit tougher.  I actually learned integration by-parts when I was in Calculus, so we had harder integration.  I'm pretty sure you guys didn't learn it, but that's okay.&lt;br /&gt;&lt;br /&gt;I was a good AP Calculus student, but I was a bad Mu Alpha Theta Calc student when it came to integration.  Mrs. Robinson would bash my head all the time whenever I missed an "easy" integration problem.  Psh!  It was easy to her.  But entering into Calculus BC, I realized it was easy.  I had more practice with integration, and I learned more advanced integration topics.  Trigonometric substition is probably my favorite type, and partial fractions is always a fun puzzle to solve at times.&lt;br /&gt;&lt;br /&gt;For you juniors this year who will be in Calculus BC next year, don't worry!  Calculus BC is about 75% Calculus AB, so I became better at what I learned in AB through BC.  You will basically learn a little more advanced ways of doing what you've already learned in AB.&lt;br /&gt;&lt;br /&gt;If there is one thing in Calculus BC that I have to say, it is SEQUENCES AND SERIES ARE BEAST!  The basic concepts of sequences and series are not too bad, but once you get to Taylor and MacClaurin Series, your mind will be baffled.  For AP, you should memorize your series, which I didn't do until right before my AP test lol, but for working out of the book, your work will look ridiculous on paper, but you will be okay.&lt;br /&gt;&lt;br /&gt;Anyway, I hope many of you can take a lot of things out of Calculus AB and bring it into Calculus BC or into college.  For those of you who may not have understood things so much, or those of you who may have spent more time complaning than learning, or those of you who were too lazy to do some of the work, I apologize.  Not on your behalf, but I apologize to you that you may not have gotten the best out of one of the best AP courses in the area.&lt;br /&gt;&lt;br /&gt;Take care all!  It was a pleasure.  Good luck!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5024678777490621618?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5024678777490621618/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/alex-tirs-final-reflection.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5024678777490621618'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5024678777490621618'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/alex-tirs-final-reflection.html' title='Alex Tir&apos;s Final Reflection'/><author><name>Archimedes</name><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='23' height='32' src='http://1.bp.blogspot.com/_M51pVgQmAi4/TD5ngKLMfFI/AAAAAAAAABQ/kvu77fl4cIg/S220/DSC_0210.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5476115946547279383</id><published>2010-05-11T09:00:00.003-05:00</published><updated>2010-05-11T09:35:28.920-05:00</updated><title type='text'>Final Reflection</title><content type='html'>Wow.  We're done with Calculus. (well, for the sole 6 juniors, maybe not, but you get the picture)  Who would've ever thought that we'd make it alive, through all the blood and glory (I know! so melodramatic...).  But, seriously!  This year has been, in one word, CRAZY.&lt;br /&gt;&lt;br /&gt;So, in all things math related, we learned some concepts that might actually help us later on in life.  For example, I'm pretty sure everyone will remember that rate=derivative=slope.  Also, an integral is the area under a curve! PLUS, the derivative of 2 is ZEEERROOOO!!!   Hopefully, you remember that.&lt;br /&gt;&lt;br /&gt;Okay, so for the semi-sad part.  All of the seniors are leaving.  How depressing.  The 2009-2010 Calculus class. &lt;br /&gt;&lt;br /&gt;Tir-...do I have to say anything?&lt;br /&gt;&lt;br /&gt;John John-the go to guy for everything and anything (moderately loud)&lt;br /&gt;&lt;br /&gt;Mamie &amp; Chelsea (because you can't have one without the other-John's co-conspirators/enemies in the Calculus realm; also, mamie=insne; chelsea=insane on certain days&lt;br /&gt;&lt;br /&gt;Ryne-the loud one&lt;br /&gt;&lt;br /&gt;Gonzales-the semi-stupid one (maybe just the stupid one)&lt;br /&gt;&lt;br /&gt;Aimee-"I didn't do that."&lt;br /&gt;&lt;br /&gt;Ricky-DROID (seriously, that's the one thing I will remember...the "best phone ever")&lt;br /&gt;&lt;br /&gt;Kaitlyn-no chocolate...&lt;br /&gt;&lt;br /&gt;Trina-Ms. Stress over calculus forever. (really and truly dedicated)&lt;br /&gt;&lt;br /&gt;Ellie-almost the same as Trina, except she's a little...more...outspoken?  ahmm...&lt;br /&gt;&lt;br /&gt;Dylan-another loud one...when he talked&lt;br /&gt;&lt;br /&gt;Jessie-quiet one most of the time (I think it was too early) and the sickly one.&lt;br /&gt;&lt;br /&gt;And I almost forgot Mher---the door closer who gave me pencils.  How trivial..&lt;br /&gt;&lt;br /&gt;And then of course there's all the Juniors...Sarah, Steph, Abbey, ME, Ashley, Milky...and that's about it.&lt;br /&gt;&lt;br /&gt;OH! the most important part...B-ROB:&lt;br /&gt;&lt;br /&gt;who left us, unfortunately, to have Sarah-Rachelle (during which time we had Mr. St. Pierre and had a proper going away party for). She's probably the best Calculus teacher on the planet because although we hated doing the work, she really did prepare us for the AP..which is the final goal..&lt;br /&gt;&lt;br /&gt;SO, to the Seniors: have a nice life, and don't forget to visit.&lt;br /&gt;&lt;br /&gt;to Brandi: Thanks.&lt;br /&gt;&lt;br /&gt;to the Juniors: we have another year left...gosh.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5476115946547279383?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5476115946547279383/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/final-reflection_11.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5476115946547279383'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5476115946547279383'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/final-reflection_11.html' title='Final Reflection'/><author><name>MalPal</name><uri>http://www.blogger.com/profile/12947933001973909637</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1075872878739449622</id><published>2010-05-10T22:26:00.002-05:00</published><updated>2010-05-10T22:47:58.552-05:00</updated><title type='text'>Ryans Final Reflection</title><content type='html'>This year has been a very upside and backwards year, not just including Calculus.  Graduating is a stepping stone in life that is very complicated filled with mixed feelings.  Although I am ready to move on and begin a new chapter in my life, i know that i will miss high school and all my friends very much so.  Looking back on the year and class as a whole, I have some regrets but who wouldn't.  I also have some memories that will last a lifetime.  Late night study groups where we all freaked out and crammed the night before a test, John and Tir getting aggravated and trying not to show it while they help us,and People falling flat on there face in the middle of class are all memories that stick out in my mind.  (Lets not hope anybody has spitwater in there ear when i said that because i ment it) Ill remember all of my classmates, they all have something different and wonderful to contribute to the group. Although this class has been some what over barring and grueling at times I am thankful for having taken it as i know it has prepared me for the next level.    &lt;br /&gt;&lt;br /&gt;One concept that i am taking away from calculus is that an intergral is the area under the curve.  This is very useful.  lol &lt;br /&gt;&lt;br /&gt;To all reading this pick up on intergrating as quick as possible it will help in the long run... and watch out for particle problems.  &lt;br /&gt;&lt;br /&gt;well I said what I had to say.  Whether it was getting me ready for college or yes even finding me a girlfriend ;) calculus has helped.  I wish all of my classmates, whether you have another year left or you are begging a whole new chapter, the best of luck.  Turn the book one page at a time.  Stay persistent in reaching your goals and I wish everybody the best of success in whatever it is they are doing.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1075872878739449622?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1075872878739449622/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ryans-final-reflection_10.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1075872878739449622'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1075872878739449622'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ryans-final-reflection_10.html' title='Ryans Final Reflection'/><author><name>gonzo</name><uri>http://www.blogger.com/profile/08631527109092505080</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1518431108616606220</id><published>2010-05-10T20:53:00.002-05:00</published><updated>2010-05-10T21:22:20.320-05:00</updated><title type='text'>Abbey's Final Reflection</title><content type='html'>Well I'm late on blogs...again haha.&lt;br /&gt;&lt;br /&gt;So, this year is coming to a close, and all I can say is yessss!  Calculus this year has made me laugh and cry at the same time.  I feel this year I really became close to the seniors.  Seniors I like to thank you: to Kaitlyn, I now know about a new brand of goldfish; Aimee, for using all my loose leaf paper; John, for answering my 215445585 questions about the same thing; Mamie, for letting me borrow pencils.  And obviously thanks to this calculus class I have Ryan G. now haha.  I'm really going to miss yall and the random late night study groups trying to finish corrections.  Also to the six juniors of our class...yall have become some of my closest friends.  I can't wait until next year when we are seniors!  :)&lt;br /&gt;&lt;br /&gt;Now let me tell you one concept, perhaps the only concept I knew all year long...AVERAGE VALUE.  formula: 1/b-a aSb equation &lt;br /&gt;*just plug in and if you have a calcultor it is even easier&lt;br /&gt;&lt;br /&gt;Also, I now love area, graph, and table free response questions because I know what I'm doing! haha&lt;br /&gt;&lt;br /&gt;Something that took me until two days before the ap test to learn was tram.  Mrs. Robinson had to teach me a completely different formula just because it wouldn't click!  (b-rob: thanks for answering all my questions and helping me at sixth hour!)&lt;br /&gt;So, the new formula is:  [(b2-b1/2)(h) + (b2-b1/2)(h)...]&lt;br /&gt;&lt;br /&gt;Some things I just want to point out that I always forgot...but now I know!&lt;br /&gt;1. use your calculator, it is your friend not enemy&lt;br /&gt;2. that calculator that I just talked about can actually find derivatives, integrals, intersection or bounds, and x-intercepts&lt;br /&gt;3. when the original graph is increasing the derivative will be positve, original decreasing then derivative negative&lt;br /&gt;4. if you get a graph and they want area, look to see if you can divide the graph into shapes&lt;br /&gt;&lt;br /&gt;This year flew by.  Good luck to all of the seniors and yall will be missed!&lt;br /&gt;&lt;br /&gt;Finally I just want to say:  I defintily stuggled with learing calculus, but now that we are done with the ap and I take a look back, I feel like I have learned a lot!  Bring on Calculus BC HAHAHA&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Abbey has left the building....&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1518431108616606220?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1518431108616606220/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/abbeys-final-reflection.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1518431108616606220'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1518431108616606220'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/abbeys-final-reflection.html' title='Abbey&apos;s Final Reflection'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-7087395546712633585</id><published>2010-05-10T06:12:00.002-05:00</published><updated>2010-05-10T06:30:21.762-05:00</updated><title type='text'>Mamie's final post</title><content type='html'>As you can see, it's 6:17 on Monday morning; the day after I was supposed to do my blog.  Surprise, surprise.&lt;br /&gt;&lt;br /&gt;Where should I start?  This year was a challenging one to say the least.  It was very fast paste at the beginning of the year.  We learned most things in the matter of only a few days.  Maybe even just one.  Throughout the year, though, we went over the same topics again and again, but some people--like me--still never caught on to some of the things taught until the very end.  Some things I eithe don't remember how to do or never really understood include optimization and angle of elevation.  These two are related a good bit.  I'm pretty sure I was awesome at optimization problems.  I think I also understood angle of elevation a little, but now if you'd ask me to do a problem, I couldn't.  I was also confused toward the end about natural log integrals.  I would never recognize them.  I would also never recognize the trig functions that gave us so much trouble in advanced math.&lt;br /&gt;&lt;br /&gt;I did pretty much master derivatives, though.  Some other things I'm pretty decent at include regular integrals, e integration, limit rules, area graphs and volume graphs.&lt;br /&gt;&lt;br /&gt;Many people mess up on substitution.  I do sometimes too, but I think I do well at remembering what has to be done in the process.  e integration is also really easy.  The exponent of e will always be u and its derivative will always be du.  You set it up as e^u then solve that way.  Limit rules: if the degree of the top is greater than the degree of the bottom the answer is infinity, if the degree of the top is less than the degree of the bottom the answer is 0.  If the degrees are equal, you make a fraction with the two terms infront of x.  Also sinax/bx will always be a/b.  I remembered this at the beginning of calculus, but now I find myself using quotient rule to figure it out.&lt;br /&gt;&lt;br /&gt;Area and volume graph problems are always the same.  You just integrate.  For area, it's top -bottom.  For volume it's top^2-bottom^2 times pi.&lt;br /&gt;&lt;br /&gt;This year really has been fun with all of you.  It still didn't hit me yet that my last day of high school is going to be Wednesday.  I'll really miss you all and I'll never forget any of you.  Except maybe Mher, because none of us really remembered he was in our Calculus AB calss when he was there.&lt;br /&gt;&lt;br /&gt;Bye everyone!!!!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-7087395546712633585?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/7087395546712633585/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/mamies-final-post.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7087395546712633585'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7087395546712633585'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/mamies-final-post.html' title='Mamie&apos;s final post'/><author><name>Mamie</name><uri>http://www.blogger.com/profile/10227274119948719504</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='22' height='32' src='http://1.bp.blogspot.com/_aMOKyfIJq-c/SoYLLYEnWjI/AAAAAAAAAAM/cm0UoIDuye0/S220/blog.bmp'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-4159074393083948692</id><published>2010-05-09T23:16:00.002-05:00</published><updated>2010-05-10T03:06:18.049-05:00</updated><title type='text'>Ash's 38th and Final Post..</title><content type='html'>As much as I hate to admit it, this is actually kind of depressing.  This school year seems to have flown by!  I'm really going to miss all of the seniors and I'm so glad I had those in Calculus in my class, otherwise I wouldn't have gotten to know their cooky personalities X] Good luck to all of you guys! I know all of yall will succeed in college and then later on in life, wherever that may take you =]&lt;br /&gt;&lt;br /&gt;Enough of the mushy-gushy stuff for now and onto the last bit of Calculus I'll have to type on here for a while.&lt;br /&gt;&lt;br /&gt;The concept that I felt really stupid when I finally understood was Definition of Derivatives.  I'm not exactly sure how it clicked, but I remember Mrs. Robinson sitting at her desk explaining it once again..when a light bulb when off in my head! =] I felt so accomplished after I worked that first problem understanding the concept..then I felt like an idiot for not getting it before X]&lt;br /&gt;&lt;br /&gt;But how about I go over real quick what is it? ^^&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;f'(x) = Lim  f(x+h)-f(x)&lt;br /&gt;        H-&gt;0     h&lt;br /&gt;&lt;br /&gt;You merely take the derivative of f(x+h) by following these steps :D&lt;br /&gt;First plug in 0 for H&lt;br /&gt;Then you take the derivative&lt;br /&gt;Then you simplify!&lt;br /&gt;The end!&lt;br /&gt;&lt;br /&gt;I can't believe it took me forever to grasp that X]&lt;br /&gt;&lt;br /&gt;And a final note before I say goodbye to Calculus AB...thank you guys for such a great year! I'm never going to forget this class and I'm pretty sure I never want to.  Thanks to Mrs. Robinson for having the patience to teach us, the motivation to push us and herself as hard as she could, the love and care she showed throughout the entire year.  We love you! =] &lt;br /&gt;Goodbye Calculus AB!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-4159074393083948692?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/4159074393083948692/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ashs-38th-and-final-post.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4159074393083948692'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4159074393083948692'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ashs-38th-and-final-post.html' title='Ash&apos;s 38th and Final Post..'/><author><name>DarkLies</name><uri>http://www.blogger.com/profile/06021157770462782617</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://1.bp.blogspot.com/_7d-Jre0G1a4/SpRzIPpSboI/AAAAAAAAAAM/lzDIPvGaT6Q/S220/Hairrr004-1.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6933759141764771847</id><published>2010-05-09T22:58:00.003-05:00</published><updated>2010-05-09T23:19:14.311-05:00</updated><title type='text'>John's Final Reflection</title><content type='html'>Wow.&lt;br /&gt;&lt;br /&gt;It really is amazing that the year is over.  It seems like it has gone by so fast...it's like I remember people in class asking how to take a derivative, and I feel like it was days ago...wait, that was days ago, like Monday when we reviewed. :-P.&lt;br /&gt;&lt;br /&gt;Jokes aside, I can't express how awesome this year has been for me in Calculus.  It feels like we flew right through it, and it was &lt;span style="font-weight:bold;"&gt;not&lt;/span&gt; as bad as some people made it out to be.  Calculus is pretty easy once you get the concepts down (derivatives and integrals).  It is of course taught by an amazing teacher, so I know that everyone after me will do just fine with very little effort. &lt;br /&gt;&lt;br /&gt;I first want to congratulate and acknowledge &lt;span style="font-weight:bold;"&gt;Mrs. Brandi Robinson&lt;/span&gt; for being an amazing teacher.  Not only did you have a baby and manage to keep your math classes running, but you also taught us so many things and prepared us so well for the AP exam.  I can honestly say that I took the AP exam with absolutely no studying outside of the school day, and I'm positive I did well on it.  I would not be able to say that without you, and I greatly thank you and appreciate everything you've done for me and the class especially.&lt;br /&gt;&lt;br /&gt;Second, I want to say that I could not have done so well in Calculus without all of my classmates as well.  The reason I was able to take the AP without studying is because of you guys, mainly.  You guys pestered me and annoyed me and asked as many questions as you could possibly come up with, sometimes asking the same exact question without realizing it, and I thank you all for it.  Through all of the explaining and answering of your questions, I was able to study almost every class period of my day.  You guys have made my year successful, and I want to thank you for that.  I also hope that I have helped make your year just as successful, because that would only be the fair thing :-).&lt;br /&gt;&lt;br /&gt;Last, I want to thank Chelsea specifically.  Of every classmate in Calculus, she has been the most useful to me.  Whether it was her asking a million and one questions or whether it was discussing calculus problems over AIM, she helped me in a million ways, and I owe my deepest gratitude and thanks to her for all of her help.  :-)&lt;br /&gt;&lt;br /&gt;Overall, Calculus AB has been a complete success on so many levels.  Not only will it get me credit hours in college, but it really has furthered my knowledge in mathematics and Calculus specifically, which is going to help my career (Engineering) in so many ways.&lt;br /&gt;&lt;br /&gt;P.S. I just want to thank you again, B-Rob, for being an amazing teacher and mentor.  I cannot fathom the words to describe how awesome you have been to me this year with all of your help and support.  With that being said, I hope you understand how much I truly cherish the teachings and knowledge you have passed to me.&lt;br /&gt;&lt;br /&gt;-John&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6933759141764771847?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6933759141764771847/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/johns-final-reflection.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6933759141764771847'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6933759141764771847'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/johns-final-reflection.html' title='John&apos;s Final Reflection'/><author><name>XxDohxX</name><uri>http://www.blogger.com/profile/08553932307864427605</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='23' height='32' src='http://3.bp.blogspot.com/_s-wvP590kRc/SoYHviaw8TI/AAAAAAAAACE/1_onYhrFm7c/S220/john+002.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6407299350339445760</id><published>2010-05-09T22:51:00.002-05:00</published><updated>2010-05-09T23:06:02.620-05:00</updated><title type='text'>final post</title><content type='html'>alright, well some of the things i've gotten comfortable doing this year in calculus is the first and second derivative test.  since those were basically the first things we did this year, i learned those the best, and i got kinda lazy throughout the rest of the year, so that's why i was the best at that.  &lt;br /&gt;&lt;br /&gt;For this post i am going to go into detail on the second derivative test to find all possible points of inflection and intervals of concavity. remember, points of inflection only happen where there is a change of concavity.&lt;br /&gt;&lt;br /&gt;Example: f'(x)= 6/(x^(2)+3) &lt;br /&gt;&lt;br /&gt;First, you have to take the derivative of that, and you have to use the quotient rule, so the beginning of the problem will look like, [(x^(2)+3)(0)-6(2x)]/(x^(2)+3)^(2), which simplifies to -12x/(x^(2)+3)^2 remember, that was just the first derivative. &lt;br /&gt;&lt;br /&gt;Second step is to take the derivative of the first derivative, that would make this step called taking the second derivative. &lt;br /&gt;&lt;br /&gt;Once again u need to use the quotient rule, so f''(x)={(x2+3)^2-(12)-[(-12x)2(x^(2)+3)2x} all that over (x^(2)+3)^4 then you get a bunch of stuff, then you simplify, then you cancel, so I am just going to type the end answer of the second derivative. Which is, (3)(6)(x+1)(x-1) all over (x^2+3)^3 &lt;br /&gt;&lt;br /&gt;The possible points of inflection are found in the numerator of the finished second derivative, in this case, if you look, it would be x=1, and x=-1 &lt;br /&gt;&lt;br /&gt;so then you set up your points, (-infinity, -1) u (-1, 1) u (1, infinity) &lt;br /&gt;&lt;br /&gt;then you plug in. f''(-2)= positive value f''(0)=negative value f''(2)=positive value &lt;br /&gt;&lt;br /&gt;then you know that your intervals concave up at (-infinity, -1) u (1,infinity) or x&lt;1,&gt;1 &lt;br /&gt;&lt;br /&gt;and it is concave down at (-1,1) or -1 &lt;br /&gt;and you're points of inflection are x=-1, and x=1&lt;br /&gt;&lt;br /&gt;okay, one thing i wish i would have done was do my homework at the beginning of the year, because that would have made learning all the other stuff throughout the year much easier.  and my advice to anyone else taking this class is to learn derivatives, integrals, related rates, substitution, limits, and implicit derivatives really well, because that's pretty much all the main focal points of the ap test and calculus in general.  and thanks ms robinson for teaching me so much and being such a cool teacher :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6407299350339445760?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6407299350339445760/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/final-post.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6407299350339445760'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6407299350339445760'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/final-post.html' title='final post'/><author><name>merf</name><uri>http://www.blogger.com/profile/06184639794381320730</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-7501384661375210144</id><published>2010-05-09T22:10:00.002-05:00</published><updated>2010-05-09T22:23:27.567-05:00</updated><title type='text'>Ricky's Last Blog</title><content type='html'>After all the anticipation, its finally over.  High school and more importantly CALCULUS.  I'm not a math person as many of u know but im glad i have been in advanced math and in calculus with all of you. i love each and every one of you and im glad to call yall my classmates.  From watching ryan get yelled at to making stupid math jokes. yall are amazing. i would be lost without most of you so i would like to say thanks for putting up with me sometimes. &lt;br /&gt;&lt;br /&gt;I will have many things that stick with me from calculus but lets see what i can really remember. first and second derivative tests is something i def cant forget. WE DID IT A MILLION TIMES.&lt;br /&gt;&lt;br /&gt;I will def remember limit rules.&lt;br /&gt;&lt;br /&gt;1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity&lt;br /&gt;&lt;br /&gt;2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero&lt;br /&gt;&lt;br /&gt;3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree.&lt;br /&gt;&lt;br /&gt;I know the optimization steps i just suck at doing them hahaahahah i know how to take integrals and how to substitute. i know eintegration and many other useful tools that i have learned along the way of my calculus year.&lt;br /&gt;&lt;br /&gt;One thing i have learned more than anything is to never give up and strive for something better. i didnt ahve to take calculus as a senior buti chose to and im glad i did.  the hard work will help me in college just like it will help the rest of the class.&lt;br /&gt;&lt;br /&gt;To Brandi:&lt;br /&gt;Even though i wanted to kill myself during the year im glad to have had you as a teacher these past two years.  From making you laugh to me almost crying cuz your tests were beast i had a great time.  This year was a great one w/ alot of things that will help me in the future.  thank you for being there for me and being my teacher even though im not the easiest to teach.&lt;br /&gt;&lt;br /&gt;TO THE JUNIORS:&lt;br /&gt;I LOVE YALL JUST AS IF YALL WERE SENIORS. I WISH YALLL COULD GRADUATE WITH US BUT YALL HAVE YALLS OWN THING TO DO NEXT YEAR. IMA BE AROUND NEXT YEAR BTW SO YALL CANT MISS ME TOOO MUCH :D&lt;br /&gt;&lt;br /&gt;TO THE CLASS OF 2010:&lt;br /&gt;LETS DO THIS :)  WE WORKED OUR A$$E$ OFF AND ITS GONNA PAY OFF. I LOVE YOU ALL AS IF YOU WERE MY BROTHERS AND SISTERS. IN FACT YALL ARE LIKE BROTHERS AND SISTERS. IF ANY OFYOU NEED ANYTHING IN THE FUTURE U KNOW THE NUMBER.&lt;br /&gt;&lt;br /&gt;Signing out,&lt;br /&gt;                    Ricky Johnson ;)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-7501384661375210144?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/7501384661375210144/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/rickys-last-blog.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7501384661375210144'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7501384661375210144'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/rickys-last-blog.html' title='Ricky&apos;s Last Blog'/><author><name>R.Johnson</name><uri>http://www.blogger.com/profile/16880396213810437297</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='12' height='32' src='http://1.bp.blogspot.com/__sVeTAhmH-Q/SpH5fUh01kI/AAAAAAAAAAU/ilR0mIJcsBs/S220/bustin+through+the+sign.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-8510346975179062651</id><published>2010-05-09T21:56:00.002-05:00</published><updated>2010-05-09T22:14:54.086-05:00</updated><title type='text'>Final Reflection</title><content type='html'>Well, the class we all anticipated to fail is finally over and I think most of us learned a lot of calculus this year. &lt;br /&gt;&lt;br /&gt;Some topics I caught on to right away such as tangent line and normal line since it was only taking a derivative and plugging in numbers.  I also  learned RRAM, LRAM, TRAM, and MRAM right away.  This is partly because I did both my comments on these topics weekly for almost 2 months.  It really is just plugging in formulas.  &lt;br /&gt;&lt;br /&gt;Other topics such as optimization and integration took longer to catch on too.  I did not understand optimization at all until Brob gave us steps on the board one day.  Those steps applied to every problem we saw.  I think its easier to follow steps then to have to decipher the problem by yourself, at leas it is for me.  Integration is not a hard topic, but it just took time to catch on too.  A LOT of practice problems were done especially substitution problems.  I still forget my +c sometimes though.  I never really did get into the habit of putting it.   I also did not know how to do anything in my calculator until around the last month of school.  That is when John started giving me funny looks when correcting APs because I tried to do everything by hand.  Who knew those a hundred something dollars would actually be worth it.  &lt;br /&gt;&lt;br /&gt;And finally, the topics that I still struggle with are graphs, angle of elevation, and related rates.  Graphs I never did get from day one.  I did get a little better with them since they were on EVERY AP, but I would have liked to have had a better understanding on them.  I always struggled with word problems hence my problem with angle of elevation and related rates.  Every problem differs so these cannot have a set of rules like optimization does.  I have a problem with drawing the picture correctly and picking out the given and what we are looking for.  &lt;br /&gt;&lt;br /&gt;All in all, I feel it was a pretty successful year, and I want to thank you brob for HOPEFULLY getting me out of calculus in college; although, I cannot picture next year without it.  &lt;br /&gt;Also, a thanks to all my classmates for asking plenty of questions, so I did not have too =)&lt;br /&gt;And a thanks to John for answering my thousands of questions even though many of them were repeated.&lt;br /&gt;I will miss all of you!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-8510346975179062651?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/8510346975179062651/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/final-reflection.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8510346975179062651'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8510346975179062651'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/final-reflection.html' title='Final Reflection'/><author><name>Chelsea</name><uri>http://www.blogger.com/profile/02545040999156198545</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1320361930143416343</id><published>2010-05-09T21:37:00.003-05:00</published><updated>2010-05-09T21:43:18.317-05:00</updated><title type='text'>sarah's final blog part 2.</title><content type='html'>OMG I ALMOST FORGOT!&lt;br /&gt;&lt;br /&gt;To B-ROB: you are an AMAZING teacher. I don't care what anyone says, hands-down.. best teacher at riverside academy. you have so much dedication for your students and you believe in all of us. even the annoying ones in the class... ;) you are one of the only teachers i know that will work with you in all of your extra hours UNTIL you get it. no matter how long it takes, you will make it click in our brains. which you did for most of us. i wish i could put a note on here for all incoming calc students saying that brob is willing to help as long as you're willing to put forth the effort. that is one thing that i've learned also, which i'm very glad i did. i can honestly say i think i'm speaking on behalf of the entire class when i say: WE LOVE YOU BROB. thank you for believing in us.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1320361930143416343?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1320361930143416343/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/sarahs-final-blog-part-2.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1320361930143416343'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1320361930143416343'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/sarahs-final-blog-part-2.html' title='sarah&apos;s final blog part 2.'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6581824463290162716</id><published>2010-05-09T21:18:00.002-05:00</published><updated>2010-05-09T21:35:00.420-05:00</updated><title type='text'>Kaitlyn's Final Reflection</title><content type='html'>Well after anticipating all year, it's finally here.  My last few days of high school.  First I just want to say how bittersweet this class was for me.  The class itself, i loved it!  The people, the messing around, the stressing out, the late night mini van trips, it was unforgettable.  I really enjoyed getting to know everyone and love our ability to have fun yet still get a lot of work done.  Now for the bitter part...the actually calculus.  I'm pretty sure it's safe to say i suck at it.  But hey, it's not the end of the world.  Don't worry, i don't plan on building bridges or roller coasters in my future, so y'all should be safe ;)&lt;br /&gt;&lt;br /&gt;Now onto the "concept" that never gave up on me and kept running into my skull until it finally got into my brain...&lt;br /&gt;&lt;br /&gt;what else could it be..... AVERAGE VALUE!&lt;br /&gt;&lt;br /&gt;It's just such a simple integral problem, how can anyone get this wrong.  The formula is 1/b-a bSa f(x).  You simply just plug in to the formula with what you were given.  And half the time it's in the calculator portion so you just have to plug the integral in your calculator then times it by 1/b-a.  Voila, you get a beautiful answer that you actually think is right :)  AMEN to that.&lt;br /&gt;&lt;br /&gt;Just so i don't seem like i only accomplished one thing in calculus all year, some other things i understood were average rate of change, mean value theorem, simple derivatives and integrals, HORIZONTAL TANGENTS, and i'm sure there are a couple of others.&lt;br /&gt;&lt;br /&gt;So onto my farewells:&lt;br /&gt;my seniors- We've had an amazing run.  We're all so close and we worked hard to get where we are.  No matter how mean i may come off sometimes, I love each and every one of y'all and wish every one luck in their future endeavors.  Now let's finish this year off right and make a name.&lt;br /&gt;oh let's doooo it &lt;br /&gt;&lt;br /&gt;the juniors-  i absolutely love y'all.  i never really talked to y'all much before this class but i'm so glad that we had it together.  Every one of y'all cracks me up, and i know y'all will be successful b/c of how determined you are.  Keep it up and don't slack off, and have fun in calculs bc  next year.  I might miss you guys ;)&lt;br /&gt;&lt;br /&gt;B-robbbb- thanks for putting up with me these last two years.  You believed in me and kept pushing me even when i said it was impossible to stick this what i call "garbage" in my brain.  I know i didn't catch on to a lot at first, but i did try and i truly believe no other teacher could put up with us as well as you did.  You're an awesome person and i truly admire you and all of your accomplishments.  Thanks again for everything :)&lt;br /&gt;&lt;br /&gt;Now i think it's time to close this blog, my LAST one.  &lt;br /&gt;&lt;br /&gt;Goodbye Calculus, Goodbye Riverside, Hello Life :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6581824463290162716?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6581824463290162716/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/kaitlyns-final-reflection.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6581824463290162716'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6581824463290162716'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/kaitlyns-final-reflection.html' title='Kaitlyn&apos;s Final Reflection'/><author><name>Kait</name><uri>http://www.blogger.com/profile/01366494316510049421</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5855853014889693994</id><published>2010-05-09T21:09:00.002-05:00</published><updated>2010-05-09T21:32:55.871-05:00</updated><title type='text'>sarah's final blog.</title><content type='html'>wow. I can't believe it's finally over. In a few days, i'll no longer by an underclassmen. I never realized how hard this was gonna be until it actually happened. I sound like i'm leaving the earth or something. ha, sorry. This is only my junior year and I'm already freaking out. I'm a big baby.&lt;br /&gt;&lt;br /&gt;To my seniors: I love you guys more than you know. Every last one of you. Including Mher, although he always seems to not be in class. Yall had an awesome year, I can tell. I hope my senior years proves to be just as great. You guys will be missed, EXTREMELY. especially those of you who have been at RA since I first walked in the front door when I was 4. I always looked up to you guys... cuz yall were older :) and it's so weird to think that next year, we will be the role models. It still hasn't hit me yet.&lt;br /&gt;&lt;br /&gt;To my juniors: Well, what can I say. it's been one heck of a ride.. From preschool, to middle school, to soon-to-be seniors. The five of you in the calc class are the people I am most close to out of the ENTIRE jr class. because I spend 90% of my day with you mon-fri... on weekends @ math trips, and during the summer @ practices. I love yall more than you can imagine. I can't wait for next year's calc BC class, it will be fun with just the six of us :)&lt;br /&gt;&lt;br /&gt;So i'll stop being all mushy for a moment and talk about calc this year.&lt;br /&gt;&lt;br /&gt;Most helpful thing: working in groups at the end of year to go over what you missed on the practice ap tests. one-on-one with another student that will actually make you do your work (thanks ellie) helped me more than you can understand. best idea yet B-ROB.&lt;br /&gt;&lt;br /&gt;Thing i struggled with most: integration. i honestly still am not 100% with it, idk what it is. substitution, ln, tan inverse. none of it came to me easily. more practice was needed probably? idk.&lt;br /&gt;&lt;br /&gt;Thing I picked up on the quickest: the derivative formulas. I can still say them to you probably quicker than anyone in the class. all those derivative practice sheets in the beginning were SOOOO helpful.&lt;br /&gt;&lt;br /&gt;My epiphany of the year: the riemann sum formulas!!!!!!!!! honestly, i didn't understand all that f(a) + f(2 + a) stuff. and multiply by delta x whatever. it must easier when taught that TRAM is&lt;br /&gt;b' + b'' / 2 (n) + b'' + b''' /2 (n). also the LRAM and RRAM are easier when taught that you add to x's divide by n (start all the way to left with LRAM and 2nd number with RRAM). that literally clicked in my head 2 days before the exam.&lt;br /&gt;&lt;br /&gt;so, even though I don't want it to, this year is coming to a close. and quick too. I'm going to miss all you entirely too much. I will probably cry... a lot. just because I'm about to be a senior. i'll be sad everyday in calculus when i realize how small the class is. (&amp;amp; when i realize that mher IS actually not there). we had a great year, from partying it up with mr. st pierre to re-partying it up with brob. &amp;amp; of course getting people to sneak out of school to buy us some soft drinks because ricky never brings his stuff! i will miss it all.&lt;br /&gt;&lt;br /&gt;GOOD LUCK TO YOU, seniors of 2010. you're gifted in more ways than you can imagine. embrace that.. you can do anything you want in your life. be proud of yourselves. i hope for the best for everyone and i'll remember you everytime i walk the halls of riverside academy.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5855853014889693994?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5855853014889693994/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/sarahs-final-blog.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5855853014889693994'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5855853014889693994'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/sarahs-final-blog.html' title='sarah&apos;s final blog.'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2154738815013490106</id><published>2010-05-09T20:49:00.002-05:00</published><updated>2010-05-09T20:58:13.953-05:00</updated><title type='text'>Stephanie's Final Blog</title><content type='html'>As this year comes to a close, I'd like to tell everyone how great our class was this year.  Especially with our mad covert partying skills, our ability to goof off and still get our work done, and lastly, our great personalities.  The seniors this year are great, and i'll miss every one of yall. To my juniors, LETS GET  THIS PARTY STARTED.  I'm ready to own the school, yall bout it?&lt;br /&gt;&lt;br /&gt;So, something i couldn't understand then it just smoked me in the face is substitution.  I think it took Ryan Breaud just yelling at me about how to substitute, then somehow making it involve either a mean girl quote, bonquiqui, or darrel, spelt like darrel, but pronounced like darrel.  I can't explain how it really stuck in my head besides taht. But give me something today, and i'll substitute the numbers out of it. Watch out.&lt;br /&gt;&lt;br /&gt;This was definately one of my favorite classes this year, besides the MASSIVE struggles i've had in this class, it definately taught me that i can achieve what I strive to do.  And to the seniors leaving, i'm glad i got to take this class with yall, and yall better remember the good times we had..and if anyone gets famous remember the girl who mission impossibled it to the cafeteria to get ice for our cups.&lt;br /&gt;&lt;br /&gt;To the juniors, well seniors now, lets make this year the greatest ever.  I mean i'm already great, and yall have class with me, so that makes you offliated with greatness, which means you're semi-great..and so add that together, then you get that we're all great. because of me, of course. :)&lt;br /&gt;&lt;br /&gt;Also, thanks to Mrs. Robinson for helping me this year and not giving up on me when i answer free response questions without thinking.  I know i couldn't of done it with out you.&lt;br /&gt;&lt;br /&gt;Lastly, i have to tell everyone thanks for making my first hour great, and you better not ever forget me. :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2154738815013490106?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2154738815013490106/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/stephanies-final-blog.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2154738815013490106'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2154738815013490106'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/stephanies-final-blog.html' title='Stephanie&apos;s Final Blog'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-4546937128179491620</id><published>2010-05-09T20:36:00.003-05:00</published><updated>2010-05-09T21:13:57.949-05:00</updated><title type='text'>Ellie's Final Reflection</title><content type='html'>Alright you guys!!! We've gotten through calculus and we've all made it.  When this class first started, I wasn't stressed out by the least.  Having a teacher like Mrs. Robinson the year before, I knew what I was getting into.  The first week was oddly enough, expected!  We got so many formulas to memorize and so many ways of taking a derivative a lot of us were going crazy, but not me, I was kind of like "whatever" with it.  Turns out two or three weeks into it, I got really scared, my grades weren't really slipping but I wasn't getting the new, or harder, derivatives.  After that, I buckled down and got to work.  Here's what I've learned throughout the year...&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Easy Concepts For Me:&lt;/strong&gt;&lt;br /&gt;&lt;em&gt;--How to Take a Derivative:&lt;/em&gt;&lt;br /&gt;You take the exponent and times it by the coefficient, what you get is the new coefficient and you're new exponent will be the original exponent minus one! &lt;br /&gt;&lt;em&gt;--The First Derivative Test:&lt;/em&gt;&lt;br /&gt;You have to take the derivative of the function and set it equal to zero. Then solve for the critical values (x values). Set those values up into intervals between negative infinity and infinity. Plug in numbers between these intervals into the first derivative to see if there are max or mins or if the graph is increasing or decreasing.&lt;br /&gt;&lt;em&gt;--The Second Derivative Test:&lt;/em&gt;&lt;br /&gt;You take the derivative of the function twice and set it equal to zero. Solve for the x values and set them up into intervals between negative infinity and infinity. Plug in numbers between those intervals into the second derivative to see where the graph is concave up, concave down, or where there is a point of inflection. &lt;br /&gt;&lt;em&gt;--Maximums, minimums, critical values, increasing, decreasing:&lt;/em&gt;&lt;br /&gt;All of these are related to first derivative test. it's simple. you take the derivative, set equal to 0, solve for x. Set up some intervals using these numbers. Plug in numbers and test your intervals. pos to neg is a min. neg to pos is a max. pos = increasing, neg = decreasing. simple stuff. remember it.&lt;br /&gt;&lt;em&gt;--Point of inflection, concave up, concave down:&lt;/em&gt;&lt;br /&gt;It's the second derivative test. set up intervals, if the intervals change signs, it is a point of inflection there. also, if its negative, that interval is concave down, positive is concave up.&lt;br /&gt;--&lt;em&gt;Finding Critical Values:&lt;/em&gt;&lt;br /&gt;To find critical values, first take the derivative of the function and set it equal to zero, solve for x. The answers you get for x are your critical values.&lt;br /&gt;&lt;em&gt;--Limits:&lt;/em&gt;&lt;br /&gt;If the degree on top is smaller than the degree on the bottom, the limit is zero.&lt;br /&gt;If the degree on top is bigger than the degree on the bottom, the limit is infinity.&lt;br /&gt;If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Some Concepts that...after reviewing over and over for weeks/months...it just clicked:&lt;/strong&gt;&lt;br /&gt;&lt;em&gt;--Slope of a normal line:&lt;/em&gt;&lt;br /&gt;Take derivative, plug in x. get a slope. however, make sure you use the negative reciprocal of the slope (normal means perpendicular to). use point-slope formula.&lt;br /&gt;&lt;em&gt;--Absolute Extrema:&lt;/em&gt;&lt;br /&gt;If you are given a point, plug those numbers into the original function to get another number. Alos, solve for critical values and plug those into the original function. Once you get your second numbers, you set each pair into new sets of points. The highest point is the absolute max and the smallest point is the absolute min.&lt;br /&gt;&lt;em&gt;--Average Rate of Change:&lt;/em&gt;&lt;br /&gt;f(b)-f(a)/(b-a)&lt;br /&gt;&lt;em&gt;--Average Value:&lt;/em&gt; &lt;br /&gt;This is just an integral times by 1/(b-a).&lt;br /&gt;&lt;em&gt;--Equation of a Tangent Line:&lt;/em&gt;&lt;br /&gt;Take the derivative and plug in the x value.&lt;br /&gt;If you are not given a y value, plug into the original equation to get the y value.&lt;br /&gt;then plug those numbers into point slope form: y − y1 = m(x − x1)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;And-of course-the things that I always struggled with, and hope to never encounter again even though I know I will....:&lt;/strong&gt;&lt;br /&gt;&lt;em&gt;--linearization:&lt;/em&gt;&lt;br /&gt;Identify the equation&lt;br /&gt;Use the formula f(x)+f ' (x)dx&lt;br /&gt;Determine your dx in the problem&lt;br /&gt;Then determine your x in the problem&lt;br /&gt;Plug in everything you get&lt;br /&gt;Solve the equation&lt;br /&gt;&lt;em&gt;--Riemanns Sums:&lt;/em&gt;&lt;br /&gt;LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]&lt;br /&gt;RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]&lt;br /&gt;MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]&lt;br /&gt;Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]&lt;br /&gt;delta x=b-a/number of subintervals&lt;br /&gt;&lt;em&gt;--Volume by Disks:&lt;/em&gt;The formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph&lt;br /&gt;&lt;em&gt;--Volume by Washers:&lt;/em&gt;&lt;br /&gt;The formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.&lt;br /&gt;&lt;em&gt;--OPTIMIZATION...&lt;/em&gt;&lt;br /&gt;I just can't do it  =(&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;Just a shout out:&lt;/strong&gt;&lt;br /&gt;&lt;em&gt;To the seniors in the Calculus Class of 2010&lt;/em&gt; - - I'll Miss Yall, we've grown up together basically.  I won't lie, I didn't like MANY of yall when I first came here because of different things that happened; however, we've all matured [thank God] and we've become friends.  Although some of us aren't best friends I can honestly say that I will miss each and every one of you all because yall are the people I spent all day every day with for the past 6 years.  &lt;br /&gt;&lt;em&gt;To the juniors in the Calculus Class of 2010&lt;/em&gt; - - I'll miss yall just as much.  There's many reason's why I will..some were my first friend at Riverside Academy because of the Cheerleading Squad, some have always been in my life because we grew up living next door to each other, some have been with me through the years with dancing and just being friends, some HOWEVER have become my friend in this past year and believe me I love yall just as much!&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;So back to the FARWELLS:&lt;/strong&gt;&lt;br /&gt;No, I'm not sad I'm graduating.  My parents say that it hasn't hit me yet...but most of yall know me already and I'm so happy to finally be just about OUT of Riverside Academy that nothing else matters.  Also, I'm not sad that I'm leaving any of yall.  Yall have been great friends and I'm glad we've become as close as we have or, not as close as we have in some cases, because LIFE GOES ON..We cannot dwell in the past much less live it.  I won't be back at Riverside once I'm out, there's no need....I'm finally getting my life started and I CAN'T WAIT!!  =)  Yeah, I'll miss yall when certain things come up and I realize what [so and so] would've said and they wont be there..but most of all .. I'll miss US as a class together and how we are with each other!  So to everyone - - THANKS!  And have a GREAT time either being a senior, or going out into the REAL WORLD and EXPERIENCING everything!&lt;br /&gt;&lt;br /&gt;Love Always,&lt;br /&gt;       &lt;em&gt;Ellie Kliebert&lt;/em&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-4546937128179491620?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/4546937128179491620/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ellies-final-reflection.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4546937128179491620'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/4546937128179491620'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ellies-final-reflection.html' title='Ellie&apos;s Final Reflection'/><author><name>LE2010</name><uri>http://www.blogger.com/profile/05859839473790389600</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='25' height='32' src='http://1.bp.blogspot.com/_uGqGEbJSAMo/SoMXxYdCIcI/AAAAAAAAAAs/r-0CfJZXIVE/S220/me.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1758258366695182343</id><published>2010-05-09T20:11:00.000-05:00</published><updated>2010-05-10T08:16:40.660-05:00</updated><title type='text'>steven's final reflectiondd</title><content type='html'>&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1758258366695182343?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1758258366695182343/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/stevens-final-reflectiondd.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1758258366695182343'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1758258366695182343'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/stevens-final-reflectiondd.html' title='steven&apos;s final reflectiondd'/><author><name>LilBigTymer</name><uri>http://www.blogger.com/profile/00493338842619363205</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='http://2.bp.blogspot.com/_pztXBOi2Cmo/SoS_x7B8KfI/AAAAAAAAAAM/Ixz4utPswXU/S220/pbnation+avatar.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-5437085120959226093</id><published>2010-05-09T19:44:00.002-05:00</published><updated>2010-05-09T20:00:42.558-05:00</updated><title type='text'>Dylan's last post</title><content type='html'>Well school is finally over.  The only thing that is left is graduation.  Hut there are a few things I have learned this past year that will definitely stick with me.&lt;br /&gt;&lt;br /&gt;Derivatives&lt;br /&gt;The derivative of simple equations such as x^3+9x^2+7x-3.  The derivative is 3x^2+18x+7.&lt;br /&gt;&lt;br /&gt;Integrals&lt;br /&gt;S 3x^2+4x-8&lt;br /&gt;=x^3+2x^2-8x&lt;br /&gt;&lt;br /&gt;Implicit derivatives&lt;br /&gt;All that you do is take the derivative like normal but of x's and y's but put dy/dx right after where you take the derivative of the y then sove for dy/dx.&lt;br /&gt;&lt;br /&gt;Limit rules&lt;br /&gt;1) if the highest exponent is the same on the top and bottom then the limit is the top coefficient over the bottom coefficient of the highest exponents.&lt;br /&gt;2) If the highest exponent is on the top then the limit is infinity.&lt;br /&gt;3) But if the highest exponent is on the bottom then the limit is 0.&lt;br /&gt;&lt;br /&gt;Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.&lt;br /&gt;&lt;br /&gt;Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.&lt;br /&gt;&lt;br /&gt;Those are the main things that I have learned that have stuck with me all year and will probably stay with me.  Calculus this year has been and interesting experience that will stick with me forever.  I wish everyone good luck with everything they do.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-5437085120959226093?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/5437085120959226093/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/dylans-last-post.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5437085120959226093'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/5437085120959226093'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/dylans-last-post.html' title='Dylan&apos;s last post'/><author><name>dylan b</name><uri>http://www.blogger.com/profile/11850647602537185836</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-3397339293447606497</id><published>2010-05-09T19:14:00.000-05:00</published><updated>2010-05-10T07:16:33.995-05:00</updated><title type='text'></title><content type='html'>Well here it goes.. every thing comes to an end sooner or later right?&lt;br /&gt;I think over the past couple of days I’ve understood that statement more than anything and to begin I would like to say thank you to everyone who was there for me over these past couple of days because even though it doesn’t make it hurt any less, it does make it somewhat easier and for that I am forever thankful.&lt;br /&gt;&lt;br /&gt;What can I say, calculus was calculus, we loved it and hated it at the same time.  I don’t think anyone will ever be able to understand just how if feel besides all of the people who were in there with me.  It was truly memorable and brought everyone closer together.  One person needed help and others didn’t hesitate to lend a hand which is awesome to me.  We truly excelled this year because of EACHOTHER, it was a team effort =)&lt;br /&gt;&lt;br /&gt;To our coach, haha teacher, Brandi, words  cannot express how truly thankful I am for you both as a teacher and as a person.  Without you, we surely would not have been as successful as we were this year in calculus.  You prepared us as much as humanly possible and even after Sara-Rachelle was born you got back to us as quickly as you could so that we would stay on track, that’s dedication.  You stuck with me, you didn’t give up, and you always believed in me which means the world to me, thank you so much for an amazing couple of years and you will always be remembered.&lt;br /&gt;&lt;br /&gt;OKAYY lets get to some calculusss!&lt;br /&gt;&lt;br /&gt;So some things that will stay with me forever and I actually get excited when I see them:&lt;br /&gt;Mmkay, well I realized I love derivatives. Of course the slope formula because it’s on AP free response and I loveeeeeeeee it. Limit rules and the basic integrals.&lt;br /&gt;&lt;br /&gt;For derivatives &lt;br /&gt;&lt;br /&gt;x^4 + 3x^3 + 6x&lt;br /&gt;&lt;br /&gt;multiply the coefficient in front of your variable by it's exponent, then subtract 1 from your exponent.&lt;br /&gt;&lt;br /&gt;4x^3 + 9x^2 + 6&lt;br /&gt;&lt;br /&gt;Hmmm, some things that I never really caught on to, only because I didn’t focus on them, were substitution and of course how to integrate a fraction, ahhhah I never looked at the comment on my blogs =)&lt;br /&gt;&lt;br /&gt;MY Epiphany: I FINALLY HAVE A RELATIONSHIP WITH MY CALCULATER AND IT DOES AMAZING THINGS FOR ME, I LOVE IT.&lt;br /&gt;&lt;br /&gt;To my seniors: it’s been a great run guys, I love ALL of y’all.  We’re such a close group, I consider each of you family and I don’t know where I’d be today without y’all.  It’s all coming to an end, what we’ve waited for our whole high school careers, so let’s hit up my pool these next couple weeks, walk across that stage and never look back.  I have complete confidence that everyone will be successful in what ever they chose to do. I LOVE Y’ALL!&lt;br /&gt;&lt;br /&gt;To the juniors: yall have also made my senior year one of the best.  Yall are an amazing bunch of kids with an extraordinary about of talent, don’t let any one tell y’all differently.  I hope that y’all will also have an amazing year next year and remember not to take this time for granted, or anything for that matter.  Good luck, I love every one of you.&lt;br /&gt;&lt;br /&gt;so this is it, goodbye blogggggerrr!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-3397339293447606497?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/3397339293447606497/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/well-here-it-goes.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3397339293447606497'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3397339293447606497'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/well-here-it-goes.html' title=''/><author><name>aim</name><uri>http://www.blogger.com/profile/02622890314828832713</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-7390569061958828990</id><published>2010-05-09T15:07:00.002-05:00</published><updated>2010-05-09T15:16:21.995-05:00</updated><title type='text'>Trina's Final Reflection</title><content type='html'>This year in calculus has definitely been a whirlwind for me.  I came into the class worried that everything was not going to click for me right away and guess what.... i was right.. I struggled with grasping different concepts throughout the first half of calculus but with the help of some friends.. i caught the hang of a few things.  Limit rules were the easiest to catch onto because all you had to do was follow the rules.  When we learned how to do derivatives i understood the basic derivatives but when it came to e derivatives and ln derivatives, those took a little bit longer... but with a lot of practice and a lot of studying i came to understand those too.&lt;br /&gt;&lt;br /&gt;When we were introduced to integrals i did not understand the concept at all... it took weeks of practice and help from my friends for me to finally grasp a piece of the concept.  But like derivatives, once i had worked integrals over and over again, the concept became very easy. &lt;br /&gt;&lt;br /&gt;Tangent lines were a pain at first as well.  One day Mrs. Robinson explained how to do them once more and everything just clicked in my mind.  I don't know why but it just did lol.  Whenever i would see problems like that on the AP's i would be so happy because i knew i had at least one right :). &lt;br /&gt;&lt;br /&gt;Finding volume and area on the free response portions were very easy for me.  Mrs. Robinson explained how to do those during lunch after we had taken the tests and i caught on right away.  If felt proud knowing that i could at least attempt these problems knowing what i was doing.&lt;br /&gt;&lt;br /&gt;One thing i struggled with to the very end was Riemann sums, i don't know if if was because i couldn't remember the formulas or if i just didn't know what to do with them.. but eventually i got a little bit better at them :).&lt;br /&gt;&lt;br /&gt;This class definitely helped me in preparation for any math courses i will have to be taking in college and i am glad i got the chance to take the class with my wonderful classmates.  A big thanks to everyone who helped me study and for those who supported me throughout the year.  A a huge thank you to Mrs. Robinson for helping me understand different concepts and calming my nerves throughout the year.  I will miss you all very much and good luck to the juniors who will be taking Calculus BC next year.&lt;br /&gt;&lt;br /&gt;Goodbye everyone :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-7390569061958828990?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/7390569061958828990/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/trinas-final-reflection.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7390569061958828990'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/7390569061958828990'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/trinas-final-reflection.html' title='Trina&apos;s Final Reflection'/><author><name>tlt2010</name><uri>http://www.blogger.com/profile/00560972387786673723</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='23' height='32' src='http://1.bp.blogspot.com/_ODF1ZkYNW0g/SoXrRXcgKUI/AAAAAAAAAAo/68qYTYouPt8/S220/senior2.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2044132229721427442</id><published>2010-05-09T12:10:00.003-05:00</published><updated>2010-05-09T12:24:47.471-05:00</updated><title type='text'>Ryan's Final Reflection</title><content type='html'>So I don't really know how to start this, but I guess I'll try.&lt;br /&gt;&lt;br /&gt;Well, I guess I'll have to say that Calculus was one of my favorite classes this year, not just because I liked the subject, but also because I loved my classmates.&lt;br /&gt;To all the seniors (Aimee, Jessie, Ricky, Ryan, Ryne, Chelsea, Mamie, Trina, Kaitlyn, Mher, Steven, Ellie, John, and Dylan):  I'm really going to miss you guys.&lt;br /&gt;To all of my fellow juniors (Sarah, Stephanie, Abbey, Malerie, and Ashley): Let's make our senior year the best Riverside Academy has ever seen. And just think that we're the first Calculus BC class EVER.&lt;br /&gt;(**hopefully I didn't forget anybody)&lt;br /&gt;To our teacher, Mrs. Robinson aka B-Rob, thanks for teaching us one of the "hardest subjects" in high school this year.&lt;br /&gt;&lt;br /&gt;Something I really believe I've grasped this year is taking derivatives, from just the basic x^2 to chain rules like cos(5x^3)sin(e^8x). &lt;br /&gt;&lt;br /&gt;Something I had a struggle with this year was Riemann sums.  It took me basically to the last couple of days before the AP to really understand them. &lt;br /&gt;&lt;br /&gt;To finish it off&lt;br /&gt;Class of 2010: have a great life in college, and don't forget about us :).&lt;br /&gt;Class of 2011: Let's Get It!&lt;br /&gt;&lt;br /&gt;Love you guys,&lt;br /&gt;Ryan B.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2044132229721427442?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2044132229721427442/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ryans-final-reflection.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2044132229721427442'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2044132229721427442'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ryans-final-reflection.html' title='Ryan&apos;s Final Reflection'/><author><name>ryan.is.in.calc.</name><uri>http://www.blogger.com/profile/09683413811378371672</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6920078520185877525</id><published>2010-05-07T15:26:00.002-05:00</published><updated>2010-05-07T15:49:44.096-05:00</updated><title type='text'>Jessie Green's Final Reflection</title><content type='html'>&lt;em&gt;&lt;strong&gt;WOW..school is really over for the seniors. i don't really know how i expected to feel about this but now i just feel lost. no more pulling all-nighters for calculus tests, no more getting help from john or mrs. robinson, and no more fun together as the calculus class of 2010. i guess one of the things that stuck with me the most from this class was the strive to achieve a greater goal. derivatives, slopes, optimazation, and related rates are the things i remember most that i learned. not because i hated them but because these were the things i strived to learn first. through all the hard test and page long problems we learned how much we could better ourselves and take in mass amounts of information. as we all know one needs a teacher to learn how to do all of these problems and mrs. robinson was better than anyone we could ever imagine. she made learning everything fun and easier and because of her we can go to college and be truly prepared for what is waiting for us. to the calculus class of 2010 i just want to say thank you for such a wonderful experience and to mrs. robinson thank you for being willing to teach us:)!!!&lt;br /&gt;&lt;br /&gt;the very first thing we learned this year was our derivatives:&lt;br /&gt;d/dx c=0 (c is a #)&lt;br /&gt;d/dx cu=cu' (c is #)&lt;br /&gt;d/dx cx=c (c is a #)&lt;br /&gt;d/dx u+v=u'+v'&lt;br /&gt;d/dx uv=uv'+vu'&lt;br /&gt;d/dx u/v=(vu'-uv')/v^2&lt;br /&gt;d/dx sinx=cosx(x')&lt;br /&gt;d/dx cosx=-sinx(x')&lt;br /&gt;d/dx tanx=sec^2x(x')&lt;br /&gt;d/dx secx=secxtanx(x')&lt;br /&gt;d/dx cscx=-cscxtanx(x')&lt;br /&gt;d/dx cotx=-csc^2x(x')&lt;br /&gt;d/dx lnu= 1/u(u')&lt;br /&gt;d/dx e^u=e^u(u')&lt;br /&gt;&lt;br /&gt;other things that stressed us all out!!!:&lt;br /&gt;&lt;br /&gt;linearization: The steps for working linearization problems are:&lt;br /&gt;1. Identify the equation&lt;br /&gt;2. Use the formula f(x)+f ' (x)dx&lt;br /&gt;3. Determine your dx in the problem&lt;br /&gt;4. Then determine your x in the problem&lt;br /&gt;5. Plug in everything you get&lt;br /&gt;6. Solve the equation&lt;br /&gt;&lt;br /&gt;riemann sums!!:&lt;br /&gt;The Riemanns Sums are:&lt;br /&gt;LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]&lt;br /&gt;RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]&lt;br /&gt;MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]&lt;br /&gt;Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]&lt;br /&gt;delta x=b-a/number of subintervals&lt;br /&gt;&lt;br /&gt;Equation of a tangent line!!!!:&lt;br /&gt;Take the derivative and plug in the x value.&lt;br /&gt;If you are not given a y value, plug into the original equation to get the y value.&lt;br /&gt;then plug those numbers into point slope form: y − y1 = m(x − x1)&lt;br /&gt;&lt;br /&gt;Finding critical values!!!!:&lt;br /&gt;To find critical values, first take the derivative of the function and set it equal to zero, solve for x. The answers you get for x are your critical values.&lt;br /&gt;&lt;br /&gt;Absolute extrema!!!:&lt;br /&gt;If you are given a point, plug those numbers into the original function to get another number. Alos, solve for critical values and plug those into the original function. Once you get your second numbers, you set each pair into new sets of points. The highest point is the absolute max and the smallest point is the absolute min.&lt;br /&gt;&lt;br /&gt;volume by disks!!!:&lt;br /&gt;&lt;br /&gt;the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph&lt;br /&gt;&lt;br /&gt;volume by washers!!!:&lt;br /&gt;&lt;br /&gt;the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.&lt;br /&gt;&lt;br /&gt;Limits!!!:&lt;br /&gt;&lt;br /&gt;If the degree on top is smaller than the degree on the bottom, the limit is zero.&lt;br /&gt;If the degree on top is bigger than the degree on the bottom, the limit is infinity.&lt;br /&gt;If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.&lt;br /&gt;&lt;br /&gt;First derivative test!!!:&lt;br /&gt;&lt;br /&gt;You have to take the derivative of the function and set it equal to zero. Then solve for the critical values (x values). Set those values up into intervals between negative infinity and infinity. Plug in numbers between these intervals into the first derivative to see if there are max or mins or if the graph is increasing or decreasing.&lt;br /&gt;&lt;br /&gt;Second derivative test!!!:&lt;br /&gt;&lt;br /&gt;You take the derivative of the function twice and set it equal to zero. Solve for the x values and set them up into intervals between negative infinity and infinity. Plug in numbers between those intervals into the second derivative to see where the graph is concave up, concave down, or where there is a point of inflection. &lt;br /&gt;&lt;br /&gt;Average rate of change !!!- f(b)-f(a)/(b-a).&lt;br /&gt;&lt;br /&gt;2. Rate of change !!- this is simply a derivative! plug in an x value and get a slope out, your answer&lt;br /&gt;&lt;br /&gt;3. Average value!! - this is 1/(b-a) times the integral from a to b of f(x). This is just an integral times by 1/(b-a).&lt;br /&gt;&lt;br /&gt;4. Maximums, minimums, critical values, increasing, decreasing - all this is related to first derivative test. it's simple. you take the derivative, set equal to 0, solve for x. Set up some intervals using these numbers. Plug in numbers and test your intervals. pos to neg is a min. neg to pos is a max. pos = increasing, neg = decreasing. simple stuff. remember it.&lt;br /&gt;&lt;br /&gt;5. Point of inflection, concave up, concave down - it's the second derivative test. set up intervals, if the intervals change signs, it is a point of inflection there. also, if its negative, that interval is concave down, positive is concave up. &lt;br /&gt;&lt;br /&gt;6. Slope of a normal line - take derivative, plug in x. get a slope. however, make sure you use the negative reciprocal of the slope (normal means perpendicular to). use point-slope formula.&lt;br /&gt;&lt;br /&gt;GOOD LUCK CLASS OF 2011!!!!!!!!!!!!! and may your senior year be filled with amazing times and fun:)&lt;br /&gt;&lt;br /&gt;Love, Jessie Green:):):):)!!!!!!!!!!&lt;/strong&gt;&lt;/em&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6920078520185877525?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6920078520185877525/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/jessie-greens-final-reflection.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6920078520185877525'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6920078520185877525'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/jessie-greens-final-reflection.html' title='Jessie Green&apos;s Final Reflection'/><author><name>jessie</name><uri>http://www.blogger.com/profile/00161955224234045105</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_4NPNxsaqHvw/So228Ks0qtI/AAAAAAAAAAU/bNmUZcNRtLU/S220/DSCN1625.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-109827517515654213</id><published>2010-05-02T23:48:00.003-05:00</published><updated>2010-05-03T01:41:02.796-05:00</updated><title type='text'>Ash's 37th Post</title><content type='html'>Ahh, so this week is the AP test and I'm terrified!! &gt;.&lt;&lt;br /&gt;Break a leg everyone!! &lt;br /&gt;Not literally...&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Hmm...this always makes me double guess myself! Limits!! -.-&lt;br /&gt;&lt;br /&gt;Limit Rules:&lt;br /&gt;&lt;br /&gt;1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity&lt;br /&gt;&lt;br /&gt;2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero&lt;br /&gt;&lt;br /&gt;3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;I found the steps for Linearization, &lt;span style="font-weight:bold;"&gt;but how do you identify one?&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;The Steps:&lt;br /&gt;Step One: Figure out the equation&lt;br /&gt;Step Two: Take the derivative&lt;br /&gt;Step Three: Add the derivative and originals [f(x)+f'(x)]&lt;br /&gt;Step Four: Solve for dx&lt;br /&gt;Step Five: Solve for x&lt;br /&gt;Step Six: Plug in everything &lt;span style="font-weight:bold;"&gt;((to what??))&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;I'm just really pumped for the 2 days of review..The AP exam is terrifying me and I feel like I'm going to freak out even more! &gt;.&lt;&lt;br /&gt;Oh, if anyone has tips when *NOT* using a calculator, I'd be forever in your debt :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-109827517515654213?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/109827517515654213/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ashs-37th-post.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/109827517515654213'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/109827517515654213'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/ashs-37th-post.html' title='Ash&apos;s 37th Post'/><author><name>DarkLies</name><uri>http://www.blogger.com/profile/06021157770462782617</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://1.bp.blogspot.com/_7d-Jre0G1a4/SpRzIPpSboI/AAAAAAAAAAM/lzDIPvGaT6Q/S220/Hairrr004-1.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-9010967567870267329</id><published>2010-05-02T22:43:00.002-05:00</published><updated>2010-05-02T22:50:43.997-05:00</updated><title type='text'>post 37</title><content type='html'>this week we just took more ap practice tests, and this wednesday we are all finally going to take the real ap test!  anyways, i'm going to do this blog on implicit derivatives because that's a super easy topic. :)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Implicit derivatives involve both x's and y's, unlike normal derivatives. &lt;br /&gt;1: So, first you have to take the derivative of whatever they give you as you normally would.&lt;br /&gt;2: Whenever you take the derivative of y, you have to note it with dy/dx.&lt;br /&gt;3: Solve for dy/dx&lt;br /&gt;(if you want to find slope plug in an x and y value)&lt;br /&gt;&lt;br /&gt;example: y^3+y^2-5y-x^2=-4&lt;br /&gt;First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0&lt;br /&gt;&lt;br /&gt;Then you have to solve for dy/dx, so you get:&lt;br /&gt;dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)&lt;br /&gt;&lt;br /&gt;and that's it for that problem, it's done.&lt;br /&gt;&lt;br /&gt;Another example:&lt;br /&gt;Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)&lt;br /&gt;&lt;br /&gt;First you take the derivative, which involves all kinds of product and exponent rule...&lt;br /&gt;6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))&lt;br /&gt;&lt;br /&gt;then, you need to foil it n stuff, so you get:&lt;br /&gt;12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)&lt;br /&gt;&lt;br /&gt;then, as usual, you would have to solve for dy/dx:&lt;br /&gt;dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)&lt;br /&gt;&lt;br /&gt;after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)&lt;br /&gt;&lt;br /&gt;that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before. &lt;br /&gt;&lt;br /&gt;and ummm, i have trouble with those questions where it gives you a bunch of x and y values, and it will give an interval or something, like 0&lt;x&lt;5 or something, and it will say take the derivative or something like that.  but i don't know how to take the derivative because it's not actually giving a function like it normally would.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-9010967567870267329?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/9010967567870267329/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_6350.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/9010967567870267329'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/9010967567870267329'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_6350.html' title='post 37'/><author><name>merf</name><uri>http://www.blogger.com/profile/06184639794381320730</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2503201533252368655</id><published>2010-05-02T22:23:00.002-05:00</published><updated>2010-05-02T22:41:57.677-05:00</updated><title type='text'>Posting...#37</title><content type='html'>&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;Normal lines:&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;Normal lines have the same steps and tangent lines but when you find the slope you have to take the reciprocal.&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;br /&gt;Example:&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;An equation of the line normal to the graph of y = (3x^2 +2x)^1/2 at (2,4) is&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;1. Since you already have a point, you do not need to plug in the x to find the y.&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;br /&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;2. Derivative: 1/2 (3x^2 + 2x) ^ -1/2 (6x +2)&lt;br /&gt;Plug in x to find slope: 6(2) +2 / 2 (3(2)^2 + 2(2)) ^1/2&lt;br /&gt;14/8 = 7/4&lt;br /&gt;Since this is normal, we need the negative reciprocal of the slope, which is -4/7&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;3. Now we can plug in: y-4 = -4/7 (x-2)&lt;br /&gt;&lt;br /&gt;Than change what you got and find your awnser choice&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;br /&gt;7y - 28 = -4 (x-2)&lt;br /&gt;7y - 28 = -4x + 8&lt;br /&gt;7y = -4x + 36&lt;br /&gt;4x + 7y = 36&lt;br /&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;br /&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;What i'm having trouble with is:&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;br /&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span"    style="font-family:Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif;font-size:100%;color:#204063;"&gt;&lt;span class="Apple-style-span" style="font-size: 13px;"&gt;Someone help me find AREA,&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span"    style="font-family:Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif;font-size:100%;color:#204063;"&gt;&lt;span class="Apple-style-span" style="font-size: 13px;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span"    style="font-family:Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif;font-size:100%;color:#204063;"&gt;&lt;span class="Apple-style-span" style="font-size: 13px;"&gt;and... Trapazoidal&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2503201533252368655?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2503201533252368655/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/posting.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2503201533252368655'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2503201533252368655'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/posting.html' title='Posting...#37'/><author><name>LilBigTymer</name><uri>http://www.blogger.com/profile/00493338842619363205</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='http://2.bp.blogspot.com/_pztXBOi2Cmo/SoS_x7B8KfI/AAAAAAAAAAM/Ixz4utPswXU/S220/pbnation+avatar.jpg'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6822346907145378351</id><published>2010-05-02T21:41:00.002-05:00</published><updated>2010-05-02T21:58:47.654-05:00</updated><title type='text'>Post #37</title><content type='html'>AP this week =0.&lt;br /&gt;&lt;br /&gt;I'll explain optimization because I am presenting it tomorrow.&lt;br /&gt;Steps:&lt;br /&gt;1. Determine everything you are given. Determine your primary and secondary equation.  Your primary equation is the one you are maximizing or minimizing.  Your secondary equation is usually using the other information given to you in the problem. &lt;br /&gt;&lt;br /&gt;2. Solve the secondary equation for one variable&lt;br /&gt;&lt;br /&gt;3.  Plug the solved secondary equation into the primary equation and simplify.&lt;br /&gt;&lt;br /&gt;4. Take the derivative of that equation.&lt;br /&gt;&lt;br /&gt;5. Solve that equation for the remaining variable.&lt;br /&gt;&lt;br /&gt;6.  Plug that variable back into the secondary equation to find the other variable(s). &lt;br /&gt;&lt;br /&gt;Example (some of you will see this problem on the white board tomorrow): &lt;br /&gt;We want to construct a box whose base length is 3 times the base width.  The material used to build the top and bottom costs $10/ft^2 and the material used to build the sides cost $6/ft^2.  If the box must have a volume of 50ft^3 determine the dimensions that will minimize the cost to build the box.  &lt;br /&gt;&lt;br /&gt;1. You are looking for the dimensions to minimize the box so your primary equation will be &lt;br /&gt;c= 10(2lw) + 6(2wh + 2lh) &lt;br /&gt;Note: you multiply by 10 because of the $10/ft^s for the top and bottom, and 6 for the $6/ft^2 for the sides.  &lt;br /&gt;Once distributed in, that equation simplifies to 60w^2+48wh&lt;br /&gt;Your secondary equation will deal with volume since you are given the restraint of v=50ft^3&lt;br /&gt;The volume of a box is lwh so 50 = lwh&lt;br /&gt;Since l is 3 times the width, l=3w&lt;br /&gt;The secondary equation is 50 = 3w^2h&lt;br /&gt;&lt;br /&gt;2. Solve for h: h=50/3w^2&lt;br /&gt;&lt;br /&gt;3. Plug into primary: 60w^2 + 48w (50/3w^2)&lt;br /&gt;Simplified is 60w^2 + 800/w&lt;br /&gt;&lt;br /&gt;4. Derivative: 120w - 800w^-2 &lt;br /&gt;which equals: 120w^3 - 800/ w^2&lt;br /&gt;&lt;br /&gt;5.  Solve for w by setting the top of the equation equal to 0.&lt;br /&gt;120w^3 -800 = 0&lt;br /&gt;w= 1.8821&lt;br /&gt;&lt;br /&gt;6.  Plug into equations to find other 2 variables:&lt;br /&gt;50 = 3w^2 h &lt;br /&gt;50= 3(1.8821)^2 h&lt;br /&gt;h=4.7050&lt;br /&gt;&lt;br /&gt;l=3w&lt;br /&gt;l=3(1.8821)&lt;br /&gt;l= 5.6463&lt;br /&gt;&lt;br /&gt;Hope this helps.&lt;br /&gt;&lt;br /&gt;I have questions on the invertible problems and the related rate problems with surface area such as number 13 on one of the last aps we took.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6822346907145378351?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6822346907145378351/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_9487.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6822346907145378351'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6822346907145378351'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_9487.html' title='Post #37'/><author><name>Chelsea</name><uri>http://www.blogger.com/profile/02545040999156198545</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-8155821060131594561</id><published>2010-05-02T20:52:00.001-05:00</published><updated>2010-05-02T20:54:13.401-05:00</updated><title type='text'>Post #37</title><content type='html'>5 days left :) Yay Yay Yay. But i'm still calculus illiterate.  &lt;br /&gt;&lt;br /&gt;Limits:&lt;br /&gt;&lt;br /&gt;If the bottom is 0 don’t assume the limit does not exist right away, first factor and cancel or use your calculator.&lt;br /&gt;&lt;br /&gt;Related Rates:&lt;br /&gt;&lt;br /&gt;1. Identify all variables and equations.&lt;br /&gt;2. Identify what you are looking for.&lt;br /&gt;3. Make a sketch and label.&lt;br /&gt;4. Write an equations involving your variables.&lt;br /&gt;*You can only have one unknown so a secondary equation may be given&lt;br /&gt;5. Take the derivative with respect to time.&lt;br /&gt;6. Substitute in derivative and solve.&lt;br /&gt;&lt;br /&gt;For example, The variables x and y are differentiable functions of t and are related by the equation y = 2x^3 - x + 4. When x = 2, dx/dt = -1. Find dy/dt when x = 2.Since everything is given you can skip straight to the derivative.dy/dt=6x^2dx/dt - dx/dtNow plug in all your givens in order to find dy/dt.dy/dt=6(2)^2(-1) - (-1)dy/dt= -23&lt;br /&gt;&lt;br /&gt;Volume by disks:&lt;br /&gt;The formula is pi S[r(x)]^2dx&lt;br /&gt;&lt;br /&gt;Volume by washers is the same as volume by disks except it has a hole in it. That means it is bounded by two graphs. The formula for volume by washers is Pi S top^2 - bottom^2 dx.&lt;br /&gt;&lt;br /&gt;Umm my ap score seems to be dropping every single test, so I need help.&lt;br /&gt;&lt;br /&gt;Mainly with ln and e integration and derivatives, related rates, substitution integration, remembering all the area/volume formulas, and stuff I need to figure out on my own. Even when I know what to do and know the way I have to do it I have a lot of trouble applying it and never get the answer. SO HELP.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-8155821060131594561?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/8155821060131594561/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_3050.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8155821060131594561'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8155821060131594561'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_3050.html' title='Post #37'/><author><name>Kait</name><uri>http://www.blogger.com/profile/01366494316510049421</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>3</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-9033600097150210545</id><published>2010-05-02T20:42:00.002-05:00</published><updated>2010-05-02T20:51:09.420-05:00</updated><title type='text'>thirty seven</title><content type='html'>alright, ap this week! finally all of our hard work will pay off. haha i can't wait to take the test and stop stressing about it. i am really nervous though... cuz i'm soooooo close to passing, :( oh well, we'll see what happens!&lt;br /&gt;here's some tips for taking the tests:&lt;br /&gt;&lt;br /&gt;when given huge word problems with the words, position (original function), velocity, and acceleration. USE YOUR pva CHART!&lt;br /&gt;&lt;br /&gt;Position&lt;br /&gt;Velocity&lt;br /&gt;Accelration&lt;br /&gt;&lt;br /&gt;when moving down the chart, take the derivative. (derivative - down) when moving up the chart, take the integral. (up - integrate) &lt;br /&gt;what i mean by down is, it will give you a position problem &amp;amp; ask for acceleration =&gt; take the derivative TWICE. etc.&lt;br /&gt;&lt;br /&gt;also, calculators are very useful. make sure you use them on the calculator portion. there's a reason why it says a calculator is REQUIRED for this part of the test. if you don't use it on EVERY problem, something is wrong.&lt;br /&gt;&lt;br /&gt;your calculator can integrate for you, it can find x-intercepts, and it can graph anything. always use itttttt.&lt;br /&gt;&lt;br /&gt;alsooooo, a VERY HELPFUL HINT mrs. robinson told me the other day. THIS IS CALCULUS CLASS, you should either be taking the derivative or integrating on EVERY SINGLE PROBLEM. some problems may look easy, and you may say OH i can solve this with basic algebra. but you are wrong.  the multiple choice section will even try to trick you like that. but just remember, ALWAYS integrate/derive. you need to be able to figure out which of those to do though&lt;br /&gt;&lt;br /&gt;good luck to everyone on wednesday :)&lt;br /&gt;can someone go over concavity/pts of inflection.&lt;br /&gt;you know all that first &amp;amp; second derivative stuff.&lt;br /&gt;THANKS.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-9033600097150210545?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/9033600097150210545/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/thirty-seven.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/9033600097150210545'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/9033600097150210545'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/thirty-seven.html' title='thirty seven'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6300078234029270710</id><published>2010-05-02T20:38:00.002-05:00</published><updated>2010-05-02T20:47:36.551-05:00</updated><title type='text'>post 37</title><content type='html'>Well the AP test is here on Wednesday.  So to review for it i will go over some old things we have learned this year.&lt;br /&gt;&lt;br /&gt;The steps for optimization are:&lt;br /&gt;1. Identify all quantities&lt;br /&gt;2. Write an equation&lt;br /&gt;3. Reduce equation&lt;br /&gt;4. Determine domain of equation&lt;br /&gt;5. Determine max/min values&lt;br /&gt;&lt;br /&gt;The limit rules are:&lt;br /&gt;1.  If the degree on top is smaller than the degree on the bottom, the limit is zero.&lt;br /&gt;2.  If the degree on top is bigger than the degree on the bottom, the limit is infinity.&lt;br /&gt;3.  If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.&lt;br /&gt;&lt;br /&gt;The steps of the First Derivative Test are:&lt;br /&gt;1. Take the derivative of the original problem.&lt;br /&gt;2. Set the first derivative equal to Zero.&lt;br /&gt;3. Solve for x.&lt;br /&gt;4. Create intervals for x&lt;br /&gt;5. Pick a number in the intervals then plug that number in the first derivative for x.&lt;br /&gt;6. Solve.&lt;br /&gt;&lt;br /&gt;The steps of the Second Derivative Test are:&lt;br /&gt;1. Take the derivative of the first derivative.&lt;br /&gt;2. Set the second derivative equal to Zero.&lt;br /&gt;3. Solve for x.&lt;br /&gt;4. Create intervals for x.&lt;br /&gt;5. Pick a number in the intervals then plug that number in the second derivative for x.&lt;br /&gt;6. Solve. &lt;div&gt; &lt;/div&gt;&lt;div&gt;The steps to finding absolute maxs and mins:&lt;/div&gt;&lt;div&gt;1. First derivative test&lt;/div&gt;&lt;div&gt;2. Plug critical values into the original function to get y-values&lt;/div&gt;&lt;div&gt;3. Plug endpoints into the original function to get y-values&lt;/div&gt;&lt;div&gt;4. The highest y-value is the absolute maximum&lt;/div&gt;&lt;div&gt;5. The lowest y-value is the absolute minimum&lt;/div&gt;&lt;div&gt; &lt;/div&gt;&lt;div&gt;For questions I need help with the table problems and graph problems on free response.&lt;/div&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6300078234029270710?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6300078234029270710/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_5528.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6300078234029270710'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6300078234029270710'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_5528.html' title='post 37'/><author><name>dylan b</name><uri>http://www.blogger.com/profile/11850647602537185836</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-3508938102967513501</id><published>2010-05-02T20:36:00.002-05:00</published><updated>2010-05-02T20:47:32.301-05:00</updated><title type='text'>post 37</title><content type='html'>Well the ap test is really close and I so need to study!&lt;br /&gt;&lt;br /&gt;Optimization:&lt;br /&gt;1. Identify all quantities&lt;br /&gt;2. Write an equation&lt;br /&gt;3. Reduce equation&lt;br /&gt;4. Determine domain of equation&lt;br /&gt;5. Determine max/min values&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;substitution:&lt;br /&gt;1. Find u and du&lt;br /&gt;2. set u equal to whatever isn't the derivative&lt;br /&gt;3. take the derivative of u&lt;br /&gt;4. substitute back in&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;LRAM- left hand approximation x[f(a)+f(a+x)+...f(b)]&lt;br /&gt;RRAM- right hand approximation x[f(a+x)+...f(b)]&lt;br /&gt;MRAM- approximation from the middle x[f(mid)+f(mid)+...]&lt;br /&gt;Trapezoidal- x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;average rate of change:&lt;br /&gt;just a slope; f(b)-f(a)/(b-a)&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;linearization:&lt;br /&gt;1. Pick out the equation&lt;br /&gt;2. f(x)+f`(x)dx&lt;br /&gt;3. Figure out your dx&lt;br /&gt;4. Figure out your x&lt;br /&gt;5. Plug in everything you get&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;slope field stuff from free response:&lt;br /&gt;(what to draw)&lt;br /&gt;positive slopes is /&lt;br /&gt;negative slope is \&lt;br /&gt;for a zero slope is a horizontal line&lt;br /&gt;for an undefined slope is a vertical line &lt;br /&gt;&lt;br /&gt;I sometimes just mess up on taking the derivative or integrating wrong.  Any tips for integration?&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-3508938102967513501?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/3508938102967513501/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_3724.html#comment-form' title='2 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3508938102967513501'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3508938102967513501'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_3724.html' title='post 37'/><author><name>gonzo</name><uri>http://www.blogger.com/profile/08631527109092505080</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>2</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-2926295894260159017</id><published>2010-05-02T20:16:00.002-05:00</published><updated>2010-05-02T20:35:40.355-05:00</updated><title type='text'>Post #37?</title><content type='html'>Well I was actually just studying..so I'm going to go over some random things.  &lt;br /&gt;&lt;br /&gt;Area formula: aSb top - bottom&lt;br /&gt;Volume formula: (pi) aSb (top)^2  - (bottom)^2&lt;br /&gt;&lt;strong&gt;QUESTION: when do you use aSb (top-bottom)^2?&lt;/strong&gt;&lt;br /&gt;&lt;br /&gt;limit rules as x--&gt; infinity are really simple:&lt;br /&gt;top degree = bottom degree--&gt;divide coeffient&lt;br /&gt;top degree &lt; bottom degree--&gt;0&lt;br /&gt;top degree &gt; bottom degree--&gt;infinity&lt;br /&gt;&lt;br /&gt;remember:&lt;br /&gt;&lt;br /&gt;increasing/decreasing, max/min is FIRST derivative test&lt;br /&gt;concave up/down, pt of inflection is SECOND derivative test&lt;br /&gt;&lt;br /&gt;if original graph is postive then the deivative graph is increasing&lt;br /&gt;if original graph is negative then the deivative graph is decreasing&lt;br /&gt;&lt;strong&gt;QUESTION: how do you know if it is concave up/down?&lt;/strong&gt;&lt;br /&gt;&lt;br /&gt;for a piecewise the derivatives must equal in order for it to be differentiable&lt;br /&gt;&lt;br /&gt;something I just learned:&lt;br /&gt;&lt;br /&gt;d/dx (integral from 0 to x^2 (sin(t^3) dt)&lt;br /&gt;= x^2[sinx^6]&lt;br /&gt;*pull the top bound out to the front and plug in t^3 for x&lt;br /&gt;&lt;br /&gt;tangent lines:&lt;br /&gt;take derivative&lt;br /&gt;set = to 0&lt;br /&gt;solve for x&lt;br /&gt;plug into point slope formula&lt;br /&gt;*linearzation is much like tangent lines&lt;br /&gt;&lt;br /&gt;average value:&lt;br /&gt;1/b-a (aSb) equation&lt;br /&gt;&lt;br /&gt;Mean Value Theorem:&lt;br /&gt;plug in points to original equation&lt;br /&gt;take those points you just found and the point given in problem to find slope&lt;br /&gt;take derivative of orginial equation and set = to the slope&lt;br /&gt;&lt;br /&gt;Position, Velocity, Acceleration:&lt;br /&gt;P--&gt; Postion&lt;br /&gt;V--&gt; Velocity&lt;br /&gt;A--&gt; Acceleration&lt;br /&gt;*Going down take derivative; going up take integral&lt;br /&gt;&lt;br /&gt;Calculator:&lt;br /&gt;change direction-zeros on graph&lt;br /&gt;bounds for integral-intersection on graph&lt;br /&gt;graph stuff!&lt;br /&gt;&lt;br /&gt;and:&lt;br /&gt;a corner IS continuous but NOT differentiable&lt;br /&gt;don't forget +C&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;I'm still worried about substitution &amp; ln integration!&lt;/strong&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-2926295894260159017?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/2926295894260159017/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_8969.html#comment-form' title='1 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2926295894260159017'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/2926295894260159017'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_8969.html' title='Post #37?'/><author><name>Falgoust</name><uri>http://www.blogger.com/profile/17830322899637763500</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_vMLqwS7hx8A/SpHqaEpcjUI/AAAAAAAAAAM/T3iVFzoKGrM/S220/1.jpg'/></author><thr:total>1</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-8665834479436967496</id><published>2010-05-02T18:28:00.002-05:00</published><updated>2010-05-02T18:36:08.118-05:00</updated><title type='text'>Post #37</title><content type='html'>Okay, so as i'm super nervous that wednesday is the AP test, i'm going to say a few things i think i'm actually getting the hang of.&lt;br /&gt;&lt;br /&gt;&lt;strong&gt;But first, when you figure out the lram, rram, and tram from a graph..which way do you draw the thing? taht's really confusing, but hopefully someone knows what i mean.&lt;/strong&gt;&lt;br /&gt;&lt;strong&gt;&lt;/strong&gt;&lt;br /&gt;So, when you're given a table on free response, don't freak out, i finally know what to do! It means sometime in that question you are going to have to find slope. and to find slope, you do the y-values - y-value divided by x-value - x-value.  SUPER EASY POINTS.&lt;br /&gt;&lt;br /&gt;Now, it'll prob ask you tangent line since it just made you find slope..and taht's easy too, it'll be the points you just used!&lt;br /&gt;&lt;br /&gt;Also, when they give you a wierd graph, all you have to do to find area is divide it into shapes and add all the areas.  And if it asks you for total distance traveled, its the absolute value of all the areas added together. And, if it asks where someone turned around, or something like that, its teh negative parts of the graph.&lt;br /&gt;&lt;br /&gt;Now, the question i always get wrong, when it says something about a graph's volume being area perpendicular to the x-axis, squared...all you do is solve the equations for x and then plug them into the volume formula and square the whole thing.&lt;br /&gt;&lt;br /&gt;As the AP comes closer i get more nervous and anxious at the same time.  I just hope i pass! And i hope everyone else does too, of course!&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-8665834479436967496?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/8665834479436967496/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_5385.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8665834479436967496'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8665834479436967496'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_5385.html' title='Post #37'/><author><name>Hymel</name><uri>http://www.blogger.com/profile/01618962517968489947</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://1.bp.blogspot.com/_X9p-SjoEvlU/SpChibxpMFI/AAAAAAAAAAM/jIBI61eUTB4/S220/1lead.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1608395049115399441</id><published>2010-05-02T15:08:00.001-05:00</published><updated>2010-05-02T15:11:37.083-05:00</updated><title type='text'>post 37</title><content type='html'>&lt;span style="font-weight: bold;font-size:85%;" &gt;&lt;span style="color: rgb(0, 0, 0);"&gt;last week :) &lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;The terms for the First Derivative Test:&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;1. Increasing&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;2. Decreasing&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;3. Horizontal Tangent&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;4. Min/Max&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;linearization:&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;1. Pick out the equation&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;2. f(x)+f`(x)dx&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;3. Figure out your dx&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;4. Figure out your x&lt;/span&gt;&lt;br /&gt;&lt;span style="color: rgb(0, 0, 0);"&gt;5. Plug in everything you get&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Optimization:&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;1. Identify all quantities&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;2. Write an equation&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;3. Reduce equation&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;4. Determine domain of equation&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;5. Determine max/min values&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Finding absolute max/min:&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;1. First derivative test&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;2. Plug critical values into the original function to get y-values&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;3. Plug endpoints into the original function to get y-values&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;4. The highest y-value is the absolute maximum&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;5. The lowest y-value is the absolute minimum&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Limit Rules:&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;1. if the degree of the top is bigger than the degree of the bottom, the limit is infinity.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;2. if the degree of the top is smaller than the degree of the bottom, the limit is 0.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;3. if the degree of the top is equal to the degree of the bottom, the limit is the coefficient of the leading term of the top divided by the coefficient of the leading term of the bottom equation.&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold;"&gt;Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]&lt;/span&gt;&lt;br /&gt;&lt;br /&gt;&lt;span style="font-weight: bold;font-size:85%;" &gt;&lt;span style="color: rgb(0, 0, 0);"&gt;&lt;/span&gt;i hate substiituion and i forget stuff about it.&lt;br /&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica,Arial,Verdana,'Trebuchet MS',sans-serif; font-size: 13px; color: rgb(0, 0, 0);"&gt;&lt;span class="Apple-style-span" style="font-family: arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;br /&gt;&lt;/span&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1608395049115399441?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1608395049115399441/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_02.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1608395049115399441'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1608395049115399441'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37_02.html' title='post 37'/><author><name>R.Johnson</name><uri>http://www.blogger.com/profile/16880396213810437297</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='12' height='32' src='http://1.bp.blogspot.com/__sVeTAhmH-Q/SpH5fUh01kI/AAAAAAAAAAU/ilR0mIJcsBs/S220/bustin+through+the+sign.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-1906364896951005247</id><published>2010-05-02T14:46:00.003-05:00</published><updated>2010-05-02T14:48:26.072-05:00</updated><title type='text'>37th post</title><content type='html'>&lt;em&gt;&lt;strong&gt;The steps for working linearization problems are:&lt;br /&gt;1. Identify the equation&lt;br /&gt;2. Use the formula f(x)+f ' (x)dx&lt;br /&gt;3. Determine your dx in the problem&lt;br /&gt;4. Then determine your x in the problem&lt;br /&gt;5. Plug in everything you get&lt;br /&gt;6. Solve the equation&lt;br /&gt;&lt;br /&gt;Example problem:(sr=square root) Use differentiability to approximate sr(4.5)&lt;br /&gt;f(x)=sr(x) sr(4)+(1/2 sr(4) )(.5)=1.125sr(x)+(1/2 sr(x) )&lt;br /&gt;dx error=.005&lt;br /&gt;dx=.5&lt;br /&gt;x=4&lt;br /&gt;&lt;br /&gt;The next topic I will talk about is integration. Integration finds the area under a curve. The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:&lt;br /&gt;LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]&lt;br /&gt;RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]&lt;br /&gt;MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]&lt;br /&gt;Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]&lt;br /&gt;delta x=b-a/number of subintervals&lt;br /&gt;&lt;br /&gt;Example problem: Find the area of f(x)x-3 on the interval [0,2] with 4 subintervals.&lt;br /&gt;&lt;br /&gt;delta x=2-0/4=1/2&lt;br /&gt;LRAM=1/2[f(0)+f(1/2)+f(1)+f(3/2)]1/2[-3+-5/2+-2+-3/2]1/2[-9]= -9/2&lt;br /&gt;RRAM=1/2[f(1/2)+f(1)+(3/2)+f(2)]1/2[-5/2+-2+-3/2+-1]1/2[-7]= -7/2&lt;br /&gt;MRAM=1/2[f(1/4)+f(3/4)+f(5/4)+f(7/4)]1/2[-11/4+-9/4+-7/4+-5/4]1/2[-8]=4&lt;br /&gt;Trapezoidal=1/4[f(0)+2f(1/2)+2f(1)+2f(7/2)+f(2)]1/4[-3+-10/2+-4+-6/2+-1]1/4[-16]= 4&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Equation of a tangent line:&lt;br /&gt;Take the derivative and plug in the x value.&lt;br /&gt;If you are not given a y value, plug into the original equation to get the y value.&lt;br /&gt;then plug those numbers into point slope form: y − y1 = m(x − x1)&lt;br /&gt;&lt;br /&gt;Finding critical values:&lt;br /&gt;To find critical values, first take the derivative of the function and set it equal to zero, solve for x. The answers you get for x are your critical values.&lt;br /&gt;&lt;br /&gt;Absolute extrema:&lt;br /&gt;If you are given a point, plug those numbers into the original function to get another number. Alos, solve for critical values and plug those into the original function. Once you get your second numbers, you set each pair into new sets of points. The highest point is the absolute max and the smallest point is the absolute min.&lt;br /&gt;&lt;br /&gt;volume by disks:&lt;br /&gt;&lt;br /&gt;the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph&lt;br /&gt;&lt;br /&gt;volume by washers:&lt;br /&gt;&lt;br /&gt;the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;okay so i am super confused about all those table problems on free response ....help!!!!!&lt;/strong&gt;&lt;/em&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-1906364896951005247?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/1906364896951005247/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/37th-post_02.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1906364896951005247'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/1906364896951005247'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/37th-post_02.html' title='37th post'/><author><name>jessie</name><uri>http://www.blogger.com/profile/00161955224234045105</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://4.bp.blogspot.com/_4NPNxsaqHvw/So228Ks0qtI/AAAAAAAAAAU/bNmUZcNRtLU/S220/DSCN1625.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-8820917129423448577</id><published>2010-05-02T12:15:00.002-05:00</published><updated>2010-05-02T12:22:47.121-05:00</updated><title type='text'>37th post</title><content type='html'>I can't believe this is the final blog before Wednesday's AP.  Good luck to everyone and i hope you all do well.  Here are some basic facts:&lt;br /&gt;&lt;br /&gt;Limits:&lt;br /&gt;&lt;br /&gt;If the degree on top is smaller than the degree on the bottom, the limit is zero.&lt;br /&gt;If the degree on top is bigger than the degree on the bottom, the limit is infinity.&lt;br /&gt;If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.&lt;br /&gt;&lt;br /&gt;First derivative test:&lt;br /&gt;&lt;br /&gt;You have to take the derivative of the function and set it equal to zero.  Then solve for the critical values (x values).  Set those values up into intervals between negative infinity and infinity.  Plug in numbers between these intervals into the first derivative to see if there are max or mins or if the graph is increasing or decreasing.&lt;br /&gt;&lt;br /&gt;Second derivative test:&lt;br /&gt;&lt;br /&gt;You take the derivative of the function twice and set it equal to zero.  Solve for the x values and set them up into intervals between negative infinity and infinity.  Plug in numbers between those intervals into the second derivative to see where the graph is concave up, concave down, or where there is a point of inflection. &lt;br /&gt;&lt;br /&gt;Tangent lines:&lt;br /&gt;&lt;br /&gt;You will be given a function and a x value.  If no y value is given, plug the x value into the original function and solve for y.  Then take the derivative of the function and plug in the x value to get a slope.  Then plug everything into point-slope form (y-y1=slope(x-x1)).&lt;br /&gt;&lt;br /&gt;Normal lines:&lt;br /&gt;&lt;br /&gt;You do the same thing as a tangent line except you take the negative reciprocal of the slope and plug in into point slope form.&lt;br /&gt;&lt;br /&gt;Some things i still don't quite understand:&lt;br /&gt;velocity, acceleration, and position problems&lt;br /&gt;f inverse problems&lt;br /&gt;related rates and angle of elevation&lt;br /&gt;optimization&lt;br /&gt;&lt;br /&gt;Have a great week everyone :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-8820917129423448577?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/8820917129423448577/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/37th-post.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8820917129423448577'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/8820917129423448577'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/37th-post.html' title='37th post'/><author><name>tlt2010</name><uri>http://www.blogger.com/profile/00560972387786673723</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='23' height='32' src='http://1.bp.blogspot.com/_ODF1ZkYNW0g/SoXrRXcgKUI/AAAAAAAAAAo/68qYTYouPt8/S220/senior2.JPG'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-9212810890835735031</id><published>2010-05-02T09:56:00.004-05:00</published><updated>2010-05-02T10:11:21.733-05:00</updated><title type='text'>Second to last...</title><content type='html'>It's so close guys :-D.  All of our hard work and doing these blogs and doing the APs will finally be paid off this Wednesday, the day of our AP. :-)&lt;br /&gt;&lt;br /&gt;So, on the final blog before the AP, what should I post about? Well...hm.&lt;br /&gt;&lt;br /&gt;I know, I'll do like a final study guide type ordeal.  I'll cover as much as I can remember and put it in one short post.&lt;br /&gt;&lt;br /&gt;1. Average rate of change - this is simple, don't confuse this with average value or rate of change.  This is just a slope.  f(b)-f(a)/(b-a).&lt;br /&gt;&lt;br /&gt;2. Rate of change - this is simply a derivative! plug in an x value and get a slope out, your answer&lt;br /&gt;&lt;br /&gt;3. Average value - this is 1/(b-a) times the integral from a to b of f(x). This is just an integral times by 1/(b-a).&lt;br /&gt;&lt;br /&gt;4. Maximums, minimums, critical values, increasing, decreasing - all this crap is related to first derivative test.  it's simple.  you take the derivative, set equal to 0, solve for x.  Set up some intervals using these numbers.  Plug in numbers and test your intervals.  pos to neg is a min.  neg to pos is a max. pos = increasing, neg = decreasing.  simple stuff. remember it.&lt;br /&gt;&lt;br /&gt;5. Point of inflection, concave up, concave down - it's the second derivative test.  set up intervals, if the intervals change signs, it is a point of inflection there.  also, if its negative, that interval is concave down, positive is concave up. &lt;br /&gt;&lt;br /&gt;6. Slope of a normal line - take derivative, plug in x.  get a slope. however, make sure you use the negative reciprocal of the slope (normal means perpendicular to). use point-slope formula.&lt;br /&gt;&lt;br /&gt;7. Equation of a tangent line - take derivative, plug in x.  get a slope, use point-slope formula.&lt;br /&gt;&lt;br /&gt;8. linearization - do an equation of a tangent line.  then plug in the decimal number they gave you.  then find y.&lt;br /&gt;&lt;br /&gt;9. finite limits - just plug in the number and get a value.  try to factor out before hand if possible.&lt;br /&gt;&lt;br /&gt;10. infinite limits - degreetop &gt; degreebottom = infinity, degreetop &lt; degreebottom = 0.  degreetop = degreebottom = degreetop/degreebottom&lt;br /&gt;&lt;br /&gt;11. Vertical asymptote - set bottom = to 0.  make sure you have factored and canceled anything beforehand.  &lt;br /&gt;&lt;br /&gt;12. Removables - factor top and bottom of the fraction.  make cancellations.  if something canceled, that factor is a removable. &lt;br /&gt;&lt;br /&gt;13. Horizontal asymptote - same as infinite limits. see 10.&lt;br /&gt;&lt;br /&gt;14. Area if only one equation is given - integrate the equation of and plug in two x values&lt;br /&gt;&lt;br /&gt;15. Area between two equations - find your bounds, then do the integral from a to b (your bounds) of top-bottom&lt;br /&gt;&lt;br /&gt;16. Volume by disks - find bounds, then do pi times integral from a to b of (top-bottom)^2&lt;br /&gt;&lt;br /&gt;17. Volume by washers - (has a hole in the graph when you rotate about an axis) find bounds, then do pi times the integral from a to b of (top)^2 - (bottom)^2.&lt;br /&gt;&lt;br /&gt;18. Cross sections - use the area of the cross section, and integrate that, but plug in your normal top-bottom for the variable.  for instance, s^2 is a square crosssection.  so you would do integral of s^2 which is (top-bottom)^2.&lt;br /&gt;&lt;br /&gt;Enjoy. &lt;br /&gt;&lt;br /&gt;:-)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-9212810890835735031?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/9212810890835735031/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/second-to-last.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/9212810890835735031'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/9212810890835735031'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/second-to-last.html' title='Second to last...'/><author><name>XxDohxX</name><uri>http://www.blogger.com/profile/08553932307864427605</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='23' height='32' src='http://3.bp.blogspot.com/_s-wvP590kRc/SoYHviaw8TI/AAAAAAAAACE/1_onYhrFm7c/S220/john+002.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-235512479471979291</id><published>2010-05-01T13:34:00.002-05:00</published><updated>2010-05-01T13:54:38.116-05:00</updated><title type='text'>Post #37</title><content type='html'>HEY EVERYONE...JUST ONE MORE WEEK OF CALCULUS...YEYYY!!!&lt;br /&gt;We've Made It!&lt;br /&gt;But first, don't forget we have to take the AP Calculus Test on Wednesday and we have to be there for 7:15AM&lt;br /&gt;&lt;br /&gt;Here's the Basics that we should know:&lt;br /&gt;&lt;br /&gt;First Derivative Test:&lt;br /&gt;1. Take the derivative of the original problem.&lt;br /&gt;2. Set the first derivative equal to Zero.&lt;br /&gt;3. Solve for x.&lt;br /&gt;4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)&lt;br /&gt;5. Pick a number in the intervals then plug that number in the first derivative for x.&lt;br /&gt;6. Solve.&lt;br /&gt;&lt;br /&gt;Second Derivative Test:&lt;br /&gt;1. Take the derivative of the first derivative.&lt;br /&gt;2. Set the second derivative equal to Zero.&lt;br /&gt;3. Solve for x.&lt;br /&gt;4. Create intervals for x. i.e. (-∞, 1) (1, 4) (4, ∞)&lt;br /&gt;5. Pick a number in the intervals then plug that number in the second derivative for x.&lt;br /&gt;6. Solve.&lt;br /&gt;&lt;br /&gt;limits:&lt;br /&gt;Rule #1 - When the degree (exponent) of the bottom is GREATER than the degree of the top, the limit is Zero.&lt;br /&gt;Rule #2 - When the degree (exponent) of the bottom is SMALLER than the degree of the top, the limit is infinity. (positive or negative) &lt;br /&gt;Rule #3 - When the degrees are equal, the limit is the coeffecients. &lt;br /&gt;&lt;br /&gt;linierazation:&lt;br /&gt;1. Pick out the equation&lt;br /&gt;2. f(x)+f`(x)dx&lt;br /&gt;3. Figure out your dx&lt;br /&gt;4. Figure out your x&lt;br /&gt;5. Plug in everything you get&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;implicit derivatives for 1st and 2nd derivatives:&lt;br /&gt;&lt;br /&gt;First Derivative:&lt;br /&gt;1. take the derivative of both sides&lt;br /&gt;2. everytime you take the derivative of y note it with dy/dx or y^1&lt;br /&gt;3. solve for dy/dx&lt;br /&gt;&lt;br /&gt;Second Derivative:&lt;br /&gt;first you find the first derivative and solve it for dy/dx by using the steps for the first derivative steps.&lt;br /&gt;you then take the second derivative of the solved equation. Plugging in d^2y/d^2x everytime you take the derivative of y again. and where you have dy/dx you plug in your solved equation for that.&lt;br /&gt;once you have everything plugged in and ready to go you then solve for d^2y/d^2x&lt;br /&gt;&lt;br /&gt;Intermediate Value Theorem:&lt;br /&gt;1. if f is continuous on [a,b] and k is any number between f(a)and f(b), then there is at least 1 number c when f(c)=k.&lt;br /&gt;&lt;br /&gt;HOW TO FIND THE EQUATION OF A TANGENT LINE: [this one always tricks me!!!!]&lt;br /&gt;1. take f1(x)&lt;br /&gt;2. plug x in to find your slope m&lt;br /&gt;3. plug x into f(x)to get y&lt;br /&gt;4. using m and (x,y) plug it into the equation (y-y1)=m(x-x1). &lt;br /&gt;&lt;br /&gt;&lt;br /&gt;WELL I HOPE ALL OF THIS HELPS!!&lt;br /&gt;~ElliE&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-235512479471979291?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/235512479471979291/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37.html#comment-form' title='0 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/235512479471979291'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/235512479471979291'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/05/post-37.html' title='Post #37'/><author><name>LE2010</name><uri>http://www.blogger.com/profile/05859839473790389600</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='25' height='32' src='http://1.bp.blogspot.com/_uGqGEbJSAMo/SoMXxYdCIcI/AAAAAAAAAAs/r-0CfJZXIVE/S220/me.jpg'/></author><thr:total>0</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6100343771072303135</id><published>2010-04-27T11:21:00.002-05:00</published><updated>2010-04-27T11:28:01.623-05:00</updated><title type='text'>post 36</title><content type='html'>MRS ROBINSON: sorry this is late! my internet went out last friday night when that bad storm passed through &amp;amp; my dad fixed it last night, but then i went to do my blog &amp;amp; it said i needed to "enable cookies on my browser" to go to blogger.com. so i just did it during 4th hour.. sorry!&lt;br /&gt;&lt;br /&gt;Product rule is used when anything two things are being multiplied together.Product rule states copy the first times the derivative of the second + copy the second times the derivative of the first or uv' + vu'&lt;br /&gt;&lt;br /&gt;Chain rule is used when you are finding the derivative with exponents. To do chain rule, work from the outside in.&lt;br /&gt;&lt;br /&gt;Ln Integration is only used with fractions and when the top is the derivative of the bottom&lt;br /&gt;&lt;br /&gt;To find the equation of a normal line, find the perpendicular slope and plug into the point slope formula.   First step is to take the derivative and plug in x to find the slope.  Now take the negative reciprocal. Since you are given a point and now have your slope, just plug in.&lt;br /&gt;&lt;br /&gt;The terms for the First Derivative Test:&lt;br /&gt;1. Increasing&lt;br /&gt;2. Decreasing&lt;br /&gt;3. Horizontal Tangent&lt;br /&gt;4. Min/Max&lt;br /&gt;&lt;br /&gt;linearization:&lt;br /&gt;1. Pick out the equation&lt;br /&gt;2. f(x)+f`(x)dx&lt;br /&gt;3. Figure out your dx&lt;br /&gt;4. Figure out your x&lt;br /&gt;5. Plug in everything you get&lt;br /&gt;&lt;br /&gt;can someone go over slope field for me? i don't remember all the little numbers you add together to figure out which way the lines are going to face. that's probably hard to understand how i described it, but try your best. haha :)&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6100343771072303135?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6100343771072303135/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/04/post-36_27.html#comment-form' title='5 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6100343771072303135'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6100343771072303135'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/04/post-36_27.html' title='post 36'/><author><name>Sarah</name><uri>http://www.blogger.com/profile/15779976441965969395</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='15' height='32' src='http://2.bp.blogspot.com/_pmHoFNFy9GI/SpHjCFxqqFI/AAAAAAAAAAM/QRF5BHPUbXM/S220/yum.jpg'/></author><thr:total>5</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6853104632171919664</id><published>2010-04-26T07:19:00.002-05:00</published><updated>2010-04-26T07:21:13.971-05:00</updated><title type='text'>post 36</title><content type='html'>yeah, this week all we did was take a bunch of ap exams, again.. but it's okay because i am getting better at them.  for this blog i am going to write about riemann summs.&lt;br /&gt;&lt;br /&gt;The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. You will need to know this because each of the next formulas require that you know what x is.&lt;br /&gt;&lt;br /&gt;LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]&lt;br /&gt;RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]&lt;br /&gt;MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]&lt;br /&gt;Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]&lt;br /&gt;&lt;br /&gt;i still have trouble with related rates, in my opinion, that is the hardest part of the ap tests we take, and it is on almost every free response section, so some help on that would be appreciated.&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6853104632171919664?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6853104632171919664/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/04/post-36_26.html#comment-form' title='5 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6853104632171919664'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6853104632171919664'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/04/post-36_26.html' title='post 36'/><author><name>merf</name><uri>http://www.blogger.com/profile/06184639794381320730</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='16' height='16' src='http://img2.blogblog.com/img/b16-rounded.gif'/></author><thr:total>5</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-6627202477126805134</id><published>2010-04-25T23:25:00.004-05:00</published><updated>2010-04-26T03:20:58.239-05:00</updated><title type='text'>Ash's 36th Post</title><content type='html'>This week consisted of APs again and Friday a clicker quiz! My gosh, I cannot believe how close the AP is!&lt;br /&gt;&lt;br /&gt;Let's see...&lt;br /&gt;&lt;br /&gt;How about some quick tips?&lt;br /&gt;&lt;br /&gt;1. Limit from h-&gt;0 (DON'T LOOK AT THE COEFFS!!)-- it's a definition of a derivative&lt;br /&gt;&lt;br /&gt;2. When you're asked about the Average Rate Of Change, you simply take the derivative.&lt;br /&gt;&lt;br /&gt;3. When they ask you about relative maximum, you take the derivative, solve for x, and check intervals.&lt;br /&gt;&lt;br /&gt;4. When you're asked about maximum value, take the derivative, solve for x, then plug in to get y.&lt;br /&gt;&lt;br /&gt;5. Maximum acceleration = slope = derivative&lt;br /&gt;&lt;br /&gt;6. When asked to find the vertical tangents, take the derivative.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;Some things I'm confused about:&lt;br /&gt;1. Keywords in free response - I never know what to do...oh and for slope field, 0 is a straight line, -1 is bottom of the line points to [left or right?] while +1 is where the bottom of the line points [left or right?]...also, what do you do if the number is bigger than 1?&lt;br /&gt;&lt;br /&gt;2. Acceleration, velocity, and [[the other word that I cannot remember]] - what is the trick to remembering what to do?&lt;br /&gt;&lt;br /&gt;3. Related Rates - how do I recognize them?&lt;br /&gt;&lt;br /&gt;4. TRAM - It goes like thisish right? delta/2[f(a)+2f(a+2(delta))+2f(a+3(delta))+2f(a+4(delta))+2f(a+5(delta))+f(b)] &lt;br /&gt;&lt;br /&gt;5. e! or ln! I cannot grasp these two..each time I see them in a problem I skip it or save it for last...&lt;br /&gt;&lt;br /&gt;6. OH!!! Calculator help! I'm still amazed at what it can do, but there's still lots I don't know about...also, do we ever use degrees?&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-6627202477126805134?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/6627202477126805134/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/04/ashs-36th-post.html#comment-form' title='7 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6627202477126805134'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/6627202477126805134'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/04/ashs-36th-post.html' title='Ash&apos;s 36th Post'/><author><name>DarkLies</name><uri>http://www.blogger.com/profile/06021157770462782617</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='24' height='32' src='http://1.bp.blogspot.com/_7d-Jre0G1a4/SpRzIPpSboI/AAAAAAAAAAM/lzDIPvGaT6Q/S220/Hairrr004-1.jpg'/></author><thr:total>7</thr:total></entry><entry><id>tag:blogger.com,1999:blog-3478091816824629985.post-3467087171975564856</id><published>2010-04-25T22:18:00.001-05:00</published><updated>2010-04-25T22:28:14.093-05:00</updated><title type='text'>posting...#36</title><content type='html'>&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;Linearization:&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;1.Identify the equation&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;2. Use the formula f(x)+f ' (x)dx&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;3. Determine your dx in the problem&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;4. Then determine your x in the problem&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;5. Plug in everything you get&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;6. Solve the equation&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;Related Rates:&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;1. Identify all of the variables and equations&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;2. Identify the things that you are looking for&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;3. Sketch a graph and then label that graph&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;4. Create and write an equation using all of the variables&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;5. Take the derivative of this equation with respect to time&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;span class="Apple-style-span"  style="font-family:arial;"&gt;&lt;span class="Apple-style-span" style="font-size: small;"&gt;6. Substitute everything back in&lt;/span&gt;&lt;/span&gt;&lt;br /&gt;&lt;span style="font-weight: bold; "&gt;7. Solve the equation&lt;/span&gt;&lt;/span&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;span style="font-weight: bold; "&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;span style="font-weight: bold; "&gt;&lt;span class="Apple-style-span" style="font-weight: normal; "&gt;Tangent line Example:&lt;br /&gt;&lt;br /&gt;Find the line tangent to the graph y=2x2+4x+6 at x=1.&lt;br /&gt;&lt;br /&gt;1. Identify the equation and point of tangency. If not given a y value, plug the x value into the original equation.&lt;br /&gt;&lt;br /&gt;y=2x2+4x+6 y=2(1)2&lt;br /&gt;&lt;br /&gt;2. Differentiate you equation.&lt;br /&gt;&lt;br /&gt;dy/dx=4x+4&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;3. Plug in x value then solve for dy/dx.&lt;br /&gt;&lt;br /&gt;&lt;br /&gt;dy/dx=4(1)+4=8&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;span style="font-weight: bold; "&gt;&lt;span class="Apple-style-span" style="font-weight: normal; "&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;span style="font-weight: bold; "&gt;&lt;span class="Apple-style-span" style="font-weight: normal; "&gt;&lt;br /&gt;&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;div&gt;&lt;span class="Apple-style-span" style="font-family: Helvetica, Arial, Verdana, 'Trebuchet MS', sans-serif; font-size: 13px; color: rgb(32, 64, 99); "&gt;&lt;span style="font-weight: bold; "&gt;&lt;span class="Apple-style-span" style="font-weight: normal; "&gt;i need help with alot but lets start with intergration&lt;/span&gt;&lt;/span&gt;&lt;/span&gt;&lt;/div&gt;&lt;div class="blogger-post-footer"&gt;&lt;img width='1' height='1' src='https://blogger.googleusercontent.com/tracker/3478091816824629985-3467087171975564856?l=br0910apcalc.blogspot.com' alt='' /&gt;&lt;/div&gt;</content><link rel='replies' type='application/atom+xml' href='http://br0910apcalc.blogspot.com/feeds/3467087171975564856/comments/default' title='Post Comments'/><link rel='replies' type='text/html' href='http://br0910apcalc.blogspot.com/2010/04/posting36.html#comment-form' title='3 Comments'/><link rel='edit' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3467087171975564856'/><link rel='self' type='application/atom+xml' href='http://www.blogger.com/feeds/3478091816824629985/posts/default/3467087171975564856'/><link rel='alternate' type='text/html' href='http://br0910apcalc.blogspot.com/2010/04/posting36.html' title='posting...#36'/><author><name>LilBigTymer</name><uri>http://www.blogger.com/profile/00493338842619363205</uri><email>noreply@blogger.com</email><gd:image rel='http://schemas.google.com/g/2005#thumbnail' width='26' height='32' src='http://2.bp.blogspot.com/_pztXBOi2Cmo/SoS_x7B8KfI/AAAAAAAAAAM/Ixz4utPswXU/S220/pbnation+avatar.jpg'/></author><thr:total>3</thr:total></entry></feed>
