the second week of calculus has come to a close. I understand the simeple concepts of the derivitives like the quotient rule vu^1-uv^1/v^2 and the product rule uv^1 + vu^1 and i even understand the sin, cos, tan, etc. rules as well. My biggest problem is that when you have a big problem and you have to do multiple derivitives i get a little lost. For example one problem says ln 3square root x-1/x+1. My first thought would be to treat it as a natural log 1/u X u^1. So i would make it look like 1/(x-1/x+2)^1/3. but i don't understand what u^1 would be or if i need to look at it some other way.
Another one i didn't really understand was -square root of x^2 +4 /2x^2 - 1/4 ln (2+ square root of x^2 +4/x). So i see from looking at it, it is the quotient rule and there is a square root so you have to make it into a fraction.. what confuses me if when i get to the ln part. Simplifying this was also very hard and i catch trouble with that as well. I just need some help breaking down the problem and looking at everything that needs to be done. I know the formulas and how to set them up, i just don't know exactly when to set them up, if that makes any sense.
Other than that, i have a good grasp on the limit stuff and the vertical and horizontal asymptotes for the test wednesday, i just need some help going over the big derivitive problems before we take the test. The only other thing about derivitives that i don't have a big grasp on yet is the e problems. y= e^x-e^-x/2. I see that the quotient rule is here but i don't know what needs to be done with the e's.
I think i will have a handle on the derivitives once someone helps me understand how to break down the problems and know everything that needs to be done in the problem. So all in all, my second week of calculus was not too bad, but i'm just a little confused at times.
Saturday, August 29, 2009
Friday, August 28, 2009
Post #2
So it's finally the end of week number two of school, and of Calculus. I won't even joke with you guys, I've been completely lost at some point during this week. I think I'm pretty set on all the different derivatives now. The most important thing that I continue to screw up though, however, is when there is a lot of canceling out to do. I tend to be lazy and try to find shortcuts in skipping steps but it ends up biting me in the behind anyway. :-(
So, things that completely confused me until Thursday...
When you take the derivative of un it is nun-1. Well, I didn't realize that say you have...
(x2-1)2 you actually have to use the formula nxn-1 times u'. So you get 2(x2-1)(2x) so ultimately we get 4x(x2-1).
Hope everyone remembers to do this...
After realizing that every single derivative so far that we've had is times by u' I don't think this will be hard to remember.
Anyway, I think that the test on Wednesday will be really hard...I know it covers limits so I figured I'd help everyone out.
When you take the limit as x goes to infinity, the rules are as follows
1) If the degree of the top is larger than the degree of the bottom, its infinity.
2) If the degree of the top is equal to the degree of the bottom, it is the leading coefficient of the bottom divided by the leading coefficient of the bottom.
3) If the degree of the top is smaller than the degree of the bottom, it is 0.
Other things include vertical asymptotes. To do this, you set the bottom part of the fraction equal to 0. Asymptotes occur at these x-values.
Horizontal Asymptotes follow the rules of limits. (the 1-3 of above). The asymptotes will either not occur (rule 1), at y=coeffi/coeffi (rule 2) or at y=0 (rule 3).
I won't explain them here but don't forget to study removeables, jumps, and infinite discontinuities :-p.
So, things that completely confused me until Thursday...
When you take the derivative of un it is nun-1. Well, I didn't realize that say you have...
(x2-1)2 you actually have to use the formula nxn-1 times u'. So you get 2(x2-1)(2x) so ultimately we get 4x(x2-1).
Hope everyone remembers to do this...
After realizing that every single derivative so far that we've had is times by u' I don't think this will be hard to remember.
Anyway, I think that the test on Wednesday will be really hard...I know it covers limits so I figured I'd help everyone out.
When you take the limit as x goes to infinity, the rules are as follows
1) If the degree of the top is larger than the degree of the bottom, its infinity.
2) If the degree of the top is equal to the degree of the bottom, it is the leading coefficient of the bottom divided by the leading coefficient of the bottom.
3) If the degree of the top is smaller than the degree of the bottom, it is 0.
Other things include vertical asymptotes. To do this, you set the bottom part of the fraction equal to 0. Asymptotes occur at these x-values.
Horizontal Asymptotes follow the rules of limits. (the 1-3 of above). The asymptotes will either not occur (rule 1), at y=coeffi/coeffi (rule 2) or at y=0 (rule 3).
I won't explain them here but don't forget to study removeables, jumps, and infinite discontinuities :-p.
POST #2
OH MY GOSH...THE SECOND WEEK OF CALCULUS IS OVER WITH...
well...i can honestly say that i just cannot grasp the concept of the whole thing. I understand the derivatives, and velocity, and i'm even getting the hang of it in Mu Alpha Theta. But this week in Calculus, it just didn't click.
So monday, was good, IT WAS A REVIEW DAY!...but then tuesday came [well i wasn't in class because i was absent] and when i got the notes, i was completely lost and couldn't do my homework. I sat down for hours to try and figure it out. I'm totally confused on what to do for arccos and arcsin. I know that i got the examples, but once i get to a problem i just freak out i guess. how would you do sin(arccos t) ????
Wednesday-Friday we reviewed from the beginning and even from Advanced Math and of course we had to use BASIC algebra, which seems soo hard because we never had to do THAT much algebra at once. We reviewed logs and exponets. I understood logs, like to find x when you have log2(1/8) you'll take the 2 and put the x as it's exponet and solve for x when it's set equal to (1/8). It'll give you -3!!
Hopefully I'm doing that right!
Then the determinates come back again when we have y=4 x
...it's the ULNA!..ha
And now i'm on this weekends homework and i'm already having trouble. I think it's the "basic" algebra that's killing me. I'M NOT SURE ON EXACTLY HOW TO START the problemS.
Oh and don't forget that we have a test Wednesday!...i'm scared!
well...i can honestly say that i just cannot grasp the concept of the whole thing. I understand the derivatives, and velocity, and i'm even getting the hang of it in Mu Alpha Theta. But this week in Calculus, it just didn't click.
So monday, was good, IT WAS A REVIEW DAY!...but then tuesday came [well i wasn't in class because i was absent] and when i got the notes, i was completely lost and couldn't do my homework. I sat down for hours to try and figure it out. I'm totally confused on what to do for arccos and arcsin. I know that i got the examples, but once i get to a problem i just freak out i guess. how would you do sin(arccos t) ????
Wednesday-Friday we reviewed from the beginning and even from Advanced Math and of course we had to use BASIC algebra, which seems soo hard because we never had to do THAT much algebra at once. We reviewed logs and exponets. I understood logs, like to find x when you have log2(1/8) you'll take the 2 and put the x as it's exponet and solve for x when it's set equal to (1/8). It'll give you -3!!
Hopefully I'm doing that right!
Then the determinates come back again when we have y=4 x
...it's the ULNA!..ha
And now i'm on this weekends homework and i'm already having trouble. I think it's the "basic" algebra that's killing me. I'M NOT SURE ON EXACTLY HOW TO START the problemS.
Oh and don't forget that we have a test Wednesday!...i'm scared!
post #2
Okay so the second week of calculus has past and it's getting better. I'm starting to understand derivatives alot more and both product and quotient rules. All the homework sheets have really helped me to understand the various types of complications one can come across in product and quotient rule problems. I have also learned that some of these problems are really ugly and that you can not cancel out everything if all the sides dont have something in common.
One problem I had trouble with was..y=ln(x(sqrt)x^2-1))..but thankfully mrs.robinson did it on the board and i got it. It's to messy to type it in this blog but the answer after its worked out is (2x^2-1)(x(sqrt.)x^2-1) all over (sqrt of x^2-1). But not everything was messy most of the natural logs were pretty easy. For example, ln2/3= ln^2-ln^3. So basically, its not that bad like i thought it would be. Thanks alot to milky who helped me to understand how to not be confused by all the numbers everywhere on some of the problems on homework. Oh, and listing the rules and stuff before you do the problem makes it so much easier. b-rob helped me alot when she recommended that because it really works for me. Just listing the simple stuff like the exponent rule, the inside exponent, and quotient rule makes it so much easier to work with! and also with a big problem, dont forget to work from the outside-in.
One problem I had trouble with was..y=ln(x(sqrt)x^2-1))..but thankfully mrs.robinson did it on the board and i got it. It's to messy to type it in this blog but the answer after its worked out is (2x^2-1)(x(sqrt.)x^2-1) all over (sqrt of x^2-1). But not everything was messy most of the natural logs were pretty easy. For example, ln2/3= ln^2-ln^3. So basically, its not that bad like i thought it would be. Thanks alot to milky who helped me to understand how to not be confused by all the numbers everywhere on some of the problems on homework. Oh, and listing the rules and stuff before you do the problem makes it so much easier. b-rob helped me alot when she recommended that because it really works for me. Just listing the simple stuff like the exponent rule, the inside exponent, and quotient rule makes it so much easier to work with! and also with a big problem, dont forget to work from the outside-in.
Thursday, August 27, 2009
CONFUSED!
i really don't know why...but i just can't seem to get the homework at all ever since i've been absent..
its okay at the beginning but once A LOT of algebra is involved..i'm not sure what to do!
its okay at the beginning but once A LOT of algebra is involved..i'm not sure what to do!
Tuesday, August 25, 2009
Homework
For number 44 and 45 I'm a little confused.
The arcsec formula what exactly is this IuI ?
Also, how do you solve arctan(x/a)?
Can someone help?
The arcsec formula what exactly is this IuI ?
Also, how do you solve arctan(x/a)?
Can someone help?
#'s 52-60 on homework
once i hit number 52 i was completely lost.. can someone explain what the steps are in how about doing these problems because i don't even know where to start. Thanks :)
what did we do today in Calc & MAO?
sorry i missed class today [i didn't want to but with 103 fever i kinda couldn't go to school]
but i did go to the doctor so hopefully it helps...
i don't want to be lost when i get back [whenever that is] so please let me know what i missed!
THANKS!
ellie
but i did go to the doctor so hopefully it helps...
i don't want to be lost when i get back [whenever that is] so please let me know what i missed!
THANKS!
ellie
Monday, August 24, 2009
#73-80 on homework of 8/24
I was doing fine until I got to these problems. I don't know if I over-mathed myself or if I'm just super slow. I completely lost where I was going with any of them. I even lost how to begin. I can't even get to where Kait got. Can someone re-explain how to do one of those problems in simple terms? Please and thank you!!
can anyone help me?
in the homework we had tonight, i thought i was doing them right until i got to 73-76. i worked a few and forgot to put everything over bottom squared while doing the product rule, and i don't know how to work it when i do that. for example in number 73i got x^2 - 2x all over (x - 1)^2. i factored an x out of the top and got x (x-2) all over (x-1)^2. How do i set all of that equal to zero and then solve to find the point(s) of the horizontal tangent line?
QUESTION for Mrs. Robinson
do i get credit for two for the two that i did for mabile
or does that count as one...?
or does that count as one...?
i posted my blog on thursday
okay i dont know if it matters but i posted my blog on thursday because i couldnt get on the computer this weekend...so hopefully b-rob gets this.
Sunday, August 23, 2009
merf's 1st post
This week in calculus, the learning began with average speed, which is basically a slope. To find the average speed you use the formula m=(y2-y1)/(x2-x1) , but first you have to find your y's and x's. For example, an anvil falls off the roof onto Ryan G's head, what is its average speed during the first two seconds if y=16t^2 describes the fall. Alright, to begin, the points you have to use are 0 and 2 because the length is the first two seconds, which would be the starting point (0) and the other point (2). So you plug both of those into the formula given in the problem, and so that's how you find your y's. So if you plug your x's and y's into the formula m= (y2-y1)/(x2-x1), you will get m=64/2, which is 32. so your final answer is 32 length per second. For Instantaneous speed you do the same thing, except you use the formula lim h->0 (f(x+h)-f(x))/(h). The example problem given to use for that one is extremely long, so I'm not even going to go there... Another thing I learned and know how to do pretty well (other than taking derivatives) is using the quotent rule, the formula is (vu^1 x uv^1)/v^2, but the way I learned it is, you take the derivative of the top, multiply it by the bottom, minus defivative of the bottom, times the top. For example, let's say you were given the problem y=(4x^(3/2))/(x), first you have to take the bottom, and multiply it by the derivative of the top. So the begginning of the problem would like like, x(6x^(1/2)). After that, you must subtract the derivative of the bottom, multiplyed by the top. So the second part of the problem would look like, -1(4x^(3/2)). So the full problem would now look like x(6x^(1/2))-1(4x^(3/2)). After you multiply all that out, you get (6x^(3/2))-4x^(3/2)), after you further simplify that, the final answer should be 2x^(3/2). THE END. ( I think this is like 330 something words)
mher out...
mher out...
Ashley's First Post
Okay, wow. First off, let me just clarify two things:
1) This class is not as terrifying as I thought.
2) I cannot explain things very well at all, but I’ll try.
In my opinion, Average Speed and Instantaneous Speed are both pretty easy. Since I probably can’t explain in 300 words how to do just one of those, I’ll explain both.
Let’s start with an example problem for Average Speed. (This is probably not accurate, by the way) A wolf runs at y=93+2.3(t)^2 miles in t hours. What is it’s average speed during the first 5 seconds?
Now, break this down into steps, it’s a lot simpler.
1) State the slope formula: (y2-y1)/(x2-x1)
2) Put the time you need to find into parenthesis or brackets: [0, 5] These are your two x’s.
3) Plug x1 into your formula: y=93+2.3(0)^2 = 93 93 is your y1
4) Plug x2 into your formula: y=93+2.3(5)^2=150.5 150.5 is your y2
5) Plug back into your slope formula: (150.5-93)/(5-0)
6) Simplify: 11.5 miles per hour
Now for Instantaneous Speed. Let’s use the same problem. A wolf runs at y=93+2.3(t)^2 miles per hour. What is the instantaneous speed for t=3?
1) State the Instantaneous Speed formula:
Lim f(t-h)-f(t)
H->0 h
2) In the original problem, replace y with f(t): f(t)=93+2.3(t)2
3) Plug in everything.
Lim 93+2.3(3-h)^2-93+2.3(3)^2
H->0 h
4) Foil the parenthesis.
Lim 93+2.3(9-6h+h^2)-93+2.3(9)
H->0 h
5) Start simplifying.
Lim (93+20.7 –13.8h+2.3h^2) – (93 + 20.7)
H->0 h
6) Simplify even more.
Lim 2.3h^2-13.8h+41.4
H->0 h
7) Cancel some h’s.
Lim 2.3h-13.8+41.4
H->0
8) Simplify as much as possible and then plug in 0 for h.
Lim 2.3(0)+27.6
H->0
9) Your answer is what is left: 27.6 miles per hour.
Sorry for all of the decimals! I didn't think that there would be that many when I created the problem. I think all of that is right. I HOPE all of that is right. If someone finds a mistake, other than typing error (I typed this in word and tried to transfer....fail), please feel free to make me look like an idiot! =]
Also, for what I don’t understand: Fractions in Derivatives. When there is a radical, do you make that into a fractional exponent? And when there is a fraction as the exponent, what do you do? In general, fractions as exponents confuse me…badly. Can someone break it down into 3 year old terms for me? Please and thank you!!
~Ashley
1) This class is not as terrifying as I thought.
2) I cannot explain things very well at all, but I’ll try.
In my opinion, Average Speed and Instantaneous Speed are both pretty easy. Since I probably can’t explain in 300 words how to do just one of those, I’ll explain both.
Let’s start with an example problem for Average Speed. (This is probably not accurate, by the way) A wolf runs at y=93+2.3(t)^2 miles in t hours. What is it’s average speed during the first 5 seconds?
Now, break this down into steps, it’s a lot simpler.
1) State the slope formula: (y2-y1)/(x2-x1)
2) Put the time you need to find into parenthesis or brackets: [0, 5] These are your two x’s.
3) Plug x1 into your formula: y=93+2.3(0)^2 = 93 93 is your y1
4) Plug x2 into your formula: y=93+2.3(5)^2=150.5 150.5 is your y2
5) Plug back into your slope formula: (150.5-93)/(5-0)
6) Simplify: 11.5 miles per hour
Now for Instantaneous Speed. Let’s use the same problem. A wolf runs at y=93+2.3(t)^2 miles per hour. What is the instantaneous speed for t=3?
1) State the Instantaneous Speed formula:
Lim f(t-h)-f(t)
H->0 h
2) In the original problem, replace y with f(t): f(t)=93+2.3(t)2
3) Plug in everything.
Lim 93+2.3(3-h)^2-93+2.3(3)^2
H->0 h
4) Foil the parenthesis.
Lim 93+2.3(9-6h+h^2)-93+2.3(9)
H->0 h
5) Start simplifying.
Lim (93+20.7 –13.8h+2.3h^2) – (93 + 20.7)
H->0 h
6) Simplify even more.
Lim 2.3h^2-13.8h+41.4
H->0 h
7) Cancel some h’s.
Lim 2.3h-13.8+41.4
H->0
8) Simplify as much as possible and then plug in 0 for h.
Lim 2.3(0)+27.6
H->0
9) Your answer is what is left: 27.6 miles per hour.
Sorry for all of the decimals! I didn't think that there would be that many when I created the problem. I think all of that is right. I HOPE all of that is right. If someone finds a mistake, other than typing error (I typed this in word and tried to transfer....fail), please feel free to make me look like an idiot! =]
Also, for what I don’t understand: Fractions in Derivatives. When there is a radical, do you make that into a fractional exponent? And when there is a fraction as the exponent, what do you do? In general, fractions as exponents confuse me…badly. Can someone break it down into 3 year old terms for me? Please and thank you!!
~Ashley
post #1
To begin, even though this past week has basically only focused on about three things, I must say I was extremely overwhelmed in the beginning. The first worksheet we did I completely screwed up because I used the product rule on basically everything except for on those which required use of the quotient rule, it didn’t click in my head that there was a formula for addition and subtraction until it was pointed out to me. Once I got over that frustration, everything became a lot easier. Average speed and instantaneous speed were easy concepts for me to grasp although when they used the term velocity it threw me off a little bit. The shortcuts made my day even better. My only problem with instantaneous speed is the h’s. when I plug into the formula..f(x+h) - f(x) all over h, I somehow end up with either too little or too many h's, hopefully if i keep practicing i'll get it sooner or later, preferably before the test.
Thankfully I seem to be remembering the newly introduced formulas quite well, including the dirivative formulas. Although I still cant remember the double and half angle formulas, yes, I am the reason we will be taking that quiz over and over again, sorry guys.
Ohhhhh, quotient rule. I get confused on whether or not I’m supposed to use it on a problem within a problem, for example: #27 on the worksheet we had to work on over the weekend,
x(1-4/(x+3))
..it probably isn't as complicated as i'm thinking it is, but i'm stuck.
p.s. anybody else gets mixed up between calc and statistics? i'm all backwards mann.
Thankfully I seem to be remembering the newly introduced formulas quite well, including the dirivative formulas. Although I still cant remember the double and half angle formulas, yes, I am the reason we will be taking that quiz over and over again, sorry guys.
Ohhhhh, quotient rule. I get confused on whether or not I’m supposed to use it on a problem within a problem, for example: #27 on the worksheet we had to work on over the weekend,
x(1-4/(x+3))
..it probably isn't as complicated as i'm thinking it is, but i'm stuck.
p.s. anybody else gets mixed up between calc and statistics? i'm all backwards mann.
Post #1
In Calculus class last week we learned about average speed, instantaneous speed, and Product rule and the Quotient Rule. I basically understood dervatives and I understand the Product and quotient rule; but I am not good at doing INstantaneous speed and average speed. I understand Average speed up to the m=y2-y1/x2-x1 then on instantaneous speed i don't even know where to start
P-zost #1
Alright so calculus is not all that I thought it would be. Its not that hard, if you can grasp the basic concepts. It uses your basic algebra skills and techniques that you picked up in algebraI and algebraII. Finding derivitives is not that hard, and quotient and product rule is pretty basic also. I find these so basic because it reliys on those algebra skills. The only thing that I need with these concepts is more work and play so that I can fully understand them and all the minor details that they involve, which I'm sure we will all get. For instance, I can find the simple derivitive of x^2 which = 2x. And I can find the derivitves of things such as (9x-2)/2x by using quotient rule. This concept is not hard. Something that is a little more confusing to me is x(1-(4x-2/2x)). This problem was something like one that was on the worksheet that we got on friday. I understand that it is product rule, but with that hole thing...(4x-2/2x) is messing with me. Its like quotient rule inside of product rule. lol ahhh!!! Like I have no clue where to start. Also I need more help on the shortcuts that we learned in class. I can pretty much get them besides like the ones that are like (2x-2/3). Instead of using qoutient rule cant i just make the top multiplied by the three raised to a negative 1 power (3^-1)???? idk i think that was one of the shortcuts.
Also I need more work with the shortcuts of finding average speed and instantanious speed. I get you find the derivative or something but idk where to start for the most part. Need work on those concepts. Im going to just through this out there too...I need to remember my junk from adv. math that I forgot.
Also I need more work with the shortcuts of finding average speed and instantanious speed. I get you find the derivative or something but idk where to start for the most part. Need work on those concepts. Im going to just through this out there too...I need to remember my junk from adv. math that I forgot.
Post 1
This week in calculus, we went over a lot. We had to remember finite limits, infinite limits, and points of discontinuity from last year. We also went over some new material. The new things we went over were average speed (or slope), instantaneous speed, and derivatives. We were given derivative formulas to help us and give us shortcuts for the lengthy formulas we learned for average speed and instantaneous speed.
I learned that when working average speed, you will be given a word problem, and in that word problem there will be either two numbers or a hint (in the first two seconds) to two numbers. You will also have an equation at the end of the word problem (y=16t^2). These numbers in the word problem will become the x variables in your formula. To get the y variables, you plug both the x variables into the equation to get two separate ys. Once that is done, you plug your xs and ys into the midpoint formula (because midpoint finds slope/average speed). The midpoint formula is y2-y1/x2-x1. Once you get your number, you are to write the units next to it (unit length/unit time).
I also understand the product rule formula in our derivative formulas. If you are asked to take the derivative of a product you are to copy down the first term being multiplied times the second term's derivative plus copy down the second term and times that by the first term's derivative.
For the most part, I believe I understand all the concepts of derivatives so far. I just get mixed up when a formula has a fraction, such as the quotient rule, or if a problem ends up having a negative exponent. When I look at the problem, I always want to put the variable with the negative exponent under everything, instead of just what it's supposed to be over. I also confuse myself when I see a number minus a number times something else. Such as 5-3sinx. When I see this, I automatically think I can combine these two numbers into one.
I learned that when working average speed, you will be given a word problem, and in that word problem there will be either two numbers or a hint (in the first two seconds) to two numbers. You will also have an equation at the end of the word problem (y=16t^2). These numbers in the word problem will become the x variables in your formula. To get the y variables, you plug both the x variables into the equation to get two separate ys. Once that is done, you plug your xs and ys into the midpoint formula (because midpoint finds slope/average speed). The midpoint formula is y2-y1/x2-x1. Once you get your number, you are to write the units next to it (unit length/unit time).
I also understand the product rule formula in our derivative formulas. If you are asked to take the derivative of a product you are to copy down the first term being multiplied times the second term's derivative plus copy down the second term and times that by the first term's derivative.
For the most part, I believe I understand all the concepts of derivatives so far. I just get mixed up when a formula has a fraction, such as the quotient rule, or if a problem ends up having a negative exponent. When I look at the problem, I always want to put the variable with the negative exponent under everything, instead of just what it's supposed to be over. I also confuse myself when I see a number minus a number times something else. Such as 5-3sinx. When I see this, I automatically think I can combine these two numbers into one.
Week #1
First week of Calculus finally done and with little blood shed. After hours of doing Calculus homework and helping people with Calculus, I notice that some people did not understand some questions on the first packet. An example question: "Supposed w'(-1)=18 and w(-1)=0. Find g(-1) assuming (wg)'(-1)=32." After staring at it for a good 10 minutes I finally understood it.
First off, you should recognize that they are using the product rule (uv)'=u(v') + v(u'). (w=u and g=v)
Secondly, if they -1 is confusing anyone just ignore it because they all plug -1 in for x, meaning that they are all the same.
To start this problem, write out the fully expanded product rule. Then, just start plugging in numbers where they belong. You should get (0)(g') + (g)(18)=36. Simplify the problem, which leaves you with 18g=36. Just divide 36 by 18 you find that g(-1)=2.
This also applies to problems that use the quotient rule also, so do not be scared because it turns out to be relatively simple.
One thing I really did not grasp until recently, though, was being able to manipulate the problem algebraically before taking the derivate. A good example of this is -2/(x^3). Besides the fact that it is still a simple problem, you can really save time by just moving the x^3 up, giving you -2(x^(-3)), which makes it much easier that doing the quotient rule. This little trick can save you a lot of time on any test, including an ACT or Mu Alpha Theta competition test.
Another little side note...If you understand derivatives and want to start trying to gain speed while doing them, stop and make sure you do the math correctly first because if your going too fast to realize that you did not subtract your fractional exponents correctly, you will have to go back and redo half of the problem or just not notice and get all of the problems wrong.
First off, you should recognize that they are using the product rule (uv)'=u(v') + v(u'). (w=u and g=v)
Secondly, if they -1 is confusing anyone just ignore it because they all plug -1 in for x, meaning that they are all the same.
To start this problem, write out the fully expanded product rule. Then, just start plugging in numbers where they belong. You should get (0)(g') + (g)(18)=36. Simplify the problem, which leaves you with 18g=36. Just divide 36 by 18 you find that g(-1)=2.
This also applies to problems that use the quotient rule also, so do not be scared because it turns out to be relatively simple.
One thing I really did not grasp until recently, though, was being able to manipulate the problem algebraically before taking the derivate. A good example of this is -2/(x^3). Besides the fact that it is still a simple problem, you can really save time by just moving the x^3 up, giving you -2(x^(-3)), which makes it much easier that doing the quotient rule. This little trick can save you a lot of time on any test, including an ACT or Mu Alpha Theta competition test.
Another little side note...If you understand derivatives and want to start trying to gain speed while doing them, stop and make sure you do the math correctly first because if your going too fast to realize that you did not subtract your fractional exponents correctly, you will have to go back and redo half of the problem or just not notice and get all of the problems wrong.
we learned a bunch this week like instantaneous speed and average speed and taking the derivative of many equations. i prettty much know how to take derivatives of like easy stuff. this week wasn't as hard as i thought it would be. especially when we first got all those formulas. im getting almost everything but i forget some small part of a problem every once in a while. i sometimes forget to subtract the one when taking the derivative and that makes everything wrong. when product rule and quotient rule come into play i either mess up with my signs or i do something dumb and get the entire thing wrong. i always find myslef messing up foil or something easy and stupid. i can use some help with knowing when to use quotient rule and when not to. when there is like 1/4x^4 i think im doing it right but im not sure. average speed and instantaneous speed is pretty easy once i learned how to do it the fast way. that part of calculus is easy. quotient rule is the most confusing for me with all the algebra involved. i make small mistakes that cost me. if anybody can help me solve some of this i'd appreciate it.
Ok, let me just say that I was somewhat scared when I walked in on Monday not because I didn't think I could do the work, but jsut because it was CALCULUS. However, I stressed myself out over nothing as usual.
To me, calculating the average speed was the easiest thing I've learned so far because it's just basic algebra with the slope formula. First, you identify the time you start and the time you end [t1, t2]. Each t value is the equivalent of a x value in the basic slope formula (y2 - y1)/(x2 - x1). Then you plug each t value in as your x in the equation given to describe the fall. The outputs are your y values. After plugging everything into your equation for slope, you get your average speed.
B-Rob stressed to us that we had to remember the Long way to find instantaneous speed (rate of change, definition of derivative), and I feel confident that I can do this becaus plugging into a formula is my thing. I like things thatare set in stone, so there's no question in my mind that I'll have a problem with this. You just take the limit as h goest to 0 of the function plus h minues the function all over h. So, this is it:
lim f(x + h) - f(x)
h-0 h
Then you just manipulate algebra to get the instantaneous speed or the derivative.
I really have a true understanding of everything, even after thirty-six formulas for derivatives were thrown on the board. For instance, the product rule states that when multiplying two things, you take one original times the derivative of the other and add that to the other switched up (this is better: (uv1 * vu1). I also understood how the quotient rule works and among other things (except, I keep forgetting the bottom of the quotient rule--my problem!).
The only things I'm unsure about do not actually involve doing the steps and formulas, just how to end the problem.
1. Like, after I take a derivative using the quotient rule and get a fraction, do I leave the bottom squared or foil it out?
2. Also, if I have a fractional exponent in the denominator, do I have to change it to a root and multiply by the conjugate to get rid of the radical on the bottom?
3. On the Section 2.3 sheet # 35, I didn't know what to do with three, so if anyone could help me, that would be great!
4. On the big packet (one with like everything), I didn't know how to do the ones where it started at a certain height. Could anyone explain to me an example (like #34?)? Thanks!
So overall, I think I had a good first week in Calculus. I learned a lot and am so psyched about the possiblity of getting college credit by taking the AP exam.
--Malerie
To me, calculating the average speed was the easiest thing I've learned so far because it's just basic algebra with the slope formula. First, you identify the time you start and the time you end [t1, t2]. Each t value is the equivalent of a x value in the basic slope formula (y2 - y1)/(x2 - x1). Then you plug each t value in as your x in the equation given to describe the fall. The outputs are your y values. After plugging everything into your equation for slope, you get your average speed.
B-Rob stressed to us that we had to remember the Long way to find instantaneous speed (rate of change, definition of derivative), and I feel confident that I can do this becaus plugging into a formula is my thing. I like things thatare set in stone, so there's no question in my mind that I'll have a problem with this. You just take the limit as h goest to 0 of the function plus h minues the function all over h. So, this is it:
lim f(x + h) - f(x)
h-0 h
Then you just manipulate algebra to get the instantaneous speed or the derivative.
I really have a true understanding of everything, even after thirty-six formulas for derivatives were thrown on the board. For instance, the product rule states that when multiplying two things, you take one original times the derivative of the other and add that to the other switched up (this is better: (uv1 * vu1). I also understood how the quotient rule works and among other things (except, I keep forgetting the bottom of the quotient rule--my problem!).
The only things I'm unsure about do not actually involve doing the steps and formulas, just how to end the problem.
1. Like, after I take a derivative using the quotient rule and get a fraction, do I leave the bottom squared or foil it out?
2. Also, if I have a fractional exponent in the denominator, do I have to change it to a root and multiply by the conjugate to get rid of the radical on the bottom?
3. On the Section 2.3 sheet # 35, I didn't know what to do with three, so if anyone could help me, that would be great!
4. On the big packet (one with like everything), I didn't know how to do the ones where it started at a certain height. Could anyone explain to me an example (like #34?)? Thanks!
So overall, I think I had a good first week in Calculus. I learned a lot and am so psyched about the possiblity of getting college credit by taking the AP exam.
--Malerie
Post 1
In the first class in calculus that we learned something new, I was completely lost. I was mad because I thought "I'm going to fail because I'll be lost the whole year." And Mrs. Robinson telling us we needed to know EVERYTHING from Advanced Math didn't help either. But once I got the hang of it I realized that it's not really all that bad. I guess I was getting nervous about the whole AP college credit thing. Now I'm kind of anxious to see how the rest of the school year plays out in first hour.
Something that I understood well this week was most things to do with the Derivative Formulas. I guess it just clicked in my head? Anyway I think it is very easy because one you establish what your u and v are then you can simply plug in to one (or more) of the formulas. Although I was a little confused about how to solve a derivative with multiple formulas in it in the beginning, but I'm pretty sure I got it now. And of course probably the easiest thing is that the derivative of any number is 0.
Something I am still a little confused about is average speed and instantaneous speeed. I get it for the most part but I get confused about the [f(x+h) - f(x)]/h (or instantaneous speed) formula. I always get confused on how to plug in a number into a formula that you have to plug into another formula.
One problem that I didn't quite know how to solve was a problem on the worksheet we had to do for homework this weekend (pg. 126 - chapter 2).
Number 37:
(x^2 + c^2) / (x^2 - c^2), c is a constant.
I get that you have to plug into the quotient rule. But I get stuck whenever you have to take the prime of (x^2 + c^2). I get that the derivative of x^2 is 2x. I do not know what to do with the c^2 however. Do you treat is like a number or like x? If treated like a number the derivative of it would be 0, but if treated like x the derivative would be 2c. It's racking my brain on what to do.
In conclusion to the first week, I think that as a class we are on our way to success on the AP. We can only achieve this goal if we keep up the hard work all year though. So everybody be ready to learn tomorrow!
-Ryan B.
Something that I understood well this week was most things to do with the Derivative Formulas. I guess it just clicked in my head? Anyway I think it is very easy because one you establish what your u and v are then you can simply plug in to one (or more) of the formulas. Although I was a little confused about how to solve a derivative with multiple formulas in it in the beginning, but I'm pretty sure I got it now. And of course probably the easiest thing is that the derivative of any number is 0.
Something I am still a little confused about is average speed and instantaneous speeed. I get it for the most part but I get confused about the [f(x+h) - f(x)]/h (or instantaneous speed) formula. I always get confused on how to plug in a number into a formula that you have to plug into another formula.
One problem that I didn't quite know how to solve was a problem on the worksheet we had to do for homework this weekend (pg. 126 - chapter 2).
Number 37:
(x^2 + c^2) / (x^2 - c^2), c is a constant.
I get that you have to plug into the quotient rule. But I get stuck whenever you have to take the prime of (x^2 + c^2). I get that the derivative of x^2 is 2x. I do not know what to do with the c^2 however. Do you treat is like a number or like x? If treated like a number the derivative of it would be 0, but if treated like x the derivative would be 2c. It's racking my brain on what to do.
In conclusion to the first week, I think that as a class we are on our way to success on the AP. We can only achieve this goal if we keep up the hard work all year though. So everybody be ready to learn tomorrow!
-Ryan B.
post 1
We have learned many things this past week. We learned from average speed to instantaneous speed, also we learned derivatives and their formulas. And within these things we have learned I understande some where on the other side I am having some problems with some.
The thing I understand the best is how to take derivatives of some different things. The reason is that the formulas on the derivatives are not hard to understand. One of the formulas is the product rule. This rule is one of the easiest to me. It is because you just follow the formula and its easy algebra. All you do is write the first take the derivative of the second and add that to where you write the second and take the derivative of the first. Also the derivatives any number. This is easy because the derivative of any number is equal to 0. Another thing is to take a derivative of anything with an x raised to a number. All that has to be done is multiply the coefficient by the exponent and then subtract the exponent by one. An example of this is 3x^3+4x^2-5x-5. The derivative is 9x^2+8x-5.
Also I understand how to do the average speed and instantaeous speed even though I did not really understand this at all at the beginning. The reason I started to understand this is because i realize that average has to do with over a time period and slope and instantaneous is has to do with now. This helps me do these problems.
One thing i do not understand fully is how to simplify a derivative of something like 1/3 x -3 . The reason is that i dont know how fix it when you have to make a fraction to get rid of the negative exponent.
Also I am having some problems with taking the derivative of a number to the square root, cube root, fourth root, etc. I can start it off but can't ever finish it correctly.
The thing I understand the best is how to take derivatives of some different things. The reason is that the formulas on the derivatives are not hard to understand. One of the formulas is the product rule. This rule is one of the easiest to me. It is because you just follow the formula and its easy algebra. All you do is write the first take the derivative of the second and add that to where you write the second and take the derivative of the first. Also the derivatives any number. This is easy because the derivative of any number is equal to 0. Another thing is to take a derivative of anything with an x raised to a number. All that has to be done is multiply the coefficient by the exponent and then subtract the exponent by one. An example of this is 3x^3+4x^2-5x-5. The derivative is 9x^2+8x-5.
Also I understand how to do the average speed and instantaeous speed even though I did not really understand this at all at the beginning. The reason I started to understand this is because i realize that average has to do with over a time period and slope and instantaneous is has to do with now. This helps me do these problems.
One thing i do not understand fully is how to simplify a derivative of something like 1/3 x -3 . The reason is that i dont know how fix it when you have to make a fraction to get rid of the negative exponent.
Also I am having some problems with taking the derivative of a number to the square root, cube root, fourth root, etc. I can start it off but can't ever finish it correctly.
Post 1
When I first walked in calculus class I was like "ahhh, help" and sadly I still am. I do understand how to do the product and quotient rule. I think I get the concept of everything, but I always seem to get stuck in the middle of each problem.
The only thing I'm fully confident in doing is finding the derivatives of -4x^3+2x^2-3x+1. For each term, you take the exponent and multiply it by the constant and subtract one from the exponent. Also, the derivative of any number is 0. for -4x^3: 3(-4)= -12 3-1=2 for 2x^2: 2(2)=4 2-1=1 for -3x: -3(1) = -1 1-1=0 So, then you have -12x^2+4x-3.
Also, I think I'm getting average speed, but can anyone tell me if I'm doing this right: Given the position equation s(t)=3t^2-3t-4, find the anverage velocity from t=2 to t=4. So, I ended up getting (2,4). I then plugged in 2 and then 4 to the equation 3t^2-3t-4. When I plugged in I got 2 and 32. After that I plugged (2,4) and 2, 32 in to the slope formula. For my final answer I got 15.
I think I'm having trouble more with the algebra kind of stuff. for example: [(-3x^2+5x-5)(sinx)] I understand copy the first, derivative of second, minus, copy the second, derivative of first (-3x^2+5x-5)(cosx)-(sinx)(-6x+5) but then what's the next step to solving?
Also, I don't understand problems like x(1-4/x+3) Do I use the product rule or no? I just don't know where to start for this one.
Yeah I'm kind of confused on a lot, but hopefully someone can help me before the test. (pleaseeeee haha)
The only thing I'm fully confident in doing is finding the derivatives of -4x^3+2x^2-3x+1. For each term, you take the exponent and multiply it by the constant and subtract one from the exponent. Also, the derivative of any number is 0. for -4x^3: 3(-4)= -12 3-1=2 for 2x^2: 2(2)=4 2-1=1 for -3x: -3(1) = -1 1-1=0 So, then you have -12x^2+4x-3.
Also, I think I'm getting average speed, but can anyone tell me if I'm doing this right: Given the position equation s(t)=3t^2-3t-4, find the anverage velocity from t=2 to t=4. So, I ended up getting (2,4). I then plugged in 2 and then 4 to the equation 3t^2-3t-4. When I plugged in I got 2 and 32. After that I plugged (2,4) and 2, 32 in to the slope formula. For my final answer I got 15.
I think I'm having trouble more with the algebra kind of stuff. for example: [(-3x^2+5x-5)(sinx)] I understand copy the first, derivative of second, minus, copy the second, derivative of first (-3x^2+5x-5)(cosx)-(sinx)(-6x+5) but then what's the next step to solving?
Also, I don't understand problems like x(1-4/x+3) Do I use the product rule or no? I just don't know where to start for this one.
Yeah I'm kind of confused on a lot, but hopefully someone can help me before the test. (pleaseeeee haha)
week #1
In calculus last week, we learned about derivatives, instantaneous speed, and average speed. We learned thirty-six formulas for derivatives, and also learned some for instantaneous speed and average speed. Although we learned many formulas, we only used about ten last week for derivatives. The two most important formulas that we learned, in my opinion, were quotient rule and product rule.
The thing that I am most comfortable with is average speed. The difference between instantaneous and average speed, is that instantaneous speed occurs only at one instant, while average speed occurs over a certain period of time. For average speed, you’re given a problem where an anvil falls off the roof onto Ryan G.’s head and then you are asked to find the average speed during the first 2 seconds of falling if y=16t^2 describes the fall. Your x’s in the problem would be (0, 2) because you are asked to find the average speed in the first two seconds. Then, to find your y’s, you would simply plug in your first x to the problem y=16t^2, which is 0, to find your first y. (y=16t^2, y=16(0)^2, y=0) After that, you plug in your second x to find your second y. (y=16t^2, y=16(2)^2, y=16(4), y=64) Therefore, your y’s in the problem would be (0, 64). After that, you use slope formula to find your answer. (y2-y1/x2-x1, 64-0/2-0, 64/2=32) So, your average speed would be 32 m/s.
The thing that we learned in calculus that I was most confused with was simplifying the derivatives. For example, after using the quotient rule or product rule to solve a derivative, after I plugged in everything I wouldn’t know what to do. I know that you take the derivative of the bottom times the top minus the derivative of the top times the bottom all over the bottom squared, but after I finish that first step I get really mixed up with all the cancellation rules and how to simplify it. Also, I do not know when the problem is completely simplified. I know that the easiest way to become comfortable with it is just by practicing, but if anyone can help that would also be nice. :)
The thing that I am most comfortable with is average speed. The difference between instantaneous and average speed, is that instantaneous speed occurs only at one instant, while average speed occurs over a certain period of time. For average speed, you’re given a problem where an anvil falls off the roof onto Ryan G.’s head and then you are asked to find the average speed during the first 2 seconds of falling if y=16t^2 describes the fall. Your x’s in the problem would be (0, 2) because you are asked to find the average speed in the first two seconds. Then, to find your y’s, you would simply plug in your first x to the problem y=16t^2, which is 0, to find your first y. (y=16t^2, y=16(0)^2, y=0) After that, you plug in your second x to find your second y. (y=16t^2, y=16(2)^2, y=16(4), y=64) Therefore, your y’s in the problem would be (0, 64). After that, you use slope formula to find your answer. (y2-y1/x2-x1, 64-0/2-0, 64/2=32) So, your average speed would be 32 m/s.
The thing that we learned in calculus that I was most confused with was simplifying the derivatives. For example, after using the quotient rule or product rule to solve a derivative, after I plugged in everything I wouldn’t know what to do. I know that you take the derivative of the bottom times the top minus the derivative of the top times the bottom all over the bottom squared, but after I finish that first step I get really mixed up with all the cancellation rules and how to simplify it. Also, I do not know when the problem is completely simplified. I know that the easiest way to become comfortable with it is just by practicing, but if anyone can help that would also be nice. :)
Post #1
I started this week off really confused but by Friday I was only slightly confused. The thing I understand the most is quotient rule. 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. 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. From there, it is just simple algebra. The answer comes out to (cos x) (x-1)-sin x/(x-1) ^2. You do not use quotient rule when there is only an x on the bottom. You just bring the x to the top and make the exponent negative then use the formula U^n.
The thing I am most confused about is when to or when not to use the product rule. I understand you use it when you’re multiplying obviously, but do you only use it when you’re multiplying quadratics and trig functions? I know you don’t use it when you are multiplying 8 cos x so I’m thinking if there is a number in the problem you do not use product rule but does that apply to all problems? I also get confused when there is a problem such as x (quadratic). I don’t know if I have to distribute the x or use product rule and take the derivative of the x. I also struggled with problems like x^2 x^-5^3+1/2x^4pi^3 or number 45 on our first packet, whether I can simplify an answer or not, and the wording on some problems, but I think I will eventually catch on with more practice. All in all this week has been overwhelming but I’m looking forward to spending the year with everyone.
The thing I am most confused about is when to or when not to use the product rule. I understand you use it when you’re multiplying obviously, but do you only use it when you’re multiplying quadratics and trig functions? I know you don’t use it when you are multiplying 8 cos x so I’m thinking if there is a number in the problem you do not use product rule but does that apply to all problems? I also get confused when there is a problem such as x (quadratic). I don’t know if I have to distribute the x or use product rule and take the derivative of the x. I also struggled with problems like x^2 x^-5^3+1/2x^4pi^3 or number 45 on our first packet, whether I can simplify an answer or not, and the wording on some problems, but I think I will eventually catch on with more practice. All in all this week has been overwhelming but I’m looking forward to spending the year with everyone.
Post Number One
This week we learned many different things, including average speed, instantaneous speed, and derivatives. Who knew average speed actually meant slope. I have to admit i was a bit overwhelmed when we had to copy all of those derivative formulas down, but of course once we were taught what they were I felt much better.
The first thing we learned was average speed, and i'm pretty comfortable with that. What I am not comfortable with is instantaneous speed. I get the point and know the formula, but I get mixed up while plugging everything in..as in the example Find the instantaneous speed at t=2. y=16t2 as lim of h goes to 0. Since i have the notes on this problem, I can do it without a problem. But when I need to do a different problem without my notes, I get confused with what and where to plut into the formula as lim h goes to 0 f(x+h) - f(x) all over h. Everytime I work it i never get the right amount of h's in the end, and I get completely frustrated after that. I think I just need to practice more without getting frustrated, then I might get it. The other thing I don't always understand is the quotient rule. I know that the formula is vu1 - uv1 all over v2. All that means is copy bottom(derivative of top) - top(derivative of bottom) all over bottom squared. This is easy for me to explain but when simplifying I always try to cross out too much. I need someone to verify what I can and cannot cross out, or my whole answer is screwed up. One more thing I don't understand is when we have to find the derivative of a number raised to something then that is raised to something else. Example: x raised to pi squared. Do you just bring pi in front of x and subract one from pi like any other polynomial?
Well, atleast I understood something throughout the week in Calculus, better than nothing right? Now for what I AM comfortable doing. First, I finally get product rule. It's so simple, I don't understand why i never got it in the beginning. Simply put, it's copy top(derivative of bottom) + copy bottom(derivative of top). The only thing you need to remember is the different derivative formulas that will come up in the problems. Though the next thing I understood was completely easy and everyone understood it, it's still something so i'll say it anyway: derivatives of polynomials. If I had a test of just that, I may actually pass it.
Next week, I think i'll probably catch on a lot quicker. I was a little emotional and stressed out this week due to different reasons, but I'm better now so hopefully i'm ready even though we're moving on. We have a good class, and i think we're ready for whatever we are going to face.
Thank youuuu :)
The first thing we learned was average speed, and i'm pretty comfortable with that. What I am not comfortable with is instantaneous speed. I get the point and know the formula, but I get mixed up while plugging everything in..as in the example Find the instantaneous speed at t=2. y=16t2 as lim of h goes to 0. Since i have the notes on this problem, I can do it without a problem. But when I need to do a different problem without my notes, I get confused with what and where to plut into the formula as lim h goes to 0 f(x+h) - f(x) all over h. Everytime I work it i never get the right amount of h's in the end, and I get completely frustrated after that. I think I just need to practice more without getting frustrated, then I might get it. The other thing I don't always understand is the quotient rule. I know that the formula is vu1 - uv1 all over v2. All that means is copy bottom(derivative of top) - top(derivative of bottom) all over bottom squared. This is easy for me to explain but when simplifying I always try to cross out too much. I need someone to verify what I can and cannot cross out, or my whole answer is screwed up. One more thing I don't understand is when we have to find the derivative of a number raised to something then that is raised to something else. Example: x raised to pi squared. Do you just bring pi in front of x and subract one from pi like any other polynomial?
Well, atleast I understood something throughout the week in Calculus, better than nothing right? Now for what I AM comfortable doing. First, I finally get product rule. It's so simple, I don't understand why i never got it in the beginning. Simply put, it's copy top(derivative of bottom) + copy bottom(derivative of top). The only thing you need to remember is the different derivative formulas that will come up in the problems. Though the next thing I understood was completely easy and everyone understood it, it's still something so i'll say it anyway: derivatives of polynomials. If I had a test of just that, I may actually pass it.
Next week, I think i'll probably catch on a lot quicker. I was a little emotional and stressed out this week due to different reasons, but I'm better now so hopefully i'm ready even though we're moving on. We have a good class, and i think we're ready for whatever we are going to face.
Thank youuuu :)
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