Saturday, January 23, 2010
Blogs
Remember to use the blog to go over questions from the practice APs. Even though we went through them in class you will find looking at them later you may have a question about why something was done or what happened between steps. If this is the case I prefer that you use that as your questions. Also, you can post as many questions as you want but remember to at least explain one question/concept. Try to pick one that we focused on while going over the APs. I have been very impressed with many of the recent comments and blogs. Many of you are getting very good at explaining concepts and how to work problems to each other. Don't stress about the exam now as it is still 100 plus days away (see the countdown at the bottom). There is still lots of time to work on the concepts and the way questions are formatted.
Friday, January 22, 2010
Post #23
so this week was reviewing the AP which..i defidently needed
i think i understand the non-calculator portion now..but since i missed thursday because of BETA Convention [which was soo much fun =) i'm glad i went] i missed out on all the notes that were taken in class. however, thanks to steph, i did get them, but it's not the same as going through them with mrs robinson explaining every detail, so my question is how to do most of the calculator portion so if you wanta explain any of those from 30-41 you would like to explain would be great!
lets go over a few things shall we:
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
Finding critical values:
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.
Absolute extrema:
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.
ON THE NON-CALCULATOR PORTION:
number four---i didn't understand how to use the table until we sat down and reviewed it---this is how you work this problem:
your key word[s] is average rate of change on [1,4]. since we were given a table, and the interval of function f, we look at the table to find the answer. in order to do this, we plug in [f(b)-f(a)] divided by [b-a]. Therefore, you look at the table and f(b)=6 and f(a)=2 so that's 6-2 divided by 4-1 which is 4/3.
number five---also with the table---this is how you work this problem:
your key word[s] is h'(3)= we know we need to take the derivativoe of h which is g'(f(x))times f'(x). then we look at the tabele to find what is needed. f(x) = 4 and then f'(x) is 2. g'(4) = 3 so when you put this together it gives you 1!
hope this helps!
~ElliE~
i think i understand the non-calculator portion now..but since i missed thursday because of BETA Convention [which was soo much fun =) i'm glad i went] i missed out on all the notes that were taken in class. however, thanks to steph, i did get them, but it's not the same as going through them with mrs robinson explaining every detail, so my question is how to do most of the calculator portion so if you wanta explain any of those from 30-41 you would like to explain would be great!
lets go over a few things shall we:
Equation of a tangent line:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
Finding critical values:
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.
Absolute extrema:
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.
ON THE NON-CALCULATOR PORTION:
number four---i didn't understand how to use the table until we sat down and reviewed it---this is how you work this problem:
your key word[s] is average rate of change on [1,4]. since we were given a table, and the interval of function f, we look at the table to find the answer. in order to do this, we plug in [f(b)-f(a)] divided by [b-a]. Therefore, you look at the table and f(b)=6 and f(a)=2 so that's 6-2 divided by 4-1 which is 4/3.
number five---also with the table---this is how you work this problem:
your key word[s] is h'(3)= we know we need to take the derivativoe of h which is g'(f(x))times f'(x). then we look at the tabele to find what is needed. f(x) = 4 and then f'(x) is 2. g'(4) = 3 so when you put this together it gives you 1!
hope this helps!
~ElliE~
Wednesday, January 20, 2010
Academic...blah blah blah #3
Non-Calculator Portion Question 19
Question 19
The equation of the curve is y=(4)/(1+x^2). Find the area of the region formed by the line y=4 and the curve on the interval [-1,1].
1. Set up your definite integral.
(equation of top)-(equation of bottom)
!!!(can't really create a real integral problem using alt codes)!!!
(4-(4/(1+x^2)))
2. Integrate
(4-(4/(1+x^2)))
4x-...
4/(1+x^2) is apparently a arctan (or tan^(-1)) function, which is u'/(1+u^2).
4x-4(arctan(x)
3. Plug in your interval. Since the picture of the graph showed a symmetric region, you can use [0,1] as your a and b but just multiply your final answer by 2.
4(1)-4(arctan(1)) - [4(0)-4(arctan(0))]
4-4(π/4) - [0-4(0)]
4-π-0
2(4-π)=8-2π
8-2π is answer B and that is how you find it. It may look difficult but it is actually simple algebra.
Question 19
The equation of the curve is y=(4)/(1+x^2). Find the area of the region formed by the line y=4 and the curve on the interval [-1,1].
1. Set up your definite integral.
(equation of top)-(equation of bottom)
!!!(can't really create a real integral problem using alt codes)!!!
(4-(4/(1+x^2)))
2. Integrate
(4-(4/(1+x^2)))
4x-...
4/(1+x^2) is apparently a arctan (or tan^(-1)) function, which is u'/(1+u^2).
4x-4(arctan(x)
3. Plug in your interval. Since the picture of the graph showed a symmetric region, you can use [0,1] as your a and b but just multiply your final answer by 2.
4(1)-4(arctan(1)) - [4(0)-4(arctan(0))]
4-4(π/4) - [0-4(0)]
4-π-0
2(4-π)=8-2π
8-2π is answer B and that is how you find it. It may look difficult but it is actually simple algebra.
Academic Detention #2 /yawn
Non-Calculator AP exam Questions 15 and 24
Question 15
Using the Table, estimate f'(2.1).
f(2.0)=1.39
f(2.2)=1.73
f(2.4)=2.10
f(2.6)=2.48
f(2.8)=2.88
f(3.0)=3.30
This problem is really easy if you remember what a derivative is, which is a slope. All you need to do for this problem is find the slope of two points nearest to 2.1. After picking the two points, 2.0 and 2.2, you must use the slope formula, which is f(b)-f(a) over b-a or (f(b)-f(a))/(b-a). Using the chart, you plug in f(b)=1.73 and f(a)=1.39, while b=2.2 and a=2.0.
(f(b)-f(a))/(b-a)
(f(2.2)-f(2.0)/(2.2-2.0)
(1.73 - 1.39)/(.2)
(.34)/(.2)
slope= 1.70
Question 24
At what vale of h is the rate of increase of √h twice the rate of increase of h?
First, pick out the important phrase in this question, which is rate of increase. You know that rate of increase deals with the first derivative and √h is twice as much as h in terms of the derivative, so you can create the equation √h=2h. Then you take the derivative and solve for h.
√h=2h
(1/2)(h^(-(1/2)))=2
1. Multiply (√h) over.
(1/2)=2(√h)
2. Divide 2 over.
(1/4)=(√h)
3. Square both sides.
(1/16)=h
So after a few short steps and simple algebra, you found your answer.
Question 15
Using the Table, estimate f'(2.1).
f(2.0)=1.39
f(2.2)=1.73
f(2.4)=2.10
f(2.6)=2.48
f(2.8)=2.88
f(3.0)=3.30
This problem is really easy if you remember what a derivative is, which is a slope. All you need to do for this problem is find the slope of two points nearest to 2.1. After picking the two points, 2.0 and 2.2, you must use the slope formula, which is f(b)-f(a) over b-a or (f(b)-f(a))/(b-a). Using the chart, you plug in f(b)=1.73 and f(a)=1.39, while b=2.2 and a=2.0.
(f(b)-f(a))/(b-a)
(f(2.2)-f(2.0)/(2.2-2.0)
(1.73 - 1.39)/(.2)
(.34)/(.2)
slope= 1.70
Question 24
At what vale of h is the rate of increase of √h twice the rate of increase of h?
First, pick out the important phrase in this question, which is rate of increase. You know that rate of increase deals with the first derivative and √h is twice as much as h in terms of the derivative, so you can create the equation √h=2h. Then you take the derivative and solve for h.
√h=2h
(1/2)(h^(-(1/2)))=2
1. Multiply (√h) over.
(1/2)=2(√h)
2. Divide 2 over.
(1/4)=(√h)
3. Square both sides.
(1/16)=h
So after a few short steps and simple algebra, you found your answer.
Academic Detention...again #1
On the non-calculator portion of our recent AP exam, I found that many people did not know how to work number 23.
Question 23
This question asks for you to approximate the value of y at x=3.1, given that the curve x^3+xtan(y)=27 through (3,0).
1. Implicit differentiation.
x^3+xtan(y)=27
3x^2 + xsec^2(y)dy + tan(y) = 0
2. Separate dy from the rest of your function.
3x^2+xsec^2(y)dy+tan(y)=0
xsec^2(y)dy=(-3x^2)-(tan(y))
dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))
3. Plug in your x and y values.
dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))
dy=((-3(3^2))-(tan(0)))/(3(sec^2(0)))
dy=((-3(9))-(0))/(3(1))
dy=(-27)/3
dy=-9
4. Create your equation for the tangent line.
(y-0)=(-9)(x-3)
5. Plug in 3.1
(y-0)=(-9)(x-3)
y=(-9)((3.1)-3)
y=(-9)(.1)
y=(-0.9)
And there it is. The answer for number 23 on the non-calculator portion of our AP exam.
Question 23
This question asks for you to approximate the value of y at x=3.1, given that the curve x^3+xtan(y)=27 through (3,0).
1. Implicit differentiation.
x^3+xtan(y)=27
3x^2 + xsec^2(y)dy + tan(y) = 0
2. Separate dy from the rest of your function.
3x^2+xsec^2(y)dy+tan(y)=0
xsec^2(y)dy=(-3x^2)-(tan(y))
dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))
3. Plug in your x and y values.
dy=((-3x^2)-(tan(y)))/(x(sec^2(y)))
dy=((-3(3^2))-(tan(0)))/(3(sec^2(0)))
dy=((-3(9))-(0))/(3(1))
dy=(-27)/3
dy=-9
4. Create your equation for the tangent line.
(y-0)=(-9)(x-3)
5. Plug in 3.1
(y-0)=(-9)(x-3)
y=(-9)((3.1)-3)
y=(-9)(.1)
y=(-0.9)
And there it is. The answer for number 23 on the non-calculator portion of our AP exam.
Tuesday, January 19, 2010
Post 22
Ok, so today is Tuesday. I totally forgot to do my blog over the weekend. Although my lateness might affect my grade on the blog, I hope it makes up for it in content. If I would have done my blog on Sunday, I probably would have looked up something old from my notes and just wrote it on here, but between yesterday night and today in class, I learned a few new things. The concepts of the things I learned may not exactly be new, but I learned new ways of looking at them. For the last couple of weeks, we have been working on the AP tests. We have covered mostly all the material given on the test, but many of the questions are difficult to answer; especially the calculator portion (I got a zero).
Anyway, a few things that have refreshed my memory include infinity rules, limits that go to zero, second implicit derivatives, average rate of change, relative maximum, vertical tangents, and some other stuff.
Some people think infinity rules are very easy, but many forget. For a limit approaching infinity you have three rules:
1. If degree of the top is greater than the degree of the bottom, the limit is approaching infinity
2. If the degree of the top is less than the degree of the bottom, the limit is approaching zero
3. If the degrees of the top and bottom are the same, you just make a fraction out of the coefficients in front of the variables.
Limits that go to zero:
I used to forget what to do with these. Today I learned that if a limit is going to zero, you are thinking about taking a derivative. If you see a +h, this means to definitely take a derivative.
A second derivative is just taking two derivatives, but taking a second implicit derivative can be tricky. When you are taking the first derivative of an implicit, you solve for dy/dx and that’s your answer. For the second derivative, you take the derivative of what dy/dx is equal to. When you further have more dy/dx’s in your second derivative, you plug the first dy/dx in to get your answer.
Average rate of change is not 1/b-a. It is f(b)-f(a)/b-a.
A vertical tangent is only found when a slope is undefined. First you need to find the derivative of the function and set the bottom equal to zero. You set the bottom equal to zero because zero at the bottom of any fraction means it is undefined.
These things I understand very well. These things are also on the non-calculator portion of the AP. I’m still a little confused oh now to use functions in my calculator. I forgot how to take a derivative in my calculator. So someone can help me with that.
Anyway, a few things that have refreshed my memory include infinity rules, limits that go to zero, second implicit derivatives, average rate of change, relative maximum, vertical tangents, and some other stuff.
Some people think infinity rules are very easy, but many forget. For a limit approaching infinity you have three rules:
1. If degree of the top is greater than the degree of the bottom, the limit is approaching infinity
2. If the degree of the top is less than the degree of the bottom, the limit is approaching zero
3. If the degrees of the top and bottom are the same, you just make a fraction out of the coefficients in front of the variables.
Limits that go to zero:
I used to forget what to do with these. Today I learned that if a limit is going to zero, you are thinking about taking a derivative. If you see a +h, this means to definitely take a derivative.
A second derivative is just taking two derivatives, but taking a second implicit derivative can be tricky. When you are taking the first derivative of an implicit, you solve for dy/dx and that’s your answer. For the second derivative, you take the derivative of what dy/dx is equal to. When you further have more dy/dx’s in your second derivative, you plug the first dy/dx in to get your answer.
Average rate of change is not 1/b-a. It is f(b)-f(a)/b-a.
A vertical tangent is only found when a slope is undefined. First you need to find the derivative of the function and set the bottom equal to zero. You set the bottom equal to zero because zero at the bottom of any fraction means it is undefined.
These things I understand very well. These things are also on the non-calculator portion of the AP. I’m still a little confused oh now to use functions in my calculator. I forgot how to take a derivative in my calculator. So someone can help me with that.
Monday, January 18, 2010
Post #22
Alrighty another blog..
So we are still going over ap tests, great. I have been getting straight zeroes! Yeah, so not encouraging. I tried doing corrections, and I still don't know what I'm doing. I think the ap tests are freaking me out. It's like I forget everything we have learned when I'm taking the test.
But I am going to go over some things that I learned:
1. If y=e^-x^2, then y^11(0) equals?
y^11(o)--->clue words to take derivative, take second derivative, plug in 0
e^-x^2(-2x)
e^-x^2(-2)+(-2x)e^-x^2(-2x)
e^-0^2(-2)+(-2(0))e^-0^2(-2(0))
1(-2)= -2
2. Rate of change--->SLOPE
3. lim (ln(2+h)-ln2)/h
h->0
All you have to do is take the derivative of ln2, but for this you need to treat 2 like an x. Then at the end plug 2 back in.
ln2= lnx
lnx= 1/x
1/x= 1/2
4. and just to review:
delta x=b-a/number of subintervals
LRAM-left hand approximation
delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-right hand approximation
delta x[f(a+delta x)+...f(b)]
MRAM-riddle approximation
delta x[f(mid)+f(mid)+...]
Trapezoidal
delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
DON'T KNOW:
Well I'm still having trouble with e integration, substitution, and working my calculator. Can someone describe to me how to plug in to the derivative or integral function in the calculator? Oh and can someone go over acceleration or velocity, like does that mean I'm dealing with derivatives or integrals?
So we are still going over ap tests, great. I have been getting straight zeroes! Yeah, so not encouraging. I tried doing corrections, and I still don't know what I'm doing. I think the ap tests are freaking me out. It's like I forget everything we have learned when I'm taking the test.
But I am going to go over some things that I learned:
1. If y=e^-x^2, then y^11(0) equals?
y^11(o)--->clue words to take derivative, take second derivative, plug in 0
e^-x^2(-2x)
e^-x^2(-2)+(-2x)e^-x^2(-2x)
e^-0^2(-2)+(-2(0))e^-0^2(-2(0))
1(-2)= -2
2. Rate of change--->SLOPE
3. lim (ln(2+h)-ln2)/h
h->0
All you have to do is take the derivative of ln2, but for this you need to treat 2 like an x. Then at the end plug 2 back in.
ln2= lnx
lnx= 1/x
1/x= 1/2
4. and just to review:
delta x=b-a/number of subintervals
LRAM-left hand approximation
delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-right hand approximation
delta x[f(a+delta x)+...f(b)]
MRAM-riddle approximation
delta x[f(mid)+f(mid)+...]
Trapezoidal
delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
DON'T KNOW:
Well I'm still having trouble with e integration, substitution, and working my calculator. Can someone describe to me how to plug in to the derivative or integral function in the calculator? Oh and can someone go over acceleration or velocity, like does that mean I'm dealing with derivatives or integrals?
22
LIMIT RULES:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
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.
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
i dont understand lram,
I dont understand rram
i dont understand trapiziodal
and m ram
can somebody help me with these i need help with these. Somebody help!!! It will be much appreciated.
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
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.
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
i dont understand lram,
I dont understand rram
i dont understand trapiziodal
and m ram
can somebody help me with these i need help with these. Somebody help!!! It will be much appreciated.
Week of 1/11 - 1/15
This week we took some more AP tests, so I'll do some problems from them.
Example 1:
Find the limit as x goes to infinity of (20x^2 - 13x + 5)/(5 - 4x^3)
The answer is zero because you use you're limit of infinity rules:
1) if the degree of the top equals the degree of the bottom, then the limit is the coeffients of the two made into a fraction
2) if the degree of the top is greater than the degree of the bottom then the answer is plus or minus infinity (depending on the coeffients given).
3) if the degree of the top is less than the degree of the bottom then the answer is always zero (perfect example is this problem).
Example 2:
Find the limit as h goes to zero of (ln(2+h) - ln(2))/(h).
This problem is the definition of a derivative. So you would just take the derivative of ln(2), which using the formula dy/dx ln(x) = 1/x the problem would equal 1/2.
Example 3:
If y=e^(-x^2), then y''(0) equals:
So you take the first derivative:
e^(-x^2) * (-2x)
Then take the second derivative:
e^(-x^2) * (-2x) * (-2)
Then plug in 0 for x.
e^0 * (-2)(0) * (-2)
Equals:
Zero
Linearization:
I have no idea what this is. I probably do know, I just don't know what to do with it when I see it in a problem.
Example:
And example of this would be like number 33 on AB Practice Exam 1 - Part B (calculator).
It tells you to look at a graph and then find the local linerearization of H(x) near x=3.
But in all honesty, I have no clue what that means. Help?
Example 1:
Find the limit as x goes to infinity of (20x^2 - 13x + 5)/(5 - 4x^3)
The answer is zero because you use you're limit of infinity rules:
1) if the degree of the top equals the degree of the bottom, then the limit is the coeffients of the two made into a fraction
2) if the degree of the top is greater than the degree of the bottom then the answer is plus or minus infinity (depending on the coeffients given).
3) if the degree of the top is less than the degree of the bottom then the answer is always zero (perfect example is this problem).
Example 2:
Find the limit as h goes to zero of (ln(2+h) - ln(2))/(h).
This problem is the definition of a derivative. So you would just take the derivative of ln(2), which using the formula dy/dx ln(x) = 1/x the problem would equal 1/2.
Example 3:
If y=e^(-x^2), then y''(0) equals:
So you take the first derivative:
e^(-x^2) * (-2x)
Then take the second derivative:
e^(-x^2) * (-2x) * (-2)
Then plug in 0 for x.
e^0 * (-2)(0) * (-2)
Equals:
Zero
Linearization:
I have no idea what this is. I probably do know, I just don't know what to do with it when I see it in a problem.
Example:
And example of this would be like number 33 on AB Practice Exam 1 - Part B (calculator).
It tells you to look at a graph and then find the local linerearization of H(x) near x=3.
But in all honesty, I have no clue what that means. Help?
Sunday, January 17, 2010
Post 22!
Okay, so this week we just reviewed and took AP practice exams and what not... but I was not at school friday so I do not know what we did that day. And since we have not been going over any new material, I am pretty much posting stuff that we learned a long time ago, so I sure hope I am not repeating myself here... but here is my blog anyway.
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Example: the variables x and y are functions of t and related by the equation y=2x^3-x+4 when x=2, dy/dt=-1. Find dy/dt when x=2
alright, so you put down the equation, y=2x^3-x+4.
Then you take the derivative of that, so you get dy/dt=6x^2(dx/dt)-(dx/dt)
then you plug in to find that dy/dt=6(2)^2(-1)-(-1)
and that is further simplified to, dy/dt=-23.
Linearization:
f(x)=f(c)+f'(c)(x-c)
example: Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
example 2: use differentials to approximate: sq root(16.5)
steps:
1: identify an equation--- f(x)=sq root(x)
2:f(x)+f;(x)dx--- sqrt(x)+ (1/(2sqrt(x)))(dx)
3:determine dx-- .5
4:determine x--- 16
5:plug in--- sqrt(16)+(1/2sqrt(16))(.5)= 4.0625
error= .0005
so yeah, I was not at school friday, did we take another practice ap test or what? Since we took the no calculator portion thursday I guess everyone took the calculator allowed part friday, but I was not there so I do not know. Someone fill me in!! and umm, I suck at integrating fractions!!! could someone tell me how to do that? thanks
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Example: the variables x and y are functions of t and related by the equation y=2x^3-x+4 when x=2, dy/dt=-1. Find dy/dt when x=2
alright, so you put down the equation, y=2x^3-x+4.
Then you take the derivative of that, so you get dy/dt=6x^2(dx/dt)-(dx/dt)
then you plug in to find that dy/dt=6(2)^2(-1)-(-1)
and that is further simplified to, dy/dt=-23.
Linearization:
f(x)=f(c)+f'(c)(x-c)
example: Approximate the tangent line to y=x^2 at x=1
you find all the different values: dy/dx=2x dy/dx=2 y=(1)^2=1
then you plug into the formula to get: f(x)=1+2(x-1)
example 2: use differentials to approximate: sq root(16.5)
steps:
1: identify an equation--- f(x)=sq root(x)
2:f(x)+f;(x)dx--- sqrt(x)+ (1/(2sqrt(x)))(dx)
3:determine dx-- .5
4:determine x--- 16
5:plug in--- sqrt(16)+(1/2sqrt(16))(.5)= 4.0625
error= .0005
so yeah, I was not at school friday, did we take another practice ap test or what? Since we took the no calculator portion thursday I guess everyone took the calculator allowed part friday, but I was not there so I do not know. Someone fill me in!! and umm, I suck at integrating fractions!!! could someone tell me how to do that? thanks
Post Number Twenty Two
This week in Calculus we took AP tests. I did okay on the first one, but of course horrible at the calculator portion. Go figure. I can honestly say I know most of the stuff, it’s just applying it that gives me trouble. I’m not sure what can help me with this problem but if anyone does know come talk to me!
One thing I know all the steps to but can’t do the actual problem is related rates!
1. Pick out all variables.
2. Pick out all equations.
3. Pick out what you are looking for.
4. Sketch a graph and label.
5. Make an equation with your variables.
6. Take the derivative with respect to time.
7. SUBSTITUTE back into to derivative.
8. SOLVE.
Sounds easy in steps, right? Well I still can’t do the problems. Yay me
I have a feeling limits will be on every AP test so here’s some tips.
Rules for limit of x approaching infinity:
When the top degree is the same as the bottom degree, the limit is the top coefficient over the bottom coefficient.
When the top degree is larger than the bottom degree, the limit is infinity.
When the top degree is smaller than the bottom degree, the limit is zero.
Something I may need help with is tangent lines. I always knew how to do them but now I think I psych myself up and forget how during the test. So this is an easy question someone can help me out with .
Also, I still suck at moving from original to derivative or second derivative and what not. PLEASE HELPPPP!
One thing I know all the steps to but can’t do the actual problem is related rates!
1. Pick out all variables.
2. Pick out all equations.
3. Pick out what you are looking for.
4. Sketch a graph and label.
5. Make an equation with your variables.
6. Take the derivative with respect to time.
7. SUBSTITUTE back into to derivative.
8. SOLVE.
Sounds easy in steps, right? Well I still can’t do the problems. Yay me
I have a feeling limits will be on every AP test so here’s some tips.
Rules for limit of x approaching infinity:
When the top degree is the same as the bottom degree, the limit is the top coefficient over the bottom coefficient.
When the top degree is larger than the bottom degree, the limit is infinity.
When the top degree is smaller than the bottom degree, the limit is zero.
Something I may need help with is tangent lines. I always knew how to do them but now I think I psych myself up and forget how during the test. So this is an easy question someone can help me out with .
Also, I still suck at moving from original to derivative or second derivative and what not. PLEASE HELPPPP!
Post #22?
Calculus Week #22
Okay so this week we didn't have school monday...or wednesday...so not too much happened. We did, however, take two practice AP tests...a calculator and a non-calculator portion.
On behalf on our class, I want to say...don't get discouraged early. Even if you got a 0, there's room for improvement and a lot of these problems were just simple tricks...I hope I can go over a few of the little tricks they keep throwing at us to help us get through this better as a class.
First things first...okay for the limit thing where it's a definition of derivative...I think a lot of us are figuring out that it's definition of a derivative but we are missing a few things about it...for example, if it doesn't have an x in...what that means is they just replaced x with a number. So when you take the derivative of the part behind the negative, you need to put that x back in...then after you take derivative, plug back in your number.
so if it says ln(3 + h) - ln 3 all over h. We know that we need an x...so the thing behind the minus is ln 3. Replace the 3 with an x because that's the only place it can go and you now have ln x to take the derivative of and then plug back in 3. Same thing goes for things like sqrt(3 + h) - sqrt(3) all over h. You would take the derivative of sqrt(x) then plug back in 3. Pretty easy concept--easily missed.
Next thing...don't get confused about what they are asking.. I know a few people and myself included are still giving them x values when they ask us for the value of or absolute max or min. If they say anything like that, they are talking about the y value. They WILL put in the x value in the answer to throw you off so try to catch this.
Also...don't forget your simple horizontal tangent rules because it will be an easy question to immediately get right...
1) if the degree of the top is equal to the degree of the bottom, the asymptote is at y= top/bottom and the limit is the same thing.
2) if the degree of the top is bigger than the degree of the bottom, the asymptote doesn't exist and the limit is infinity
3) if the degree of the top is smaller than the degree of the bottom, the asymptote is at y=0 and the limit is 0.
These are simple rules that should be remembered.
Anyway, that's all for now.
Okay so this week we didn't have school monday...or wednesday...so not too much happened. We did, however, take two practice AP tests...a calculator and a non-calculator portion.
On behalf on our class, I want to say...don't get discouraged early. Even if you got a 0, there's room for improvement and a lot of these problems were just simple tricks...I hope I can go over a few of the little tricks they keep throwing at us to help us get through this better as a class.
First things first...okay for the limit thing where it's a definition of derivative...I think a lot of us are figuring out that it's definition of a derivative but we are missing a few things about it...for example, if it doesn't have an x in...what that means is they just replaced x with a number. So when you take the derivative of the part behind the negative, you need to put that x back in...then after you take derivative, plug back in your number.
so if it says ln(3 + h) - ln 3 all over h. We know that we need an x...so the thing behind the minus is ln 3. Replace the 3 with an x because that's the only place it can go and you now have ln x to take the derivative of and then plug back in 3. Same thing goes for things like sqrt(3 + h) - sqrt(3) all over h. You would take the derivative of sqrt(x) then plug back in 3. Pretty easy concept--easily missed.
Next thing...don't get confused about what they are asking.. I know a few people and myself included are still giving them x values when they ask us for the value of or absolute max or min. If they say anything like that, they are talking about the y value. They WILL put in the x value in the answer to throw you off so try to catch this.
Also...don't forget your simple horizontal tangent rules because it will be an easy question to immediately get right...
1) if the degree of the top is equal to the degree of the bottom, the asymptote is at y= top/bottom and the limit is the same thing.
2) if the degree of the top is bigger than the degree of the bottom, the asymptote doesn't exist and the limit is infinity
3) if the degree of the top is smaller than the degree of the bottom, the asymptote is at y=0 and the limit is 0.
These are simple rules that should be remembered.
Anyway, that's all for now.
Post # 22
So, i'm looking through my notes and limits are something i need to glue in my brain..so i'll take this time to re-type the instructions and rules so hopefully they stay there.
Limits are used to find where an x value is going on a graph. There are two different kinds of limits, limits that appraoch infinity or negative infinity or limits that approach a number.
If you are solving a limit approaching infinity, you do these things:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
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 degrees.
Limits approaching numbers can be solved/found in a few different ways.
First, you can plug the number x is going to into the x values of the limit and see what it comes out to be. Sometimes you can do this and come out with an actual number, but other times, the bottom comes out to zero. Many people, like myself, think this means the limit is undefined, but this is not always correct. If the limit comes out to be zero, you have to use other methods to solve it. Another easy way to solve a limit is to try to break up the limit and cancle what you can, then plug in your x value. If this does not work, then you would have to plug in the limit into your calculator. If you are working with the definition of a derivative, all you have to do is take the derivative of the last term, and that is your limit!
So, hopefully this is something that people can remember..mainly myself because it always seems to slip my mind.
Now, something i'm struggling with..is it odd taht i still can't quite understand substitution. I get as far as finding u and du then i know if something's missing waht to do..but i don't understand how to plug in du into the equation..can anyone help?
Limits are used to find where an x value is going on a graph. There are two different kinds of limits, limits that appraoch infinity or negative infinity or limits that approach a number.
If you are solving a limit approaching infinity, you do these things:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
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 degrees.
Limits approaching numbers can be solved/found in a few different ways.
First, you can plug the number x is going to into the x values of the limit and see what it comes out to be. Sometimes you can do this and come out with an actual number, but other times, the bottom comes out to zero. Many people, like myself, think this means the limit is undefined, but this is not always correct. If the limit comes out to be zero, you have to use other methods to solve it. Another easy way to solve a limit is to try to break up the limit and cancle what you can, then plug in your x value. If this does not work, then you would have to plug in the limit into your calculator. If you are working with the definition of a derivative, all you have to do is take the derivative of the last term, and that is your limit!
So, hopefully this is something that people can remember..mainly myself because it always seems to slip my mind.
Now, something i'm struggling with..is it odd taht i still can't quite understand substitution. I get as far as finding u and du then i know if something's missing waht to do..but i don't understand how to plug in du into the equation..can anyone help?
Posting...#22
This week we didn't have much school and i was sick so i only took one of the practice ap
So this is how you do Optimization:
1. Identify primary and secondare equations
(primary is the one you are maximizing or minimizing)
(secondary is the other equation)
2.Solve secondary equation for one variable and plug into the primary
3.Take the derivative of primary equation, set equal to zero, then solve for x.
4. Plug into secondary equation to find the other value.
Example:
Find the length and width of a rectangle with a perimeter of 60 meters and a maximum area.
1. since your maximizing the area, the area of a rectangle formula will be the primary. A=lw
The perimeter formula would be the secondary equation because you give the perimeter 60=2l+2w
2. Solve secondary equation for one variable.
60=2l+2w
60-2w=2l
l=60-2w/2
Now plug into your primary.
A=60-2w/2(w)
3. Multiply the w in to simplify equation.
60w-2w^2/2=30w-w^2
then take the derivateve 30-2w
4. Solve for w.
20=2w
w=15
PLug in you secondary
60=2l+2(15)
60=2l+30
30+2l
l=15
I'm still having trouble with implicit derivatives so if anyone can help me that would be great thanks
So this is how you do Optimization:
1. Identify primary and secondare equations
(primary is the one you are maximizing or minimizing)
(secondary is the other equation)
2.Solve secondary equation for one variable and plug into the primary
3.Take the derivative of primary equation, set equal to zero, then solve for x.
4. Plug into secondary equation to find the other value.
Example:
Find the length and width of a rectangle with a perimeter of 60 meters and a maximum area.
1. since your maximizing the area, the area of a rectangle formula will be the primary. A=lw
The perimeter formula would be the secondary equation because you give the perimeter 60=2l+2w
2. Solve secondary equation for one variable.
60=2l+2w
60-2w=2l
l=60-2w/2
Now plug into your primary.
A=60-2w/2(w)
3. Multiply the w in to simplify equation.
60w-2w^2/2=30w-w^2
then take the derivateve 30-2w
4. Solve for w.
20=2w
w=15
PLug in you secondary
60=2l+2(15)
60=2l+30
30+2l
l=15
I'm still having trouble with implicit derivatives so if anyone can help me that would be great thanks
post 22
This past week we only had school 3 days and I missed half of one of those three. And it just so happened Mrs. Robinson was not there either so I missed the right day. But the other two days we took another practice AP test. Some things that were on it were:
Related rates were on there. The steps for related rates are:
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
The limit rules were on there too. They are:
1. if the top and bottom exponents are the same the limit is the top coefficient over the bottom coefficient.
2. if the top exponent is bigger than the bottom exponent the limit is infinity.
3. if the top is less than the bottom, it goes the limit is 0.
The Riemans sum was also on there and it is:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals
Tangent lines were on there too and the steps for finding the tangent line are:
1. Take the derivative of the equation like normal
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
For the things I am having trouble with is mainly tables and graphs. I used to know how to do them now I guess I forgot. And the biggest problem is that I need to study. If I could fix that problem I will be all right.
Related rates were on there. The steps for related rates are:
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
The limit rules were on there too. They are:
1. if the top and bottom exponents are the same the limit is the top coefficient over the bottom coefficient.
2. if the top exponent is bigger than the bottom exponent the limit is infinity.
3. if the top is less than the bottom, it goes the limit is 0.
The Riemans sum was also on there and it is:
LRAM-Left hand approximation=delta x[f(a)+f(a+delta x)+...f(b-delta x)]
RRAM-Right hand approximation=delta x[f(a+delta x)+...f(b)]
MRAM-Middle approximation=delta x[f(mid)+f(mid)+...]
Trapezoidal-delta x/2[f(a)+2f(a+delta x)+2f(a+2 delta x)+...f(b)]
*delta x=b-a/number of subintervals
Tangent lines were on there too and the steps for finding the tangent line are:
1. Take the derivative of the equation like normal
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
For the things I am having trouble with is mainly tables and graphs. I used to know how to do them now I guess I forgot. And the biggest problem is that I need to study. If I could fix that problem I will be all right.
Post #22
This week I learned that I need a lot more practice if I plan to pass the AP.
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation
5. Take the derivative with respect to time
6. Substitute in derivative and solve.
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/dx=-1. find dy/dt when x=2.
1. Given: x=2 dx=dt=-1 y=2x^3-x+4
2. Unknown: dy/dt = ?
3. You are given the equation: y=2x^3-x+4
4. Derivative with respect to time: dy/dt=6x^2 dx/dt - dx/dt
5. Plug in given: dy/dt= 6(2)^2 (-1) - (-1)
dy/dt = -23
Example 2:
Air is being pumped into a spherical balloon at a rate of 4.5 cubic ft/min. Find the rate of change of the radius when the radius is 2ft.
1. Given: r=2 dv/dt = 4.5 ft^3/min
2. Unknown: dr/dt = ?
3. Equation will be volume of a sphere: v=4/3 pi r^3
4. Derivative: dv/dt = 4/3 pi 3 r^2 dr/dt
5. Plug in: 4.5 ft^3/min = 4/3 pi 3 (2ft)^2 dr/dt
4.5ft^3/min = 16 pi ft^2 dr/dt
dr/dt = 9/32 pi ft/min
Substitution:
It takes the place of the derivative rules.
1. Find a derivative inside the integral
2. Set u=the non-derivative
3. Take derivative of u
4. Substitute back in
Example:
Integrate (x^2+1) (2x) dx
Since you can't integrate product rule, you know you have to use substitution.
u=x^2 + 1 du= 2x dx
Integrate: u du
1/2 u ^2
Plug u back in
1/2 ( x^2+1)^2 + c
Example 2:
Integrate x (x^2+1)^2 dx
u= x^2 +1 du= 2x
Since you are missing a 2 in the problem, you have to add a 1/2 to get rid of the 2
1/2 S u^2 du
1/2 (1/3) u^3
1/6 (x^2 +1) ^3 + c
I need help on numbers 12, 17-20, 22, 24-26, and 28 on non-calculator portion and 29, 31-34, 37-45 on the calculator portion. I know it's a lot.
Post #22
After reading Trina's post, I've decided to post my own blog and answer a few of her questions.
1. Table Problems
Questions 4 and 5 on the non-calculator portion are questions that use a table, which is given to you.
Question 4 asks what the average rate of change of the function F on [1,4] is. To solve this problem just take your values for F(1) and F(4) and plug them into the slope formula, (F(b)-F(a))/(b-a). Using your graph you find that F(1)=2 and F(4)=6.
Plug 4 in for b and 1 in for a, giving you (6-2)/(4-1)= 4/3
Question 5 asks you to find H'(3) when H(x)=G(F(x)). All you do is differentiate, or take the derivative, of H(x). In order to do this though, you need to realize that G(F(x)) can be differentiated similar to sin(2x). You differentiate sin(2x) by taking the derivative of the outside, sin, and keeping the same inside, then you multiply everything by the derivative of the inside, which would give you cos(2x)*(2), or 2cos(2x). Now that you realize this, you can differentiate H(x). First derivative of outside, G'(f(x)), then you multiply by the derivative of your inside, G'(F(x))*F'(x). After all of this you plug in 3 for x. G'(F(x))*F'(x); G'(F(3))*F'(3); G'(4)*(2); (1/2)*(2)=1.
3. Graph Problems
Question 10 asks at what time does the object attain it's maximum acceleration, giving the graph of the object's velocity. The key word here is the object's velocity. Since a derivative is just the slope of a graph, your actually looking for where on the graph is the slope the steepest, or highest y value. After looking at the graph, you notice that at 8 < t < 9 is where the slope is the greatest because it goes from -4 to 0 in one second, which gives you a slope=4.
Question 11 asks at what time is the object farthest from the starting point. To answer this question, you need to use the rules of interpreting graphs. Knowing that position is the original function, velocity the first derivative, and acceleration the second derivative. Now that you know your going from your first derivative to the original graph, you just draw it by using your rules: 1. Any part of the first derivative graph that is above the x axis is increasing on the original graph; 2. Any part of the first derivative graph that is below the x axis is decreasing on the original graph. So from 0 to 6 the original graph is increasing, and from 6 to 9 the original graph is decreasing. This creates a negative parabola negative absolute value graph. Once done drawing just find the maximum of that graph and that is your answer, which is 6.
Question 12 states that at t=8, the object was at position x=10 at t=5, the object's position was x=?. To be honest i just guessed on this question, knowing that my position graph was a parabola and at t=6 that was the farthest possible point that the object could travel from the starting point, so that ruled out answer choice E. Next i saw that at t=8, x=10, and since 8 is farther away from 6 than 5 was, I just chose 13, answer choice D, because it made sense, being that it was greater that 10, but I did not want to choose 15, choice E, because that seemed like it would too big.
I hope I can help with this blog Trina.
1. Table Problems
Questions 4 and 5 on the non-calculator portion are questions that use a table, which is given to you.
Question 4 asks what the average rate of change of the function F on [1,4] is. To solve this problem just take your values for F(1) and F(4) and plug them into the slope formula, (F(b)-F(a))/(b-a). Using your graph you find that F(1)=2 and F(4)=6.
Plug 4 in for b and 1 in for a, giving you (6-2)/(4-1)= 4/3
Question 5 asks you to find H'(3) when H(x)=G(F(x)). All you do is differentiate, or take the derivative, of H(x). In order to do this though, you need to realize that G(F(x)) can be differentiated similar to sin(2x). You differentiate sin(2x) by taking the derivative of the outside, sin, and keeping the same inside, then you multiply everything by the derivative of the inside, which would give you cos(2x)*(2), or 2cos(2x). Now that you realize this, you can differentiate H(x). First derivative of outside, G'(f(x)), then you multiply by the derivative of your inside, G'(F(x))*F'(x). After all of this you plug in 3 for x. G'(F(x))*F'(x); G'(F(3))*F'(3); G'(4)*(2); (1/2)*(2)=1.
3. Graph Problems
Question 10 asks at what time does the object attain it's maximum acceleration, giving the graph of the object's velocity. The key word here is the object's velocity. Since a derivative is just the slope of a graph, your actually looking for where on the graph is the slope the steepest, or highest y value. After looking at the graph, you notice that at 8 < t < 9 is where the slope is the greatest because it goes from -4 to 0 in one second, which gives you a slope=4.
Question 11 asks at what time is the object farthest from the starting point. To answer this question, you need to use the rules of interpreting graphs. Knowing that position is the original function, velocity the first derivative, and acceleration the second derivative. Now that you know your going from your first derivative to the original graph, you just draw it by using your rules: 1. Any part of the first derivative graph that is above the x axis is increasing on the original graph; 2. Any part of the first derivative graph that is below the x axis is decreasing on the original graph. So from 0 to 6 the original graph is increasing, and from 6 to 9 the original graph is decreasing. This creates a negative parabola negative absolute value graph. Once done drawing just find the maximum of that graph and that is your answer, which is 6.
Question 12 states that at t=8, the object was at position x=10 at t=5, the object's position was x=?. To be honest i just guessed on this question, knowing that my position graph was a parabola and at t=6 that was the farthest possible point that the object could travel from the starting point, so that ruled out answer choice E. Next i saw that at t=8, x=10, and since 8 is farther away from 6 than 5 was, I just chose 13, answer choice D, because it made sense, being that it was greater that 10, but I did not want to choose 15, choice E, because that seemed like it would too big.
I hope I can help with this blog Trina.
POST 22
Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
linierazation:
The steps:
1. Determine the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything
LRAM formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
EXAMPLE: calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x equals: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
..plug into original equation.
RRAM formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)].. plug in to original equation.
MRAM formula is:
delta x [ f(mid) + f(mid) + .... ]
In order to find the midpoints, add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal: multiply by delta x/2-
The formula: delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
umm.. for what i don't understand, i really dont get what to do when i have to integrete a fraction, i keep forgetting.. i know i can't do quotient rule, but i just don't remember what i can do.
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
linierazation:
The steps:
1. Determine the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything
LRAM formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
EXAMPLE: calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x equals: -1+3 /2 = 2/2 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
..plug into original equation.
RRAM formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)].. plug in to original equation.
MRAM formula is:
delta x [ f(mid) + f(mid) + .... ]
In order to find the midpoints, add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1
-3 + -2/ 2 = -5/2 and -2 + -1 / 2 = -3/2
so 1[f(-5/2) + f(-3/2)] and the plug in
Trapezoidal: multiply by delta x/2-
The formula: delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
umm.. for what i don't understand, i really dont get what to do when i have to integrete a fraction, i keep forgetting.. i know i can't do quotient rule, but i just don't remember what i can do.
22nd post
This week we took a practice AP exam.. unfortunately i did not do very well on them but i will try to explain some concepts i have got down pact.
For limit problems:
if you have a problem that asks you to find the limit as x goes to infinity, then you must use your limit rules.
1. If the degree on the top is smaller than the degree on the bottom, the answer is 0.
2. If the degree on the top is bigger than the degree on the bottom, then the answer is infinity.
3. If the degree on the top is the same as the degree on the botton, you divide the coefficients to get the answer.
(the degree of the number is the highest exponent x is raised to)
so for example.. if they gave you this problem and asked you to find the limit as x goes to infinity.
(4x^2 + 3x)/(x^2+2).. the degrees are the same so you would divide the coefficients which are 4 and 1.. so the answer would be 4.
Another thing i have a pretty good handle on is taking the derivative of e.
If you have e raised to something, you bring what e is raised to to the front and e raised to whatever stays the same (NOTE: if e is raised to the zero power, it will always equal 1).
Some things i do not know:
on the AP test, i need help on the following things,
1. the table problems: i do not understand how to use the table on questions 4 and 5
2. number 9, where you have to trace those little lines on that graph.
3.questions 10-12, where they are dealing with velocity, acceleration, and time.. i have no idea how to start those and how to find values on the graph.
4. Integrating with trig functions (probably my biggest problem)
5. number 15, with that other table, i tried to take an educated guess but i'm not sure how to get the right answer.
6. Number 16, with the Riemann sums
7.number 19, i do not understand how to get an equation and what to plug in.
8. Number 20, i know you have to do average value, but again i'm at a loss on how to start
9. LOCAL LINEARIZATION ( trig yet again on number 23)
10. number 24, figuring out what the value of h is?
Well thats about it folks, i'm going to be sitting here most of the day trying to work my corrections, if anyone could help me... i would be forever grateful lol.. bye for now!
For limit problems:
if you have a problem that asks you to find the limit as x goes to infinity, then you must use your limit rules.
1. If the degree on the top is smaller than the degree on the bottom, the answer is 0.
2. If the degree on the top is bigger than the degree on the bottom, then the answer is infinity.
3. If the degree on the top is the same as the degree on the botton, you divide the coefficients to get the answer.
(the degree of the number is the highest exponent x is raised to)
so for example.. if they gave you this problem and asked you to find the limit as x goes to infinity.
(4x^2 + 3x)/(x^2+2).. the degrees are the same so you would divide the coefficients which are 4 and 1.. so the answer would be 4.
Another thing i have a pretty good handle on is taking the derivative of e.
If you have e raised to something, you bring what e is raised to to the front and e raised to whatever stays the same (NOTE: if e is raised to the zero power, it will always equal 1).
Some things i do not know:
on the AP test, i need help on the following things,
1. the table problems: i do not understand how to use the table on questions 4 and 5
2. number 9, where you have to trace those little lines on that graph.
3.questions 10-12, where they are dealing with velocity, acceleration, and time.. i have no idea how to start those and how to find values on the graph.
4. Integrating with trig functions (probably my biggest problem)
5. number 15, with that other table, i tried to take an educated guess but i'm not sure how to get the right answer.
6. Number 16, with the Riemann sums
7.number 19, i do not understand how to get an equation and what to plug in.
8. Number 20, i know you have to do average value, but again i'm at a loss on how to start
9. LOCAL LINEARIZATION ( trig yet again on number 23)
10. number 24, figuring out what the value of h is?
Well thats about it folks, i'm going to be sitting here most of the day trying to work my corrections, if anyone could help me... i would be forever grateful lol.. bye for now!
Ash's 22nd Post
Okay, can I just say I'm going crazy?
I'm relating everything to math and saw this "The Gaga Law (RAH)^2 (AH)^3 + RO (MA + MAMA) + (GA)^2 + OOH(LA)^2 = Bad Romance" I thought it was the most amazing thing ever....I'm scaring myself
Now that the fun stuff is over, let's move on to deeper, darker subjects.
These AP tests are mentally freaking me out
I'm forgetting everything...when I mean everything...I mean EVERYTHING!!!
I forgot how to take the derivative of e!! -.-
I forgot what delta equals!
*sigh*
I can't really explain anything, again, because, I'm at a loss
The tables...examples: 4, 5, 15, 22....HOW do you use those???
Graphs: 9: Hoow do you solve that? 10: How do you change velocity to acceleration? I'm so lost on this process 11 and 12: what are they asking?? 20: WHAT is the answer and how/why??? I'm SO confused!! 25: they're asking for.....the x value right, NOT the y value?
I think one of my main problems is not knowing what they're asking!! I don't understand these questions because of the way they are written!! I sit there and ponder the true meanings of these things as if they were the questions of life!
Yes, I need extreme help
I know this
I do not deny this fact
I don't know how to get it -.-
This was not meant as a whining post, just a post where I'm trying to get questions answered....hopefully someone can help =/
I'm relating everything to math and saw this "The Gaga Law (RAH)^2 (AH)^3 + RO (MA + MAMA) + (GA)^2 + OOH(LA)^2 = Bad Romance" I thought it was the most amazing thing ever....I'm scaring myself
Now that the fun stuff is over, let's move on to deeper, darker subjects.
These AP tests are mentally freaking me out
I'm forgetting everything...when I mean everything...I mean EVERYTHING!!!
I forgot how to take the derivative of e!! -.-
I forgot what delta equals!
*sigh*
I can't really explain anything, again, because, I'm at a loss
The tables...examples: 4, 5, 15, 22....HOW do you use those???
Graphs: 9: Hoow do you solve that? 10: How do you change velocity to acceleration? I'm so lost on this process 11 and 12: what are they asking?? 20: WHAT is the answer and how/why??? I'm SO confused!! 25: they're asking for.....the x value right, NOT the y value?
I think one of my main problems is not knowing what they're asking!! I don't understand these questions because of the way they are written!! I sit there and ponder the true meanings of these things as if they were the questions of life!
Yes, I need extreme help
I know this
I do not deny this fact
I don't know how to get it -.-
This was not meant as a whining post, just a post where I'm trying to get questions answered....hopefully someone can help =/
post 22
here is some stuff we can go over.
Related Rates:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation involving your variables
5. Take the derivative with respect to time
6. Substitute in derivative and solve.
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)]
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)]
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)+...]
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)]
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
im not too good at doing most of the little thigns thats involved w/ every problem. I sometimes forget a negative or i forget to put a certain number. One thing im not too goodd at is angles of elevation and most optimization problems.
Related Rates:
1. Identify all variables and equations
2. Identify what you are looking for
3. Make a sketch and label
4. Write an equation involving your variables
5. Take the derivative with respect to time
6. Substitute in derivative and solve.
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)]
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)]
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)+...]
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)]
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
im not too good at doing most of the little thigns thats involved w/ every problem. I sometimes forget a negative or i forget to put a certain number. One thing im not too goodd at is angles of elevation and most optimization problems.
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