The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
Example problem:(sr=square root) Use differentiability to approximate sr(4.5)
f(x)=sr(x) sr(4)+(1/2 sr(4) )(.5)=1.125sr(x)+(1/2 sr(x) )
dx error=.005
dx=.5
x=4
The next topic I will talk about is integration. Integration finds the area under a curve. The Riemann sum approximates the area using the rectangles or trapezoids. The Riemanns Sums are:
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
Example problem: Find the area of f(x)x-3 on the interval [0,2] with 4 subintervals.
delta x=2-0/4=1/2
LRAM=1/2[f(0)+f(1/2)+f(1)+f(3/2)]1/2[-3+-5/2+-2+-3/2]1/2[-9]= -9/2
RRAM=1/2[f(1/2)+f(1)+(3/2)+f(2)]1/2[-5/2+-2+-3/2+-1]1/2[-7]= -7/2
MRAM=1/2[f(1/4)+f(3/4)+f(5/4)+f(7/4)]1/2[-11/4+-9/4+-7/4+-5/4]1/2[-8]=4
Trapezoidal=1/4[f(0)+2f(1/2)+2f(1)+2f(7/2)+f(2)]1/4[-3+-10/2+-4+-6/2+-1]1/4[-16]= 4
For things I do not know I forgot how to do problems that ask to find acceleration and velocity when all that is given is postitions and an equation of t(x). This is what my wiki was on and failed it horribly because I have no idea what I am doing.
Sunday, January 10, 2010
21?
okay so i think everyone will agree that the first week back to calculus was extremely roughhhh. I suck at practice AP's but hopefully i'll get better at them.
i'm just going to review some things that we all may need to know.
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.
LRAM- left hand approximation.
x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation.
x[f(a+x)+...f(b)]
MRAM- approximation from the middle.
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)]
Example: Find the area of f(x)x-3 on the interval [0,2] with 4 subintervals.
delta x=2-0/4=1/2
LRAM=1/2[f(0)+f(1/2)+f(1)+f(3/2)]
1/2[-3+-5/2+-2+-3/2]
1/2[-9]= -9/2
RRAM=1/2[f(1/2)+f(1)+(3/2)+f(2)]
1/2[-5/2+-2+-3/2+-1]
1/2[-7]= -7/2
MRAM=1/2[f(1/4)+f(3/4)+f(5/4)+f(7/4)]
1/2[-11/4+-9/4+-7/4+-5/4]
1/2[-8]= 4
Trapezoidal=1/4[f(0)+2f(1/2)+2f(1)+2f(7/2)+f(2)]
1/4[-3+-10/2+-4+-6/2+-1]
1/4[-16]= 4
as far as questions go.. i have trouble understanding what exactly it is i'm being asked to do on the AP test. umm other things.. substitution, it'd be great if anyone would like to review that for me.
i'm just going to review some things that we all may need to know.
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.
LRAM- left hand approximation.
x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation.
x[f(a+x)+...f(b)]
MRAM- approximation from the middle.
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)]
Example: Find the area of f(x)x-3 on the interval [0,2] with 4 subintervals.
delta x=2-0/4=1/2
LRAM=1/2[f(0)+f(1/2)+f(1)+f(3/2)]
1/2[-3+-5/2+-2+-3/2]
1/2[-9]= -9/2
RRAM=1/2[f(1/2)+f(1)+(3/2)+f(2)]
1/2[-5/2+-2+-3/2+-1]
1/2[-7]= -7/2
MRAM=1/2[f(1/4)+f(3/4)+f(5/4)+f(7/4)]
1/2[-11/4+-9/4+-7/4+-5/4]
1/2[-8]= 4
Trapezoidal=1/4[f(0)+2f(1/2)+2f(1)+2f(7/2)+f(2)]
1/4[-3+-10/2+-4+-6/2+-1]
1/4[-16]= 4
as far as questions go.. i have trouble understanding what exactly it is i'm being asked to do on the AP test. umm other things.. substitution, it'd be great if anyone would like to review that for me.
Posting...#21
I accidently hit enter so i posted twice my mistake. I can't say what we did in the week no more because all we do now is review.
So ill go over somehitn called the First derivative test
Steps: 1st take a derivative
2: set it =0
3: solve for x=>max mins, extrema , horizontal tangents
4: Set up intervals using step 3
5: plug in 1st derivative
6: If your finding absolute max/min plug values from 5 into original fuction
Example:
F(x)=(1/2)-cos(x) Find the extrema for f(x) on the (0,2pie)
f ' (x)=1/2-cos(x)
1/2-cos(x)=0
-cos(x)=1/2
x=cos^-1 (1/2)
x= pie/3, 5pie/3
b) 0,pie/3 pie/3 , 5pie/3 5pie/3 , 2 pie
then plug invalues b/w these sets
f '(pie/4)= 1/2-cos (pie/4)
which is decresing
b/w pie/3 and 5pie/3 it is increasing and b/w 5pie/3 and 2pie its decreasings
so pie/3 is a min
and 5pie/3 is a max
And im having trouble with disks and washers so if someone could help me thanks
So ill go over somehitn called the First derivative test
Steps: 1st take a derivative
2: set it =0
3: solve for x=>max mins, extrema , horizontal tangents
4: Set up intervals using step 3
5: plug in 1st derivative
6: If your finding absolute max/min plug values from 5 into original fuction
Example:
F(x)=(1/2)-cos(x) Find the extrema for f(x) on the (0,2pie)
f ' (x)=1/2-cos(x)
1/2-cos(x)=0
-cos(x)=1/2
x=cos^-1 (1/2)
x= pie/3, 5pie/3
b) 0,pie/3 pie/3 , 5pie/3 5pie/3 , 2 pie
then plug invalues b/w these sets
f '(pie/4)= 1/2-cos (pie/4)
which is decresing
b/w pie/3 and 5pie/3 it is increasing and b/w 5pie/3 and 2pie its decreasings
so pie/3 is a min
and 5pie/3 is a max
And im having trouble with disks and washers so if someone could help me thanks
Post Number Twenty One
First thing’s first, NO SCHOOL TOMORROW (:
This week we did practice ap exams. We’ll be doing this for the rest of the year, which scares me because I all of a sudden know NOTHING. I’m not even sure what to do my blog on, that’s how bad off I am. I have a lot of trouble just recognizing what I need to do for the problem. Hopefully as the weeks go on and the more practice tests we do the better I get. So if anyone can help me out with anything, come. I’m begging you, seriously!
Formulas that I know:
Average Value : 1/b – a bSa f(a)
Area between curves : bSa (top equation – bottom equation)
“e integration”: Se^x = e^x + C
ln integration: S 1/x dx = ln x + C
*don’t change to x^-1
Volume by disks: pi bSa [R(x)]^2 dx
Volume by washers: pi bSa (top)^2 – (bottom)^2
Quotient rule: vu’ – uv’/v^2
Product Rule: uv’ + vu’
Substitution steps:
1. Find a derivative inside the integral
2. Set u = to the non-derivative
3. Take the derivative of u
4. Substitute back in
If only I can actually work the problem..
Calculus Graph Vocabulary incase people forgot
1. Increasing – original
2. Decreasing - original
3. Positive Slope – first
4. Negative Slope – first
5. Concave up – second
6. Concave down – second
7. Horizontal tangent
8. Slope = 0
9. Derivative above axis – related to positive slope and concave up
10.Derivative below axis – related to negative slope and concave down
11. Zero of derivative
12. Maximum – original
13. Minimum – original
14. Point of inflection – second
Unfortunately the only one I actually know how to do from the practice test is number one, so that is going to be my example.
Lim x>infinity (3x^2 – 4)/(2 – 7x – x^2)
The answer is -3 because it follows the limit rules. If the degrees match, take the top coefficient over the bottom coefficient, thus giving you the answer.
Obviously I need a lot of help. I need it with everything, so please do so!
This week we did practice ap exams. We’ll be doing this for the rest of the year, which scares me because I all of a sudden know NOTHING. I’m not even sure what to do my blog on, that’s how bad off I am. I have a lot of trouble just recognizing what I need to do for the problem. Hopefully as the weeks go on and the more practice tests we do the better I get. So if anyone can help me out with anything, come. I’m begging you, seriously!
Formulas that I know:
Average Value : 1/b – a bSa f(a)
Area between curves : bSa (top equation – bottom equation)
“e integration”: Se^x = e^x + C
ln integration: S 1/x dx = ln x + C
*don’t change to x^-1
Volume by disks: pi bSa [R(x)]^2 dx
Volume by washers: pi bSa (top)^2 – (bottom)^2
Quotient rule: vu’ – uv’/v^2
Product Rule: uv’ + vu’
Substitution steps:
1. Find a derivative inside the integral
2. Set u = to the non-derivative
3. Take the derivative of u
4. Substitute back in
If only I can actually work the problem..
Calculus Graph Vocabulary incase people forgot
1. Increasing – original
2. Decreasing - original
3. Positive Slope – first
4. Negative Slope – first
5. Concave up – second
6. Concave down – second
7. Horizontal tangent
8. Slope = 0
9. Derivative above axis – related to positive slope and concave up
10.Derivative below axis – related to negative slope and concave down
11. Zero of derivative
12. Maximum – original
13. Minimum – original
14. Point of inflection – second
Unfortunately the only one I actually know how to do from the practice test is number one, so that is going to be my example.
Lim x>infinity (3x^2 – 4)/(2 – 7x – x^2)
The answer is -3 because it follows the limit rules. If the degrees match, take the top coefficient over the bottom coefficient, thus giving you the answer.
Obviously I need a lot of help. I need it with everything, so please do so!
Post 21
Ok, so this whole week we've been doing practice AP tests (twice a day if you also have stats). These tests are really hard, and I'm horrible at them. Most of the time I have no clue what is ever going on. Other times, I think I know what's going on and eliminate answers I believe are not correct so I can make an educated guess, and my guesses are always wrong. I have learned that guessing is bad on these tests, especially if you are as horrible as I am with the whole guessing game. I have also learned you can make negative points on these tests if you do guess incorrectly. This week's really been crazy, but B-rob says it'll get better when we have more practice on the AP tests. She says the ask the same questions on every test, so with practice we'll eventually do well. It's the eventually part that bothers me because we have our first test tomorrow.
Anyway, this week was my side of the class's week to answer the problems on the wiki page. I picked number three which was a problem asking for the area beneath a curve and the volume of a solid. I picked this question because I knew I liked area under the curve, and volume wasn't much different. My problem actually had area by washers, which is very similar to area beneath the curve. So I picked this problem without really looking at how I'd solve it. I thought it would be easy enough. When I looked at the problem, I immediatly saw I didn't know how to solve for the bounds, or even integrate for that matter. When I asked for help, I found the problem could be done very easily in my calculator. The problem was I wasn't very sure how to use my calculator to do this problem. When I figured it out, it was a very easy problem to do. When I finished the problem; however, I doubted myself because I came out with crazy decimals. I was told to be confident in my answer because AP questions would not be very pretty. So for this blog, I'll explain how to do area under a curve and volume by disks and washers.
Area under a curve is very simple. In these problems, you will have two graphs that will yield two equations. You will then plug these equations in to the area under a curve formula. The formula is:
Integral from a-b (top eq-bottom eq)
Sometimes the problem will give you bounds, but if not they're easy to find. To find your bounds, you set your equations equal to each other and solve them. Once you find your bounds, just plug into the formula and integrate as you normally would for a definite integral.
For volume, you are usually given a graph, or an equation that you are asked to revolve around something to make it a solid, and you are asked to then find the volume of that solid. Volume can be found in two different ways by doing this, it can be found by using "disks" or "washers."
For example, you would use disks when you only have one equation. When this one equation (one graph) is rotated about a line it will make a solid object with no hole. If a problem asks you to rotate a graph around a horizontal line that is not the x-axis, you are to subtract that line from the equation and solve. For disks the formula is:
pi * Integral from a-b (eq)^2
Ex: Rotate the graph y=x^4 + x^2 from [0,2] about the line y=1
Ok, so for this one, it's asking you to rotate the graph about a horizontal line. This cannot be done, so you will have to subtract it from your equation once it's plugged in, like this.
pi* Integral from 0-2 (x^4 +x^2 -1)^2
Then integrate as you would in a normal definite integral
For disks, you would use the same process except you would have two equations. The formula for disks would be the same as it is for area under the curve, except the two equations would be squared as they are in disks and multiplied by pi as they are in disks.
Ok, some things I don't understand. My problem this week is I don't remember a thing I learned. Every time I went to make an educated guess it was wrong, and I always eliminated the possible answers. If anyone is good at these AP things, can they please help me?
Anyway, this week was my side of the class's week to answer the problems on the wiki page. I picked number three which was a problem asking for the area beneath a curve and the volume of a solid. I picked this question because I knew I liked area under the curve, and volume wasn't much different. My problem actually had area by washers, which is very similar to area beneath the curve. So I picked this problem without really looking at how I'd solve it. I thought it would be easy enough. When I looked at the problem, I immediatly saw I didn't know how to solve for the bounds, or even integrate for that matter. When I asked for help, I found the problem could be done very easily in my calculator. The problem was I wasn't very sure how to use my calculator to do this problem. When I figured it out, it was a very easy problem to do. When I finished the problem; however, I doubted myself because I came out with crazy decimals. I was told to be confident in my answer because AP questions would not be very pretty. So for this blog, I'll explain how to do area under a curve and volume by disks and washers.
Area under a curve is very simple. In these problems, you will have two graphs that will yield two equations. You will then plug these equations in to the area under a curve formula. The formula is:
Integral from a-b (top eq-bottom eq)
Sometimes the problem will give you bounds, but if not they're easy to find. To find your bounds, you set your equations equal to each other and solve them. Once you find your bounds, just plug into the formula and integrate as you normally would for a definite integral.
For volume, you are usually given a graph, or an equation that you are asked to revolve around something to make it a solid, and you are asked to then find the volume of that solid. Volume can be found in two different ways by doing this, it can be found by using "disks" or "washers."
For example, you would use disks when you only have one equation. When this one equation (one graph) is rotated about a line it will make a solid object with no hole. If a problem asks you to rotate a graph around a horizontal line that is not the x-axis, you are to subtract that line from the equation and solve. For disks the formula is:
pi * Integral from a-b (eq)^2
Ex: Rotate the graph y=x^4 + x^2 from [0,2] about the line y=1
Ok, so for this one, it's asking you to rotate the graph about a horizontal line. This cannot be done, so you will have to subtract it from your equation once it's plugged in, like this.
pi* Integral from 0-2 (x^4 +x^2 -1)^2
Then integrate as you would in a normal definite integral
For disks, you would use the same process except you would have two equations. The formula for disks would be the same as it is for area under the curve, except the two equations would be squared as they are in disks and multiplied by pi as they are in disks.
Ok, some things I don't understand. My problem this week is I don't remember a thing I learned. Every time I went to make an educated guess it was wrong, and I always eliminated the possible answers. If anyone is good at these AP things, can they please help me?
Post #21
Since I been having trouble remembering linearization, I'll explain that.
The key word in linearization is approximate. Also, if there is a decimal in the problem, its another hint that you will be doing linearization.
The steps of linearization are
1. Identify and equation
2. f(x)+f'(x) dx
3. Determine dx (the decimal)
4. Determine x (the whole number)
5. Plug in
Examples:
Use differentials to approximate the square root of 16.5.
1. The equation would be f(x) = the square root of x
2. Take the derivative of f(x) which we found to be the square root of x
x^1/2 = 1/2 x^-1/2 dx or 1/2 square root of x dx
Then at f(x) and f'(x): the square root of x + 1/2 square root of x dx
3. dx = .5
4. x = 16
5. Plug in: the square root of 16 + 1/2 the square root of 16 (.5)
4+1/((2)(4) (.5) = 4 + 1/8 (.5) = 4.0625
The example we been seeing on the practice AP is
If f(1)=2 and f'(1)=5, use the equation of the line tangent to the graph of f at x=1 to approximate f(1.2).
For this problem, all you have to do is find the equation of the tangent line then plug in 1.2 for x.
To find the equation of a tangent line, you need a point and a slope.
You are given that f(1)=2 so your x is 1 and your y is 2 (1,2) is your point.
Then they give you that f'(1)=5 so your derivative (slope) is 5.
Now plug in: y-2=5(x-1)
Next plug in 1.2 for x
y-2=5(1.2-1)
y-2=5(.2)
y=1+2 = 3
I can use help on related rate problems because I get confused about what is given and what is not. Also, I need to learn how to work my calculator for the calculator portion. Other than that, I think I just need to study big time tonight because I am quite terrified for tomorrow.
Post #21
Calculus Week #21
So this week in Calculus we came back from the holidays, and we focused on doing some review by doing AP Calculus Practice Examinations.
I just want to say that these were a real eye opener to me because I sucked so hard on these...like... I can't even say how bad it is.
Anyway, some things I've learned that I'll share :-)
If you take the derivative of f^2(x) you treat it as if it was like sin. You do 2f(x) times f'(x) times 1. So its almost as if you have (f(x))^2 and you use the chain rule. Bring two to the front, subtract one. times by the derivative of the inside...well the derivative of f(x) is f'(x) times the derivative of x, which is 1. I never realized that you could apply all of those rules to function notation and it really seemed to help on a lot of problems.
Another thing that is new... say you have
x^2-2
-----
4-x^2 as the limit approaches 2
For this, you end up getting 2/0. For finite limits, if you get a # over 0, it is always infinity. That's a really nice little trick.
Also, all of the problems that deal with velocity and acceleration...like I knew before you had to use derivatives and integrals but I never realized the whole solve for c thing. So anyway, with those I find it easiest to just find all your 3 equations...position, velocity, and acceleration by whatever way possible. Then use what they give you...they will tell you enough information to plug in and solve the equations to be able to get the right thing they want. These have become really easy...
Another thing that's sort of new to me...whenever you have equations solved for x and are doing the area and stuff...like just because you turn the calculator sideways...you still put the one that was on top originally first when you do the integral.
Also, say you have an integral from like 2 to 3x-1... the 3x-1 will change basically like everything. Like if before the integral it had like 1...it would become 2.
On problems that tell you like... x=2t-1 is the substitution used for some integral...how is it changed etc... you use that to find your new bounds and you can eliminate so many answers its amazing. Also, make sure you solve this for t, take its derivative and make sure whatever you have is accounted for in the problem.
If you get confused using fnInt and nDeriv...you can do it another way. Graph the function you want to take the integral of or do a derivative. Then press SECOND, then CALC, and go down to dy/dx for the derivative or Sf(x)dx for integral. If you are doing the dy/dx it will just ask you for a value. You put in the x value that you want to know the derivative at. For the integral, it will ask you your lower bound, you type the number and press enter and then it will ask you for the upper bound...press in the number and press enter. I like this way because it shows you visually where the area you are calculating which is sometimes useful. Also I like this way because you don't have to worry about messing up parenthesizes. A downfall to this method is you will have to write down the decimal version of the number if you want to use it late on whereas if you use fnInt, you can just SECOND ANS. Anyway, just letting you guys know.
Something I still keep forgetting is that say you have the integral of 3x...you can just put it as 3 times the integral of x. This proves very useful...
Something I need to remember is that a secant line is the line between two points...usually the two points they give you.
Anyway, that was a bit of what I was noticing as I studied my notes...I'm off to take the practice online thing...
So this week in Calculus we came back from the holidays, and we focused on doing some review by doing AP Calculus Practice Examinations.
I just want to say that these were a real eye opener to me because I sucked so hard on these...like... I can't even say how bad it is.
Anyway, some things I've learned that I'll share :-)
If you take the derivative of f^2(x) you treat it as if it was like sin. You do 2f(x) times f'(x) times 1. So its almost as if you have (f(x))^2 and you use the chain rule. Bring two to the front, subtract one. times by the derivative of the inside...well the derivative of f(x) is f'(x) times the derivative of x, which is 1. I never realized that you could apply all of those rules to function notation and it really seemed to help on a lot of problems.
Another thing that is new... say you have
x^2-2
-----
4-x^2 as the limit approaches 2
For this, you end up getting 2/0. For finite limits, if you get a # over 0, it is always infinity. That's a really nice little trick.
Also, all of the problems that deal with velocity and acceleration...like I knew before you had to use derivatives and integrals but I never realized the whole solve for c thing. So anyway, with those I find it easiest to just find all your 3 equations...position, velocity, and acceleration by whatever way possible. Then use what they give you...they will tell you enough information to plug in and solve the equations to be able to get the right thing they want. These have become really easy...
Another thing that's sort of new to me...whenever you have equations solved for x and are doing the area and stuff...like just because you turn the calculator sideways...you still put the one that was on top originally first when you do the integral.
Also, say you have an integral from like 2 to 3x-1... the 3x-1 will change basically like everything. Like if before the integral it had like 1...it would become 2.
On problems that tell you like... x=2t-1 is the substitution used for some integral...how is it changed etc... you use that to find your new bounds and you can eliminate so many answers its amazing. Also, make sure you solve this for t, take its derivative and make sure whatever you have is accounted for in the problem.
If you get confused using fnInt and nDeriv...you can do it another way. Graph the function you want to take the integral of or do a derivative. Then press SECOND, then CALC, and go down to dy/dx for the derivative or Sf(x)dx for integral. If you are doing the dy/dx it will just ask you for a value. You put in the x value that you want to know the derivative at. For the integral, it will ask you your lower bound, you type the number and press enter and then it will ask you for the upper bound...press in the number and press enter. I like this way because it shows you visually where the area you are calculating which is sometimes useful. Also I like this way because you don't have to worry about messing up parenthesizes. A downfall to this method is you will have to write down the decimal version of the number if you want to use it late on whereas if you use fnInt, you can just SECOND ANS. Anyway, just letting you guys know.
Something I still keep forgetting is that say you have the integral of 3x...you can just put it as 3 times the integral of x. This proves very useful...
Something I need to remember is that a secant line is the line between two points...usually the two points they give you.
Anyway, that was a bit of what I was noticing as I studied my notes...I'm off to take the practice online thing...
post 21
well another week of calculus comes to a close and all this reviewing is helping me a goodd bit.
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
limits:
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
related rates:
-Write down the given
-figure out which formula they want you to use
-plug in the given
-take the derivative
-solve for the unknown
im not too good at the lram rram mram stuff even though its super easy. i can never remember how to do it and when i finally to i forget.
volume by disks:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
limits:
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
related rates:
-Write down the given
-figure out which formula they want you to use
-plug in the given
-take the derivative
-solve for the unknown
im not too good at the lram rram mram stuff even though its super easy. i can never remember how to do it and when i finally to i forget.
Post #...1/10/2010
Instead of balantly typing the steps for the first and second derivatives on here, I have decided to do a couple of problems from the AP exams and explain them in terms I'm pretty sure you will understand as well (although this remains to be seen).
The following problem falls into the LINEARIZATION category:
**When the local linearization of f(x) = sqrt(9 + sin2x) near 0 is used, an estimate of 0.06 is...
Linearization is just the equation of a tangent line with the x value they give you plugged in.
f'(x) = cos2x / sqrt(9 + sin2x)
Since youknow that your x value is 0 (near 0), plug it in to find your slope.
This then gives you 1/sqrt(9), whcih simplifies to 1/3.
This is your slope of the tangent line. Now the only thing you are missing is your y value. You find this by plugging your x value into your original which gives you the point (0,3)
So now you have a point and a slope..plug in to point slope form.
y-3 =1/3(x-0)
Now plug in your 0.06 (since you are approximating to that) into your equation of tangent line. This then yields 3.02
Now for an easy problem:
If y = sin^3(1-2x), then dy/dx is...
As you can tell, the problem is just telling you to take the derivative. Now this part may be a little tricky because it is indeed a chain rule that involves multiple steps, thus possibly having a negative effect on your final answer.
Ok. so first deal witht the exponent:
dy/dx = 3sin^2(1-2x)
Now the derivative of sin:
dy/dx = 3sin^2(1-2x)cos(1-2x)
Now the derivative of the inside:
dy/dx = 3sin^2(1-2x)cos(1-2x)(-2)
Which then simplifies further to:
dy/dx = -6sin^2(1-2x)cos(1-2x)
Ok. Thats enough examples for now..will do more in the future!!
~mal
The following problem falls into the LINEARIZATION category:
**When the local linearization of f(x) = sqrt(9 + sin2x) near 0 is used, an estimate of 0.06 is...
Linearization is just the equation of a tangent line with the x value they give you plugged in.
f'(x) = cos2x / sqrt(9 + sin2x)
Since youknow that your x value is 0 (near 0), plug it in to find your slope.
This then gives you 1/sqrt(9), whcih simplifies to 1/3.
This is your slope of the tangent line. Now the only thing you are missing is your y value. You find this by plugging your x value into your original which gives you the point (0,3)
So now you have a point and a slope..plug in to point slope form.
y-3 =1/3(x-0)
Now plug in your 0.06 (since you are approximating to that) into your equation of tangent line. This then yields 3.02
Now for an easy problem:
If y = sin^3(1-2x), then dy/dx is...
As you can tell, the problem is just telling you to take the derivative. Now this part may be a little tricky because it is indeed a chain rule that involves multiple steps, thus possibly having a negative effect on your final answer.
Ok. so first deal witht the exponent:
dy/dx = 3sin^2(1-2x)
Now the derivative of sin:
dy/dx = 3sin^2(1-2x)cos(1-2x)
Now the derivative of the inside:
dy/dx = 3sin^2(1-2x)cos(1-2x)(-2)
Which then simplifies further to:
dy/dx = -6sin^2(1-2x)cos(1-2x)
Ok. Thats enough examples for now..will do more in the future!!
~mal
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