This week in calculus we reviewed some of the concepts we did not fully understand. Let's go over some of the things we may already know:
First derivative test:
You take the derivative of the function and set it equal to zero. Then solve for the x values (critical points). Then you set the x values up into intervals between negative infinity and infinity. After that, you plug in numbers between the intervals into the first derivative to solve to see if there is a max or min or to see if the function is increasing or decreasing.
Second derivative test:
You take the derivative of the function twice and set it equal to zero. Solve for the x values once more and set them up into intervals between negative infinity and infinity. Then plug in numbers between those intervals and plug them into the second derivative to see where the graph is concave or, concave down, or where there is a point of inflection.
Tangent lines:
The problem will give you a function and an x value. If no y value is given, plug the x value into the original function and solve for y. Once that is done, you take the derivative of the function, plug in the x value and solve to find the slope. Then set everything up into point-slope form (y-y1=slope(x-x1)).
Limit rules:
If the degree on the top is bigger than the degree on the bottom, the limit is infinity
If the degree on the top is smaller than the degree on the bottom, the limit is zero.
If the degree on the top is the same as the degree on the bottom, divide the coefficients to find the limit.
Things i'm still having trouble with:
optimization
angles of elevation and related rates
nasty looking integrals
Have a great Easter vacation :)
Saturday, April 3, 2010
Friday, April 2, 2010
Post #33
SPRING BREAK...FINALLY!!!! Just a few more weeks to go!
This week mrs. robinson answered questions that we needed help with. Monday and tuesday helped me a lot! Wednesday I went to NIcholls Orientation, and Thursday I went to the St. Charles Drunk Driving Skit. Today's Good Friday and it offically starts our Spring Break!!!
So i'll go over implicit derivatives because I used to understand them, then after learning some other things i got them confused and intertwined with one another. So, here it goes...
steps to IMPLICIT DERIVATIVES:
1. Take the derivative like normal.
2. For each y-term, you put y' or dy/dx behind it.
3. Solve for dy/dy or y'.
For Example:
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
the Second Implicit Derivative...basically the same way!
steps to SECOND IMPLICIT DERIVATIVES
1. Take the derivative of the first derivative
2. Put d^2y/dx^2 for dy/dx
3. Simplify
4. Plug in values
so an example using the same equation is:
we have to take the first implicit derivative FIRST...[durr]
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x <-- that's our first implicit derivative
now in order to take the second implicit derivative we just take the second derivative of dy/dx = -18x..
dy/dx = -18x
dy/dx = -18
so since we know that dy/dx is -18x we can plug that in
-18x = -18
when solving for x we get 1!
NOW, i understand and grasp that...but for what i really don't get is still Angle of Elevation..and Optimization..and I know there's other things, but most of them are either formula's to memorize [which i'm working on] or things that i understand and just make careless mistakes!
~ElliE~
This week mrs. robinson answered questions that we needed help with. Monday and tuesday helped me a lot! Wednesday I went to NIcholls Orientation, and Thursday I went to the St. Charles Drunk Driving Skit. Today's Good Friday and it offically starts our Spring Break!!!
So i'll go over implicit derivatives because I used to understand them, then after learning some other things i got them confused and intertwined with one another. So, here it goes...
steps to IMPLICIT DERIVATIVES:
1. Take the derivative like normal.
2. For each y-term, you put y' or dy/dx behind it.
3. Solve for dy/dy or y'.
For Example:
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
the Second Implicit Derivative...basically the same way!
steps to SECOND IMPLICIT DERIVATIVES
1. Take the derivative of the first derivative
2. Put d^2y/dx^2 for dy/dx
3. Simplify
4. Plug in values
so an example using the same equation is:
we have to take the first implicit derivative FIRST...[durr]
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x <-- that's our first implicit derivative
now in order to take the second implicit derivative we just take the second derivative of dy/dx = -18x..
dy/dx = -18x
dy/dx = -18
so since we know that dy/dx is -18x we can plug that in
-18x = -18
when solving for x we get 1!
NOW, i understand and grasp that...but for what i really don't get is still Angle of Elevation..and Optimization..and I know there's other things, but most of them are either formula's to memorize [which i'm working on] or things that i understand and just make careless mistakes!
~ElliE~
Monday, March 29, 2010
32 post
So the blog wouldn't let me sign in yesterday, but I finally got it working today.
Well, I think I have talked about the first derivative test way too much. So, I guess I will focus on the second derivative test this time.
PROBLEM: Finding the points of inflection for x^3-6x^2+12x.
1. take the derivative:
3x^2-12x+12
2. take the second derivative:
6x-12
3. set = to 0:
6x-12=0 x=2 --->gives possible pts. of inflection
4. set up intervals:
(-infinity,2) u (2,infinity)
5. plug in to SECOND DERIVATIVE:
6(0)-12= -ve --->concave down
6(3)-12= +ve --->concave up
So, x=2 is the point of inflection.
Also, I thought I might review implicit derivatives.
PROBLEM: y^3+y^2-5y-x^2=-4
1. take derivative of both sides: (implicit derivatives have an = sign)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
*remember every time you take the derivative of y, you have to note it by dy/dx or y^1
3. solve for dy/dx: (you are going to have to take out a dy/dx when solving)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)=2x
dy/dx(3y^2+2y-5)=2x
dy/dx=2x/3y^2+2y-5
*Also, if you want the slope you must plug in a x and y-value.
So, I would just like to point out that I understand most (definitely not all) things we have learned so far in Calculus. I just seem to forget a step, mess up a derivative, or the problem is just simply tough haha
Well, I think I have talked about the first derivative test way too much. So, I guess I will focus on the second derivative test this time.
PROBLEM: Finding the points of inflection for x^3-6x^2+12x.
1. take the derivative:
3x^2-12x+12
2. take the second derivative:
6x-12
3. set = to 0:
6x-12=0 x=2 --->gives possible pts. of inflection
4. set up intervals:
(-infinity,2) u (2,infinity)
5. plug in to SECOND DERIVATIVE:
6(0)-12= -ve --->concave down
6(3)-12= +ve --->concave up
So, x=2 is the point of inflection.
Also, I thought I might review implicit derivatives.
PROBLEM: y^3+y^2-5y-x^2=-4
1. take derivative of both sides: (implicit derivatives have an = sign)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
*remember every time you take the derivative of y, you have to note it by dy/dx or y^1
3. solve for dy/dx: (you are going to have to take out a dy/dx when solving)
3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)=2x
dy/dx(3y^2+2y-5)=2x
dy/dx=2x/3y^2+2y-5
*Also, if you want the slope you must plug in a x and y-value.
So, I would just like to point out that I understand most (definitely not all) things we have learned so far in Calculus. I just seem to forget a step, mess up a derivative, or the problem is just simply tough haha
32 post
First derivative test:
-take the derivative of the original function
-solve for x (the values will be your critical values)
-set those values up into intervals between negative infinity and infinity
- plug in numbers between the intervals into the function
-this will show you when the function is increasing, decreasing, and you will find max's and mins.
Second derivative test:
-take the derivative of the original function twice
-solve for x values(critical values)
-set up into intervals between infinity and negative infinity
-plug in values between the intervals into the function
-this will show you where the graph is concave up and down, and where there is a point of inflection.
Another concept i have a decent handle on is 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
one thing that i need a lot more help with is intergration. and intergration of trig functions expecially.
-take the derivative of the original function
-solve for x (the values will be your critical values)
-set those values up into intervals between negative infinity and infinity
- plug in numbers between the intervals into the function
-this will show you when the function is increasing, decreasing, and you will find max's and mins.
Second derivative test:
-take the derivative of the original function twice
-solve for x values(critical values)
-set up into intervals between infinity and negative infinity
-plug in values between the intervals into the function
-this will show you where the graph is concave up and down, and where there is a point of inflection.
Another concept i have a decent handle on is 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
one thing that i need a lot more help with is intergration. and intergration of trig functions expecially.
Sunday, March 28, 2010
Ash's 32nd Post
So...it's the 32nd week of school...wait 9X4=36....4 more weeks left? What?? I think my math is off...
So, those substitution packets really kicked my butt!! I found out that I have no idea how to do it! I tried working on them this weekend and got nowhere! I learned it, and I used to be good at it...I'm just horrible now. Along with everything else...did anyone else forget everything? =/
I'm going to try to explain something...but don't take my word for it! =]
Product Rule.
Formula: (Derivative of the first)(the original second)+(The original first)(Derivative of the second)
Step 1: Obviously, you do the formula.
Step 2: Simplify! :)
Example: 3xy+z^2 (let's ignore the dx/dx, dy/dx, and dz/dx for simplicity, kay?)
Step 1: (3)(y)+(3x)(1)+2z
Step 2: 3y+3x+2z
Quotient Rule.
Formula: (Derivative of the top)(the original bottom)-(the original top)(derivative of the bottom)/(bottom)^2
Step 1: You do the formula.
Step 2: Simplify!
Example: 3x^2-x/4x
Step 1: (6x-1)(4x)-(3x^2)(4)/(4x)^2
Step 2: (24x^2-4x-12x^2)/(4x^2)
(12x^2-4x)/(4x^2)
Or you can simplify it completely:
3-1/x
A common mistake is to change the + and the - on those two! Don't get them mixed up!!!
For things I need help on:
Just about everything...I think I'm just going to go through all of the blogs a thousand times and hope that something clicks =/
So, those substitution packets really kicked my butt!! I found out that I have no idea how to do it! I tried working on them this weekend and got nowhere! I learned it, and I used to be good at it...I'm just horrible now. Along with everything else...did anyone else forget everything? =/
I'm going to try to explain something...but don't take my word for it! =]
Product Rule.
Formula: (Derivative of the first)(the original second)+(The original first)(Derivative of the second)
Step 1: Obviously, you do the formula.
Step 2: Simplify! :)
Example: 3xy+z^2 (let's ignore the dx/dx, dy/dx, and dz/dx for simplicity, kay?)
Step 1: (3)(y)+(3x)(1)+2z
Step 2: 3y+3x+2z
Quotient Rule.
Formula: (Derivative of the top)(the original bottom)-(the original top)(derivative of the bottom)/(bottom)^2
Step 1: You do the formula.
Step 2: Simplify!
Example: 3x^2-x/4x
Step 1: (6x-1)(4x)-(3x^2)(4)/(4x)^2
Step 2: (24x^2-4x-12x^2)/(4x^2)
(12x^2-4x)/(4x^2)
Or you can simplify it completely:
3-1/x
A common mistake is to change the + and the - on those two! Don't get them mixed up!!!
For things I need help on:
Just about everything...I think I'm just going to go through all of the blogs a thousand times and hope that something clicks =/
Post 32
wow, post 32, we have been in school forreevvverrr... but anyway, the school year is almost over and this is just gonna be one less blog i have to do.
for this blog i'm going to go over volume and area of disks. a disk is a solid object, and to find the volume of it, you just solve for x or y of the given equation. you can tell by the problem which axis its about. so you draw the equation, then you reflect it. the formula is S r^2 dx, and the radius is the equation that is given. and the formula for the area of a disk is the same thing but the radius is not squared.
Example: Find the volume of the solid obtained by rotating about the x-axis the region under the curve from -2 to 1: y=sqroot(-2x^2-10x+48)
so, since the formula needs to you to square the equation, the square root just disappears. then, you have to integrate the equation, then you do top-bottom, and multiple by pi. so after you integrate it looks like this: -2/3(x)^3-5x^2+48x
then after you plug in everything and do top-bottom it looks like this: -2/3(1)^3-5(1)^2+48(1)-(-2/3(-2)^3-5(-2)^2+48(-2))
then you can simplify it on your own even more, or plug it all into your calculator. i chose to simplify it more by hand first because with that many numbers it is pretty easy to make a mistake in your calculator. So i further simplified it to -(2/3)-5+48+(16/3)+20+96 which equals: 491/3.
so then, you have to multiply that by pi, like i mentioned above, and the final answer is 491(pi)/3.
since we have to post things we have trouble with, so other people can comment and help us out, i am going to go ahead and say that i suck at integrating things with ln, e, or sin and cos n stuff, but after doing that packet on integration, i can do the simple integrating, just not the stuff i mentioned above.
for this blog i'm going to go over volume and area of disks. a disk is a solid object, and to find the volume of it, you just solve for x or y of the given equation. you can tell by the problem which axis its about. so you draw the equation, then you reflect it. the formula is S r^2 dx, and the radius is the equation that is given. and the formula for the area of a disk is the same thing but the radius is not squared.
Example: Find the volume of the solid obtained by rotating about the x-axis the region under the curve from -2 to 1: y=sqroot(-2x^2-10x+48)
so, since the formula needs to you to square the equation, the square root just disappears. then, you have to integrate the equation, then you do top-bottom, and multiple by pi. so after you integrate it looks like this: -2/3(x)^3-5x^2+48x
then after you plug in everything and do top-bottom it looks like this: -2/3(1)^3-5(1)^2+48(1)-(-2/3(-2)^3-5(-2)^2+48(-2))
then you can simplify it on your own even more, or plug it all into your calculator. i chose to simplify it more by hand first because with that many numbers it is pretty easy to make a mistake in your calculator. So i further simplified it to -(2/3)-5+48+(16/3)+20+96 which equals: 491/3.
so then, you have to multiply that by pi, like i mentioned above, and the final answer is 491(pi)/3.
since we have to post things we have trouble with, so other people can comment and help us out, i am going to go ahead and say that i suck at integrating things with ln, e, or sin and cos n stuff, but after doing that packet on integration, i can do the simple integrating, just not the stuff i mentioned above.
Post #32
So, i haven't been to school in a few days..so i'm going to explain a few throwback things that noone has talked about in a while.
Lets start with chain rule:
outside to inside is the major hint..
such as cos(2s)
first derivative of cosine, then the derivative of 2s, multiply them togeter..and presto.
Now, lets talk about product rule with chain rule:
Make sure if you have xcos2x, you use product rule. How do you know to use product rule? You are multiplying two terms with variables.
(derivative of first)(copy second) + (derivative of second)(copy first)
*when you take the derivative of cos2x, make sure to include the chain rule!
Things i always misunderstand
*tangent lines, i can't decide waht to do for these.
*related rates
*optimization..
Lets start with chain rule:
outside to inside is the major hint..
such as cos(2s)
first derivative of cosine, then the derivative of 2s, multiply them togeter..and presto.
Now, lets talk about product rule with chain rule:
Make sure if you have xcos2x, you use product rule. How do you know to use product rule? You are multiplying two terms with variables.
(derivative of first)(copy second) + (derivative of second)(copy first)
*when you take the derivative of cos2x, make sure to include the chain rule!
Things i always misunderstand
*tangent lines, i can't decide waht to do for these.
*related rates
*optimization..
post 32
ok so this week we went over ap's from last week. we didn't take any this week, i'm pretty sure. i don't really remember because everyone was pretty much focused on state this weekend. haha
which congrats to alex, ryne, and john! great job. oh & i won a doorprize. WOOOOHHHHH. haha
Chain Rule:
you start from outside in. let's say you have cos(2x)
first take derivative of cos
then of multiply that by 2x
Product Rule:
copy the first x derivative of the second + copy the second x derivative of the first
Quotient Rule:
copy the bottom x derivative of the top - copy the top x derivative of the bottom/bottom ^2
so those are some derivative rules.
can someone go over the rules for simple derivatives.
like what is derivative of tan?
and sec?
csc?
cot?
all those weird ones that are rarely used. i get them mixed up a lot
like i think tan is sec^2
or sec is tan^2
i'm not really sure?
and csc is something like cscxtanx?
idk. someone please go over that for me
which congrats to alex, ryne, and john! great job. oh & i won a doorprize. WOOOOHHHHH. haha
Chain Rule:
you start from outside in. let's say you have cos(2x)
first take derivative of cos
then of multiply that by 2x
Product Rule:
copy the first x derivative of the second + copy the second x derivative of the first
Quotient Rule:
copy the bottom x derivative of the top - copy the top x derivative of the bottom/bottom ^2
so those are some derivative rules.
can someone go over the rules for simple derivatives.
like what is derivative of tan?
and sec?
csc?
cot?
all those weird ones that are rarely used. i get them mixed up a lot
like i think tan is sec^2
or sec is tan^2
i'm not really sure?
and csc is something like cscxtanx?
idk. someone please go over that for me
Post 32
Well the last nine weeks are here and the AP test is coming soon. This week we reviewed practice AP tests and did a integration worksheet.
Some old stuff I will go over are related rates. The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
Next I will talk about Riemann Sums. 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
Next I will talk about stuff from the beginning of school the product rule and quotient rule.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
For questions, I need help with area volume problems. If someone could refresh my memory that would be great.
Some old stuff I will go over are related rates. The steps for related rates are:
1. Identify all of the variables and equations
2. Identify the things that you are looking for
3. Sketch a graph and then label that graph
4. Create and write an equation using all of the variables
5. Take the derivative of this equation with respect to time
6. Substitute everything back in
7. Solve the equation
Next I will talk about Riemann Sums. 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
Next I will talk about stuff from the beginning of school the product rule and quotient rule.
Product rule is copy the first times the derivative of the last plus copy the first times the derivative of the second.
Quotient rule is copy the bottom times derivative of the top minus copy the top times the derivative of the bottom all divided by the bottom squared.
For questions, I need help with area volume problems. If someone could refresh my memory that would be great.
post 32
calculus for another week yeerddd me.
Disks
Formula: piS(equation)^2dx
Find Bounds (How?)
Integrate using Definite Integration
Washers
Formula: piS[(top equation)^2-(bottom equation)^2]dx
Find Bounds
Integrate using Definite Integration
First derivative test:
The problem will give u a function and you have to take the derivative of that function and solve for the x values. Then you plug those x values into intervals between negative infinity and infinity. Then plug in numbers between those intervals into the derivative to find if the function is a max, min, increasing or decreasing.
Second derivative test:
You take the derivative of the function twice and solve for the x values once more. Set those values up into intervals between negative infinity and infinity. Plug in a number between these intervals into the second derivative to see if it is concave up, concave down, or a point of inflection.
Product Rule:
You take the derivative of the first and then leave the second as is, then added to the original first and take the derivative of the second.
Related rates-
1.Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
i suck at tangent lines
Disks
Formula: piS(equation)^2dx
Find Bounds (How?)
Integrate using Definite Integration
Washers
Formula: piS[(top equation)^2-(bottom equation)^2]dx
Find Bounds
Integrate using Definite Integration
First derivative test:
The problem will give u a function and you have to take the derivative of that function and solve for the x values. Then you plug those x values into intervals between negative infinity and infinity. Then plug in numbers between those intervals into the derivative to find if the function is a max, min, increasing or decreasing.
Second derivative test:
You take the derivative of the function twice and solve for the x values once more. Set those values up into intervals between negative infinity and infinity. Plug in a number between these intervals into the second derivative to see if it is concave up, concave down, or a point of inflection.
Product Rule:
You take the derivative of the first and then leave the second as is, then added to the original first and take the derivative of the second.
Related rates-
1.Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
i suck at tangent lines
Posting...#32
Some example problems i found for area and volume that might help if your confused.
y=-4x^2+41x+94 AND y=x-2 inbetween 7 and 1
so graph and then put the top equation over the bottom equation
the integral of (-4x^2+41x+94)-(x-2)dx
simplify: the integral of (-4x^2+40x+96) dx
((-4/3)(x)^3+20(x)^2+96(x)) of 1 and 7
f(7)=(3584/3) f(1)=(344/3)
ANSWER: 1080
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
so graph and then put take the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: (153 PI)
Also i need someone to help me at school with something so just tell me if you can. thanks.
Post Number Thirty Two
This week in Calculus we did reviewed mostly, and then we also got an integration worksheet. I never really knew how to do substitution but I started to get a hang of it finally. Hopefully I remember it from now on.. probably not though.
Some easy things that I sometimes don’t always remember and may help:
Product Rule:
You take the derivative of the first and then leave the second as is, then added to the original first and take the derivative of the second.
First derivative test:
The problem will give u a function and you have to take the derivative of that function and solve for the x values. Then you plug those x values into intervals between negative infinity and infinity. Then plug in numbers between those intervals into the derivative to find if the function is a max, min, increasing or decreasing.
Chain Rule:
For a chain rule, you will have a function raised to a power. For example, lets say you have (2x)^3, You have to bring the 3 out in front and then take the derivative of the inside. So you end up with 6(2x)^2.
Tangent line:
If no y value is given, plug the x value into the original function to find the y value. Then take the derivative of the function and plug in the x value to find the slope. Once this is done, plug everything into point-slope form y-y1=slope(x-x1).
I don’t really know how to find area or volume. If anyone can show me an example please do so.
Also I don’t know how to do the problems with K’s.
All help would be appreciated.
Some easy things that I sometimes don’t always remember and may help:
Product Rule:
You take the derivative of the first and then leave the second as is, then added to the original first and take the derivative of the second.
First derivative test:
The problem will give u a function and you have to take the derivative of that function and solve for the x values. Then you plug those x values into intervals between negative infinity and infinity. Then plug in numbers between those intervals into the derivative to find if the function is a max, min, increasing or decreasing.
Chain Rule:
For a chain rule, you will have a function raised to a power. For example, lets say you have (2x)^3, You have to bring the 3 out in front and then take the derivative of the inside. So you end up with 6(2x)^2.
Tangent line:
If no y value is given, plug the x value into the original function to find the y value. Then take the derivative of the function and plug in the x value to find the slope. Once this is done, plug everything into point-slope form y-y1=slope(x-x1).
I don’t really know how to find area or volume. If anyone can show me an example please do so.
Also I don’t know how to do the problems with K’s.
All help would be appreciated.
Post #32
This week in calculus, we went over questions from old APs and had a worksheet on integration, so I will review that.
The integration that I seem to have the most trouble with are all the inverse integrations. The first step in these problems is to recognize it as tan inverse or sin inverse.
The derivative of arcsin is 1/ the square root of 1-u^2
acrsin is -1/ the square root of 1-u^2
acrtan is 1/ 1+ u^2
Example
22. Determine the integral of 1/the square root of 4-t^2 dt on [1,2]
A. pi/6 B. pi/2 C. pi D. pi/4 E. pi/3
This should first be recognized as acrsin.
Next you need to know that the formula of 1/ the square root of a^2 - x^2 is sin -1 (x/a)
Now plug into the formula: sin-1 (t/2)
From there, plug in the bounds and simplify.
sin-1 (1) - sin-1 (1/2)
pi/2 - pi/6 = pi/3
The answer is E.
7. Compute the integral of 4/ 1+4t^2 dt on [0,1/2]
A. 2pi B. pi/2 C. 3pi/2 D. pi E. 0
Factor out the four, so you are left with 4 integral of dt/1+4t^2
This can be recognized and arctan because of the +1 in the bottom.
The 4t^2 at the bottom can also be (2t)^2
Now you can make your u= 2t and your du=2
Since there is no 2 in the problem, you have to multiply by 1/2 to get rid of it.
4(1/2) Integral of dt/ 1+ u^2 on [0,1/2]
Now that we have the formula needed for arctan, all we have to do is integrate then plug in and simplify.
2 tan-1 (u)
2 tan-1(2t) on [0,1/2]
2 ( tan (2)(1/2)) - 2 (tan (2)(0))
2 (pi/4) - 2(0)
pi/2 - 0 = pi/2
The answer is B.
Integration from the worksheet
8. Integrate the (the square root of x - 1) ^2/ the square root of x
u= the square root of x -1 du = 1/2 x^-1/2
since there is a negative exponent, the x^-1/2 will go to the bottom becoming x^1/2, which is what is needed
but there is an extra 2 at the bottom. To get rid of the 2, you will have to multiply by 2
2 integral u^2
2(1/3) u ^3
2/3 u^3
2/3 (the square root of x -1)^3 +C
13. The integral of (1 + 1/t)^3 (1/t^2)
u= 1+ 1/t du= -1/t^2
Multiply the integral by a negative since there is not any in the problem
- integral u^3
- 1/4 u^4
-1/4 (1+ 1/t)^4 +c
Questions:
I can still use help and examples on related rates, area problems, and volume problems.
The integration that I seem to have the most trouble with are all the inverse integrations. The first step in these problems is to recognize it as tan inverse or sin inverse.
The derivative of arcsin is 1/ the square root of 1-u^2
acrsin is -1/ the square root of 1-u^2
acrtan is 1/ 1+ u^2
Example
22. Determine the integral of 1/the square root of 4-t^2 dt on [1,2]
A. pi/6 B. pi/2 C. pi D. pi/4 E. pi/3
This should first be recognized as acrsin.
Next you need to know that the formula of 1/ the square root of a^2 - x^2 is sin -1 (x/a)
Now plug into the formula: sin-1 (t/2)
From there, plug in the bounds and simplify.
sin-1 (1) - sin-1 (1/2)
pi/2 - pi/6 = pi/3
The answer is E.
7. Compute the integral of 4/ 1+4t^2 dt on [0,1/2]
A. 2pi B. pi/2 C. 3pi/2 D. pi E. 0
Factor out the four, so you are left with 4 integral of dt/1+4t^2
This can be recognized and arctan because of the +1 in the bottom.
The 4t^2 at the bottom can also be (2t)^2
Now you can make your u= 2t and your du=2
Since there is no 2 in the problem, you have to multiply by 1/2 to get rid of it.
4(1/2) Integral of dt/ 1+ u^2 on [0,1/2]
Now that we have the formula needed for arctan, all we have to do is integrate then plug in and simplify.
2 tan-1 (u)
2 tan-1(2t) on [0,1/2]
2 ( tan (2)(1/2)) - 2 (tan (2)(0))
2 (pi/4) - 2(0)
pi/2 - 0 = pi/2
The answer is B.
Integration from the worksheet
8. Integrate the (the square root of x - 1) ^2/ the square root of x
u= the square root of x -1 du = 1/2 x^-1/2
since there is a negative exponent, the x^-1/2 will go to the bottom becoming x^1/2, which is what is needed
but there is an extra 2 at the bottom. To get rid of the 2, you will have to multiply by 2
2 integral u^2
2(1/3) u ^3
2/3 u^3
2/3 (the square root of x -1)^3 +C
13. The integral of (1 + 1/t)^3 (1/t^2)
u= 1+ 1/t du= -1/t^2
Multiply the integral by a negative since there is not any in the problem
- integral u^3
- 1/4 u^4
-1/4 (1+ 1/t)^4 +c
Questions:
I can still use help and examples on related rates, area problems, and volume problems.
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
The integral of (1 + 1/t)^3 (1/t^2)
u= 1+ 1/t du= -1/t^2
Multiply the integral by a negative since there is not any in the problem
- integral u^3
- 1/4 u^4
-1/4 (1+ 1/t)^4 +c
Find the volume of the solid obtained by rotating about the x-axis the region under the curve from -2 to 1: y=sqroot(-2x^2-10x+48)
the sq root dissapearas and you then have to integrate the equation.
after you intergrate you plug in : top- bottom times pi.
-2/3(x)^3-5x^2+48x
top-bottom: -2/3(1)^3-5(1)^2+48(1)-(-2/3(-2)^3-5(-2)^2+48(-2))
=491/3.
=491(pi)/3.
for some reason i just can't do the tangent line problems on the ap test, idk where i'm going wrong.
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
The integral of (1 + 1/t)^3 (1/t^2)
u= 1+ 1/t du= -1/t^2
Multiply the integral by a negative since there is not any in the problem
- integral u^3
- 1/4 u^4
-1/4 (1+ 1/t)^4 +c
Find the volume of the solid obtained by rotating about the x-axis the region under the curve from -2 to 1: y=sqroot(-2x^2-10x+48)
the sq root dissapearas and you then have to integrate the equation.
after you intergrate you plug in : top- bottom times pi.
-2/3(x)^3-5x^2+48x
top-bottom: -2/3(1)^3-5(1)^2+48(1)-(-2/3(-2)^3-5(-2)^2+48(-2))
=491/3.
=491(pi)/3.
for some reason i just can't do the tangent line problems on the ap test, idk where i'm going wrong.
Post #32
Finding the Area and volume under a curve
Int (Top Equation)-(Bottom Equation) on the interval [a,b]
You can find a and b by setting the equations equal to each other and solving because a and b are their intersections.
Two rules needed to know: If the area is on the y then a and b need to be y values and solved for x. If the area is on the x then a and b need to be x values and solved for y.
Example:
Find the area of the rigion bounded by f(x)=2-x^2 and g(x)=x.
2-x^2=x 2-x^2-x=0 (-x-1)(x+2) x=-1 x=2
int (Top)-(Bottom) dx [-1,2]
To find the top and bottom equation just graph them on your graphing calculator. You'll see that 2-x^2 is on top with x on the bottom.
int (2-x^2)-(x) dx [-1,2]
int (2-x^2-x) = 2x-(1/3)x^3-(1/2)x^2
Solve like an ordinary definite integral.
2(2)-(1/3)((2)^3)-(1/2)((2)^2)-[2(-1)-(1/3)((-1)^3)-(1/2)((-1)^2)]=(3/2)
Volume is little different because there are two ways to find the volume of a region, depending on the region itself. The two methods are discs and washers.
Discs: (π)int [R(x)]² dx [a,b]
Example:
(π)int √(sinx)² dx [0,π]
(π)int sinx dx [0,π]
(π)(-cosx) [0,π] -cos(π)-(-cos(0))
π(1+1)=2π
Washers: (π)int (Top equation)²-(Bottom equation)² dx [a,b]
Example:
√(x) and (x²)
(π)int (√(x))² - ((x²))² [0,1]
((1/2)x^2) - ((1/5)x^5) 1/2(1)-(1/5)(1) - [1/2(0) -(1/5)(0)]= 3/10
(π)(3/10)= (3π)/10
Int (Top Equation)-(Bottom Equation) on the interval [a,b]
You can find a and b by setting the equations equal to each other and solving because a and b are their intersections.
Two rules needed to know: If the area is on the y then a and b need to be y values and solved for x. If the area is on the x then a and b need to be x values and solved for y.
Example:
Find the area of the rigion bounded by f(x)=2-x^2 and g(x)=x.
2-x^2=x 2-x^2-x=0 (-x-1)(x+2) x=-1 x=2
int (Top)-(Bottom) dx [-1,2]
To find the top and bottom equation just graph them on your graphing calculator. You'll see that 2-x^2 is on top with x on the bottom.
int (2-x^2)-(x) dx [-1,2]
int (2-x^2-x) = 2x-(1/3)x^3-(1/2)x^2
Solve like an ordinary definite integral.
2(2)-(1/3)((2)^3)-(1/2)((2)^2)-[2(-1)-(1/3)((-1)^3)-(1/2)((-1)^2)]=(3/2)
Volume is little different because there are two ways to find the volume of a region, depending on the region itself. The two methods are discs and washers.
Discs: (π)int [R(x)]² dx [a,b]
Example:
(π)int √(sinx)² dx [0,π]
(π)int sinx dx [0,π]
(π)(-cosx) [0,π] -cos(π)-(-cos(0))
π(1+1)=2π
Washers: (π)int (Top equation)²-(Bottom equation)² dx [a,b]
Example:
√(x) and (x²)
(π)int (√(x))² - ((x²))² [0,1]
((1/2)x^2) - ((1/5)x^5) 1/2(1)-(1/5)(1) - [1/2(0) -(1/5)(0)]= 3/10
(π)(3/10)= (3π)/10
Post
So this week basically all I did was go over the tests and do the problems... most of which I already knew... so there's not much for me to post this week that comes to my head...
I'll just explain something I guess...
I find a lot of people still are having issues with integration...like, in particular...change of variable (I think it's called this anyway).
Basically this happens whenever you set your u equal to something and when you take the du, you don't have enough x's that you need...Not sure how to explain it other wise...
Basically let's say the problem was...
x sqrt(x+1)
To integrate this, you would set your u equal to x+1...so
u=x+1
du=1 dx
Now, usually the du will have an x in it because the original problem had an x...but this one doesn't. So what we do, is solve the "u=x+1" for x...so we get that x=u-1.
Now we can substitute back in.
(u-1) (for x) times sqrt(u).
So this is (u-1)(u^(1/2)). Now you can distribute the u^(1/2) into the u-1 and then integrate powers like normal...
Once you learn this change of variable, a lot of problems become way easier.
Another thing...
Let's say the problem says h(x) = f^-1(x) or the inverse of f(x). It then gives you f(x) and asks you to find h'(x)...the formula for this is
h'(x) = 1/(f'(f^-1(x)))
And if you don't know how to do f^-1(x) you should probably look that up from advanced math...and while speaking of that, make sure to look over your trig identities from there as well.
Other than that, you people are starting to learn more :-P. Keep up the good work.
I'll just explain something I guess...
I find a lot of people still are having issues with integration...like, in particular...change of variable (I think it's called this anyway).
Basically this happens whenever you set your u equal to something and when you take the du, you don't have enough x's that you need...Not sure how to explain it other wise...
Basically let's say the problem was...
x sqrt(x+1)
To integrate this, you would set your u equal to x+1...so
u=x+1
du=1 dx
Now, usually the du will have an x in it because the original problem had an x...but this one doesn't. So what we do, is solve the "u=x+1" for x...so we get that x=u-1.
Now we can substitute back in.
(u-1) (for x) times sqrt(u).
So this is (u-1)(u^(1/2)). Now you can distribute the u^(1/2) into the u-1 and then integrate powers like normal...
Once you learn this change of variable, a lot of problems become way easier.
Another thing...
Let's say the problem says h(x) = f^-1(x) or the inverse of f(x). It then gives you f(x) and asks you to find h'(x)...the formula for this is
h'(x) = 1/(f'(f^-1(x)))
And if you don't know how to do f^-1(x) you should probably look that up from advanced math...and while speaking of that, make sure to look over your trig identities from there as well.
Other than that, you people are starting to learn more :-P. Keep up the good work.
Post #32
This week in calculus we went over some of the concepts we did not know very well. Let's go over some simple concepts.
Product Rule:
When taking the derivative, sometimes you have to use the product rule. You take the derivative of the first and then leave the second as is, then add a plus sign and leave the first as is and take the derivative of the second.
Chain Rule:
For a chain rule, you will have a function raised to a power. For example, lets say you have (2x)^3, You have to bring the 3 out in front and then take the derivative of the inside. So you end up with 6(2x)^2.
First derivative test:
The problem will give u a function and you have to take the derivative of that function and solve for the x values. Then you plug those x values into intervals between negative infinity and infinity. Then plug in numbers between those intervals into the derivative to find if the function is a max, min, increasing or decreasing.
Second derivative test:
You take the derivative of the function twice and solve for the x values once more. Set those values up into intervals between negative infinity and infinity. Plug in a number between these intervals into the second derivative to see if it is concave up, concave down, or a point of inflection.
Tangent line:
if no y value is given, plug the x value into the original function to find the y value. Then take the derivative of the function and plug in the x value to find the slope. Once this is done, plug everything into point-slope form y-y1=slope(x-x1).
Things i still have trouble with:
-area problems where u have to find the bounds
- When they give you a graph of acceleration and you have to find where the initial position is
-Problems where they want you to find the increase in value between years (2005-2007).
Have a great weekend :)
Product Rule:
When taking the derivative, sometimes you have to use the product rule. You take the derivative of the first and then leave the second as is, then add a plus sign and leave the first as is and take the derivative of the second.
Chain Rule:
For a chain rule, you will have a function raised to a power. For example, lets say you have (2x)^3, You have to bring the 3 out in front and then take the derivative of the inside. So you end up with 6(2x)^2.
First derivative test:
The problem will give u a function and you have to take the derivative of that function and solve for the x values. Then you plug those x values into intervals between negative infinity and infinity. Then plug in numbers between those intervals into the derivative to find if the function is a max, min, increasing or decreasing.
Second derivative test:
You take the derivative of the function twice and solve for the x values once more. Set those values up into intervals between negative infinity and infinity. Plug in a number between these intervals into the second derivative to see if it is concave up, concave down, or a point of inflection.
Tangent line:
if no y value is given, plug the x value into the original function to find the y value. Then take the derivative of the function and plug in the x value to find the slope. Once this is done, plug everything into point-slope form y-y1=slope(x-x1).
Things i still have trouble with:
-area problems where u have to find the bounds
- When they give you a graph of acceleration and you have to find where the initial position is
-Problems where they want you to find the increase in value between years (2005-2007).
Have a great weekend :)
Post #32
alright so this past week we went over all sorts of things and what i didn't grasp right away i'll go over...but there is some simple stuff that everyone should know
Let’s see what I remember shall we.. if you take the derivative of f^2(x) you treat it as if it was like sine. You do 2f(x) times f’(x) times 1. So it’s almost as if you have (f(x))^2 and you use the chain rule. Bring the 2 to the front, subtract 1, times by the derivative of the inside. And since the derivative of f(x) is f’(x) times the derivative of x, which is 1. These rules seem to help me A LOT!
Lets say you have:
x^2-2 DIVIDED BY 4-x^2 AS THE LIMIT APPROACHES 2
soo..
for this you end up getting two divided by zero. For finite limits though, if you get a number over zero, it’s always infinity. Therefore the answer is INFINITY!
Remember:
If the top exponent is greater than the bottom it’s the limit as it approaches infinity, and if the top exponent is less than the bottom it’s the limit as it approaches zero!
DON’T GET THEM CONFUSED LIKE I DID!
Here’s a trick:
Remember if you’re divided a number by another number and you have a bigger number on top it’s usually not zero. Also, if you’re dividing a number by another number and you have a smaller number on top it’s usually not a big number. Therefore, when the bigger number is on top since it’s a bigger number than zero it’s INFINITY and when the bigger number is on the bottom that means that it’s being lessened so it’s ZERO!
volume by disks
so you know that the formula is pi times the integral of the [function given] squared times dx. well then you gotta know what to do right 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...oh, REMEMBER TO GRAPH..
just look at what they give you...they're should be numbers that they want them inbetween
IF NOT..WHAT DO YOU DO???
volume by washers
so you know that 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 inbetween 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. REMEMBER TO GRAPH!
just a reminder if it's for the x-axis then you solve for y and if it's on the y-axis then you solve for x..and if you plug the y-axis one's in your calculator to graph then you'll have to turn your calculator sideways with the screen on the right to see how it would really look.
area...it's worked the same just they'll give you two equations and if you don't have numbers for the inbetween then you'll set them equal to get them, but if you do then don't worry and keep going. for area there is no pi and no squaring. so you put the integral of the first equation minus the second equation and simplify and solve. then take the integral of it and plug in the numbers inbetween!
examples:
AREA:
y=-4x^2+41x+94 AND y=x-2 inbetween 7 and 1
so graph and then put the top equation over the bottom equation
the integral of (-4x^2+41x+94)-(x-2)dx
simplify: the integral of (-4x^2+40x+96) dx
((-4/3)(x)^3+20(x)^2+96(x)) of 1 and 7
f(7)=(3584/3) f(1)=(344/3)
ANSWER: 1080
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
so graph and then put take the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: (153 PI)
hope this helps...
~ElliE~
Let’s see what I remember shall we.. if you take the derivative of f^2(x) you treat it as if it was like sine. You do 2f(x) times f’(x) times 1. So it’s almost as if you have (f(x))^2 and you use the chain rule. Bring the 2 to the front, subtract 1, times by the derivative of the inside. And since the derivative of f(x) is f’(x) times the derivative of x, which is 1. These rules seem to help me A LOT!
Lets say you have:
x^2-2 DIVIDED BY 4-x^2 AS THE LIMIT APPROACHES 2
soo..
for this you end up getting two divided by zero. For finite limits though, if you get a number over zero, it’s always infinity. Therefore the answer is INFINITY!
Remember:
If the top exponent is greater than the bottom it’s the limit as it approaches infinity, and if the top exponent is less than the bottom it’s the limit as it approaches zero!
DON’T GET THEM CONFUSED LIKE I DID!
Here’s a trick:
Remember if you’re divided a number by another number and you have a bigger number on top it’s usually not zero. Also, if you’re dividing a number by another number and you have a smaller number on top it’s usually not a big number. Therefore, when the bigger number is on top since it’s a bigger number than zero it’s INFINITY and when the bigger number is on the bottom that means that it’s being lessened so it’s ZERO!
volume by disks
so you know that the formula is pi times the integral of the [function given] squared times dx. well then you gotta know what to do right 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...oh, REMEMBER TO GRAPH..
just look at what they give you...they're should be numbers that they want them inbetween
IF NOT..WHAT DO YOU DO???
volume by washers
so you know that 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 inbetween 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. REMEMBER TO GRAPH!
just a reminder if it's for the x-axis then you solve for y and if it's on the y-axis then you solve for x..and if you plug the y-axis one's in your calculator to graph then you'll have to turn your calculator sideways with the screen on the right to see how it would really look.
area...it's worked the same just they'll give you two equations and if you don't have numbers for the inbetween then you'll set them equal to get them, but if you do then don't worry and keep going. for area there is no pi and no squaring. so you put the integral of the first equation minus the second equation and simplify and solve. then take the integral of it and plug in the numbers inbetween!
examples:
AREA:
y=-4x^2+41x+94 AND y=x-2 inbetween 7 and 1
so graph and then put the top equation over the bottom equation
the integral of (-4x^2+41x+94)-(x-2)dx
simplify: the integral of (-4x^2+40x+96) dx
((-4/3)(x)^3+20(x)^2+96(x)) of 1 and 7
f(7)=(3584/3) f(1)=(344/3)
ANSWER: 1080
VOLUME:
y=the sqr. root of (-2(x)^2-10(x)+48) inbetween 1 and -2
so graph and then put take the equation they give you into the integral and square it
PI times the integral of (-2(x)^2-10(x)+48) dx
PI [(-2/3)(x)^3-5(x)2+48(x)] of 1 and -2
f(1)=(127/3) f(-2)=(-332/3)
ANSWER: (153 PI)
hope this helps...
~ElliE~
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