Thanks to all of the help on my comments, I now understand how to integrate fractions!!
But, if they tell you to just integrate tan...all you do is sinx/cosx? (+ C if it's indefinite?)
And for substitution (I think I have this) you can alter the problem with the u and du in it to make du the derivative of u, right? If that makes any sense.
Thursday, February 11, 2010
Maternity Leave
Remember to post any questions that you have to the blog. I will try to check in every day if not every other day. It will not be quite as often right after I have the baby.
Wednesday, February 10, 2010
post 25
Ok, so my internet is finally up and I can type up and post the blog that I have had written down for days... so 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.
One thing I struggle with is related rates, if anyone has any tips for that I guess I could use them.
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.
One thing I struggle with is related rates, if anyone has any tips for that I guess I could use them.
#35
Ok. So #35 is actually not an AB topic. However, the mistake we made in class was the bounds are 0 to 2 on the x and we did 0 to 3 when we found the interesection. If you alter this to match the x-values then you will get 7.6 something which is close to 8. However, shells are much easier than washers and can only be done in a washer problem about the y-axis (for now). So you could do this entirely in your calculator MUCH faster. The formula for shells is
2pi(integral (x(top-bottom))) and you would use the bounds of the intersection regardless of the limitations given in the problem. So this would be 0 to 3. You use the x values not the ys when revolving about y and doing shells.
Nothing needs to be squared or any of that. This is Alot simpler. It is a shortcut that you learn next year but could help save time this year. So in your calculator you would type.
fnInt(x(9-x^2 - (9-3x)),x,0,3) then multiply that by 2pi.
You don't have to learn this method. It will not actually be on the AP but if it makes more sense for this method than you can try it to check your work.
Monday, February 8, 2010
Ash's 25th Post Part II
I can finally post my blog and do my wiki because: YAY!! THE VIRUS IS GONE FROM MY COMPUTER!! (Trojan)
But along with that being gone, each time I click something the "Open With" box pops up!! *throws computer against the wall* Does anyone know how to fix this?
Fun Fact of the week:
I was watching The Lion King and in Scar's song "Be Prepared" he refers to math with these lyrics: "And injustice deliciously squared"
And there's the random math fact of the week!!
Okay, so, that out of the way, let's move on to important things.
AP Tests = not fun.
How is everyone else doing? are you improving at all? I found the last test to be pretty...I dunno, out there?
The Calculator potion killed me! (nevermind the fact that my calculator decided to be stupid and not work)
I want to try to explain something, but I don't want to explain it wrong and confuse myself and others.
I need help with remembering what to do when. Okay, let me change that.
I know what to do, I just can't do it. Does anyone else have this problem?
I'm getting stuck on large derivatives and integrating fractions. Can anyone give me an example problem of integrating fractions or tips? It's on every AP test and it kills me that I can't figure it out! -.-
Please and thank you!!
But along with that being gone, each time I click something the "Open With" box pops up!! *throws computer against the wall* Does anyone know how to fix this?
Fun Fact of the week:
I was watching The Lion King and in Scar's song "Be Prepared" he refers to math with these lyrics: "And injustice deliciously squared"
And there's the random math fact of the week!!
Okay, so, that out of the way, let's move on to important things.
AP Tests = not fun.
How is everyone else doing? are you improving at all? I found the last test to be pretty...I dunno, out there?
The Calculator potion killed me! (nevermind the fact that my calculator decided to be stupid and not work)
I want to try to explain something, but I don't want to explain it wrong and confuse myself and others.
I need help with remembering what to do when. Okay, let me change that.
I know what to do, I just can't do it. Does anyone else have this problem?
I'm getting stuck on large derivatives and integrating fractions. Can anyone give me an example problem of integrating fractions or tips? It's on every AP test and it kills me that I can't figure it out! -.-
Please and thank you!!
Post 25
Gosh, I forgot to do my blog on time again this week. I need to start getting on the ball. Anyway, this week we did more ap tests. In addition to the two multiple choice portions we usually take every Monday and Tuesday, we took a free response portion. The free response portion was a slope field question. We were given one similar to it to study from prior to the test on Monday. Tuesday and Wednesday we were given first the non calculator portion and second the calculator portion. These two tests were from a different book, so we had to do corrections on them. We have to do corrections on all of them really, but this time we are getting points for the corrections.
On the non calculator portion, I thought I did well and answered a few more questions than normal. I thought I would get a perfect, but the class average was higher and I made stupid mistakes. It was the same for the calculator portion.
Upon making my corrections, I’ve learned a few things:
Normal line
How not to screw up a second derivative including a chain rule
And how useful my calculator can be on the calculator portion if I would just use it
Ok, for a normal line, I remembered I had to use the same formula for tangent line. I also remembered that in the tangent line formula, the slope given by the derivative was a perpendicular slope. A normal line is a line with the same slope, but I forgot to take the negative reciprocal of the slope given. This is all you do for a normal line.
Ok, for a chain rule with multiple derivatives. Don’t forget to bring the exponent to the front, subtract one from it, then multiply it by the inside then the derivative of the inside. I always forget the fourth step.
When doing the calculator portion, they let you have a calculator for a reason. Your calculator can help you out with everything on the test. So use it.
The problem I have is I remember B-rob saying something about when a limit is undefined it is actually going to infinity? I put this for two tests and got it wrong. Am I making something up here?
On the non calculator portion, I thought I did well and answered a few more questions than normal. I thought I would get a perfect, but the class average was higher and I made stupid mistakes. It was the same for the calculator portion.
Upon making my corrections, I’ve learned a few things:
Normal line
How not to screw up a second derivative including a chain rule
And how useful my calculator can be on the calculator portion if I would just use it
Ok, for a normal line, I remembered I had to use the same formula for tangent line. I also remembered that in the tangent line formula, the slope given by the derivative was a perpendicular slope. A normal line is a line with the same slope, but I forgot to take the negative reciprocal of the slope given. This is all you do for a normal line.
Ok, for a chain rule with multiple derivatives. Don’t forget to bring the exponent to the front, subtract one from it, then multiply it by the inside then the derivative of the inside. I always forget the fourth step.
When doing the calculator portion, they let you have a calculator for a reason. Your calculator can help you out with everything on the test. So use it.
The problem I have is I remember B-rob saying something about when a limit is undefined it is actually going to infinity? I put this for two tests and got it wrong. Am I making something up here?
idk.
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'.
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
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 take the first implicit derivative first
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
You have to identify what you are looking for and what you are given. Not only does this make it easier on you, it's kind of necessary, especially when you want those points on free response questions (or so I'm told). You also have to realize that when you are doing related rates, you have to put dy/dt or dx/dt or whatever whenever you are taking the derivative of some variable in relation to time (hence the t). So given that:
Given xy = 4
you want to know what dy/dt equals given x = 8 and dx/dt=10.
Take the derivative (product rule):
dx/dt y + dy/dt x = 0
Plug in everything:
dy/dt = -10y/8
= -5y/4
I need help on the problems with I, II, and III...you know what I mean? How do I attack the problem???
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'.
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
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 take the first implicit derivative first
36x^2 + 2y = 9
36x + 2(dy/dx) = 0
2(dy/dx) = -36x
-36x/2
dy/dx = -18x
You have to identify what you are looking for and what you are given. Not only does this make it easier on you, it's kind of necessary, especially when you want those points on free response questions (or so I'm told). You also have to realize that when you are doing related rates, you have to put dy/dt or dx/dt or whatever whenever you are taking the derivative of some variable in relation to time (hence the t). So given that:
Given xy = 4
you want to know what dy/dt equals given x = 8 and dx/dt=10.
Take the derivative (product rule):
dx/dt y + dy/dt x = 0
Plug in everything:
dy/dt = -10y/8
= -5y/4
I need help on the problems with I, II, and III...you know what I mean? How do I attack the problem???
25
Some keywords I never understood:
Local minimum- first derivative, set equal to zero, solve, set up intervals and check
Maximum speed- absolute value of velocity
Linearization- find the equation of the tangent line
How Many? – integrate
Lim h> 0 + h – definition of derivative- take derivative of what is behind the parenthesis
Maximum value – find the y- value at the max; take derivative, set equal to 0, solve for x, plug in to find y.
Maximum acceleration- where is maximum slope
Starting point-position- related to original function; zeros of velocity will give you max and mins of position
Vertical tangents- take implicit derivative
Rate- take derivative
Somethings I know how to do:
Lim
x>infintity (20x^2 – 13x + 5)/(5 – 4x^3)
The limit is 0 because of the limit rule when the top degree is smaller than the bottom degree the limit is 0. 2<3
Limh>0 (ln(2+h) – ln2)/h is 1/2 .
You take the derivative of what is outside of the parenthesis to figure out the answer. Since it is ln 2 make that 1/x and plug in 2 therefore giving you the limit at ½.
The average rate of chage of function f on [1, 4] is..
You are given a table of the values
For this problem you use f(b) – f(a)/b – a
Therefore f(4) – f(1)/3 equals 6-2/3 which equals 4/3
Local minimum- first derivative, set equal to zero, solve, set up intervals and check
Maximum speed- absolute value of velocity
Linearization- find the equation of the tangent line
How Many? – integrate
Lim h> 0 + h – definition of derivative- take derivative of what is behind the parenthesis
Maximum value – find the y- value at the max; take derivative, set equal to 0, solve for x, plug in to find y.
Maximum acceleration- where is maximum slope
Starting point-position- related to original function; zeros of velocity will give you max and mins of position
Vertical tangents- take implicit derivative
Rate- take derivative
Somethings I know how to do:
Lim
x>infintity (20x^2 – 13x + 5)/(5 – 4x^3)
The limit is 0 because of the limit rule when the top degree is smaller than the bottom degree the limit is 0. 2<3
Limh>0 (ln(2+h) – ln2)/h is 1/2 .
You take the derivative of what is outside of the parenthesis to figure out the answer. Since it is ln 2 make that 1/x and plug in 2 therefore giving you the limit at ½.
The average rate of chage of function f on [1, 4] is..
You are given a table of the values
For this problem you use f(b) – f(a)/b – a
Therefore f(4) – f(1)/3 equals 6-2/3 which equals 4/3
okayyy so everytime we have a monday off all my stuff gets messed uppp and an think that monday is sunday, etc. sorrrrrry!
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)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
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.
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
i need help with substitution and a refresher on optimization would also be nice.
ohhh and i have no idea how to work my calculator if anyone would like to go over some thingggggggssssss =)
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)
The terms for the First Derivative Test:
1. Increasing
2. Decreasing
3. Horizontal Tangent
4. Min/Max
Steps:
1. Take the derivative
2. Set it equal to zero
3. Solve for x to get the possible critical points [[can someone clarify this part???? My notebook says it equals critical points AND mins/maxs/horizontal tangents]]
4. Set up intervals with your x value(s)
5. Plug into your first derivative
6. To find an absolute extrema, plug in the values from step 5 into your original function
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.
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
i need help with substitution and a refresher on optimization would also be nice.
ohhh and i have no idea how to work my calculator if anyone would like to go over some thingggggggssssss =)
Post #25?
Well I would do specific problems from the ap test like I have been doing, but I left my binder in my locker so I guess I will just review.
Alright here are some things that keep popping up, but we just forget.
LIMIT RULES:
1. top degree is larger than the bottom degree-->the limit approaches infinity
2. bottom degree is larger than the top degree-->the limit approaches zero
3. and if the top and bottom degree are equal-->then you make a fraction out of the coefficients
FIRST DERIVATIVE TEST:
deals with increasing and decreasing/max and mins
1. derivative of the function
2. solve for x values which are critical values
3. make those numbers into intervals between -ve infinity and infinity
4. when seeing if +ve or -ve plug those numbers into the derivative
5. if -ve it is decreasing and +ve it is increasing, changing from -ve to to positive you have a min and going from +ve to -ve you have a max, absolute maxs and minx deal with the highest and lowest value
SECOND DERIVATIVE TEST:
deals with concave up and down/points of inflection
1. take the derivative of the function twice
2. set equal to 0 and solve for x (these are not really critical values, more like points of interest)
3. set up intervals just like the first derivative test
4. plug in also like the first derivative test but plug into the second derivative
5. -ve is concave down and +ve is concave up, points of inflection is where the concavity changes like going from -ve to +ve
EQUATION OF A TANGENT LINE:
1. take the derivative
2. plug in x value
3. if not given a y value, plug into the original equation to get the y value
4. then plug those numbers into point slope form: y − y1 = m(x − x1)
5. *if wanting the normal line, the slope is just the reciprocal
AND slope field stuff:
I'm just going to review how to draw it.
positive slopes is /
negative slope is \
for a zero slope is a horizontal line
for an undefined slope is a vertical line
Well I would just like to say that I made a six on the calculator part, and yes I'm still happy about that. But I am having trouble on the non calculator part. Something that I do need help on is e integration and ln integration. Thankssss
Alright here are some things that keep popping up, but we just forget.
LIMIT RULES:
1. top degree is larger than the bottom degree-->the limit approaches infinity
2. bottom degree is larger than the top degree-->the limit approaches zero
3. and if the top and bottom degree are equal-->then you make a fraction out of the coefficients
FIRST DERIVATIVE TEST:
deals with increasing and decreasing/max and mins
1. derivative of the function
2. solve for x values which are critical values
3. make those numbers into intervals between -ve infinity and infinity
4. when seeing if +ve or -ve plug those numbers into the derivative
5. if -ve it is decreasing and +ve it is increasing, changing from -ve to to positive you have a min and going from +ve to -ve you have a max, absolute maxs and minx deal with the highest and lowest value
SECOND DERIVATIVE TEST:
deals with concave up and down/points of inflection
1. take the derivative of the function twice
2. set equal to 0 and solve for x (these are not really critical values, more like points of interest)
3. set up intervals just like the first derivative test
4. plug in also like the first derivative test but plug into the second derivative
5. -ve is concave down and +ve is concave up, points of inflection is where the concavity changes like going from -ve to +ve
EQUATION OF A TANGENT LINE:
1. take the derivative
2. plug in x value
3. if not given a y value, plug into the original equation to get the y value
4. then plug those numbers into point slope form: y − y1 = m(x − x1)
5. *if wanting the normal line, the slope is just the reciprocal
AND slope field stuff:
I'm just going to review how to draw it.
positive slopes is /
negative slope is \
for a zero slope is a horizontal line
for an undefined slope is a vertical line
Well I would just like to say that I made a six on the calculator part, and yes I'm still happy about that. But I am having trouble on the non calculator part. Something that I do need help on is e integration and ln integration. Thankssss
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