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
Super Bowl Post.
So congratulations to the Saints on the win.
What can I cover...well
Volume by Discs (a disc does not have a hole in it)
pi times the integral from a to b of the equation squared.
Volume by Washers (a washer does have a hole in it)
pi times the integral from a to b of the top equation squared minus the bottom equation squared...
Area between two equations
the integral from a to b of the top equation minus the bottom equation
So...some people still do NOT realize this relationship:
Position is the integral of velocity.
Velocity is the integral of acceleration.
Acceleration is the derivative of velocity.
Velocity is the derivative of position.
If they give you the acceleration and want to know when you the particle changes directions...be smart about what to do. Think about what you need...
Well when a problem says when the particle changes...you should all know by now that you need to find the max's or min's of the position function, correct? Well how do you find max's or min's? You use the first derivative test. Well what is the derivative of position? Velocity. Wonderful.
So now that we know that we don't have to integrate all the way to position and just to velocity we just saved ourselves some time. So you would integrate the acceleration equation and solve for c then solve your equation when it is set equal to 0 so that you can find max's or min's. The max's or min's of the position function is where the particle changes directions.
Anyway, hopefully everyone is happy :-P
What can I cover...well
Volume by Discs (a disc does not have a hole in it)
pi times the integral from a to b of the equation squared.
Volume by Washers (a washer does have a hole in it)
pi times the integral from a to b of the top equation squared minus the bottom equation squared...
Area between two equations
the integral from a to b of the top equation minus the bottom equation
So...some people still do NOT realize this relationship:
Position is the integral of velocity.
Velocity is the integral of acceleration.
Acceleration is the derivative of velocity.
Velocity is the derivative of position.
If they give you the acceleration and want to know when you the particle changes directions...be smart about what to do. Think about what you need...
Well when a problem says when the particle changes...you should all know by now that you need to find the max's or min's of the position function, correct? Well how do you find max's or min's? You use the first derivative test. Well what is the derivative of position? Velocity. Wonderful.
So now that we know that we don't have to integrate all the way to position and just to velocity we just saved ourselves some time. So you would integrate the acceleration equation and solve for c then solve your equation when it is set equal to 0 so that you can find max's or min's. The max's or min's of the position function is where the particle changes directions.
Anyway, hopefully everyone is happy :-P
Sunday, February 7, 2010
super bowl sunday post
Substitution takes the position of derivative rules when integrating. The steps for substitution are:
1. Find a derivative of something else inside of the integral.
2. Set u equal to the non derivative found in the integral
3. Then take the derivative of u4. Substitute back in
5. Solve
I am going to talk about 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
Example problem:
The variables x and y are differentiable functions of t and are related by the equation y= 2x^3-x+4when x=2 dx/dt=1. Find dy/dt when x=2.
dy/dt=?
dy/dt=6x^2 dx/dt-1 dx/dt
x=2 dx/dt=1 dy/dt=6(2)^2(-1)-(-1)
y=2x^3-x+4 dy/dt=-23
The angle of elevation is pretty much the same thing as rate of change except the difference is you look for an angle of a triange instead of a rate.
Example problem: A balloon rises at a rate of 5 meters/second from a point on the ground 50 meters from the observer. Find the rate of change of the angle of elevation of the balloon to the observer when the balloon is 50 meters from the ground.
tan(theta)=4/x
tan(theta)=5/50
sec^2(theta)=1/10 d(theta)/dt=1/10 dy/dt
d(theta)/dt=cos^2(theta)=1/10(4)
1/2(1/10)(4)= 1/5 radians/second
Small errors are my main problem on the AP tests.
1. Find a derivative of something else inside of the integral.
2. Set u equal to the non derivative found in the integral
3. Then take the derivative of u4. Substitute back in
5. Solve
I am going to talk about 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
Example problem:
The variables x and y are differentiable functions of t and are related by the equation y= 2x^3-x+4when x=2 dx/dt=1. Find dy/dt when x=2.
dy/dt=?
dy/dt=6x^2 dx/dt-1 dx/dt
x=2 dx/dt=1 dy/dt=6(2)^2(-1)-(-1)
y=2x^3-x+4 dy/dt=-23
The angle of elevation is pretty much the same thing as rate of change except the difference is you look for an angle of a triange instead of a rate.
Example problem: A balloon rises at a rate of 5 meters/second from a point on the ground 50 meters from the observer. Find the rate of change of the angle of elevation of the balloon to the observer when the balloon is 50 meters from the ground.
tan(theta)=4/x
tan(theta)=5/50
sec^2(theta)=1/10 d(theta)/dt=1/10 dy/dt
d(theta)/dt=cos^2(theta)=1/10(4)
1/2(1/10)(4)= 1/5 radians/second
Small errors are my main problem on the AP tests.
Post Number Twenty Five
So i'm watching the superbowl and people oh so conveniently reminded me about the blog. So here it goes:
Max and mins are on every test:
First, find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.Second, plug both intervals as well as the critical values into your original equation. Next, be sure to keep the value after substitution with each appropriate number. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
I can tell you how to do tangent lines but for some reason I never know how to do it on the tests:
Take the derivative of the equation like normal.
Plug in the x value which gives you your slope.
Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1).
If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
Related Rates:
1. Identify all variables and equations.
2. Identify what you are looking for.
3. Make a sketch and label.
4. Write an equations involving your variables.
*You can only have one unknown so a secondary equation may be given
5. Take the derivative with respect to time.
6. Substitute in derivative and solve.
I still need help with ln and e integration and differentiation. Any volunteers?
Oh and WHO DAT (:
Posting...25
EVT: the EVT states that a continuous function on a close interval [a,b], must have both a minimum and a maximum on the interval. However, the max and min can occur at the endpoints.
Rolles theorem gives the conditions that guarantee the existance of an extrema in the interior of a closed inteval.
Rolles: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a)= f(b) then there is at least one number, “c” in (a,b) such the f '(c)=0
MVT: If f is continuous on the closed interval [a,b] and differentiable on the interval (a,b) then there exists a number c in (a,b) such that f '(c)= (f(b)-f(a))/(b-a).
Example: f(x)-x^(2)-3x+2, show that f ' (x)=0 on some interval. (Hint: if they don't give an interval, assume it's x-intercepts)
so first you factor and find that x=1 and x=2
so then, you check and see that f(1)=f(2)=0 so it's continuous and it's differentiable.
So after you've found that, you take the derivative and set it equal to zero.
F '(x)=2x-3
2x-3=0
2x=3
x=3/2
c=3/2. And that's it. You've found c.
----------------
Another example
f(x)=x^(4)-2x^(2) on [-2,2] find all values of c for which f '(c)=0
continuous: yes
differentiable: yes. So let's get started.
First, you have to plug in to see if f(-2) is equal to f(2)
so, f(-2)=(-2)^(4) -2 (-2)^(2)= 8
f(2)= (2)^(4)-2(2)^(2)=8
so that's good. They equal each other. By the way, i'm not sure if i'm typing up all my parentheses correctly, so sorry if I mess that up.
So then you take the derivative and set = to zero
f '(x)=4x^(3)-4x=0
4x(x^(2)-1)=0
x= 0, 1, -1
c= -1, 0, 1. and you're finished again.
Rolles theorem gives the conditions that guarantee the existance of an extrema in the interior of a closed inteval.
Rolles: Let f be continuous on a closed interval [a,b] and differentiable on the open interval (a,b). If f(a)= f(b) then there is at least one number, “c” in (a,b) such the f '(c)=0
MVT: If f is continuous on the closed interval [a,b] and differentiable on the interval (a,b) then there exists a number c in (a,b) such that f '(c)= (f(b)-f(a))/(b-a).
Example: f(x)-x^(2)-3x+2, show that f ' (x)=0 on some interval. (Hint: if they don't give an interval, assume it's x-intercepts)
so first you factor and find that x=1 and x=2
so then, you check and see that f(1)=f(2)=0 so it's continuous and it's differentiable.
So after you've found that, you take the derivative and set it equal to zero.
F '(x)=2x-3
2x-3=0
2x=3
x=3/2
c=3/2. And that's it. You've found c.
----------------
Another example
f(x)=x^(4)-2x^(2) on [-2,2] find all values of c for which f '(c)=0
continuous: yes
differentiable: yes. So let's get started.
First, you have to plug in to see if f(-2) is equal to f(2)
so, f(-2)=(-2)^(4) -2 (-2)^(2)= 8
f(2)= (2)^(4)-2(2)^(2)=8
so that's good. They equal each other. By the way, i'm not sure if i'm typing up all my parentheses correctly, so sorry if I mess that up.
So then you take the derivative and set = to zero
f '(x)=4x^(3)-4x=0
4x(x^(2)-1)=0
x= 0, 1, -1
c= -1, 0, 1. and you're finished again.
For what i don't understand is intergration help meh now!?!?!?
post 25
so, saints just won the superbowl :) woooooh!
this week we did the same thing, worked on ap tests, and went over them
here's some stuff:
whatever e is raised to is your u. du is derivative of u
absolute maxs and mins:
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
linearization:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
i don't understand tram and mram.
Post #25
So I'm going to explain some of the questions I missed on the non-calculator portion because some of yall may have missed them too.
An equation of the line normal to the graph of y=the square root of (3x^2+2x) at (2,4) is
A. -4x+y=20 B. 4x+7y=20 C. -7x+4y=2 D. 7x+4y=30 E. 4x+7y=30
To find an equation of a normal line, you need a point and a perpendicular slope. To find the slope, take the derivative of the equation and plug in your point. The perpendicular slope will be the negative reciprocal of the slope.
Derivative:
(3x^2+2x)^1/2
1/2(3x^2+2x)^-1/2 (6x+2)
6x+2/ 2 (3x^2 +2x)^1/2
Plug in:
6(2) + 2 / 2(3(2)^2 + 2(2)) ^1/2
14/ 8
m= 7/4
the perpendicular slope = -4/7
Now that you have your perpendicular slope, you plug into point-slope formula.
y-4 = -4/7 (x-2)
Since that answer does not appear in the choice, you need to solve the equation for an answer they give. For the equation, multiply everything by seven to get rid of the fraction first then solve
7y-28 = -4 ( x-2)
7y-28 =-4x+8
4x+7y -36 = 0
4x+7y = 36
Therefore, the answer is D.
The integral of (e^3lnx + e^3x) dx=
The key word in this problem is the integral sign. Looking at the first term, you know the ln and e cancel leaving you with x^3.
The problem is now the integral of x^3 + e^3x now integrate.
The integral of x^3 is 1/4 x^4 and the integral of e^3x is 1/3 e^3x
So the answer will be x^4/4 + e^3x/3 + c.
The average value of the function (x-1)^2 on the interval from x=1 to x=5 is
A. -16/3 B. 16/3 C. 64/3 D. 66/3 E. 256/3
To find the average value of a function, you do 1/b-a times the integral of the function on the interval given.
1/ 5-1 integral (x-1)^2 on [1,5]
To integrate this function, you first have to factor it out.
x^2 - 2x +1
Integrate: 1/3 x^3 - x^2 + x on [1,5]
Next step is to plug in and multiply by 1/b-a
1/4 [1/3 (5)^3 -(5)^2 +5] - [1/3(1)^3 - (1)^2 +1 ]
1/4 [ 1/3 (125) - 25 +5 ] - [1/3 -1 +1]
1/4 ( 125/3 - 25 + 5 - 1/3 )
1/4 (64/3 )
64/12 = 16/3 So the answer is B.
I have questions on 30, 35, 39, 42, and 44 on the calculator portion of the last AP.
An equation of the line normal to the graph of y=the square root of (3x^2+2x) at (2,4) is
A. -4x+y=20 B. 4x+7y=20 C. -7x+4y=2 D. 7x+4y=30 E. 4x+7y=30
To find an equation of a normal line, you need a point and a perpendicular slope. To find the slope, take the derivative of the equation and plug in your point. The perpendicular slope will be the negative reciprocal of the slope.
Derivative:
(3x^2+2x)^1/2
1/2(3x^2+2x)^-1/2 (6x+2)
6x+2/ 2 (3x^2 +2x)^1/2
Plug in:
6(2) + 2 / 2(3(2)^2 + 2(2)) ^1/2
14/ 8
m= 7/4
the perpendicular slope = -4/7
Now that you have your perpendicular slope, you plug into point-slope formula.
y-4 = -4/7 (x-2)
Since that answer does not appear in the choice, you need to solve the equation for an answer they give. For the equation, multiply everything by seven to get rid of the fraction first then solve
7y-28 = -4 ( x-2)
7y-28 =-4x+8
4x+7y -36 = 0
4x+7y = 36
Therefore, the answer is D.
The integral of (e^3lnx + e^3x) dx=
The key word in this problem is the integral sign. Looking at the first term, you know the ln and e cancel leaving you with x^3.
The problem is now the integral of x^3 + e^3x now integrate.
The integral of x^3 is 1/4 x^4 and the integral of e^3x is 1/3 e^3x
So the answer will be x^4/4 + e^3x/3 + c.
The average value of the function (x-1)^2 on the interval from x=1 to x=5 is
A. -16/3 B. 16/3 C. 64/3 D. 66/3 E. 256/3
To find the average value of a function, you do 1/b-a times the integral of the function on the interval given.
1/ 5-1 integral (x-1)^2 on [1,5]
To integrate this function, you first have to factor it out.
x^2 - 2x +1
Integrate: 1/3 x^3 - x^2 + x on [1,5]
Next step is to plug in and multiply by 1/b-a
1/4 [1/3 (5)^3 -(5)^2 +5] - [1/3(1)^3 - (1)^2 +1 ]
1/4 [ 1/3 (125) - 25 +5 ] - [1/3 -1 +1]
1/4 ( 125/3 - 25 + 5 - 1/3 )
1/4 (64/3 )
64/12 = 16/3 So the answer is B.
I have questions on 30, 35, 39, 42, and 44 on the calculator portion of the last AP.
Post # 25
So, when i was actually at school we did things involving taking more practice ap's..
so in attempts to teach this to myself, i'm going to explain trapezoidal rule..
so if i do anything wrong, i would appreciate if someone could just comment it and let me know..or easier tricks to learning it!
so, here it goes..
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 =
11[ f(-3) + f(-3 +1)]1[ f( -3) + f(-2)]
then plug into your equationRRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)]
and then plug into your equation...
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1-3 + -2/ 2
= -5/2 and -2 + -1 / 2
= -3/2
so 1[f(-5/2) + f(-3/2)]
and the plug in..
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
THat's really the only thing i'm confused about..my silly mistakes ruin my ap scores...
so in attempts to teach this to myself, i'm going to explain trapezoidal rule..
so if i do anything wrong, i would appreciate if someone could just comment it and let me know..or easier tricks to learning it!
so, here it goes..
LRAM is left hand approximation and the formula is:
delta x [f(a) + f( delta x +a) .... + f( delta x - b)]
Say you are asked to calculate the left Riemann Sum for -4x -5 on the interval [-3, -1] divided into 2 subintervals.
delta x would equal: -1+3 /2 = 2/2 =
11[ f(-3) + f(-3 +1)]1[ f( -3) + f(-2)]
then plug into your equationRRAM is right hand approximation and the formula is:
delta x [ f(a + delta x) + .... + f(b)]
so using the same example:
1[ f( -2) + f(-1)]
and then plug into your equation...
MRAM is to calculate the middle and the formula is:
delta x [ f(mid) + f(mid) + .... ]
To find midpoints, you would add the two numbers together then divide by two
In this problem the numbers would be: -3 , -2, -1-3 + -2/ 2
= -5/2 and -2 + -1 / 2
= -3/2
so 1[f(-5/2) + f(-3/2)]
and the plug in..
Trapezoidal is different because instead of multiplying by delta x, you multiply by delta x/2 and you also have on more term then your number of subintervals.
The formula is : delta x/2 [f(a) + 2f(a + delta x) + 2f(a+ 2 delta x) + ....f(b)]
For this problem: 1/2 [ f(-3) + 2 f(-2) + f( -1)] and then plug in.
THat's really the only thing i'm confused about..my silly mistakes ruin my ap scores...
post 25
time for some calculus before the super bowl yeeerrrrrrrrrrrdddmeeeeeeeeeeee
maxs and mins:
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Infinity rules:
1. If the degree of the top is greater than the degree of the bottom, then the limit is going to infinity.
2. If the degree at the top is less than the degree at the bottom, then the limit is going to zero.
3. If the degree of the top is the same as the degree of the bottom, set the
coefficients to a fraction.
some things i'm not too good w/ is slope fields, e integration, and lram rram and mram.
maxs and mins:
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Infinity rules:
1. If the degree of the top is greater than the degree of the bottom, then the limit is going to infinity.
2. If the degree at the top is less than the degree at the bottom, then the limit is going to zero.
3. If the degree of the top is the same as the degree of the bottom, set the
coefficients to a fraction.
some things i'm not too good w/ is slope fields, e integration, and lram rram and mram.
25th post
Tangent lines were on there too and the steps for finding the tangent line are:
1. Take the derivative of the equation like normal
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
absolute maxs and mins:
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Example:
Integrate (x^2+1) (2x) dx
Since you can't integrate product rule, you know you have to use substitution.
u=x^2 + 1 du= 2x dx
Integrate: u du
1/2 u ^2
Plug u back in
1/2 ( x^2+1)^2 + c
Example 2:
Integrate x (x^2+1)^2 dx
u= x^2 +1 du= 2x
Since you are missing a 2 in the problem, you have to add a 1/2 to get rid of the 2
1/2 S u^2 du
1/2 (1/3) u^3
1/6 (x^2 +1) ^3 + c
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
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
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
1. Take the derivative of the equation like normal
2. Plug in the x value which gives you your slope
3. Use the slope you get and the point given and plug into slope intercept form (y-y1)=slope(x-x1)
*If a point is not given and only an x value is given plug the x value into the original which will give you a y value creating a point.
absolute maxs and mins:
1. Find critical values by using the first derivative test: take the derivative set equal to zero and solve for the variable.
2. Plug both intervals as well as the critical values into your original equation
3. Be sure to keep the value after substitution with each appropriate number.
4. To determine which constitute a max and which constitute a min, you must look and point out those of which have the lowest values-min- and those of the highest value-max.
Substitution takes the place of the derivative rules for problems such as product rule and quotient rule. The steps to substitution are:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Example:
Integrate (x^2+1) (2x) dx
Since you can't integrate product rule, you know you have to use substitution.
u=x^2 + 1 du= 2x dx
Integrate: u du
1/2 u ^2
Plug u back in
1/2 ( x^2+1)^2 + c
Example 2:
Integrate x (x^2+1)^2 dx
u= x^2 +1 du= 2x
Since you are missing a 2 in the problem, you have to add a 1/2 to get rid of the 2
1/2 S u^2 du
1/2 (1/3) u^3
1/6 (x^2 +1) ^3 + c
e integration:
whatever is raised to the e power will be your u and du will be the derivative of u. For example:
e^2x-1dx
u=2x-1 du=2
rewrite the function as:
1/2{ e^u du, therefore
1/2e^2x-1+C will be the final answer.
Optimization can be used for finding the maximum/minimum amount of area of something. Steps in order to optimize anything:
1. Identify primary and secondary equations. Primary deals with the variable that is being maximized or minimized. The secondary equation is usually the other equation that ties in all the information given in the problem.
2. Solve the secondary equation for one variable and then plug that variable back into the primary. If the primary equation only have one variable you can skip this step.
3. Take the derivative of the primary equation after plugging in the variable, set it equal to zero, and then solve for the variable.
4. Plug that variable back into the secondary equation in order to solve for the last missing variable. Check endpoint if necessary to find the maximum or minimum answers
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
linearization.
The steps for solving linearization problems are:
1. Pick out the equation
2. f(x)+f`(x)dx
3. Figure out your dx
4. Figure out your x
5. Plug in everything you get
implicit derivatives:
First Derivative:
1. take the derivative of both sides
2. everytime you take the derivative of y note it with dy/dx or y^1
3. solve for dy/dx
Related Rates:
1: identify all variables in equations
2: identify what you are looking for
3: sketch and label
4: write an equation involving your variables. (you can only have one unknown so a secondary equation may be given)
5: take the derivative with respect to time.
6: substitute derivative and solve.
Questions
If ya'll will post the actual problems you are struggling with I will help with some of them. I left in such a hurry I don't have a copy of the test.
Subscribe to:
Posts (Atom)