WOW..school is really over for the seniors. i don't really know how i expected to feel about this but now i just feel lost. no more pulling all-nighters for calculus tests, no more getting help from john or mrs. robinson, and no more fun together as the calculus class of 2010. i guess one of the things that stuck with me the most from this class was the strive to achieve a greater goal. derivatives, slopes, optimazation, and related rates are the things i remember most that i learned. not because i hated them but because these were the things i strived to learn first. through all the hard test and page long problems we learned how much we could better ourselves and take in mass amounts of information. as we all know one needs a teacher to learn how to do all of these problems and mrs. robinson was better than anyone we could ever imagine. she made learning everything fun and easier and because of her we can go to college and be truly prepared for what is waiting for us. to the calculus class of 2010 i just want to say thank you for such a wonderful experience and to mrs. robinson thank you for being willing to teach us:)!!!
the very first thing we learned this year was our derivatives:
d/dx c=0 (c is a #)
d/dx cu=cu' (c is #)
d/dx cx=c (c is a #)
d/dx u+v=u'+v'
d/dx uv=uv'+vu'
d/dx u/v=(vu'-uv')/v^2
d/dx sinx=cosx(x')
d/dx cosx=-sinx(x')
d/dx tanx=sec^2x(x')
d/dx secx=secxtanx(x')
d/dx cscx=-cscxtanx(x')
d/dx cotx=-csc^2x(x')
d/dx lnu= 1/u(u')
d/dx e^u=e^u(u')
other things that stressed us all out!!!:
linearization: The steps for working linearization problems are:
1. Identify the equation
2. Use the formula f(x)+f ' (x)dx
3. Determine your dx in the problem
4. Then determine your x in the problem
5. Plug in everything you get
6. Solve the equation
riemann sums!!:
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
Equation of a tangent line!!!!:
Take the derivative and plug in the x value.
If you are not given a y value, plug into the original equation to get the y value.
then plug those numbers into point slope form: y − y1 = m(x − x1)
Finding critical values!!!!:
To find critical values, first take the derivative of the function and set it equal to zero, solve for x. The answers you get for x are your critical values.
Absolute extrema!!!:
If you are given a point, plug those numbers into the original function to get another number. Alos, solve for critical values and plug those into the original function. Once you get your second numbers, you set each pair into new sets of points. The highest point is the absolute max and the smallest point is the absolute min.
volume by disks!!!:
the formula is pi times the integral of the [function given] squared times dx. so just solve it by taking the integral of it and then pluging in the numbers they give you. just like before you'll have two numbers so whatever the answer is for the top one will be first and then you subtract the answer you get for the bottom one. then graph
volume by washers!!!:
the formla is pie times the integral of the [top function] squared minus the [bottom function] squared times dx. so to do this, if you don't have the in between number you have to set the functions equal, but if you do, then it's worked the same way as above. square the formula's that were given and simplify. then take the integral of it and plug in the numbers they give you or you found by setting the formulas equal to each other and then solve like any other one by subracting them. then graph.
Limits!!!:
If the degree on top is smaller than the degree on the bottom, the limit is zero.
If the degree on top is bigger than the degree on the bottom, the limit is infinity.
If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.
First derivative test!!!:
You have to take the derivative of the function and set it equal to zero. Then solve for the critical values (x values). Set those values up into intervals between negative infinity and infinity. Plug in numbers between these intervals into the first derivative to see if there are max or mins or if the graph 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 and set them up into intervals between negative infinity and infinity. Plug in numbers between those intervals into the second derivative to see where the graph is concave up, concave down, or where there is a point of inflection.
Average rate of change !!!- f(b)-f(a)/(b-a).
2. Rate of change !!- this is simply a derivative! plug in an x value and get a slope out, your answer
3. Average value!! - this is 1/(b-a) times the integral from a to b of f(x). This is just an integral times by 1/(b-a).
4. Maximums, minimums, critical values, increasing, decreasing - all this is related to first derivative test. it's simple. you take the derivative, set equal to 0, solve for x. Set up some intervals using these numbers. Plug in numbers and test your intervals. pos to neg is a min. neg to pos is a max. pos = increasing, neg = decreasing. simple stuff. remember it.
5. Point of inflection, concave up, concave down - it's the second derivative test. set up intervals, if the intervals change signs, it is a point of inflection there. also, if its negative, that interval is concave down, positive is concave up.
6. Slope of a normal line - take derivative, plug in x. get a slope. however, make sure you use the negative reciprocal of the slope (normal means perpendicular to). use point-slope formula.
GOOD LUCK CLASS OF 2011!!!!!!!!!!!!! and may your senior year be filled with amazing times and fun:)
Love, Jessie Green:):):):)!!!!!!!!!!
Friday, May 7, 2010
Sunday, May 2, 2010
Ash's 37th Post
Ahh, so this week is the AP test and I'm terrified!! >.<
Break a leg everyone!!
Not literally...
Hmm...this always makes me double guess myself! Limits!! -.-
Limit Rules:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree.
I found the steps for Linearization, but how do you identify one?
The Steps:
Step One: Figure out the equation
Step Two: Take the derivative
Step Three: Add the derivative and originals [f(x)+f'(x)]
Step Four: Solve for dx
Step Five: Solve for x
Step Six: Plug in everything ((to what??))
I'm just really pumped for the 2 days of review..The AP exam is terrifying me and I feel like I'm going to freak out even more! >.<
Oh, if anyone has tips when *NOT* using a calculator, I'd be forever in your debt :)
Break a leg everyone!!
Not literally...
Hmm...this always makes me double guess myself! Limits!! -.-
Limit Rules:
1. If the degree of the top is larger than the degree of the bottom, the limit approaches infinity
2. If the degree of the bottom is larger than the degree of the tip, the limit approaches zero
3. If the degree of the bottom is equal to the degree of the top, then you make a fraction out of the coefficients in front of the largest degree.
I found the steps for Linearization, but how do you identify one?
The Steps:
Step One: Figure out the equation
Step Two: Take the derivative
Step Three: Add the derivative and originals [f(x)+f'(x)]
Step Four: Solve for dx
Step Five: Solve for x
Step Six: Plug in everything ((to what??))
I'm just really pumped for the 2 days of review..The AP exam is terrifying me and I feel like I'm going to freak out even more! >.<
Oh, if anyone has tips when *NOT* using a calculator, I'd be forever in your debt :)
post 37
this week we just took more ap practice tests, and this wednesday we are all finally going to take the real ap test! anyways, i'm going to do this blog on implicit derivatives because that's a super easy topic. :)
Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)
example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
Then you have to solve for dy/dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)
and that's it for that problem, it's done.
Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)
First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))
then, you need to foil it n stuff, so you get:
12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)
then, as usual, you would have to solve for dy/dx:
dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)
after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)
that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before.
and ummm, i have trouble with those questions where it gives you a bunch of x and y values, and it will give an interval or something, like 0
Implicit derivatives involve both x's and y's, unlike normal derivatives.
1: So, first you have to take the derivative of whatever they give you as you normally would.
2: Whenever you take the derivative of y, you have to note it with dy/dx.
3: Solve for dy/dx
(if you want to find slope plug in an x and y value)
example: y^3+y^2-5y-x^2=-4
First you just take the derivative, but don't forget to not the derivatives of the y's! So you get: 3y^2(dy/dx)+2y(dy/dx)-5(dy/dx)-2x=0
Then you have to solve for dy/dx, so you get:
dy/dx(3y^2+2y-5)=2x which then is further solved for to get dy/dx=2x/(3y^2+2y-5)
and that's it for that problem, it's done.
Another example:
Okay, let's say you want to find the slope of 3(x^2+y^2)^2=100xy at the point (3,1)
First you take the derivative, which involves all kinds of product and exponent rule...
6(x^2+y^2)(2x+2y(dy/dx))=100(y+x(dy/dx))
then, you need to foil it n stuff, so you get:
12x^3+12x^2(dy/dx)+12xy^2+12y^2(dy/dx))=100y+100x(dy/dx)
then, as usual, you would have to solve for dy/dx:
dy/dx=(-12^3-12xy^2+100y)/(12x^2+12y-100x)
after you solved for dy/dx, you plug in your x and y value from the point given to get your answer, so I think the final answer would be: 1.84 (if I put it in my calculator right)
that is it for implicit derivatives, and they are really easy to identify, it is the exact same thing as a derivative pretty much, just with x's and y's. Just don't forget to plug in the point that some problems will give you at the end, I have forgotten to do it before.
and ummm, i have trouble with those questions where it gives you a bunch of x and y values, and it will give an interval or something, like 0
Posting...#37
Normal lines:
Example:
7y - 28 = -4 (x-2)
7y - 28 = -4x + 8
7y = -4x + 36
4x + 7y = 36
Normal lines have the same steps and tangent lines but when you find the slope you have to take the reciprocal.
Example:
An equation of the line normal to the graph of y = (3x^2 +2x)^1/2 at (2,4) is
1. Since you already have a point, you do not need to plug in the x to find the y.
2. Derivative: 1/2 (3x^2 + 2x) ^ -1/2 (6x +2)
Plug in x to find slope: 6(2) +2 / 2 (3(2)^2 + 2(2)) ^1/2
14/8 = 7/4
Since this is normal, we need the negative reciprocal of the slope, which is -4/7
Plug in x to find slope: 6(2) +2 / 2 (3(2)^2 + 2(2)) ^1/2
14/8 = 7/4
Since this is normal, we need the negative reciprocal of the slope, which is -4/7
3. Now we can plug in: y-4 = -4/7 (x-2)
Than change what you got and find your awnser choice
Than change what you got and find your awnser choice
7y - 28 = -4 (x-2)
7y - 28 = -4x + 8
7y = -4x + 36
4x + 7y = 36
What i'm having trouble with is:
Someone help me find AREA,
and... Trapazoidal
Post #37
AP this week =0.
I'll explain optimization because I am presenting it tomorrow.
Steps:
1. Determine everything you are given. Determine your primary and secondary equation. Your primary equation is the one you are maximizing or minimizing. Your secondary equation is usually using the other information given to you in the problem.
2. Solve the secondary equation for one variable
3. Plug the solved secondary equation into the primary equation and simplify.
4. Take the derivative of that equation.
5. Solve that equation for the remaining variable.
6. Plug that variable back into the secondary equation to find the other variable(s).
Example (some of you will see this problem on the white board tomorrow):
We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom costs $10/ft^2 and the material used to build the sides cost $6/ft^2. If the box must have a volume of 50ft^3 determine the dimensions that will minimize the cost to build the box.
1. You are looking for the dimensions to minimize the box so your primary equation will be
c= 10(2lw) + 6(2wh + 2lh)
Note: you multiply by 10 because of the $10/ft^s for the top and bottom, and 6 for the $6/ft^2 for the sides.
Once distributed in, that equation simplifies to 60w^2+48wh
Your secondary equation will deal with volume since you are given the restraint of v=50ft^3
The volume of a box is lwh so 50 = lwh
Since l is 3 times the width, l=3w
The secondary equation is 50 = 3w^2h
2. Solve for h: h=50/3w^2
3. Plug into primary: 60w^2 + 48w (50/3w^2)
Simplified is 60w^2 + 800/w
4. Derivative: 120w - 800w^-2
which equals: 120w^3 - 800/ w^2
5. Solve for w by setting the top of the equation equal to 0.
120w^3 -800 = 0
w= 1.8821
6. Plug into equations to find other 2 variables:
50 = 3w^2 h
50= 3(1.8821)^2 h
h=4.7050
l=3w
l=3(1.8821)
l= 5.6463
Hope this helps.
I have questions on the invertible problems and the related rate problems with surface area such as number 13 on one of the last aps we took.
I'll explain optimization because I am presenting it tomorrow.
Steps:
1. Determine everything you are given. Determine your primary and secondary equation. Your primary equation is the one you are maximizing or minimizing. Your secondary equation is usually using the other information given to you in the problem.
2. Solve the secondary equation for one variable
3. Plug the solved secondary equation into the primary equation and simplify.
4. Take the derivative of that equation.
5. Solve that equation for the remaining variable.
6. Plug that variable back into the secondary equation to find the other variable(s).
Example (some of you will see this problem on the white board tomorrow):
We want to construct a box whose base length is 3 times the base width. The material used to build the top and bottom costs $10/ft^2 and the material used to build the sides cost $6/ft^2. If the box must have a volume of 50ft^3 determine the dimensions that will minimize the cost to build the box.
1. You are looking for the dimensions to minimize the box so your primary equation will be
c= 10(2lw) + 6(2wh + 2lh)
Note: you multiply by 10 because of the $10/ft^s for the top and bottom, and 6 for the $6/ft^2 for the sides.
Once distributed in, that equation simplifies to 60w^2+48wh
Your secondary equation will deal with volume since you are given the restraint of v=50ft^3
The volume of a box is lwh so 50 = lwh
Since l is 3 times the width, l=3w
The secondary equation is 50 = 3w^2h
2. Solve for h: h=50/3w^2
3. Plug into primary: 60w^2 + 48w (50/3w^2)
Simplified is 60w^2 + 800/w
4. Derivative: 120w - 800w^-2
which equals: 120w^3 - 800/ w^2
5. Solve for w by setting the top of the equation equal to 0.
120w^3 -800 = 0
w= 1.8821
6. Plug into equations to find other 2 variables:
50 = 3w^2 h
50= 3(1.8821)^2 h
h=4.7050
l=3w
l=3(1.8821)
l= 5.6463
Hope this helps.
I have questions on the invertible problems and the related rate problems with surface area such as number 13 on one of the last aps we took.
Post #37
5 days left :) Yay Yay Yay. But i'm still calculus illiterate.
Limits:
If the bottom is 0 don’t assume the limit does not exist right away, first factor and cancel or use your calculator.
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.
For example, The variables x and y are differentiable functions of t and are related by the equation y = 2x^3 - x + 4. When x = 2, dx/dt = -1. Find dy/dt when x = 2.Since everything is given you can skip straight to the derivative.dy/dt=6x^2dx/dt - dx/dtNow plug in all your givens in order to find dy/dt.dy/dt=6(2)^2(-1) - (-1)dy/dt= -23
Volume by disks:
The formula is pi S[r(x)]^2dx
Volume by washers is the same as volume by disks except it has a hole in it. That means it is bounded by two graphs. The formula for volume by washers is Pi S top^2 - bottom^2 dx.
Umm my ap score seems to be dropping every single test, so I need help.
Mainly with ln and e integration and derivatives, related rates, substitution integration, remembering all the area/volume formulas, and stuff I need to figure out on my own. Even when I know what to do and know the way I have to do it I have a lot of trouble applying it and never get the answer. SO HELP.
Limits:
If the bottom is 0 don’t assume the limit does not exist right away, first factor and cancel or use your calculator.
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.
For example, The variables x and y are differentiable functions of t and are related by the equation y = 2x^3 - x + 4. When x = 2, dx/dt = -1. Find dy/dt when x = 2.Since everything is given you can skip straight to the derivative.dy/dt=6x^2dx/dt - dx/dtNow plug in all your givens in order to find dy/dt.dy/dt=6(2)^2(-1) - (-1)dy/dt= -23
Volume by disks:
The formula is pi S[r(x)]^2dx
Volume by washers is the same as volume by disks except it has a hole in it. That means it is bounded by two graphs. The formula for volume by washers is Pi S top^2 - bottom^2 dx.
Umm my ap score seems to be dropping every single test, so I need help.
Mainly with ln and e integration and derivatives, related rates, substitution integration, remembering all the area/volume formulas, and stuff I need to figure out on my own. Even when I know what to do and know the way I have to do it I have a lot of trouble applying it and never get the answer. SO HELP.
thirty seven
alright, ap this week! finally all of our hard work will pay off. haha i can't wait to take the test and stop stressing about it. i am really nervous though... cuz i'm soooooo close to passing, :( oh well, we'll see what happens!
here's some tips for taking the tests:
when given huge word problems with the words, position (original function), velocity, and acceleration. USE YOUR pva CHART!
Position
Velocity
Accelration
when moving down the chart, take the derivative. (derivative - down) when moving up the chart, take the integral. (up - integrate)
what i mean by down is, it will give you a position problem & ask for acceleration => take the derivative TWICE. etc.
also, calculators are very useful. make sure you use them on the calculator portion. there's a reason why it says a calculator is REQUIRED for this part of the test. if you don't use it on EVERY problem, something is wrong.
your calculator can integrate for you, it can find x-intercepts, and it can graph anything. always use itttttt.
alsooooo, a VERY HELPFUL HINT mrs. robinson told me the other day. THIS IS CALCULUS CLASS, you should either be taking the derivative or integrating on EVERY SINGLE PROBLEM. some problems may look easy, and you may say OH i can solve this with basic algebra. but you are wrong. the multiple choice section will even try to trick you like that. but just remember, ALWAYS integrate/derive. you need to be able to figure out which of those to do though
good luck to everyone on wednesday :)
can someone go over concavity/pts of inflection.
you know all that first & second derivative stuff.
THANKS.
here's some tips for taking the tests:
when given huge word problems with the words, position (original function), velocity, and acceleration. USE YOUR pva CHART!
Position
Velocity
Accelration
when moving down the chart, take the derivative. (derivative - down) when moving up the chart, take the integral. (up - integrate)
what i mean by down is, it will give you a position problem & ask for acceleration => take the derivative TWICE. etc.
also, calculators are very useful. make sure you use them on the calculator portion. there's a reason why it says a calculator is REQUIRED for this part of the test. if you don't use it on EVERY problem, something is wrong.
your calculator can integrate for you, it can find x-intercepts, and it can graph anything. always use itttttt.
alsooooo, a VERY HELPFUL HINT mrs. robinson told me the other day. THIS IS CALCULUS CLASS, you should either be taking the derivative or integrating on EVERY SINGLE PROBLEM. some problems may look easy, and you may say OH i can solve this with basic algebra. but you are wrong. the multiple choice section will even try to trick you like that. but just remember, ALWAYS integrate/derive. you need to be able to figure out which of those to do though
good luck to everyone on wednesday :)
can someone go over concavity/pts of inflection.
you know all that first & second derivative stuff.
THANKS.
post 37
Well the AP test is here on Wednesday. So to review for it i will go over some old things we have learned this year.
The steps for optimization are:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
The limit rules are:
1. If the degree on top is smaller than the degree on the bottom, the limit is zero.
2. If the degree on top is bigger than the degree on the bottom, the limit is infinity.
3. If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.
The steps of the First Derivative Test are:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve.
The steps of the Second Derivative Test are:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x.
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve.
The steps for optimization are:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
The limit rules are:
1. If the degree on top is smaller than the degree on the bottom, the limit is zero.
2. If the degree on top is bigger than the degree on the bottom, the limit is infinity.
3. If the degree on top is the same as the degree on the bottom, you divide the coefficients to get the limit.
The steps of the First Derivative Test are:
1. Take the derivative of the original problem.
2. Set the first derivative equal to Zero.
3. Solve for x.
4. Create intervals for x
5. Pick a number in the intervals then plug that number in the first derivative for x.
6. Solve.
The steps of the Second Derivative Test are:
1. Take the derivative of the first derivative.
2. Set the second derivative equal to Zero.
3. Solve for x.
4. Create intervals for x.
5. Pick a number in the intervals then plug that number in the second derivative for x.
6. Solve.
The steps to finding absolute maxs and mins:
1. First derivative test
2. Plug critical values into the original function to get y-values
3. Plug endpoints into the original function to get y-values
4. The highest y-value is the absolute maximum
5. The lowest y-value is the absolute minimum
For questions I need help with the table problems and graph problems on free response.
post 37
Well the ap test is really close and I so need to study!
Optimization:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
substitution:
1. Find u and du
2. set u equal to whatever isn't the derivative
3. take the derivative of u
4. substitute back in
LRAM- left hand approximation x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation x[f(a+x)+...f(b)]
MRAM- approximation from the middle x[f(mid)+f(mid)+...]
Trapezoidal- x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
average rate of change:
just a slope; f(b)-f(a)/(b-a)
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
slope field stuff from free response:
(what to draw)
positive slopes is /
negative slope is \
for a zero slope is a horizontal line
for an undefined slope is a vertical line
I sometimes just mess up on taking the derivative or integrating wrong. Any tips for integration?
Optimization:
1. Identify all quantities
2. Write an equation
3. Reduce equation
4. Determine domain of equation
5. Determine max/min values
substitution:
1. Find u and du
2. set u equal to whatever isn't the derivative
3. take the derivative of u
4. substitute back in
LRAM- left hand approximation x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation x[f(a+x)+...f(b)]
MRAM- approximation from the middle x[f(mid)+f(mid)+...]
Trapezoidal- x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
average rate of change:
just a slope; f(b)-f(a)/(b-a)
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
slope field stuff from free response:
(what to draw)
positive slopes is /
negative slope is \
for a zero slope is a horizontal line
for an undefined slope is a vertical line
I sometimes just mess up on taking the derivative or integrating wrong. Any tips for integration?
Post #37?
Well I was actually just studying..so I'm going to go over some random things.
Area formula: aSb top - bottom
Volume formula: (pi) aSb (top)^2 - (bottom)^2
QUESTION: when do you use aSb (top-bottom)^2?
limit rules as x--> infinity are really simple:
top degree = bottom degree-->divide coeffient
top degree < bottom degree-->0
top degree > bottom degree-->infinity
remember:
increasing/decreasing, max/min is FIRST derivative test
concave up/down, pt of inflection is SECOND derivative test
if original graph is postive then the deivative graph is increasing
if original graph is negative then the deivative graph is decreasing
QUESTION: how do you know if it is concave up/down?
for a piecewise the derivatives must equal in order for it to be differentiable
something I just learned:
d/dx (integral from 0 to x^2 (sin(t^3) dt)
= x^2[sinx^6]
*pull the top bound out to the front and plug in t^3 for x
tangent lines:
take derivative
set = to 0
solve for x
plug into point slope formula
*linearzation is much like tangent lines
average value:
1/b-a (aSb) equation
Mean Value Theorem:
plug in points to original equation
take those points you just found and the point given in problem to find slope
take derivative of orginial equation and set = to the slope
Position, Velocity, Acceleration:
P--> Postion
V--> Velocity
A--> Acceleration
*Going down take derivative; going up take integral
Calculator:
change direction-zeros on graph
bounds for integral-intersection on graph
graph stuff!
and:
a corner IS continuous but NOT differentiable
don't forget +C
I'm still worried about substitution & ln integration!
Area formula: aSb top - bottom
Volume formula: (pi) aSb (top)^2 - (bottom)^2
QUESTION: when do you use aSb (top-bottom)^2?
limit rules as x--> infinity are really simple:
top degree = bottom degree-->divide coeffient
top degree < bottom degree-->0
top degree > bottom degree-->infinity
remember:
increasing/decreasing, max/min is FIRST derivative test
concave up/down, pt of inflection is SECOND derivative test
if original graph is postive then the deivative graph is increasing
if original graph is negative then the deivative graph is decreasing
QUESTION: how do you know if it is concave up/down?
for a piecewise the derivatives must equal in order for it to be differentiable
something I just learned:
d/dx (integral from 0 to x^2 (sin(t^3) dt)
= x^2[sinx^6]
*pull the top bound out to the front and plug in t^3 for x
tangent lines:
take derivative
set = to 0
solve for x
plug into point slope formula
*linearzation is much like tangent lines
average value:
1/b-a (aSb) equation
Mean Value Theorem:
plug in points to original equation
take those points you just found and the point given in problem to find slope
take derivative of orginial equation and set = to the slope
Position, Velocity, Acceleration:
P--> Postion
V--> Velocity
A--> Acceleration
*Going down take derivative; going up take integral
Calculator:
change direction-zeros on graph
bounds for integral-intersection on graph
graph stuff!
and:
a corner IS continuous but NOT differentiable
don't forget +C
I'm still worried about substitution & ln integration!
Subscribe to:
Posts (Atom)