The formula for the volume of disks is S (top)^2 - (bottom)^2 dx
The formula for the area of washers is S (top) - (bottom)
The steps are:
1. Draw the graphs of the equations
2. Subtract top graph's equation by the bottom graph's equation(in disks each equation would be squared)
3. Set equations equal and solve for x to find bounds
4. Plug in the bounds and the outcome of step 2
5. Integrate
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.
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 = 1
1[ f(-3) + f(-3 +1)]
1[ f( -3) + f(-2)]
then plug into your equation
RRAM 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.
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
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.
related rates:
The steps for related rates are….
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
Friday, March 26, 2010
Wednesday, March 24, 2010
Post
These are the two I never get.
Problem: -4x -5 on the interval [-3, -1] divided into 2 subintervals.
MRAM is to calculate the middle
formula: 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.
formula: 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.
What I don't know is Substitution but these are the steps..
steps:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Example?
Problem: -4x -5 on the interval [-3, -1] divided into 2 subintervals.
MRAM is to calculate the middle
formula: 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.
formula: 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.
What I don't know is Substitution but these are the steps..
steps:
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
Example?
Post
blogggggggggyyyyyyyyyy again!
1. The graph of y=5x^4-x^5 has an inflection point or points at
y=5x^4-x^5
20x^3-5x^3
20x^2(3-x)
3-x=0
-x=-3
divide by -1
x=3
2. d/dx(integral form 0 to x^2)sin^2t(dt)
sin^2(x^2)
2xsin^2(x^2)
3. The average value of f(x)=1/x from x=1 to x=e is
(1/e-1)[ln e -ln 1] *the e and 1 are the absoule value of
(1/e-1)[1-0]
= 1/e-1
4. Find the area under the curve y=2x-x^2 from x=1 to x=2 with n=4 left-endpoint rectangles.
use LRAM
delta x=2-1/4=1/4
1/4[f(1)+(5/4)+f(6/4)+f(7/4)]
1/4(0+.4375+.75+.9375)
1/4(2.125)= 17/32
i need help on SUBSTITUION, i keep forgetting ahhh
1. The graph of y=5x^4-x^5 has an inflection point or points at
y=5x^4-x^5
20x^3-5x^3
20x^2(3-x)
3-x=0
-x=-3
divide by -1
x=3
2. d/dx(integral form 0 to x^2)sin^2t(dt)
sin^2(x^2)
2xsin^2(x^2)
3. The average value of f(x)=1/x from x=1 to x=e is
(1/e-1)[ln e -ln 1] *the e and 1 are the absoule value of
(1/e-1)[1-0]
= 1/e-1
4. Find the area under the curve y=2x-x^2 from x=1 to x=2 with n=4 left-endpoint rectangles.
use LRAM
delta x=2-1/4=1/4
1/4[f(1)+(5/4)+f(6/4)+f(7/4)]
1/4(0+.4375+.75+.9375)
1/4(2.125)= 17/32
i need help on SUBSTITUION, i keep forgetting ahhh
Monday, March 22, 2010
Ash's 31st Post Part II
So, after feeling blehhh all day, eating soup, and taking a much needed nap, I'm on to Calculus! :)
Since I cannot find my binder and half of my corrections, I'm going to try to blog about things I actually remember how to do!
1. Implicit Derivatives
Take the derivative as you would normally, inserting dy/dx and dx/dx each time you take the derivative of y or x
Solve for dy/dx
FINISHED! =]
2. Min/Max on Calc Portion
Type in your equation
Graph it
2nd->Calc
Either Min or Max
Go a bit left of what you think
Hit enter
Go a bit right of what you think
Hit enter
Go in between those two
Hit enter
VOILA!
3. Definite Integration
Integrate (opposite of a derivative)
Plug in b (bottom number on the S)
Plug in a (top number on the S)
B-A = answer!
4. Disks
Formula: piS(equation)^2dx
Find Bounds (How?)
Integrate using Definite Integration
5. Washers
Formula: piS[(top equation)^2-(bottom equation)^2]dx
Find Bounds
Integrate using Definite Integration
Questions:
~How do you find the bounds for the disks and washers?
~I'm still lost on the ln and e integrations and derivatives...they are so easy when someone goes over them with me, but, I cannot get them on the AP
Each time I see ln, e, trig, or anything remotely difficult, I skip it. I usually do not have enough time to come back to these or if there is enough time, I get so confused. I don't have a specific problem right now, because I don't have my binder, but is there a way to somehow improve my time? I don't think practicing helps me because I still get the same ones wrong even after I correct them! =/
Does anyone have any suggstions? I'd be eternally grateful!
Since I cannot find my binder and half of my corrections, I'm going to try to blog about things I actually remember how to do!
1. Implicit Derivatives
Take the derivative as you would normally, inserting dy/dx and dx/dx each time you take the derivative of y or x
Solve for dy/dx
FINISHED! =]
2. Min/Max on Calc Portion
Type in your equation
Graph it
2nd->Calc
Either Min or Max
Go a bit left of what you think
Hit enter
Go a bit right of what you think
Hit enter
Go in between those two
Hit enter
VOILA!
3. Definite Integration
Integrate (opposite of a derivative)
Plug in b (bottom number on the S)
Plug in a (top number on the S)
B-A = answer!
4. Disks
Formula: piS(equation)^2dx
Find Bounds (How?)
Integrate using Definite Integration
5. Washers
Formula: piS[(top equation)^2-(bottom equation)^2]dx
Find Bounds
Integrate using Definite Integration
Questions:
~How do you find the bounds for the disks and washers?
~I'm still lost on the ln and e integrations and derivatives...they are so easy when someone goes over them with me, but, I cannot get them on the AP
Each time I see ln, e, trig, or anything remotely difficult, I skip it. I usually do not have enough time to come back to these or if there is enough time, I get so confused. I don't have a specific problem right now, because I don't have my binder, but is there a way to somehow improve my time? I don't think practicing helps me because I still get the same ones wrong even after I correct them! =/
Does anyone have any suggstions? I'd be eternally grateful!
Post -.-
Well, since my mother decided to destroy not only my phone (by dropping it in water, purposely) but also my internet, it was unable to submit my blog by 12 on Sunday night, or rather Monday morning, So. here it is..from my laptop at school. Monday morning, Yes, I'm a tad bit angry...
Okay, since I've repeated steps possibly a thousand times and have seen them on blogs for I don't know how long, I'm just going to do a couple of example problems that actually apply the aforementioned (I LOVE that word!) steps.
Okay, so examples...:
EXAMPLE 1:Given the equation y = (x-3)/(2-5x). Find dy/dx.
Quotient Rule (you could technically do product, but I prefer things set in stone (plug into a formula nd all))
dy/dx = ((2-5x)(1) -(-5)(x-3))/(2-5x)^2
= (2-5x + 5x -15)/(2-5x)^2
= - 13/(2-5x)^2
What I did: I took the derivative of the function using the quotient rule**be sure not to mix up NEGATIVES!!
EXAMPLE 2: What is the maximum value for the following: f(x) = xe^-x
Take derivative using product rule (multiplying two things).
x(-e^-x) + e^-x(1)remember that the der. e is e^w/e times the der. of the exponent
Simplify:
-xe^-x + e^-x
You can factor out an e^-x
(-x+1)e^-x
To find the critical values (possible maxima) set (-x+1) equal to zero and solve for x.
This yields x = 1. Now plug in that one to e^-x.
This then gives you 1/e====>your maximum value.
EXAMPLE 3: The table shows the speed of an object in feet per second, during a 3 second period.
time(sec)-----0--1--2--3
speed(ft/sec)-30-22-12-0
Estimate the distance traveled using the trapezoid method.
All you have to do for this particular problem is find delta x then plug into the formula: delta x /2 [f(first one) +2f(next)...f(last one)]
b-a/n = delta x
3-0/3 = 1
1/2[30 + 2(22) + 2(12) + 0]
1/2[20 + 44 + 24]
=49
EXAMPLE 4: Which best describes the behavior of the function y = arctan(1/lnx) at x = t?
A. It has a jump discontinuity
B. It has an infinite discontinuity.
C. It has a removable discontinuity
D. It is both continuous and differentiable
E. It is continuous but not differentiable.
Now since I found this question on a calculator portion, I would first graph it...which I believe to be a very obvious course of action, but whatever. So, I know the answer, but maybe just maybe, you should try it yourself??
Okay, I've got to go. Have a nice week!
Okay, since I've repeated steps possibly a thousand times and have seen them on blogs for I don't know how long, I'm just going to do a couple of example problems that actually apply the aforementioned (I LOVE that word!) steps.
Okay, so examples...:
EXAMPLE 1:Given the equation y = (x-3)/(2-5x). Find dy/dx.
Quotient Rule (you could technically do product, but I prefer things set in stone (plug into a formula nd all))
dy/dx = ((2-5x)(1) -(-5)(x-3))/(2-5x)^2
= (2-5x + 5x -15)/(2-5x)^2
= - 13/(2-5x)^2
What I did: I took the derivative of the function using the quotient rule**be sure not to mix up NEGATIVES!!
EXAMPLE 2: What is the maximum value for the following: f(x) = xe^-x
Take derivative using product rule (multiplying two things).
x(-e^-x) + e^-x(1)remember that the der. e is e^w/e times the der. of the exponent
Simplify:
-xe^-x + e^-x
You can factor out an e^-x
(-x+1)e^-x
To find the critical values (possible maxima) set (-x+1) equal to zero and solve for x.
This yields x = 1. Now plug in that one to e^-x.
This then gives you 1/e====>your maximum value.
EXAMPLE 3: The table shows the speed of an object in feet per second, during a 3 second period.
time(sec)-----0--1--2--3
speed(ft/sec)-30-22-12-0
Estimate the distance traveled using the trapezoid method.
All you have to do for this particular problem is find delta x then plug into the formula: delta x /2 [f(first one) +2f(next)...f(last one)]
b-a/n = delta x
3-0/3 = 1
1/2[30 + 2(22) + 2(12) + 0]
1/2[20 + 44 + 24]
=49
EXAMPLE 4: Which best describes the behavior of the function y = arctan(1/lnx) at x = t?
A. It has a jump discontinuity
B. It has an infinite discontinuity.
C. It has a removable discontinuity
D. It is both continuous and differentiable
E. It is continuous but not differentiable.
Now since I found this question on a calculator portion, I would first graph it...which I believe to be a very obvious course of action, but whatever. So, I know the answer, but maybe just maybe, you should try it yourself??
Okay, I've got to go. Have a nice week!
Ash's 31st Post Part I
So, I've been sick and nasty all day and I just woke up again and realized the time. I promise to do a blog tomorrow if I'm not hugging the toilet all day and night.
Until then, goodnight
Until then, goodnight
Sunday, March 21, 2010
post 31
Alright, so this week we took another ap test, and finally b-rob is coming back! yay!! and unfortunately my binder is completely not ready, so i pray we don't have to turn those in anytime soon. So, this blog is going to be on reimann sums.
Integration uses Riemann sums, which is the approximation of area by using rectangles or trapezoids. Integration is just finding the area of something with a curve that you would not normally be able to get.
So there are four different methods of integration, LRAM, RRAM, MRAM, and trapezoidal.
The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. You will need to know this because each of the next formulas require that you know what x is.
LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
on thing i could use some touching up on are those questions that are like, "find the normal line to...."
so yeah, that would be my one question for this blog.
Integration uses Riemann sums, which is the approximation of area by using rectangles or trapezoids. Integration is just finding the area of something with a curve that you would not normally be able to get.
So there are four different methods of integration, LRAM, RRAM, MRAM, and trapezoidal.
The first formula you need to know is x=(b-a)/n [a,b] with n subintervals. You will need to know this because each of the next formulas require that you know what x is.
LRAM- left hand approximation. (this puts the rectangles used to find the area on the left side of the curve) x[f(a)+f(a+x)+...f(b)]
RRAM- right hand approximation. (this puts the rectangles used to find the area on the right side of the curve) x[f(a+x)+...f(b)]
MRAM- approximation from the middle. (this puts the rectangles right on top of the curve, so that the curve goes through the middle of each one) x[f(mid)+f(mid)+...]
Trapezoidal- this does not use squares, instead it uses trapezoids to eliminate most of the empty space inside the curve, and I think this is the most accurate. x/2[f(a)+2f(a+x)+2f(a+2x)+...f(b)]
on thing i could use some touching up on are those questions that are like, "find the normal line to...."
so yeah, that would be my one question for this blog.
Post...
Okay so it's Sunday night...I'm slightly bored/aggravated, and I have no idea what to blog about... I guess I will take a look at the previous tests we took and see what you guys often missed and see if I can explain it a little better.
Okay, one question that surprised me that people still didn't know was a question about "change in y with respect to x". Basically all this means is take the implicit derivative. If you have forgotten, all you do for implicit derivatives is take the derivative like normal except whenever you take the derivative of y, write dy/dx. After you have taken the derivative, move all like terms to one side (i.e. move the dy/dx's on one side, and the terms without dy/dx's on the other side). Factor out dy/dx and then divide by what's left on that side. You should now have dy/dx = something/something. That is your answer.
Something that some of you are still not doing is a little trick for determining local minimums... Say for instance you have a function. You take the derivative of that function, and set equal to 0 and solve. This will give you the possible points of inflection. The easiest way to determine if it is a minimum (on multiple choice) is to take the second derivative and plug in. If it comes out negative (concave up), it was a minimum. If it comes out negative (concave down), it was a maximum. Using shortcuts like this is really really useful when you need to save time on the AP multiple choice (or at least I imagine it would be).
I can not stress the following enough: simplify an integral before you do it. It's almost always easier...especially on those ones where it looks really difficult to integrate, like it was something you've never done before...Well most of the time it's just written in an odd way...for instance, the integral of 4e^(2lnx)...that's a bit annoying to integrate...you can change it to 4e^(lnx^2) which makes it a lot easier because now the e and the ln cancel, leaving you with 4x^2, which is a very simple integral. So please, just remember to simplify before you integrate.
Anyway, going to go find something to do.
Okay, one question that surprised me that people still didn't know was a question about "change in y with respect to x". Basically all this means is take the implicit derivative. If you have forgotten, all you do for implicit derivatives is take the derivative like normal except whenever you take the derivative of y, write dy/dx. After you have taken the derivative, move all like terms to one side (i.e. move the dy/dx's on one side, and the terms without dy/dx's on the other side). Factor out dy/dx and then divide by what's left on that side. You should now have dy/dx = something/something. That is your answer.
Something that some of you are still not doing is a little trick for determining local minimums... Say for instance you have a function. You take the derivative of that function, and set equal to 0 and solve. This will give you the possible points of inflection. The easiest way to determine if it is a minimum (on multiple choice) is to take the second derivative and plug in. If it comes out negative (concave up), it was a minimum. If it comes out negative (concave down), it was a maximum. Using shortcuts like this is really really useful when you need to save time on the AP multiple choice (or at least I imagine it would be).
I can not stress the following enough: simplify an integral before you do it. It's almost always easier...especially on those ones where it looks really difficult to integrate, like it was something you've never done before...Well most of the time it's just written in an odd way...for instance, the integral of 4e^(2lnx)...that's a bit annoying to integrate...you can change it to 4e^(lnx^2) which makes it a lot easier because now the e and the ln cancel, leaving you with 4x^2, which is a very simple integral. So please, just remember to simplify before you integrate.
Anyway, going to go find something to do.
Post Number Thirty One
Almost doneeeeeeeeeeeeeeeeeeeeeeeeeeeeeee with school.
All these weeks of aps and I’m finally starting to do a little better, hopefully it stays improving..
Implicit derivatives are pretty easy.
1.Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
I finally remembered mean value theorem on the last test!
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a
Related rates..i know the steps but can someone work a problem for me?
1.Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
I need help with tangent lines, normal lines, definition of a derivative, deciphering between graphs of derivatives, instantaneous speed….
Etc etc
Limit rules for the limit approaching infinity
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
Goodnight calculus!
One 9 weeks left!
Oh lehhhh do it
All these weeks of aps and I’m finally starting to do a little better, hopefully it stays improving..
Implicit derivatives are pretty easy.
1.Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
I finally remembered mean value theorem on the last test!
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a
Related rates..i know the steps but can someone work a problem for me?
1.Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
I need help with tangent lines, normal lines, definition of a derivative, deciphering between graphs of derivatives, instantaneous speed….
Etc etc
Limit rules for the limit approaching infinity
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
Goodnight calculus!
One 9 weeks left!
Oh lehhhh do it
post 31
Well the third 9 weeks have come and gone so now we are at the final stretch.
One thing I am going to talk about is 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
Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
The next two types of integration are indefinite and definite. The answer for indefinite integration is an equation. But on the other hand the answer for a definite intergration is a number.
Indefinite Integration-Sx^n(dx)={(x^n+1)/(n+1)}+C
Definine Integration-bSa f(x)dx=F(b)-F(a)=number
Example- (S=integral symbol)
Indefinite
S x+5
x^2+5x+C
Definite
1S0 x+5
1S0 x^2+5x
1^2+5(1)-0^2+5(0)
=6
For what I help with is instantaneous speed.
One thing I am going to talk about is 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
Also I am going to talk about taking implicit derivatives. The steps for taking implicit derivatives are:
1. Take the derivative of both sides like you would normally do
2. Everytime the derivative of y is taken it needs to be notated with either y ' or dy/dx
3. Solve for dy/dx or y ' as if you are solving for x.
The next two types of integration are indefinite and definite. The answer for indefinite integration is an equation. But on the other hand the answer for a definite intergration is a number.
Indefinite Integration-Sx^n(dx)={(x^n+1)/(n+1)}+C
Definine Integration-bSa f(x)dx=F(b)-F(a)=number
Example- (S=integral symbol)
Indefinite
S x+5
x^2+5x+C
Definite
1S0 x+5
1S0 x^2+5x
1^2+5(1)-0^2+5(0)
=6
For what I help with is instantaneous speed.
post 31
calculus again. geeeeeez.
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
MEAN VALUE THEOREM:
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a\
EXTREME VALUE THEOREM:
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
steps for related rates:
1. Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
Limits:
If the degree on top is bigger than the degree on the bottom, the limit is infinity
If the degree on top is smaller than the degree on the bottom, the limit is zero.
If the degree on the top is the same as the degree on the bottom, you divide the coefficients to get the limit.
If they give you a limit problem where there is any letter going to 0 and they have a huge problem with parenthesis in it, you take the derivative of what is behind the parenthesis and plug in for x if needed.
someone help me w/ tangent lines. i cannot remember them for anythinggggg
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
MEAN VALUE THEOREM:
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a\
EXTREME VALUE THEOREM:
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
steps for related rates:
1. Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
Limits:
If the degree on top is bigger than the degree on the bottom, the limit is infinity
If the degree on top is smaller than the degree on the bottom, the limit is zero.
If the degree on the top is the same as the degree on the bottom, you divide the coefficients to get the limit.
If they give you a limit problem where there is any letter going to 0 and they have a huge problem with parenthesis in it, you take the derivative of what is behind the parenthesis and plug in for x if needed.
someone help me w/ tangent lines. i cannot remember them for anythinggggg
post #31
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
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.
related rates:
The steps for related rates are….
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
limits:
Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
To find area between curves:
The formula you use is b(int)a (top eq.) - (bottom eq.).
If a and b is not already given to you, then you much set the equations equal to each other and solve.
You find which equation is top/bottom by graphing both and simply looking to see which one is on top.
If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
First derivative test:
-take the derivative of the original function
-solve for x (the values will be your critical values)
-set those values up into intervals between negative infinity and infinity
- plug in numbers between the intervals into the function
-this will show you when the function is increasing, decreasing, and you will find max's and mins.
Second derivative test:
-take the derivative of the original function twice
-solve for x values(critical values)
-set up into intervals between infinity and negative infinity
-plug in values between the intervals into the function
-this will show you where the graph is concave up and down, and where there is a point of inflection.
Implicit Derivatives
The only difference between implicit derivatives and regular derivatives is that implicit derivatives include dy or y', the actual derivative of y.
y=x+2 y'=1
In an implicit derivative, you are always asked to solve for y'.
Example:
x^2+2y=0
1. Take derivative of both sides first.
2x+2y'=0
2. Then solve for y'.
y'=(-2x)/2
Some examples include:
4x+13y^2=4 y'=(-4/26y)
cos(x)=y y'=-sin(x)
y^3+y^2-5y-x^2=4 y'=2x/((3y+5)(y-1))
Find the volume of the solid formed by revolving the region bounded by the graphs of y=squareroot of x and y=x^2
after graphing in your graphing calculater you find that you need to use washers
so you get =(pie)S(squareroot of x)^2-(x^2)^2 dxx=1 so (pie)[(1/2)-(1/5)]-03(pie)/10 is your awnser
1. Find a derivative inside the interval
2. set u = the non-derivative
3. take the derivative of u
4. substitute back in
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.
related rates:
The steps for related rates are….
1. Pick out all variables
2. Pick out all equations
3. Pick out what you are looking for
4. Sketch a graph and label
5. Create an equation with your variables
6. Take the derivative respecting time
7. Substitute back into the derivative
8. Solve
limits:
Rules for Limits:…
1. if the degree of top equals the degree of bottom, the answer is the top coefficient over bottom coefficient
2. if top degree is bigger than bottom degree, the answer is positive or negative infinity
2. if top degree is less than bottom degree, the answer is 0
To find area between curves:
The formula you use is b(int)a (top eq.) - (bottom eq.).
If a and b is not already given to you, then you much set the equations equal to each other and solve.
You find which equation is top/bottom by graphing both and simply looking to see which one is on top.
If the area is on the y-axis, then the a and b values need to be set as y-values, and the equations must be solved for x.
First derivative test:
-take the derivative of the original function
-solve for x (the values will be your critical values)
-set those values up into intervals between negative infinity and infinity
- plug in numbers between the intervals into the function
-this will show you when the function is increasing, decreasing, and you will find max's and mins.
Second derivative test:
-take the derivative of the original function twice
-solve for x values(critical values)
-set up into intervals between infinity and negative infinity
-plug in values between the intervals into the function
-this will show you where the graph is concave up and down, and where there is a point of inflection.
Implicit Derivatives
The only difference between implicit derivatives and regular derivatives is that implicit derivatives include dy or y', the actual derivative of y.
y=x+2 y'=1
In an implicit derivative, you are always asked to solve for y'.
Example:
x^2+2y=0
1. Take derivative of both sides first.
2x+2y'=0
2. Then solve for y'.
y'=(-2x)/2
Some examples include:
4x+13y^2=4 y'=(-4/26y)
cos(x)=y y'=-sin(x)
y^3+y^2-5y-x^2=4 y'=2x/((3y+5)(y-1))
Find the volume of the solid formed by revolving the region bounded by the graphs of y=squareroot of x and y=x^2
after graphing in your graphing calculater you find that you need to use washers
so you get =(pie)S(squareroot of x)^2-(x^2)^2 dxx=1 so (pie)[(1/2)-(1/5)]-03(pie)/10 is your awnser
31
yo yo yo.
we took 2 ap's this week, except it was... OUR EXAMS. oh geez. i did okay i guess. we took them wednesday and tuesday, cuz jr's had a fieldtrip monday.
anyway, let's go over stuff..
steps for linearization:
1. Identify equation
2. Use f(x)+f ' (x)dx
3. Determine your dx
4. Then determine your x
5. Plug in everything
6. Solve
MEAN VALUE THEOREM:
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a\
EXTREME VALUE THEOREM:
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
steps for related rates:
1. Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
normal lines & tangent lines & integrating trig functions! HELP NEEDED. go
we took 2 ap's this week, except it was... OUR EXAMS. oh geez. i did okay i guess. we took them wednesday and tuesday, cuz jr's had a fieldtrip monday.
anyway, let's go over stuff..
steps for linearization:
1. Identify equation
2. Use f(x)+f ' (x)dx
3. Determine your dx
4. Then determine your x
5. Plug in everything
6. Solve
MEAN VALUE THEOREM:
If f is continous on the closed interval [a,b] and differentiable on the open interval (a,b), then there exists a number c in (a,b) such that F'(c) = f(b) - f(a) / b-a\
EXTREME VALUE THEOREM:
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
steps for related rates:
1. Identify all variables
2. Identify what you are looking for
3. Sketch & label that graph
4. write an equation using all of the variables
5. Take the derivative of this equation
6. Substitute everything back in
7. Solve
normal lines & tangent lines & integrating trig functions! HELP NEEDED. go
Posting...#31
Don't forget about your tangent line andd your normal line.
Tangent Line:
1. If you are only given a x value, plug into your equation to find the y value.
2. take the derivative of the equation
3. Plug in your x value into the derivative to find the slope
4. Put in point slope
EXAMPLE: y=3x^3 + 4x^2 +5 at x=1
1. 3(1)^3+4(1)^2+5 = 12
2. 9x^2 + 8x
3. 9(1)^2 + 8(1) = 17
4. y-12 = 17 (x-1)
Normal line is the same except you take the negative reciprocal of the slope to plug in.
y-12 = -1/17 (x-1)
And also your disk.
FORMULA FOR DISKS: pi S[r(x)]^2dx, this formula is almost exactly the same as when you are asked to find the area of a disk, however, while finding the area you do not square the radius (the given equation).
FORMULA FOR WASHERS: pi S[(top)^2-(bottom)^2]dx
FORMULA FOR WASHERS: pi S[(top)^2-(bottom)^2]dx
On the AP test lately i been missing the trapezoid questions which i thought i new so if someone could help me with taht thanks.
Post #31
More APs this week.
I'll review some from the calculator portion because I have more trouble with that one.
9. What is the approximate instantaneous rate of change of the function
f(t) = the interval of cos (x) dx on [0,8t] at t= pi/3?
A. -1.333 B. -6.928 C. -.8660 D. -4 E. -.5000
Instantaneous rate of change means to take the derivative, and since there is an integral, this problem wants you to take the derivative of a integral which you do by plugging in b and multiplying that by the derivative of b.
cos (8t) (8) = 8 cos(8t)
Now plug in pi/3.
8 cos (8 pi/3)
Plug into your calculator and you get -4 D.
10. What is the error when the integral sin (pi x) dx on [0,1] is approximated by the trapezoidal rule with n=3.
A. 0.059 B. 0.051 C. 0.032 D. 0.109 E. 0.011
The formula for trapezoidal rule is delta x/ 2 [f(a) + 2f (a + delta x) + 2f(a + 2 delta x)+....... + f(b)]
delta x is found by doing b-a/n
Delta x = 1-0/3 = 1/3
Divide by 2 and you get 1/6
Plugging into the formula: 1/6[f(0) + 2f(1/3) + 2f (2/3) + f(1)] which equals .5773333333
Now you have to find the actual value of the integral. You can do this by entering into your calculator.
The exact value is .63661977
To find the error, subtract the two values: .63661977-.5773333333= .059
The answer is A. 0.059.
Jumping to the non calculator portion. This question has been on a few APs and it is really simple so I'm going to explain it.
28. Determine dx/x ln (ln (2-cos (x)))
The derivative of a natural long is just 1/whatever the natural log is of times the derivative of the inside
1/ ln (2-cos (x)) takes care of the first natural log
times that by the derivative of the inside which is 1/ (2- cos (x)
Now times that natural log by its inside (sin (x)).
Everything together is 1/ ln (2-cos (x)) (1/ (2-cos (x) (sin (x)) OR
sin (x) / ln (2-cos (x)) (2- cos (x))
That is answer choice E.
I have questions on a few problems on both portions.
Non calculator- #'s 13, 14, 18, 23, and 24.
Calculator- #'s 2 (invertible again) , 8, and 11.
I'll review some from the calculator portion because I have more trouble with that one.
9. What is the approximate instantaneous rate of change of the function
f(t) = the interval of cos (x) dx on [0,8t] at t= pi/3?
A. -1.333 B. -6.928 C. -.8660 D. -4 E. -.5000
Instantaneous rate of change means to take the derivative, and since there is an integral, this problem wants you to take the derivative of a integral which you do by plugging in b and multiplying that by the derivative of b.
cos (8t) (8) = 8 cos(8t)
Now plug in pi/3.
8 cos (8 pi/3)
Plug into your calculator and you get -4 D.
10. What is the error when the integral sin (pi x) dx on [0,1] is approximated by the trapezoidal rule with n=3.
A. 0.059 B. 0.051 C. 0.032 D. 0.109 E. 0.011
The formula for trapezoidal rule is delta x/ 2 [f(a) + 2f (a + delta x) + 2f(a + 2 delta x)+....... + f(b)]
delta x is found by doing b-a/n
Delta x = 1-0/3 = 1/3
Divide by 2 and you get 1/6
Plugging into the formula: 1/6[f(0) + 2f(1/3) + 2f (2/3) + f(1)] which equals .5773333333
Now you have to find the actual value of the integral. You can do this by entering into your calculator.
The exact value is .63661977
To find the error, subtract the two values: .63661977-.5773333333= .059
The answer is A. 0.059.
Jumping to the non calculator portion. This question has been on a few APs and it is really simple so I'm going to explain it.
28. Determine dx/x ln (ln (2-cos (x)))
The derivative of a natural long is just 1/whatever the natural log is of times the derivative of the inside
1/ ln (2-cos (x)) takes care of the first natural log
times that by the derivative of the inside which is 1/ (2- cos (x)
Now times that natural log by its inside (sin (x)).
Everything together is 1/ ln (2-cos (x)) (1/ (2-cos (x) (sin (x)) OR
sin (x) / ln (2-cos (x)) (2- cos (x))
That is answer choice E.
I have questions on a few problems on both portions.
Non calculator- #'s 13, 14, 18, 23, and 24.
Calculator- #'s 2 (invertible again) , 8, and 11.
Post #31
This week in Calculus, we took AP's and then corrected and did key words on our own..I actually did decent on these! So, to expain a few things..
I'm still having trouble with substitution..but, I'm trying..
okay, let me explain the chain rule..
that i now understand for derivatives.
If your given f(g(1)))' and they also tell you f(x) = x^3 +4x ; g(x) = 4x
you start off with plugging in the formula then taking the derivative by chain rule..
so, start with
(4x)^3 + 4(4x) Next, plug in 1 for the x's
4(1)^3 + 4(4(1))
= 64 + 8 = 72
These ahve gotten alot either to me because i finally undersatnd...
and i now know that if they ask for a derivative of that thing, you just find it like you have sin or cosine or any other kind of chain rule derivative... :)
So, lets get to the things i don't understand..
instantaneous speed..i forgot the formula, or how you do it!
and average rate of change..or mean value theorem..i can't ever remember how to do those!
I'm still having trouble with substitution..but, I'm trying..
okay, let me explain the chain rule..
that i now understand for derivatives.
If your given f(g(1)))' and they also tell you f(x) = x^3 +4x ; g(x) = 4x
you start off with plugging in the formula then taking the derivative by chain rule..
so, start with
(4x)^3 + 4(4x) Next, plug in 1 for the x's
4(1)^3 + 4(4(1))
= 64 + 8 = 72
These ahve gotten alot either to me because i finally undersatnd...
and i now know that if they ask for a derivative of that thing, you just find it like you have sin or cosine or any other kind of chain rule derivative... :)
So, lets get to the things i don't understand..
instantaneous speed..i forgot the formula, or how you do it!
and average rate of change..or mean value theorem..i can't ever remember how to do those!
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