post #32

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

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?

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

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!

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!

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

I'M A TURD

GABBA GABBA

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.

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.

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