Post #35

Review of AP questions..

3. If the integral of f(x) dx on [a,b] = a +2b, then the integral of (f(x)+5) dx on [a,b] =
The mistake I did on this problem was I forgot to integrate the 5 so my answer came out to be a+2b+5 which is an answer choice.
If you integrate the 5 you get a+2b + 5x9
If you plug in your bounds you get a+2b+5b-5a which simplifies to 7b-4a (also an answer choice and the correct answer)
You have to watch out for their tricks.

12. At what point on the graph of y=1/2x^2 is the tangent line parallel to the line 2x-4y=3?

The steps to doing these problems are to
1. Take the derivative of the equation
y' = x
2. Set that equal to the slope of the line ( slope = -a/b)
x= -2/-4 x= 1/2
3. Solve for x
x=1/2
4. Plug back into original equation to find y.
1/2 (1/2)^2
(1/2) (1/4) = 1/8
The point is (1/2,1/8)

5b (short answer). Find the particular solution y=f(x) to the differential equation with the initial condition f(-1) =1.
The equation is dy/dx = 1+y /x

The steps for particular solution are:
1. Separate variables
2. Integrate each side
3. Put +c on x side (left)
4. Do e^ = e^
5. Simplify
6. Solve for c
7. Solve for y

The easiest way to separate variables in this problem is to cross multiply.
dyx = dx (1+y)
dy x / 1+y = dx
Multiply each side by 1/x
dy/ 1+y = dx/x

2. Now integrate each side
The derivative of y is 1, which is what is in the numerator so you know this is going to be natural log
ln|1+y| = ln|x| +c

3. e^ln |1+y| = e^ |x| + e^c

4. 1+y = c|x| (Since you multiply when adding exponents)

5. You are given f(-1) = 1 so you know your point is (-1,1)
Now plug in and solve for c.
1+1 = c |-1|
2 = c (1)
c=2

6. Plug c back into the equation and solve for y.
|1+y| = 2|x|
y = 2|x| -1

Hope this helps with corrections.

I can use some help on related rate questions and questions like 84 when it says if the cross sections of S perpendicular to the x-axis and also remembering being in radians.