Question

ok this is so tragic but I need someone to explain a few steps to me. HELP

Answer 1

so I am terrible in algebra. and I was solving a differential equation for calculus. and I know how to solve a differential equation but these steps here threw me off. can someone explain to me how this happened lol.

Answer 2

I'd imagine there are earlier steps to this problem?

Answer 3

Perhaps also include the whole of the question? As they have substituted values into equations from earlier.

Answer 4

The first line is just solving for C.
\[\frac{-1}{2*2^2} = \frac{-1}{8} = \frac{1}{2}+C\]

Answer 5

\[C = \frac{-1}{8}-\frac{1}{2} = \frac{-1}{8} -\frac{4}{8} = \frac{-5}{8}\]

Answer 6

At this point, it appears they referred back to a previous equation containing C, and substituted in (-5/8)

Answer 7

And I would guess that the equation they substituted C into was
\[y^2 = \frac{1}{2C-x^2}\]

Answer 8

oh of course lol, but I was just confused here, when they isolated the y^2 it threw me off. O.o I was trying to isolate it by hand, and I did not get what was written as the answer online.

Answer 9

http://apcentral.collegeboard.com/apc/public/repository/ap10_calculus_ab_scoring_guidelines.pdf if you would like to make sense of my confusion, it is here, and it question number 6 part C

Answer 10

I understood what was going on minus the last 3 steps .

Answer 11

Ah, good =) I am glad you were able to make sense of it.

Answer 12

Here's how they isolated y^2:

Answer 13

\[-\frac{1}{2y^2} = \frac{x^2}{2} + C\]
multiply both sides by (-2y^2) to get
\[1 = -x^2y^2 + 2C*y^2\]
on the right side, factor out y^2 to obtain
\[1 = (-x^2 +2C)y^2\]
Divie both sides by (-x^2+2c) to get
\[\frac{1}{2C-x^2} = y^2\]

Answer 14

That algebra is a little tricky, I realize, but hopefully each step made sense.

Answer 15

ohh yeah that made sense :O err I wish my algebra skills were stronger. I wouldnt have gotten this right l: -sigh- But thank you this helped a lot.

Answer 16

The trick that a lot of people don't understand is that if the variable you're trying to solve for is in the denominator, your first step is to get it out of there. The problem instantly becomes easier to solve.

Answer 17

yeah that makes more sense, I wouldn't have thought about that, it is much easier. Hopefully something like this is on my AP exam and I will know what to do.

Answer 18

Hopefully =D