Another differential I can't solve

• May 10th 2011, 05:05 AM
MrJoe2000
Another differential I can't solve
The equation is y= b/a*sqrt(a^2 - x^2)

the solution apparently has a y term in it but I can't figure out where it comes from. Can anyone give me a hand?
• May 10th 2011, 05:13 AM
topsquark
Quote:

Originally Posted by MrJoe2000
The equation is y= b/a*sqrt(a^2 - x^2)

the solution apparently has a y term in it but I can't figure out where it comes from. Can anyone give me a hand?

A "y" term?

What is the derivative of $\sqrt{a^2 - x^2}$?

The only thing I can think of is that your answer has the y subbed back in to "clean" things up a bit. Your solution is (you can fill in the details)
$y' = - \frac{b}{a} \frac{x}{\sqrt{a^2 - x^2}} = - \frac{b}{a} \frac{x}{y}$

Is that what your final answer is supposed to look like? If not I have no idea what your "y" term in your question means.

-Dan
• May 10th 2011, 05:19 AM
MrJoe2000
Thanks Dan. I got up to where you did but missed that they subbed the y back into the equation. The book answer is -(b^3*x)/(a^2*y)
• May 10th 2011, 05:27 AM
amul28
you are correct Joe. You will get that answer
substitute
$\sqrt{a^2-x^2}=\frac{ay}{b}$
• May 10th 2011, 05:29 AM
MrJoe2000
How do you get the cubed and squared terms? I got to the point that Dan did but I'm not sure how to get from there to the final answer.
• May 10th 2011, 05:36 AM
amul28
taking
$\frac{1}{\sqrt{a^2 - x^2}}=\frac{b}{ay}$

$y' = - \frac{b}{a} \frac{x}{\sqrt{a^2 - x^2}} = - \frac{bx}{a} \frac{b}{ay}$

$y=-\frac{b^2x}{a^2y}$

i dont think it is $b^3$ in your answer
• May 10th 2011, 05:43 AM
MrJoe2000
Can you explain why after you obtain the -b/a*x/y answer you continue to substitute in? Why wouldn't I simply stop there?
• May 10th 2011, 05:49 AM
amul28
$y' = - \frac{b}{a} \frac{x}{\sqrt{a^2 - x^2}}$

not $y' = - \frac{b}{a} \frac{x}{y}$ i guess Dan made some mistake in substituting!!!
• May 10th 2011, 05:52 AM
MrJoe2000
OOh I see I need to substitute in for 1/y not y itself.
• May 10th 2011, 05:53 AM
amul28
im not clear with what you just said?
• May 10th 2011, 05:55 AM
MrJoe2000
I understand why you needed to substitute b/(a*y) and not just y.
• May 10th 2011, 05:57 AM
amul28
yes.
• May 10th 2011, 06:19 AM
topsquark
Quote:

Originally Posted by MrJoe2000
Thanks Dan. I got up to where you did but missed that they subbed the y back into the equation. The book answer is -(b^3*x)/(a^2*y)

Yes, sorry about that. For some reason I decided to use $y = \sqrt{a^2 - x^2}$ in the final substitution.

I mean, I was just testing you. Yeah, that was it! Uhhhh...I'm gonna go now. (slinks away into a dark alley)

-Dan