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?

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- May 10th 2011, 05:05 AMMrJoe2000Another 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 AMtopsquark
A "y" term?

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

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)

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

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 AMMrJoe2000
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 AMamul28
you are correct Joe. You will get that answer

substitute

$\displaystyle \sqrt{a^2-x^2}=\frac{ay}{b}$ - May 10th 2011, 05:29 AMMrJoe2000
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 AMamul28
taking

$\displaystyle \frac{1}{\sqrt{a^2 - x^2}}=\frac{b}{ay}$

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

so your answer should be

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

i dont think it is $\displaystyle b^3$ in your answer - May 10th 2011, 05:43 AMMrJoe2000
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 AMamul28
if you dont substitute then your answer would be

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

not $\displaystyle y' = - \frac{b}{a} \frac{x}{y}$ i guess Dan made some mistake in substituting!!! - May 10th 2011, 05:52 AMMrJoe2000
OOh I see I need to substitute in for 1/y not y itself.

- May 10th 2011, 05:53 AMamul28
im not clear with what you just said?

you are substiuting to get y in your answer - May 10th 2011, 05:55 AMMrJoe2000
I understand why you needed to substitute b/(a*y) and not just y.

- May 10th 2011, 05:57 AMamul28
yes.

- May 10th 2011, 06:19 AMtopsquark