Thread: Another differential I can't solve

1. 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?

2. 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

3. 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)

4. you are correct Joe. You will get that answer
substitute
$\sqrt{a^2-x^2}=\frac{ay}{b}$

5. 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.

6. 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

7. 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?

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

not $y' = - \frac{b}{a} \frac{x}{y}$ i guess Dan made some mistake in substituting!!!

9. OOh I see I need to substitute in for 1/y not y itself.

10. im not clear with what you just said?

11. I understand why you needed to substitute b/(a*y) and not just y.

12. yes.

13. 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