If ω & A are constant then d[ω√(A^2 - x^2)]/dt =?

A) ω/{2√(A^2 - x^2)}

B) -ω/{2√(A^2 - x^2)}

C) -ωx/{√(A^2 - x^2)}

D) None of these

Can u please explain the reasoning too?

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- Jun 29th 2008, 07:47 AMfardeen_genDifferential calculus?
**If ω & A are constant then d[ω√(A^2 - x^2)]/dt =?**

A) ω/{2√(A^2 - x^2)}

B) -ω/{2√(A^2 - x^2)}

C) -ωx/{√(A^2 - x^2)}

D) None of these

Can u please explain the reasoning too? - Jun 29th 2008, 10:00 AMReckoner
Are you differentiating with respect to $\displaystyle t$ or with respect to $\displaystyle x$?

For $\displaystyle \frac{d}{dt}\left[\omega\sqrt{A^2 - x^2}\right]$ we have

$\displaystyle \frac{d}{dt}\left[\omega\sqrt{A^2 - x^2}\right] = \frac{d}{dt}\left[\omega\left(A^2 - x^2\right)^{1/2}\right]$

$\displaystyle = \frac{\omega}2\left(A^2 - x^2\right)^{-1/2}\frac{d}{dt}\left[A^2 - x^2\right]$

$\displaystyle = \frac{\omega}2\left(A^2 - x^2\right)^{-1/2}\left(-2x\frac{dx}{dt}\right)$

Then you can simplify a bit. - Jun 29th 2008, 10:25 AMMathstud28
- Jun 29th 2008, 10:26 AMReckoner
- Jun 29th 2008, 10:29 AMMathstud28
- Jun 30th 2008, 05:46 AMfardeen_gen
Is the answer 0 then? So D is the correct option, right?

- Jun 30th 2008, 08:58 AMReckoner
We do not know for sure that $\displaystyle x$ is constant, so you should leave it in a form involving $\displaystyle \frac{dx}{dt}$. If $\displaystyle x$ happens to be constant the derivative will still work out to be 0.

So in this case, assuming you typed the problem correctly and we are differentiating with respect to $\displaystyle t$, (D) would be correct.