# Differential calculus?

• Jun 29th 2008, 07:47 AM
fardeen_gen
Differential 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 AM
Reckoner
Quote:

Originally Posted by fardeen_gen
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?

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 AM
Mathstud28
Quote:

Originally Posted by Reckoner
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.

I believe that

$\displaystyle \frac{d}{dt}\bigg[f(\omega,A,x)\bigg]=0$

Yeah?
• Jun 29th 2008, 10:26 AM
Reckoner
Quote:

Originally Posted by Mathstud28
I believe that

$\displaystyle \frac{d}{dt}\bigg[f(\omega,A,x)\bigg]=0$

Yeah?

Correct, as long as $\displaystyle x$ is constant with respect to $\displaystyle t$.
• Jun 29th 2008, 10:29 AM
Mathstud28
Quote:

Originally Posted by Reckoner
Correct, as long as $\displaystyle x$ is constant with respect to $\displaystyle t$.

.........................I know this............................I was pointing something out........in your post.
• Jun 30th 2008, 05:46 AM
fardeen_gen
Is the answer 0 then? So D is the correct option, right?
• Jun 30th 2008, 08:58 AM
Reckoner
Quote:

Originally Posted by fardeen_gen
Is the answer 0 then? So D is the correct option, right?

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.