# quick question

• August 23rd 2009, 07:29 PM
Ep!taph
quick question
y = 15 sin(ωt + a), determine dy/dt.... i'm just confused by the ω on this one
• August 23rd 2009, 07:36 PM
Prove It
Quote:

Originally Posted by Ep!taph
y = 15 sin(ωt + a), determine dy/dt.... i'm just confused by the ω on this one

It appears that $\omega$ and $a$ are arbitrary constants.

So you can use the chain rule to find this derivative.

Let $u = \omega t + a$.

Then $y = \sin{u}$.

So $\frac{du}{dt} = \omega$ and $\frac{dy}{du} = \cos{u} = \cos{(\omega t + a)}$.

Therefore

$\frac{dy}{dt} = \frac{du}{dt}\cdot\frac{dy}{du}$

$= \omega \cos{(\omega t + a)}$.
• August 23rd 2009, 07:47 PM
Ep!taph
ah thanks... just as i thought, what about this one?

v = 10(1-e^(-t/5)), determine dv/dt?
• August 23rd 2009, 08:28 PM
CaptainBlack
Quote:

Originally Posted by Ep!taph
ah thanks... just as i thought, what about this one?

v = 10(1-e^(-t/5)), determine dv/dt?

What is the problem you are having with this one?

(multiply out the bracket, do you still have a problem, what is it? Consider using the chain rule)

CB