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

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- Aug 23rd 2009, 07:29 PMEp!taphquick question
y = 15 sin(ωt + a), determine dy/dt.... i'm just confused by the ω on this one

- Aug 23rd 2009, 07:36 PMProve It
It appears that $\displaystyle \omega$ and $\displaystyle a$ are arbitrary constants.

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

Let $\displaystyle u = \omega t + a$.

Then $\displaystyle y = \sin{u}$.

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

Therefore

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

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

v = 10(1-e^(-t/5)), determine dv/dt? - Aug 23rd 2009, 08:28 PMCaptainBlack