y = 15 sin(ωt + a), determine dy/dt.... i'm just confused by the ω on this one
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
$\displaystyle \frac{dy}{dt} = \frac{du}{dt}\cdot\frac{dy}{du}$
$\displaystyle = \omega \cos{(\omega t + a)}$.