Deriving formulas used for vectors in physics.

[leafjerky]

Hey guys and gals, I'm taking an online physics course just to kind of learn the basics before I take the real thing this summer. The course is from OpenYale for those interested (oyc.yale.edu). Anyways, the professor was talking about some formulas for finding ##\vec{r}(t) = r(i(cos\omega t) + j(sin\omega t))##. He then goes on to say that ##\vec{v}(t) = \vec{r}'(t)##. So then ##\vec{v}(t) = r(i(-\omega sin\omega t) + j(\omega cos\omega t)).## I understand it all up to this point. He then loses me when he says that ##\vec{a} = \vec{v}'(t)## which he says is equal to ##-\omega ^2 \vec{r}##. Can someone walk me through how he came up with this? Sorry if the title is off as I didn't really know how to word this. Thanks

 

[A.T.]

http://en.wikipedia.org/wiki/Centripetal_force#Calculus_derivation

 

[HomogenousCow]

Just differentiate v(t) again and you'll see that it's proportional to r(t)

 

[leafjerky]

I was hoping for the step-by-step as I wasn't thinking and hadn't attempted it yet, but then I realized that it's ##\vec{r}## and not just ##r##, then I saw A.T.'s link and looked through it and it makes sense. I still haven't worked it out but I'm assuming that you factor out the ##-\omega## and what remains is just equal to ##\vec{r}##.