# Deriving formulas used for vectors in physics.

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1. May 9, 2015

### 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

2. May 9, 2015

3. May 9, 2015

### HomogenousCow

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

4. May 9, 2015

### 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 $\vec{r}$.