The position vector of a point is given by $\displaystyle \vec{r}(t) = (e^tcost, e^tsint)$. Prove that:

a) $\displaystyle \vec{a} = 2\vec{v} - 2\vec{r}$

b) the angle between the position vector $\displaystyle \vec{r}$ and the acceleration vector $\displaystyle \vec{a}$ is constant. Calculate this angle.

Answer: b) pi/2

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I tried to derive the position vector two times but it isn't working.