How to show angle between velocity and acceleration vector is constant in polar co-ordinates? [closed]

In polar co-ordinates we have

$\vec{r} = r\hat{r}$ and $\vec{v} = \dot{r}\hat{r} + r\dot{\theta}\hat{\theta}$ and $\vec{a} = (\ddot{r}-r\dot{\theta}^2)\hat{r} + 2\dot{r}\dot{\theta}\hat{\theta}$

Now suppose we had $r(t) = be^{wt}$ and $\theta(t) = wt$. Then $\vec{v} = wbe^{wt}(\hat{r} + \hat{\theta})$ and $\vec{a} = w^2be^{wt}\hat{r} + 2w^2be^{wt}\hat{\theta}$

How could I show that the angle between $\vec{v}$ and $\vec{a}$ is constant?

closed as off-topic by John Rennie, ACuriousMind♦May 18 '17 at 10:36

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• Please note that homework-like questions and check-my-work questions are generally considered off-topic here. We intend our questions to be potentially useful to a broader set of users than just the one asking, and prefer conceptual question over those just asking for a specific computation. – ACuriousMind May 18 '17 at 10:36

$$\vec{a} \cdot \vec{b} = |\vec{a}||\vec{b}|\cos \theta,$$ hence $$\cos \theta = \frac{\vec{a} \cdot \vec{b}}{|\vec{a}||\vec{b}|}.$$
Consider how you would compute the angle between two vectors in Cartesian coordinates, we would take $cos^{-1}$ of $\frac{\vec{v} \cdot \vec{a}}{|\vec{v}||\vec{a}|}$. Since $\hat{r}$ and $\hat{\theta}$ are mutually perpendicular, just like $\hat{i}$ and $\hat{j}$, we can think of the $\hat{r}$ and $\hat{\theta}$ components as a rotation of the $\hat{i}$ and $\hat{j}$ coordinates. As angles between vectors are preserved under rotation, this analysis should give the correct answer for the angle between the two vectors.