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Question

First off, I do realize that:

$$\vec{r} = r \hat{r}$$ $$\dot{\vec{r}} = \dot{r} \hat{r} + r \dot{\theta} \hat{\theta}$$ $$\ddot{\vec{r}} = \ddot{r} \hat{r} + \dot{r} \hat{r} + \dot{r} \dot{\hat{r}} + \dot{r} \dot{\theta} \hat{\theta} + r \ddot{\theta} \hat{\theta} + r \dot{\theta} \hat{\dot{\theta}}$$

However, how does one differentiate $\vec{r}$ and get $\theta$ to pop up in the equation?

After all, in my mind, when differentiating $\vec{r}$, shouldn't it look like this as shown below?

$$\dot{\vec{r}} = \dot{r} \hat{r} + r \dot{\hat{r}}$$

Any help in answering the question would be much appreciated!

Note: Not a homework question.

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  • $\begingroup$ Hi Athenian. If you haven't already done so, please take a minute to read the definition of when to use the homework-and-exercises tag, and the Phys.SE policy for homework-like problems. $\endgroup$ – Qmechanic Jan 28 at 17:46
  • $\begingroup$ @Qmechanic, Thank you for the notice. However, this is not a homework nor an exercise problem. I was looking over some sources on a vector caculus wikipedia page on how to differentiate $r$ and was simply curious how $\theta$ popped out of "nowhere". Regardless, I'll be sure to read through the policy and thank you for providing the link. $\endgroup$ – Athenian Jan 29 at 6:08
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Your final equation is correct. You just need to calculate $\frac{d\hat{r}}{dt}$.

Writing $\hat{r}$ in Cartesian form:

$$\hat{r}=(\cos\theta)\hat{x}+(\sin\theta)\hat{y}$$

We also know that $\hat{\theta}$ is defined to always be perpendicular to $\hat{r}$, and points in the direction of increasing $\theta$, so:

$$\hat{\theta}=(-\sin\theta)\hat{x}+(\cos\theta)\hat{y}$$

Taking the time derivative of $\hat{r}$:

$$\frac{d\hat{r}}{dt}=\left(-\sin\theta\frac{d\theta}{dt}\right)\hat{x}+\left(\cos\theta\frac{d\theta}{dt}\right)\hat{y}=\frac{d\theta}{dt}((-\sin\theta)\hat{x}+(\cos\theta)\hat{y})=\frac{d\theta}{dt}\hat{\theta}$$

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In polar coordinates, $\vec r$ has to be understood as a function of $r,\theta$.

Specifically, we have $$ \vec r(r,\theta) = r\,\hat r(\theta) $$ hence via product and chain rule $$ \dot{\vec r} = \frac{dr}{dt} \hat r + r \frac{d \hat r}{d\theta} \frac{d\theta}{dt} = \dot r \hat r + r \dot\theta \hat\theta $$ where we have used that $$ \frac{d \hat r}{d\theta} = \frac{d}{d\theta} \left( \cos(\theta) \,\hat x + \sin(\theta) \,\hat y\right) = -\sin(\theta) \,\hat x + \cos(\theta) \,\hat y = \hat \theta $$

Alternatively, if you're familiar with some concepts from differential geometry and know that in polar coordinates, we have $$ ds^2 = dr^2 + r^2d\theta^2 $$ then $$ \dot{\vec r} = \dot r \partial_r + \dot\theta \partial_\theta = \dot r \hat r + r \dot\theta \hat\theta $$ as $$ ds^2(\partial_r,\partial_r) = 1 \qquad ds^2(\partial_\theta,\partial_\theta) = r^2 $$ and thus $$ \partial_r = \hat r \qquad \partial_\theta = r \hat\theta $$

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