Velocity in spherical coordinates

[Another]

Why the velocity in spherical coordinates equal to $v^2 = v \dot{} v = \dot{r}^2 + \dot{r}^2\dot{\theta}^2$

maybe

$v^2 = v \dot{} v = (\hat{ \theta } \dot{ \theta } r +\hat{r} \dot{r} + \hat{ \phi } \dot{\phi } r \sin{ \theta}) \dot{} (\hat{ \theta } \dot{ \theta } r +\hat{r} \dot{r} + \hat{ \phi } \dot{\phi } r \sin{ \theta}) = \dot{\theta}^2 r^2 + \dot{r}^2 + r^2 \dot{\phi}^2 \sin^2{\theta}$

[Moderator's note: Moved from a technical forum and thus no template.]

 

[Ibix]

The movement is in a plane, so they've picked coordinates such that the plane is the equatorial plane of the coordinates - hence $\dot{\phi}=0$.

Incidentally, it should be $r^2{\dot{\theta}}^2$ in your first expression. I presume that's just a typo since it's correct in the text and your next expression.

 

[Nugatory]

The question about why the velocity in spherical coordinates takes the form it does has been answered, so any further discussion of this problem should be in a new thread in the homework forums.

This thread can be closed.