# Velocity in spherical coordinates

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.]

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Ibix
2020 Award
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.

Another and etotheipi
Nugatory
Mentor
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.