# Spherical Coordinates

 Sci Advisor HW Helper P: 3,149 Start from Cartesian coordinates in which $ds^2 = dx^2+dy^2+dz^2$ then calculate the differentials dx, dy and dz using: $$x = r \sin \theta \cos \phi$$ $$y = r \sin \theta \sin \phi$$ $$z = r \cos \theta$$ Substitute for dx, dy and dz in $ds^2 = dx^2+dy^2+dz^2$ and after a bit of algebra you should get the desired result.
 Sci Advisor HW Helper PF Gold P: 4,100 To find the velocity vector, write the position vector as $r(t) \hat r$. Then $v(t)=\frac{d}{dt} \left[ r(t) \hat r \right]$. Use the product rule. You'll have to compute $\frac{d}{dt} \hat r$, where $$\hat r= \sin\theta\cos\phi \hat\imath + \sin\theta\sin\phi \hat\jmath + \cos\theta \hat k$$. To simplify what you get, you might find it useful to know that $$\hat \theta= \cos\theta\cos\phi \hat\imath + \cos\theta\sin\phi \hat\jmath - \sin\theta \hat k$$ and $$\hat \phi= -\sin\phi \hat\imath + \cos\phi \hat\jmath$$ You can derive these expressions for the spherical-polar unit vectors if you calculate the vectorial element $$d \vec s = (dx)\hat \imath + (dy)\hat \jmath + (dz)\hat k$$ using Tide's expressions for x, y, and z. [The strategy is to group the terms in $dr$, $d\theta$, and $d\phi$.]