Deriving equations of motion in spherical coordinates OK, we've been asked to derive the equations of motion in spherical coordinates. According to the assignment, we should end up with this: 
$$
\bf \vec{v} \rm = \frac{d \bf \vec{r} \rm}{dt} = \dot{r} \bf \hat{r} \rm + r \dot{\theta}\hat{\boldsymbol \theta} \rm + r \dot{\phi}\sin \theta \bf \hat{\boldsymbol\phi}\rm
$$
In this case θ is the angle from the z axis and phi is the angle in the x-y plane. 
Now, if I take it that position $$\bf \vec{r} \rm = r \bf \hat{r}$$ and say $$ \bf \hat{r} \rm = \bf \hat{x} \rm \sin\theta \cos\phi + \bf \hat{y} \rm \sin \theta \sin \phi + \bf \hat{z}\rm \cos\theta \\
\hat{\boldsymbol\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta\\
\hat{\boldsymbol\phi} = \bf \hat{x} \rm (-\sin\phi) + \bf \hat{y} \rm \cos \phi\\
$$
now maybe I am making this more complex than it is. And maybe it's just a notation problem (I really hate the dot notation sometimes because I feel it obscures things, but I need to know it, I know). 
If we assume that when r changes, $\phi$ and $\theta$ and their unit vectors stay the same, then we can safely say that $\frac{d \hat{\boldsymbol\phi}}{dr} = 0$ and $\frac{d \hat{\boldsymbol\theta}}{d r} = 0.$ (someone please tell me if i am wrong).
If we do the same thing with changing θ and $\phi$ though, the result is different. hen we change θ, r has to change because it changes direction, and when we change $\phi$ $r$ has to change because it changes direction in that case also. 
When I take the derivative of $\hat{r}$ with respect to $\theta$, I get the following: 
$$\frac{d \bf \hat{r}}{d\theta} = \bf \hat{x} \rm \cos\theta \cos\phi + \bf \hat{y} \rm \cos \theta \sin \phi - \bf \hat{z}\rm\sin\theta $$
which as it happens also is equal to $\hat{\boldsymbol{\theta}}$
Now, if I look at $\bf \vec{r} \rm = r \bf \hat{r}$ and take the derivative w/r/t time, I should get $\frac{d \bf \vec{r} \rm}{dt} = r \frac{d\bf \hat{r}}{dt} + \frac{dr}{dt}\bf \hat{r} \rm$
I notice that this happens (and some of this is just seeing the notation):
$$
\frac{d \bf \vec{r} \rm}{dt} = r \frac{d\bf \hat{r}}{dt} + \frac{dr}{dt}\bf \hat{r} \rm = \dot{r} \bf \hat{r} \rm + r \hat{\boldsymbol \theta} \rm $$
ANd I feel like I am almost there. But I am having trouble making that last step. I am getting a bit frustrated because I can't seem to make the differentiation work the way it does in the text and I haven't found a derivation online that matches up with anything I have seen in class. Again, maybe it's just the notation used. But I am trying to understand where the $\dot{\theta}$ term comes up, and how to get that extra dimension in. 
Any help is most appreciated. Thanks.
EDIT: fixed the unit vector phi expression. 
 A: Suppose $\vec{r} = r \hat{r}$ where $ \hat{r} = \cos \phi \sin \theta \hat{x}+\sin \phi \sin \theta \hat{y}+ \cos \theta \hat{z}$. Now, this symbol represents the position of a particle at time $t$ where the spherical frame $\hat{r},\hat{\theta},\hat{\phi}$ is co-moving. On the other hand, the cartesian frame $\hat{x},\hat{y},\hat{z}$ is constant across all of three-dimensional space. So, differentiate with respect to time 
\begin{align}
\frac{d}{dt}\vec{r} 
&= \frac{dr}{dt} \hat{r}+r\frac{d\hat{r}}{dt} \\
&= \dot{r} \hat{r}+r \frac{d}{dt} \left[\cos \phi \sin \theta \hat{x}+\sin \phi \sin \theta \hat{y}+ \cos \theta \hat{z} \right] \\
&= \dot{r} \hat{r}+r \frac{d}{dt} \left[\cos \phi \sin \theta \right] \hat{x}+r\frac{d}{dt} \left[\sin \phi \sin \theta \right]\hat{y}+ r\frac{d}{dt} \left[\cos \theta \right]\hat{z}  \\
&= \dot{r} \hat{r}+r  \left[-\dot{\phi}\sin \phi \sin \theta+\dot{\theta}\cos \phi \cos \theta \right] \hat{x}+ \\
& \qquad +r \left[\dot{\phi}\cos \phi \sin \theta + \dot{\theta}\sin \phi \cos \theta \right]\hat{y}+ r \left[- \dot{\theta}\sin \theta \right]\hat{z}  \\
&= \dot{r} \hat{r}+ r\dot{\theta}\left[\cos \phi \cos \theta  \hat{x} +\sin \phi \cos \theta \hat{y}-\sin \theta \hat{z}\right] \\
&\qquad + r\sin \theta \dot{\phi} \left[ -\sin \phi \hat{x}+\cos \phi \hat{y}\right]
\end{align}
Again, the concept here is that the spherical frame rides along with the particle motion and you're trying to express the velocity in terms of the co-moving frame.  Identify the $\hat{\phi}$ and $\hat{\theta}$. As Austen was telling you, there is no $\hat{r}$ component in the $\frac{d}{dt} \hat{r}$ term because if there was a change then the length of the unit-vector would evolve... or you can just differentiate $\hat{r} \cdot \hat{r}=1$ with respect to time to see that the time-rate of change for $\hat{r}$ is infact orthogonal to $\hat{r}$ itself.
I would set-aside the questions about differentiating w.r.t. $\phi$ or $\theta$ for a moment until you complete this. 
A: Start from $\mathbf{r}=r\hat{\mathbf{r}}$ and take the derivative using the product rule
$$
\dot{\mathbf{r}}=\dot r\hat{\mathbf{r}}+r\dot{\hat{\mathbf{r}}}.
$$
The first term is the first term of the answer you seek, so his focuses attention on $\dot{\hat{\mathbf{r}}}$. The derivative of a unit vector has no radial component, so can be expressed in terms of the other two unit vectors.
At this point you can resort to explicit coordinates, but a graphical proof is more illuminating. Think about what happens when only $\dot{\theta}\neq 0$, and then when only $\dot{\phi}\neq 0$ is changing.
