# Using spherical coordinates, show that $\nabla \wedge( \cos(\theta)\nabla\phi)=\nabla(1/r)$

Having problems with how to obtain the following, I have written in my notes but can't see how the end result is formed.

Using spherical co-ordinates $(r, \theta, \phi)$ evaluate $\nabla \phi$ and show that $\nabla \wedge( \cos(\theta)\nabla\phi)=\nabla(1/r)$

Now ${\mathbf r} = x \mathbf{i}+y \mathbf{j}+z \mathbf{k}= r \sin(\theta) \cos(\phi) \mathbf{i} + r \sin(\theta) \sin(\phi) \mathbf{j} + r \cos(\theta) \mathbf{k}$

It says that $\nabla \phi$ = $\frac {1}{r\sin(\theta)} {\bf e_{\phi}}$

I'm just confused to how you get to this?

As for the $\nabla \wedge( \cos(\theta)\nabla\phi)=\nabla(1/r)$ this makes more sense to me, once the first part is obtained.

Many thanks in advance.

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First, establish what $\mathbf{e}_r$, $\mathbf{e}_\phi$ and $\mathbf{e}_\theta$ are. $$\begin{eqnarray} \mathbf{e}_r &=& \sin(\theta) \cos(\phi) \mathbf{i} + \sin(\theta) \sin(\phi) \mathbf{j} + \cos(\theta) \mathbf{k} \\ \mathbf{e}_\theta &=& \cos(\theta) \cos(\phi) \mathbf{i} + \cos(\theta) \sin(\phi) \mathbf{j} - \sin(\theta) \mathbf{k} \\ \mathbf{e}_\phi &=& - \sin(\phi) \mathbf{i} + \cos(\phi) \mathbf{j} \end{eqnarray}$$ They form orthonormal system $\langle \mathbf{e}_r, \mathbf{e}_r \rangle = \langle \mathbf{e}_\phi, \mathbf{e}_\phi \rangle = \langle \mathbf{e}_\theta, \mathbf{e}_\theta \rangle = 1$, $\langle \mathbf{e}_r, \mathbf{e}_\phi \rangle = \langle \mathbf{e}_r, \mathbf{e}_\theta \rangle = \langle \mathbf{e}_\phi, \mathbf{e}_\theta \rangle = 0$. Now $$\nabla \phi = \frac{\partial \phi}{\partial x} \mathbf{i} + \frac{\partial \phi}{\partial y} \mathbf{j} + \frac{\partial \phi}{\partial z} \mathbf{k}$$ To compute partial derivatives, use $$\begin{pmatrix} \frac{\partial r}{\partial x} & \frac{\partial r}{\partial y} & \frac{\partial r}{\partial z} \\ \frac{\partial \theta}{\partial x} & \frac{\partial \theta}{\partial y} & \frac{\partial \theta}{\partial z} \\ \frac{\partial \phi}{\partial x} & \frac{\partial \phi}{\partial y} & \frac{\partial \phi}{\partial z} \\ \end{pmatrix} = \begin{pmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} & \frac{\partial x}{\partial \phi} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} & \frac{\partial y}{\partial \phi} \\ \frac{\partial z}{\partial r} & \frac{\partial z}{\partial \theta} & \frac{\partial z}{\partial \phi} \end{pmatrix}^{-1} = \begin{pmatrix} \sin(\theta) \cos(\phi) & r \cos(\theta) \cos(\phi) & -r \sin(\theta) \sin(\phi) \\ \sin(\theta) \sin(\phi) & r \cos(\theta) \sin(\phi)& r \sin(\theta) \cos(\phi) \\ \cos(\theta) & -r \sin(\theta) & 0 \end{pmatrix}^{-1} = \begin{pmatrix} \sin(\theta) \cos(\phi) & \sin(\theta) \sin(\phi) & \cos(theta) \\ \frac{\cos(\theta) \cos(\phi) }{r} & \frac{\cos(\theta) \sin(\phi) }{r} & -\frac{\sin(\theta) }{r} \\ -\frac{\sin(\phi)}{r \cos(\theta)} & \frac{\cos(\phi)}{r \cos(\theta)} & 0 \end{pmatrix}$$ Thus $$\nabla \phi = -\frac{\sin(\phi)}{r \cos(\theta)} \mathbf{i} + \frac{\cos(\phi)}{r \cos(\theta)} \mathbf{j} = \frac{1}{r \cos(\theta)} \mathbf{e}_\phi$$