Verify equation (6.8); that is, find∇fin spherical coordinates as we did for cylindrical coordinates.

The given information is the spherical coordinates to be used in solving the question.

In mathematics and physics, scalar field or scalar-valued function referred to a scalarvalueto everypointin aspace– possiblyphysical space. The scalar may either be a (dimensionless)mathematical numberor aphysical quantity.

Let $f\left(r,\theta ,\varphi \right)$ be the gradient of a general function in a spherical coordinates.

The gradient is the directional derivative in the form of $\frac{df}{ds}$ .

The arc length is equal to which is in the direction of $e_r$ , which further leads to the derivative $\frac{df}{dr}$.

In the direction of $e_{\theta }$ the arc length is$rd\theta$ , which further leads to the derivative$\frac{df}{r\partial \theta }$ .

Since the angle $\varphi$ is obtained by rotating around k-axis. The arc length is therefore $r\sin \theta d\varphi$, which gives the directional derivatives in the direction of $e_{\theta }$as $\frac{\partial f}{r\sin \theta \partial \varphi }$ .

Therefore, the gradient is $\nabla f=e_r\frac{\partial f}{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial f}{\partial \theta }+e_z\frac{\partial f}{\partial z}$ .

Hence, the solution to this problem is $\nabla f=e_r\frac{\partial f}{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial f}{\partial \theta }+e_z\frac{\partial f}{\partial z}$.