91Ó°ÊÓ

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 $f\left(r,\mathrm{θ},\mathrm{ϕ}\right)$ 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 $e_{\mathrm{θ}}$ the arc length is$rd\mathrm{θ}$ , which further leads to the derivative$\frac{df}{r∂\mathrm{θ}}$ .

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

Therefore, the gradient is $∇f=e_r\frac{∂f}{∂r}+e_{\mathrm{θ}}\frac{1}{r}\frac{∂f}{∂\mathrm{θ}}+e_z\frac{∂f}{∂z}$ .

Hence, the solution to this problem is $∇f=e_r\frac{∂f}{∂r}+e_{\mathrm{θ}}\frac{1}{r}\frac{∂f}{∂\mathrm{θ}}+e_z\frac{∂f}{∂z}$.