Question

As in Problem 17, find the following gradients in two ways and show that your answers are equivalent $\mathbf{\nabla }\mathit{x}$.

Short answer

The solution to this problem is$\nabla x=\hat{i}$.

Step-by-step solution

Given Information.

The given information is $\nabla x$.

Definition of Scalar field.

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

Find the solution.

$$\begin{gathered} \mathrm{Use}\mathrm{the}\mathrm{equation}\nabla \mathrm{f}={\mathrm{e}}_{\mathrm{r}}\frac{\partial \mathrm{f}}{\partial \mathrm{r}}+{\mathrm{e}}_{\mathrm{\theta }}\frac{1}{\mathrm{r}}\frac{\partial \mathrm{f}}{\partial \mathrm{\theta }}+{\mathrm{e}}_{\mathrm{z}}\frac{\partial \mathrm{f}}{\partial \mathrm{z}} \\ \nabla \mathrm{x}=\nabla \mathrm{r}\cos \left(\mathrm{\theta }\right) \\ =\frac{\partial \mathrm{r}\cos \left(\mathrm{\theta }\right)}{\partial \mathrm{r}}{\mathrm{e}}^{\mathrm{r}}+\frac{\partial \mathrm{r}\cos \left(\mathrm{\theta }\right)}{\partial \mathrm{r}}\frac{1}{\mathrm{r}}{\mathrm{e}}_{\mathrm{\theta }}\nabla \mathrm{x} \\ =\cos \left(\mathrm{\theta }\right){\mathrm{e}}_{\mathrm{r}}-\sin \left(\mathrm{\theta }\right){\mathrm{e}}_{\mathrm{\theta }} \\ \mathrm{Simplify}\mathrm{further}. \\ \nabla \mathrm{x}=\cos \left(\mathrm{\theta }\right)\left[\cos \left(\mathrm{\theta }\right)\hat{\mathrm{i}}+\sin \left(\mathrm{\theta }\right)\hat{\mathrm{j}}\right]-\sin \left(\mathrm{\theta }\right)\left[-\sin \left(\mathrm{\theta }\right)\hat{\mathrm{i}}+\cos \left(\mathrm{\theta }\right)\hat{\mathrm{j}}\right] \\ =\hat{\mathrm{i}} \\ \mathrm{Hence},\mathrm{the}\mathrm{solution}\mathrm{to}\mathrm{this}\mathrm{problem}\mathrm{is}\nabla \mathrm{x}=\hat{\mathrm{i}}. \end{gathered}$$