Question

As in Problem 17, find the following gradients in two ways and show that your answers are equivalent $\mathbf{\nabla }{\mathit{r}}^{\mathbf{2}}$

Short answer

$\nabla r^2=2r{e}_r$

Step-by-step solution

Given Information

The given information is $\nabla r^2$ and $r=\sqrt{x^2+y^2}$.

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.

Finding the gradient

Using the equation,

$\begin{array}{rcl}\nabla \varphi & =& grad\varphi \\ & =& i\frac{\partial \varphi }{\partial x}+j\frac{\partial \varphi }{\partial y}+k\frac{\partial \varphi }{\partial z}\end{array}$

Find the value of gradient.

$\begin{array}{rcl}{r}^2& =& x^2+y^2\\ \nabla r^2& =& 2x\hat{i}+2y\hat{j}\end{array}$

$\begin{array}{rcl}\nabla r^2& =& 2x\times \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\hat{i}+2y\times \frac{\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}\hat{j}\\ & =& 2\sqrt{x^2+y^2}\left[\frac{x}{\sqrt{x^2+y^2}}\hat{i}+\frac{y}{\sqrt{x^2+y^2}}\hat{j}\right]\\ & =& 2r\left[\cos \left(\theta \right)\hat{i}+sin\left(\theta \right)\hat{j}\right]\\ & =& 2r{e}_r\end{array}$

Hence, the solution to this problem is$\nabla r^2=2r{e}_r$ .