Question

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

Short answer

$\nabla y=\hat{j}$

Step-by-step solution

Given Information

The given information is $\nabla y$.

Definition of Scalar field

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.

Find the gradients

Use the equation .$\nabla f=e_r\frac{\partial f}{\partial r}+e_{\theta }\frac{1}{r}\frac{\partial f}{\partial \theta }+e_z\frac{\partial f}{\partial z}$

$\begin{array}{rcl}\nabla y& =& \nabla rsin\left(\theta \right)\\ & =& \frac{\partial rsin\left(\theta \right)}{\partial r}{e}_r+\frac{\partial rsin\left(\theta \right)}{\partial r}\frac{1}{r}{e}_{\theta }\nabla y\\ & =& sin\left(\theta \right)e_r+\cos \left(\theta \right)e_{\theta }\end{array}$

Simplify further.

$\begin{array}{rcl}\nabla y& =& sin\left(\theta \right)\left[\cos \left(\theta \right)\hat{i}+sin\left(\theta \right)\hat{j}\right]+\cos \left(\theta \right)\left[-sin\left(\theta \right)\hat{i}+\cos \left(\theta \right)\hat{j}\right]\\ & =& \hat{j}\end{array}$

Hence, the solution to this problem is .$\nabla y=\hat{j}$