Question

For Problem 11,

(a) Find the magnitude and direction of the electric field at (2,1).

(b) Find the direction in which the temperature is decreasing most rapidly at(-3,2)

(c) Find the rate of change of temperature with distance at (1,2)in the direction

Short answer

(a) Magnitude:$2\sqrt{5}$

Direction:$4\hat{i}-2\hat{j}$

(b) Direction:$6\hat{i}+4\hat{j}$

(c) Rate of change of temperature with distance at$(1,2):\sqrt{10}$

Step-by-step solution

Given Information.

The gradient is$\nabla \Phi =2x\hat{i}-2y\hat{j}$

Definition of gradient.

Gradient is defined by the equation mentioned below.

$\mathbf{\nabla }\mathit{z}\mathbf{=}\mathit{i}\frac{\widehat{\mathbf{\partial }\mathbf{z}}}{\mathbf{\partial }\mathbf{x}}\mathbf{+}\mathit{j}\frac{\widehat{\mathbf{\partial }\mathbf{z}}}{\widehat{\mathbf{\partial }\mathbf{y}}}$

Find electric field by taking the negative of electric field.

Find the electric field using the formula.

$\begin{array}{rcl}E& =& -\nabla \varphi \\ -\nabla \varphi & =& -2x\hat{i}+2y\hat{j}\\ E& =& -4\hat{i}+2\hat{j}\\ \left|E\right|& =& 2\sqrt{5}\end{array}$

Take the opposite direction of the gradient.

At (3,2), the direction of maximum decrease of temperature is in the opposite direction of its gradient.

$\begin{array}{rcl}\nabla \Phi & =& 2(-3)\hat{i}-2\left(2\right)\hat{j}\\ \overrightarrow{v}& =& -\nabla \Phi \\ \overrightarrow{v}& =& 6\hat{i}+4\hat{j}\end{array}$

Take the vector dot product to find the rate of change of temperature.

In the direction $3\hat{i}-\hat{j}$, the rate of change of temperature with distance (1,2) is given by the dot product.

$\begin{array}{rcl}\frac{d\varphi }{du}& =& (\nabla \varphi )·u\\ \hat{u}& =& \frac{1}{\sqrt{10}}(3,-1)\\ \nabla \varphi & =& 2\hat{i}-4\hat{j}\\ \frac{d\varphi }{du}& =& \frac{1}{\sqrt{10}}(3,-1)·(2,-4)\\ \frac{d\varphi }{du}& =& \sqrt{10}\end{array}$