Repeat Problem 14b for the following points and directions.

$$\begin{gathered} \mathbf{\left(}\mathit{a}\mathbf{\right)}\mathbf{}\mathbf{(}\mathbf{4}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{,}\hat{\mathbf{i}}\mathbf{+}\hat{\mathbf{j}} \\ \mathbf{\left(}\mathit{b}\mathbf{\right)}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{3}\mathbf{,}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{4}\hat{\mathbf{i}}\mathbf{+}\mathbf{3}\hat{\mathbf{j}} \\ \mathbf{\left(}\mathit{c}\mathbf{\right)}\mathbf{}\mathbf{(}\mathbf{2}\mathbf{,}\mathbf{2}\mathbf{)}\mathbf{,}\mathbf{-}\mathbf{3}\hat{\mathbf{i}}\mathbf{+}\hat{\mathbf{j}} \\ \mathbf{\left(}\mathit{d}\mathbf{\right)}\mathbf{}\mathbf{(}\mathbf{-}\mathbf{4}\mathbf{,}\mathbf{-}\mathbf{1}\mathbf{)}\mathbf{,}\mathbf{4}\hat{\mathbf{i}}\mathbf{-}\mathbf{3}\hat{\mathbf{j}} \end{gathered}$$

The given equation is$\nabla \varphi =-2x\hat{i}-8y\hat{j}-\hat{k}$

Gradient is definedby the equation mentioned below.

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

At $\overrightarrow{u}=(1,1,0)$, calculate the function.

$\begin{array}{rcl}\varphi & =& 32-x^2-4y^2-z\\ & =& 0\end{array}$

Find the directional derivate of the function.

$\begin{array}{rcl}\frac{d\varphi }{d\hat{u}}& =& (\nabla \varphi )·\hat{u}\\ \hat{u}& =& \frac{1}{\sqrt{2}}(1,1,0)\end{array}$

Calculate the gradient of the function.

$\nabla \varphi =-8\hat{i}+16\hat{j}-\hat{k}$

Substitute the values to find the directional derivate.

$\begin{array}{rcl}\frac{d\varphi }{d\hat{u}}& =& \frac{1}{\sqrt{2}}(1,1,0)·(-8,16,-1)\\ & =& 4\sqrt{2}\end{array}$

From the sign, deduce that the movement is in the upward direction.

The second point is $\overrightarrow{u}=(-3,1,0)$$\overrightarrow{u}=(-3,1,0)$.

Calculate the unit vector

$\hat{u}=\frac{1}{5}(4,3,0)$

Calculate the gradient of the function.

$\nabla \varphi =6\hat{i}-8\hat{j}-\hat{k}$

Substitute the values to find the directional derivate.

$\begin{array}{rcl}\frac{d\varphi }{d\hat{u}}& =& \frac{1}{5}(4,3,0)·(6,-8,-1)\\ & =& 0\end{array}$

From the sign, deduce that the movement is neither upwards nor downwards but in the same direction.

The third point is .$\overrightarrow{u}=(2,2,0)$

Calculate the unit vector

$\hat{u}=\frac{1}{\sqrt{10}}(-3,1,0)$

Calculate the gradient of the function

$\nabla \varphi =-4\hat{i}-16\hat{j}-\hat{k}$

Substitute the values to find the directional derivate.

$\begin{array}{rcl}\frac{d\varphi }{d\hat{u}}& =& \frac{1}{\sqrt{10}}(-3,1,0)·(-4,-16,-1)\\ & =& \frac{-4}{\sqrt{10}}\end{array}$

From the sign, deduce that the movement is in the downward direction.

The third point is .$\overrightarrow{u}=(-4,-1,0)$

Calculate the unit vector

$\hat{u}=\frac{1}{5}(4,-3,0)$

Calculate the gradient of the function

$\nabla \varphi =-8\hat{i}+24\hat{j}-\hat{k}$

Substitute the values to find the directional derivate

$\begin{array}{rcl}\frac{d\varphi }{d\hat{u}}& =& \frac{1}{5}(4,-3,0)·(8,8,-1)\\ & =& \frac{8}{5}\end{array}$

From the sign, deduce that the movement is in the upward direction.