(a) Given $\mathit{\varphi }\mathbf{=}{\mathit{x}}^{\mathbf{2}}\mathbf{-}{\mathit{y}}^{\mathbf{2}}\mathit{z}$, find $\mathit{\Delta }\mathit{\varphi }$ at $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$.

(b) Find the directional derivative of φ at $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$ in the direction $\mathit{i}\mathbf{}\mathbf{-}\mathbf{}\mathbf{2}\mathit{j}\mathbf{}\mathbf{+}\mathbf{}\mathit{k}$.

(c)Find the equations of the normal line to the surface ${\mathit{x}}^{\mathbf{2}}\mathbf{-}{\mathit{y}}^{\mathbf{2}}\mathit{z}\mathbf{=}\mathbf{0}$ at $\mathbf{(}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{,}\mathbf{1}\mathbf{)}$.

The equation of the surface is$\varphi =x^2-y^{2}z$, vector $i-2j+k$, and point $(1,1,1)$.

The directional derivative is the rate of change along a unit vector.

The formulae for the directional derivative is$\frac{\mathbf{d}\mathbf{\varphi }}{\mathbf{d}\mathbf{u}}$.

Part(a)

For Equation of gradient.

Formula states the equation mentioned below.

$\begin{array}{c}\Delta F=\frac{\partial f}{\partial x}i+\frac{\partial f}{\partial y}j+\frac{\partial f}{\partial z}k\\ \Delta F=\left(2x_0\right)i-2y_0{z}_{0}j-{y^2}_{0}k\end{array}$

Put$(1,1,1)$ in the above equation.

$\Delta F=2i-2j-k$

Hence the Gradient is $\Delta \varphi =2i-2j-k$.

Part (b)

For Directional derivative

Put the values mentioned below in the formula.

$\begin{array}{c}\varphi =(2,-2,-1)\\ u=\frac{1}{\sqrt{6}}(1,1,1)\end{array}$

The equation becomes as follows.

$\begin{array}{c}\frac{d\varphi }{du}=(2,-2,-1)\times \frac{1}{\sqrt{6}}(1,1,1)\\ \frac{d\varphi }{du}=\frac{5}{\sqrt{6}}\end{array}$

Hence the directional derivative is $\frac{5}{\sqrt{6}}$ .

Part (c)

Equation of normal line at point$(1,1,1)$ .

$\begin{array}{c}\frac{x-x_0}{a}+\frac{y-y_0}{b}=\frac{z-z_0}{c}\\ \frac{x-1}{5}-\frac{y-1}{2}=\frac{z-1}{-1}\end{array}$

Hence the equation of normal line to the surface$\frac{x-1}{2}-\frac{y-1}{2}=\frac{z-1}{-1}$.