91影视

The equation of the surface is$蠒=x^2-y^{2}z$, 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$\frac{\mathbf{d}\mathbf{蠒}}{\mathbf{d}\mathbf{u}}$.

Part(a)

For Equation of gradient.

Formula states the equation mentioned below.

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

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

$螖F=2i-2j-k$

Hence the Gradient is $螖蠒=2i-2j-k$.

Part (b)

For Directional derivative

Put the values mentioned below in the formula.

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

The equation becomes as follows.

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

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$\frac{x-1}{2}-\frac{y-1}{2}=\frac{z-1}{-1}$.