91影视

The equation of the surface is $蠒=x^2+\sin y-xz$,

vector $i+2j-2k,蠒=5$, and point $(1,\frac{蟺}{2},-3)$.

A tangent plane is a plane in which all the lines are tangent to a particular point. Equation of tangent plane is $\frac{\mathbf{鈭�}\mathbf{F}}{\mathbf{鈭�}\mathbf{x}}\mathbf{(}\mathit{x}\mathbf{-}{\mathit{x}}_{\mathbf{0}}\mathbf{)}\mathbf{+}\frac{\mathbf{鈭�}\mathbf{F}}{\mathbf{鈭�}\mathbf{y}}\mathbf{(}\mathit{y}\mathbf{-}{\mathit{y}}_{\mathbf{0}}\mathbf{)}\mathbf{+}\frac{\mathbf{鈭�}\mathbf{F}}{\mathbf{鈭�}\mathbf{z}}\mathbf{(}\mathit{z}\mathbf{-}{\mathit{z}}_{\mathbf{0}}\mathbf{)}\mathbf{=}\mathbf{0}$.

Find equation of gradient.

The following formula states the given equation.

$\begin{array}{c}螖F=\frac{鈭�f}{鈭�x}i+\frac{鈭�f}{鈭�y}j+\frac{鈭�f}{鈭�z}k\\ 螖F=(2x_0-z_0)i+\cos y_{0}j-x_{0}k\end{array}$

Put $(1,\frac{蟺}{2},-3)$in the above equation.

$螖F=5i-k$

Find directional derivative.

Put the values mentioned below in the formula.

$\begin{array}{c}蠒=(5,0,-1)\\ u=\frac{1}{3}(1,\frac{蟺}{2},-3)\end{array}$

The equations become as follows:

$\begin{array}{c}\frac{d蠒}{du}=(5,0,-1)脳\frac{1}{3}(1,\frac{蟺}{2},-3)\\ \frac{d蠒}{du}=\frac{7}{3}\end{array}$

Hence the directional derivative is $\frac{7}{3}$.

The formula states the equation mentioned below.

$\begin{array}{c}\frac{鈭�F}{鈭�x}(x-x_0)+\frac{鈭�F}{鈭�y}(y-y_0)+\frac{鈭�F}{鈭�z}(z-z_0)=0\\ (2x_0-z_0)(x-x_0)+\cos y_0(y-y_0)+x_0(z-z_0)=0\end{array}$

Put the values mentioned below in the above formula.

$\begin{array}{c}{x}_0=1,\\ y_0=\frac{蟺}{2},\\ z_0=-3\end{array}$

The equation becomes as follow:

$\begin{array}{c}5(x-1)+0(y-\frac{蟺}{2})-(z+3)=0\\ 5x-5-z-3=0\\ 5x-z-8=0\\ 5x-z=8\end{array}$

Hence the equation of the tangent plane is $5x-z=8$.

Equation of normal line at point $(1,\frac{蟺}{2},-3)$.

$\begin{array}{c}\frac{x-x_0}{a}+\frac{y-y_0}{b}=\frac{z-z_0}{c}\\ \frac{x-1}{5}+\frac{z+3}{-1},y=\frac{蟺}{2}\end{array}$

Hence the equation of normal line to the surface is $\frac{x-1}{5}=\frac{z+3}{-1},y=\frac{蟺}{2}$.