91影视

$f(x,y)=x^2鈭�y^2$

At point $P=\left(x_0,y_0\right)=(2,3)\text{and}u=(伪,尾)=\frac{3}{5}i鈭�\frac{4}{5}j$

The equation of line tangent

$\begin{array}{r}{f}_x(2,3)(x鈭�2)+f_y(2,3)(y鈭�3)=z鈭�f(2,3)\end{array}$ $\left(1\right)$

Consider

localid="1650356170652" $f_x\left(x_0,y_0\right)={\left.\frac{d}{dx}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dx}\left(x^2鈭�y^2\right)\right|}_{\left(x_0,y_0\right)}$

localid="1650356178311" $f_x(2,3)={\left.2x\right|}_{(2,3)}$

localid="1650356183164" $f_x(2,3)=4$ localid="1650356189103" $\left(2\right)$

$\begin{array}{r}{f}_y\left(x_0,y_0\right)={\left.\frac{d}{dy}f(x,y)\right|}_{\left(x_0,y_0\right)}={\left.\frac{d}{dy}\left(x^2鈭�y^2\right)\right|}_{\left(x_0,y_0\right)}\end{array}$

$\begin{array}{r}{f}_y(2,3)=鈭�{\left.2y\right|}_{(2,3)}\end{array}$

$\begin{array}{r}{f}_y(2,3)=鈭�4\end{array}$ $\left(3\right)$

$\begin{array}{r}f\left(x_0,y_0\right)=f(2,3)=\left(2^2鈭�3^2\right)\end{array}$

$\begin{array}{r}f(2,3)=鈭�5\end{array}$ $\left(4\right)$

Substituting $\left(2\right),\left(3\right),\left(4\right)$in $\left(1\right)$

$\begin{array}{r}4(x鈭�2)鈭�6(y鈭�3)=z+5\end{array}$

$\begin{array}{r}4x鈭�8鈭�6y+18=z+5\end{array}$

$\begin{array}{r}4x鈭�6y鈭�z=5鈭�18+8\end{array}$

$\begin{array}{r}4x鈭�6y鈭�z=鈭�5\end{array}$

The equation of normal line is

Now

$x=x_0+伪t,y=y_0+尾t,z=z_0+纬t$

Where

$z_0=f\left(x_0,y_0\right)\text{and}纬=D_{w}f\left(x_0,y_0\right)$

Consider directional derivative

$\begin{array}{r}{D}_{v}f\left(x_0,y_0\right)={\mathrm{Lim}}_{h鈫�0}鈦�\frac{f\left(x_0+伪h,y_0+尾h\right)鈭�f\left(x_0,y_0\right)}{h}\end{array}$

$\begin{array}{r}{D}_{v}f(2,3)={\mathrm{Lim}}_{h鈫�0}鈦�\frac{f\left(2+\frac{3}{5}h,3鈭�\frac{4}{5}h\right)鈭�f(2,3)}{h}\end{array}$ $\left(5\right)$

Therefore,

$\begin{array}{r}f\left(2+\frac{3}{5}h,3鈭�\frac{4}{5}h\right)={\left(2+\frac{3}{5}h\right)}^2鈭�{\left(3鈭�\frac{4}{5}h\right)}^2\end{array}$

$\begin{array}{r}=\left(4+\frac{12}{5}h+\frac{9}{25}{h}^2\right)鈭�\left(9鈭�\frac{24}{5}h+\frac{16}{25}{h}^2\right)\end{array}$

$\begin{array}{r}=鈭�5+\frac{36}{5}h鈭�\frac{7}{25}{h}^2\end{array}$ $\left(6\right)$

And

$f\left(x_0,y_0\right)=2^2鈭�3^2=鈭�5$ $\left(7\right)$

Substituting $\left(6\right),\left(7\right)$in $\left(1\right)$

$\begin{array}{r}{D}_{w}f(2,3)={\mathrm{Lim}}_{h鈫�0}鈦�\frac{鈭�5+\frac{36}{5}h鈭�\frac{7}{25}{h}^2+5}{h}\end{array}$

$\begin{array}{r}={\mathrm{Lim}}_{h鈫�0}鈦�\frac{36}{5}鈭�\frac{7}{25}h\end{array}$

$\begin{array}{r}{D}_{v}f(2,3)=\frac{36}{5}\end{array}$

$x=2+\frac{3}{5}t,y=3鈭�\frac{4}{5}t,z=鈭�5+\frac{36}{5}t$