91影视

Given $f(x,y)=x^2鈭�y^2$

At point$P=\left(x_0,y_0\right)=(2,3)\text{and}u=(伪,尾)=\left(\frac{{\displaystyle \sqrt{2}}}{{\displaystyle 2}},\frac{{\displaystyle \sqrt{2}}}{{\displaystyle 2}}\right)$

The equation of line of tangent is

$\begin{array}{r}{f}_x\left(x_0,y_0\right)\left(x鈭�x_0\right)+f_y\left(x_0,y_0\right)\left(y鈭�y_0\right)=z鈭�f\left(x_0,y_0)\right.\end{array}$

$f_x(2,3)(x鈭�2)+f_y(2,3)(y鈭�3)=z鈭�f(2,3)$

Assume

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

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

$f_x(2,3)=4$ $\left(2\right)$

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

$f_y(2,3)=鈭�{\left.2y\right|}_{(2,3)}$

$f_y(2,3)=鈭�6$ $\left(3\right)$

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

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

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

$4(x鈭�2)鈭�6(y鈭�3)=z+5$

$4x鈭�8鈭�6y+18=z+5$

$4x鈭�6y鈭�z=5鈭�18+8$

$4x鈭�6y鈭�z=鈭�5$

Here

Consider the equation of normal line is

Now

$x=x_0+伪t,y=y_0+尾t,z=z_0+纬t$

Where

$z_0=f\left(x_0+伪t\right),y=y_0+尾z=z_0+纬t$

Consider directional derivative

$D_{n}f\left(x_0,y_0\right)={\mathrm{Lim}}_{h鈫�0}鈦�\frac{f\left(x_0+伪h,y_0+尾h\right)鈭�f\left(x_0,y_0\right)}{h}$

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

therefore

$f\left(2+\frac{\sqrt{2}}{2}h,3+\frac{\sqrt{2}}{2}h\right)={\left(2+\frac{\sqrt{2}}{2}h\right)}^2鈭�{\left(3+\frac{\sqrt{2}}{2}h\right)}^2$

$=\left(4+2\sqrt{2}h+\frac{1}{2}{h}^2\right)鈭�\left(9+3\sqrt{2}h+\frac{1}{2}{h}^2\right)$

$=鈭�5鈭�\sqrt{2}h$

$f\left(x_0,y_0\right)=2^2鈭�3^2=鈭�5$

Substituting

$D_{k}f(2,3)={\mathrm{Lim}}_{h鈫�0}鈦�\frac{鈭�5鈭�\sqrt{2}h+5}{h}$

$={\mathrm{Lim}}_{h鈫�0}鈭�\sqrt{2}$

$D_{n}f(2,3)=鈭�\sqrt{2}$

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