접평면과 선형근사

(a) The specified graph is $z=f(x,y)=x^2-y^2$ or $F(x,y,z)=x^2-y^2-z=0$. Observe that $f(-2,1)=3$. The vector

$$\begin{aligned} { \boldsymbol{\nabla} } F(-2,1,3) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< 2x,-2y,-1\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< -4,-2,-1\right> \end{aligned}$$

is a normal vector to the graph at $(-2,1,3)$. So the tangent plane is

$$-4(x+2)-2(y-1)-(z-3)=0\text{ or } 4x+2y+z=-3$$

and the normal line is

$$\left< x,y,z\right>=\left< -2,1,3\right>+t\left< 4,2,1\right>$$

(b) The specified graph is $z=f(x,y)=e^{xy}$ or $F(x,y,z)=e^{xy}-z=0$. Observe that $f(2,0)=1$. The vector

$$\begin{aligned} { \boldsymbol{\nabla} } F(2,0,1) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< ye^{xy},xe^{xy},-1\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< 0,2,-1\right> \end{aligned}$$

is a normal vector to the graph at $(2,0,1)$. So the tangent plane is

$$0(x-2)+2(y-0)-(z-1)=0\text{ or } 2y-z=-1$$

and the normal line is

$$\left< x,y,z\right>=\left< 2,0,1\right>+t\left< 0,2,-1\right>$$