접평면과 선형근사

(a) The function $z(x,y)$ obeys

$$z(x,y)^3-z(x,y)+2xy-y^2=0$$

for all $(x,y)$ near $(2,4)$. Differentiating with respect to $x$ and $y$

$$\begin{aligned} 3z(x,y)^2\frac{\partial z}{\partial x}(x,y) -\frac{\partial z}{\partial x}(x,y)+2y&=0 \\ 3z(x,y)^2\frac{\partial z}{\partial y}(x,y) -\frac{\partial z}{\partial y}(x,y)+2x-2y&=0 \end{aligned}$$

Substituting in $x=2$, $y=4$ and $z(2,4)=1$ gives

$$\begin{aligned} 3\frac{\partial z}{\partial x}(2,4) -\frac{\partial z}{\partial x}(2,4)+8&=0 & &\iff & \frac{\partial z}{\partial x}(2,4) & = -4 \\ 3\frac{\partial z}{\partial y}(2,4) -\frac{\partial z}{\partial y}(2,4)+4-8&=0 & &\iff & \frac{\partial z}{\partial y}(2,4) & = 2 \end{aligned}$$

The linear approximation is

$$\begin{aligned} z(x,y)&\approx z(2,4)+z_x(2,4)(x-2)+z_y(2,4)(y-4)= 1-4(x-2)+2(y-4) \\ &=1-4x+2y \end{aligned}$$

(b) Substituting in $x=2.02$ and $y=3.96$ gives

$$\begin{aligned} z(2.02,3.96)\approx 1-4\times0.02+2\times(-0.04) =0.84 \end{aligned}$$