접평면과 선형근사

(a) The linear approximation to $f(x,y)$ at $(1,3)$ is

$$\begin{aligned} f(x,y) \approx f(1,3) + f_x(1,3)\,(x-1) +f_y(1,3)\,(y-3) = 1 + 3(x-1) -2(y-3) \end{aligned}$$

So the change is $z$ is approximately

$$\begin{aligned} 3(1.2-1) -2(2.6-3) = 1.4 \end{aligned}$$

(b) The equation of the tangent plane is

$$\begin{aligned} z = f(1,3) + f_x(1,3)\,(x-1) +f_y(1,3)\,(y-3) = 1 + 3(x-1) -2(y-3) \end{aligned}$$

or

$$3x-2y-z = -4$$

(a) The surface has equation $G(x,y,z) = x^2 z^3 + y \sin(\pi x) + y^2 =0$. So a normal vector to the surface at $(1,1-1)$ is

$$\begin{aligned} { \boldsymbol{\nabla} } G(1,1,-1) &= \big[ \big(2x z^3 +\pi y\cos(\pi x)\big)\hat{\boldsymbol{\imath}} +\big(\sin(\pi x) +2y\big)\hat{\boldsymbol{\jmath}} +3z^2 x^2\,\hat{\mathbf{k}}\big]_{(x,y,z)=(1,1,-1)} \\ &= \big(-2-\pi\big)\hat{\boldsymbol{\imath}} +2\,\hat{\boldsymbol{\jmath}} +3\,\hat{\mathbf{k}} \end{aligned}$$

So the equation of the tangent plane is

$$\begin{aligned} \big(-2-\pi\big)(x-1) + 2(y-1) +3 (z+1) =0\quad\text{or}\quad -(2+\pi)x +2y +3z = -\pi-3 \end{aligned}$$

(b) The functions $z(x,y)$ obeys

$$\begin{aligned} x^2 z(x,y)^3 +y\sin(\pi x) +y^2 = 0 \end{aligned}$$

for all $x$ and $y$. Differentiating this equation with respect to $x$ gives

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

Evaluating at $(1,1,-1)$ gives

$$\begin{aligned} -2 + 3 \frac{\partial z}{\partial x}(1,1) -\pi =0 \implies \frac{\partial z}{\partial x}(1,1) = \frac{\pi+2}{3} \end{aligned}$$

(c) Using the linear approximation about $(x,y)=(1,1)$,

$$\begin{aligned} z(x,1) \approx z(1,1) +\frac{\partial z}{\partial x}(1,1)\ (x-1) \end{aligned}$$

gives

$$\begin{aligned} z(0.97,1) \approx -1 +\frac{\pi+2}{3}\ (-0.03) = -1 -\frac{\pi+2}{100} = -\frac{\pi+102}{100} \end{aligned}$$

(a) The function $F(y,z)$ obeys $F(y,z)^4+y^4+z^4+F(y,z)yz=17$ for all $y$ and $z$ near $y=1$, $z=2$. Applying the derivatives $\frac{\partial }{\partial y}$ and $\frac{\partial }{\partial z}$ to this equation gives

$$\begin{aligned} 4F(y,z)^3F_y(y,z)+4y^3+F_y(y,z)yz +F(y,z)z&=0 \\ 4F(y,z)^3F_z(y,z)+4z^3+F_z(y,z)yz +F(y,z)y&=0 \end{aligned}$$

Substiututing $F(1,2)=0$, $y=1$ and $z=2$ gives

$$\begin{aligned} 4+2F_y(1,2)&=0 & &\ \implies\ & &F_y(1,2)=-2 \\ 32+2F_z(1,2)&=0 & &\ \implies\ & & F_z(1,2)=-16 \end{aligned}$$

(b) Using the tangent plane to $x=F(y,z)$ at $y=1$ and $z=2$, which is

$$x \approx F(1,2) +F_y(1,2)\,(y-1) +F_z(1,2)\,(z-2)$$

with $y=1.01$ and $z=1.98$ gives

$$\begin{aligned} x=F(1.01, 1.98) &\approx F(1,2)+F_y(1,2)(1.01-1)+F_z(1,2)(1.98-2) \\ &=0-2(.01)-16(-0.02) =0.3 \end{aligned}$$