접평면과 선형근사

Let $(x_0,y_0,z_0)$ be any point on the surface. A vector normal to the surface at $(x_0,y_0,z_0)$ is

$$\begin{aligned} &{ \boldsymbol{\nabla} }\Big(xy e^{-(x^2+y^2)/2}-z\Big)\bigg|_{(x_0,y_0,z_0)} \\& =\left< y_0 e^{-(x_0^2+y_0^2)/2}-x_0^2y_0 e^{-(x_0^2+y_0^2)/2}, x_0 e^{-(x_0^2+y_0^2)/2}-x_0y_0^2 e^{-(x_0^2+y_0^2)/2},-1\right> \end{aligned}$$

The tangent plane to the surface at $(x_0,y_0,z_0)$ is horizontal if and only if this vector is vertical, which is the case if and only if its $x$- and $y$-components are zero, which in turn is the case if and only if

$$\begin{aligned} &y_0(1-x_0^2)=0\text{ and }x_0(1-y_0^2)=0\\ &\iff\big\{y_0=0\text{ or }x_0=1\text{ or }x_0=-1\big\} \text{ and }\big\{x_0=0\text{ or }y_0=1\text{ or }y_0=-1\big\}\\ &\iff (x_0,y_0)=(0,0)\text{ or }(1,1)\text{ or }(1,-1) \text{ or }(-1,1)\text{ or }(-1,-1) \end{aligned}$$

The values of $z_0$ at these points are $0$, $e^{-1}$, $-e^{-1}$, $-e^{-1}$ and $e^{-1}$, respectively. So the horizontal tangent planes are $z=0$, $z=e^{-1}$ and $z=-e^{-1}$. At the highest and lowest points of the surface, the tangent plane is horizontal. So the largest and smallest values of $z$ are $e^{-1}$ and $-e^{-1}$, respectively.

(a) A normal vector to the surface at $(0,2,1)$ is

$$\begin{aligned} { \boldsymbol{\nabla} }\big(xy-2x+yz+x^2+y^2+z^3-7\big)\big|_{(0,2,1)} &=\left< y-2+2x\,,\,x+z+2y\,,\,y+3z^2\right>\big|_{(0,2,1)}\\ &=\left< 0, 5, 5\right> \end{aligned}$$

So the tangent plane is

$$\begin{aligned} 0(x-0)+5(y-2)+5(z-1)=0\text{ or }y+z=3 \end{aligned}$$

The vector parametric equations for the normal line are

$$\mathbf{r}(t)=\left< 0,2,1\right>+t\left< 0,5,5\right>$$

(b) Differentiating

$$xy-2x+y\,z(x,y)+x^2+y^2+z(x,y)^3=7$$

gives

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

In particular, at $(0,2,1)$, $z_y(0,2)=-\frac{4+1}{2+3}=-1$.

(c) Differentiating $z_x$ with respect to $y$ gives

$$\begin{aligned} z_{xy}(x,y) &=-\frac{1}{y+3z(x,y)^2} -\frac{2-2x-y}{{[y+3z(x,y)^2]}^2}\big(1+6z(x,y)z_y(x,y)\big)\\ &=-\frac{1}{y+3z(x,y)^2} -\frac{2-2x-y}{{[y+3z(x,y)^2]}^2} \left(1-6z(x,y)\frac{x+2y+z(x,y)}{y+3z(x,y)^2}\right) \end{aligned}$$

As an alternate solution, we could also differentiate $z_y$ with respect to $x$. This gives

$$\begin{aligned} z_{yx}(x,y) &=-\frac{1+z_x(x,y)}{y+3z(x,y)^2} +\frac{x+2y+z(x,y)}{{[y+3z(x,y)^2]}^2}6z(x,y)z_x(x,y)\\ &=-\frac{1}{y+3z(x,y)^2}\left(1+\frac{2-2x-y}{y+3z(x,y)^2}\right) +\frac{x+2y+z(x,y)}{{[y+3z(x,y)^2]}^2}6z(x,y)\frac{2-2x-y}{y+3z(x,y)^2} \end{aligned}$$

(a) A vector perpendicular to $x^2+z^2=10$ at $(1,1,3)$ is

$${ \boldsymbol{\nabla} }(x^2+z^2)\big|_{(1,1,3)} =(2x\hat{\boldsymbol{\imath}}+2z\hat{\mathbf{k}})\big|_{(1,1,3)} =2\hat{\boldsymbol{\imath}}+6\hat{\mathbf{k}}\text{ or } \frac{1}{2} \left< 2,0,6\right>=\left< 1,0,3\right>$$

(b) A vector perpendicular to $y^2+z^2=10$ at $(1,1,3)$ is

$${ \boldsymbol{\nabla} }(y^2+z^2)\big|_{(1,1,3)} =(2y\hat{\boldsymbol{\jmath}}+2z\hat{\mathbf{k}})\big|_{(1,1,3)} =2\hat{\boldsymbol{\jmath}}+6\hat{\mathbf{k}}\text{ or }\frac{1}{2} \left< 0,2,6\right>=\left< 0,1,3\right>$$

A vector is tangent to the specified curve at the specified point if and only if it perpendicular to both $(1,0,3)$ and $(0,1,3)$. One such vector is

$$\left< 0,1,3\right>\times\left<1,0,3\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 0 & 1 & 3 \\ 1 & 0 & 3 \end{matrix}\right] =\left< 3,3,-1\right>$$

(c) The specified tangent line passes through $(1,1,3)$ and has direction vector $\left< 3,3,-1\right>$ and so has vector parametric equation

$$\mathbf{r}(t)=\left< 1,1,3\right>+t\left< 3,3,-1\right>$$

$\mathbf{r}(t)=\left< x(t)\,,\,y(t)\,,\,z(t)\right>$ intersects $z^3 + xyz -2 = 0$ when

$$z(t)^3+x(t)\,y(t)\,z(t)-2=0\iff \big(t^2\big)^3 + \big(t^3)(t)\big(t^2\big)-2=0 \iff 2t^6=2\iff t=1$$

since $t$ is required to be positive. The direction vector for the curve at $t=1$ is

$$\mathbf{r}'(1)=3\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+2\,\hat{\mathbf{k}}$$

A normal vector for the surface at $\mathbf{r}(1)=\left< 1,1,1\right>$ is

$${ \boldsymbol{\nabla} }(z^3+xyz)\big|_{(1,1,1)}=[yz\hat{\boldsymbol{\imath}}+xz\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}}]_{(1,1,1)} =\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+4\hat{\mathbf{k}}$$

The angle $\theta$ between the curve and the normal vector to the surface is determined by

$$\begin{aligned} \big|\left< 3,1,2\right>\big|\,\big|\left< 1,1,4\right>\big|\cos\theta =\left< 3,1,2\right> \cdot\left< 1,1,4\right> &\iff \sqrt{14}\sqrt{18}\cos\theta=12 \\ &\iff \sqrt{7\times 36}\cos\theta=12 \\ &\iff \cos\theta=\frac{2}{\sqrt{7}} \\ &\iff \theta=40.89^\circ \end{aligned}$$

The angle between the curve and the surface is $90-40.89=49.11^\circ$ (to two decimal places).

(a) We are told that

$$z(x,y)^3 - xy\,z(x,y)^2 - 4x = 0$$

for all $(x,y)$ (sufficiently near $(1,1)$). Differentiating this equation with respect to $x$ gives

$$\begin{aligned} & 3z(x,y)^2\,\frac{\partial z}{\partial x}(x,y) -y\, z(x,y)^2 -2xy\,z(x,y)\frac{\partial z}{\partial x}(x,y) - 4=0 \\ \implies &\frac{\partial z}{\partial x} = \frac{4+yz^2}{3z^2-2xyz} \end{aligned}$$

and differentiating with respect to $y$ gives

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

(b) When $(x,y,z)=(1,1,2)$,

$$\begin{aligned} \frac{\partial z}{\partial x}(1,1) = \frac{4+(1)(2)^2}{3(2)^2-2(1)(1)(2)} = 1\qquad \frac{\partial z}{\partial y}(1,1) = \frac{(1)(2)^2}{3(2)^2-2(1)(1)(2)} = \frac{1}{2} \end{aligned}$$

(c) Under the linear approximation at $(1,1)$

$$z(x,y) \approx z(1,1) + z_x(1,1)\,(x-1) + z_y\,(1,1)\,(y-1) = 2 + (x-1) +\frac{1}{2}(y-1)$$

So errors of $\pm 0.03$ in $x$ and $\pm 0.02$ in $y$ leads of errors of about

$$\pm\left[0.03 + \frac{1}{2}(0.02)\right] =\pm 0.04$$

in $z$.

(d) By the chain rule

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}\theta} z\big(x(\theta),y(\theta)\big) &= z_x\big(x(\theta),y(\theta)\big)\,x'(\theta) +z_y\big(x(\theta),y(\theta)\big)\,y'(\theta) \\ &=-z_x\big(1 + \cos\theta,\sin\theta\big)\,\sin\theta +z_y\big(1 + \cos\theta,\sin\theta\big)\,\cos\theta \end{aligned}$$

At $A$, $x=2$, $y=0$, $z=2$ (since $z^3-(2)(0)z^2-4(2)=0$) and $\theta=0$, so that

$$\begin{aligned} \frac{\partial z}{\partial x}(2,0) = \frac{4+(0)(2)^2}{3(2)^2-2(2)(0)(2)} = \frac{1}{3}\qquad \frac{\partial z}{\partial y}(2,0) = \frac{(2)(2)^2}{3(2)^2-2(2)(0)(2)} = \frac{2}{3} \end{aligned}$$

and

$$\begin{aligned} \frac{\mathrm{d}z}{\mathrm{d}\theta} &=-\frac{1}{3}\sin(0) +\frac{2}{3}\cos(0) =\frac{2}{3} \end{aligned}$$

At $B$, $x=1$, $y=1$, $z=2$ and $\theta=\frac{\pi}{2}$, so that, by part (b),

$$\begin{aligned} \frac{\partial z}{\partial x}(1,1) = 1\qquad \frac{\partial z}{\partial y}(1,1) = \frac{1}{2} \end{aligned}$$

and

$$\begin{aligned} \frac{\mathrm{d}z}{\mathrm{d}\theta} &=-\sin\frac{\pi}{2} +\frac{1}{2}\cos\frac{\pi}{2} = -1 \end{aligned}$$