다변수 미분 전체 문제

Let $C(A,B,\theta)=\sqrt{A^2+B^2-2AB\cos\theta}$. Then $C\big(3,4,\frac{\pi}{2}\big)=5$. Differentiating $C^2=A^2+B^2-2AB\cos\theta$ gives

$$\begin{aligned} 2C\frac{\partial C}{\partial A}(A,B,\theta)&=2A-2B\cos\theta &\quad&\implies\quad & 10\frac{\partial C}{\partial A}\big(3,4,\tfrac{\pi}{2}\big)&=6\\ 2C\frac{\partial C}{\partial B}(A,B,\theta)&=2B-2A\cos\theta & &\implies & 10\frac{\partial C}{\partial B}\big(3,4,\tfrac{\pi}{2}\big)&=8\\ 2C\frac{\partial C}{\partial \theta}(A,B,\theta)&=2AB\sin\theta & &\implies & 10\frac{\partial C}{\partial \theta}\big(3,4,\tfrac{\pi}{2}\big)&=24\\ \end{aligned}$$

Hence the approximate maximum error in the computed value of $C$ is

$$\begin{aligned} |\Delta C|&\approx \left| \frac{\partial C}{\partial A}\big(3,4,\tfrac{\pi}{2}\big)\Delta A +\frac{\partial C}{\partial B}\big(3,4,\tfrac{\pi}{2}\big)\Delta B +\frac{\partial C}{\partial \theta}\big(3,4,\tfrac{\pi}{2}\big)\Delta \theta\right| \\ &\le(0.6)(0.1)+(0.8)(0.1)+(2.4)\frac{\pi}{180}\\ &=\frac{\pi}{75}+0.14\le 0.182 \end{aligned}$$

Substituting $(x_0,y_0)=(3,4)$ and $(x,y)=(3.02,3.96)$ into

$$f(x,y)\approx f(x_0,y_0)+f_x(x_0,y_0)(x-x_0)+f_y(x_0,y_0)(y-y_0)$$

gives

$$\begin{aligned} f(3.02,3.96)&\approx f(3,4)+0.02f_x(3,4)-0.04f_y(3,4)\\ &=60+0.02\left(20+\frac{36}{5}\right)-0.04\left(15+\frac{48}{5}\right)\\ &=59.560 \end{aligned}$$

since

$$f_x(x,y)=y\sqrt{x^2+y^2}+\frac{x^2y}{\sqrt{x^2+y^2}}\qquad f_y(x,y)=x\sqrt{x^2+y^2}+\frac{xy^2}{\sqrt{x^2+y^2}}$$

(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

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

(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

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

(a) We are told that for all $x,y$ (with $(x,y,z)$ near $(2,-1/2,1)$), the function $z(x,y)$ obeys

$$\begin{aligned} z(x,y)^4 -xy^2 z(x,y)^2 +y=0 \tag{$*$}\end{aligned}$$

Differentiating $(*)$ with respect to $x$ gives

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

Similarly, differentiating this equation with respect to $y$ gives

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

(b) Substituting $(x, y, z) = (2, -1/2, 1)$ into the results of part (a) gives

$$\begin{aligned} \frac{\partial z}{\partial x}(2,-1/2) &= \frac{1/4}{4-1} =\frac{1}{12}\\ \frac{\partial z}{\partial y}(2,-1/2) &= \frac{-2-1}{4-1}=-1 \end{aligned}$$

(c) Under the linear approximation about $(2,-1/2)$,

$$\begin{aligned} f(x,y) &\approx f(2,-1/2) + f_x(2,-1/2)\,(x-2) + f_y(2,-1/2)\,(y+1/2) \\ &= 1 +\frac{1}{12}(x-2) - (y+0.5) \end{aligned}$$

In particular

$$\begin{aligned} f(1.94,-0.4) \approx 1 -\frac{0.06}{12}-0.1 \end{aligned}$$

so that

$$\begin{aligned} f(1.94,-0.4) - 1 \approx -0.105 \end{aligned}$$

(d) The tangent plane is

$$\begin{aligned} z&=f(2,-1/2) + f_x(2,-1/2)\,(x-2) + f_y(2,-1/2)\,(y+1/2) \\ &= 1 +\frac{1}{12}(x-2) - (y+0.5) \end{aligned}$$

or

$$\begin{aligned} \frac{x}{12} -y -z = -\frac{1}{3} \end{aligned}$$