다변수 미분 전체 문제

Think of the volume as being the function $V(p,T)$ of pressure and temperature that is determined implicitly (at least for $p\approx 1$, $T\approx 5$ and $V\approx 2$) by the equation

$$\big(pV(p,T)^2 + 16\big)\big(V(p,T) - 1\big) = T V(p,T)^2 \tag{$*$}$$

To determine the approximate change in $V$, we will use the linear approximation to $V(p,T)$ at $p=1$, $T=5$. So we will need the partial derivatives $V_p(1,5)$ and $V_T(1,5)$. As the equation $(*)$ is valid for all $p$ near $1$ and $T$ near 5, we may differentiate $(*)$ with respect to $p$, giving

$$\begin{aligned} &\big(V^2 +2 pV V_p\big)\big(V - 1\big) + \big(pV ^2 + 16\big)V_p = 2T V V_p \end{aligned}$$

and we may also differentiate $(*)$ with respect to $T$, giving

$$\begin{aligned} &\big(2 pV V_T\big)\big(V - 1\big) + \big(pV ^2 + 16\big)V_T = V^2 + 2T V V_T \end{aligned}$$

In particular, when $p=1$, $V=2$, $T=5$,

$$\begin{aligned} \big(4 +4 V_p(1,5)\big)\big(2 - 1\big) + \big(4 + 16\big)V_p(1,5) &= 20 V_p(1,5) &\quad\implies V_p(1,5)&=-1 \\ 4 V_T(1,5)\big(2 - 1\big) + \big(4 + 16\big)V_T(1,5) &= 4 + 20 V_T(1,5) &\quad\implies V_T(1,5)&=1 \end{aligned}$$

so that the change in $V$ is

$$\begin{aligned} V(1.2\,,\,5.3)-V(1,5) \approx V_p(1,5)\,(0.2) + V_T(1,5)\,(0.3) = -0.2+0.3 = 0.1 \end{aligned}$$

We have

$${ \boldsymbol{\nabla} } h(x,y)= -200 e^{-(x^2+2y^2)}\left< x,2y\right>\text{ and, in particular, } { \boldsymbol{\nabla} } h(3,2)= -200 e^{-17}\left< 3,4\right>$$

(a) At $(3,2)$ the dune slopes downward the most steeply in the direction opposite ${ \boldsymbol{\nabla} } h(3,2)$, which is (any positive multiple of) $\left< 3,4\right>$.

(b) The rate is $D_{\hat{\boldsymbol{\jmath}}} h(3,2)={ \boldsymbol{\nabla} } h(3,2)\cdot\hat{\boldsymbol{\jmath}}=-800 e^{-17}$.

(c) To remain at the same height, you should walk perpendicular to ${ \boldsymbol{\nabla} } h(3,2)$. So you should walk in one of the directions $\pm\big(\frac{4}{5},-\frac{3}{5}\big)$.

(d) Suppose that you are walking along a steepest descent curve. Then the direction from $(x,y)$ to $(x+\mathrm{d}x, y+\mathrm{d}y)$, with $(\mathrm{d}x,\mathrm{d}y)$ infinitesmal, must be opposite to ${ \boldsymbol{\nabla} } h(x,y)= -200 e^{-(x^2+2y^2)}(x,2y)$. Thus $(\mathrm{d}x,\mathrm{d}y)$ must be parallel to $(x,2y)$ so that the slope

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2y}{x} \implies \frac{\mathrm{d}y}{y}=2\frac{\mathrm{d}x}{x} \implies \ln y=2\ln x+C \end{aligned}$$

We must choose $C$ to obey $\ln 2=2\ln 3+C$ in order to pass through the point $(3,2)$. Thus $C=\ln\frac{2}{9}$ and the curve is $\ln y=2\ln x+\ln\frac{2}{9}$ or $y=\frac{2}{9}x^2$.

(a) The maximum and minimum can occur either in the interior of the disk or on the boundary of the disk. The boundary of the disk is the circle $x^2+y^2=4$.

• Any absolute max or min in the interior of the disk must also be a local max or min and, since there are no singular points, must also be a critical point of $h$. Since $T_x=-8x$ and $T_y=-2y$, the only critical point is $(x,y)=(0,0)$, where $T=20$. Since $4x^2+y^2\ge 0$, we have $T(x,y)=20-4x^2-y^2\le 20$. So the maximum value of $T$ (even in $\mathbb{R}^2$) is $20$.
• At each point of $x^2+y^2=4$ we have $T(x,y)=20-4x^2-y^2=20 -4x^2-(4-x^2)=16-3x^2$ with $-2\le x\le 2$. So $T$ is a minimum when $x^2$ is a maximum. Thus the minimum value of $T$ on the disk is $16-3(\pm 2)^2=4$.

So all together, the maximum and minimum values of $T(x,y)$ in $x^2+y^2\le 4$ are $20$ (at $(0,0)$) and $4$ (at $(\pm 2,0)$), respectively.

(b) To increase its temperature as quickly as possible, the ant should move in the direction of the temperature gradient ${ \boldsymbol{\nabla} } T(1,1)=\left< -8x,-2y\right>\Big|_{(x,y)=(1,1)}=\left< -8,-2\right>$. A unit vector in that direction is $\frac{1}{\sqrt{17}}\left< -4,-1\right>$.

(c) The ant's rate f increase of temperature (per unit time) is

$$\begin{aligned} { \boldsymbol{\nabla} } T(1,1)\cdot\left< -2,-1\right> =\left< -8,-2\right>\cdot\left< -2,-1\right> =18 \end{aligned}$$

(d) We are being asked to find the $(x,y)=(x,2-x^2)$ which maximizes

$$\begin{aligned} T\big(x,2-x^2\big) =20 -4x^2-\big(2-x^2\big)^2 = 16-x^4 \end{aligned}$$

The maximum of $16-x^4$ is obviously $16$ at $x=0$. So the ant should go to $\big(0,2-0^2\big)=(0,2)$.