다변수 미분 전체 문제

(a) By inspection, when $y=0$, then $f(x,y)=1$ as long as $x \neq 0$. So, if we follow the $x$-axis in towards the origin, $f(x,y)=1$ along this route. (b) Also by inspection, when $x=0$, then $f(x,y)=0$ as long as $y \neq 0$. So, if we follow the $y$-axis in towards the origin, $f(x,y)=0$ along this route. (c) Since two different directions give us different values as we approach the origin, $\lim\limits_{(x,y)\to(0,0)}f(x,y)$ does not exist.

If $f_y(0,0)<0$, then $f(0,y)$ decreases as $y$ increases from $0$. Thus moving in the positive $y$ direction takes you downhill. This means you aren't at the lowest point in a valley, since you can still move downhill. On the other hand, as $f_y(0,0)<0$, $f(0,y)$ also decreases as $y$ increases towards $0$ from slightly negative values. Thus if you move in the negative $y$-direction from $y=0$, your height $z$ will increase. So you are not at a locally highest point—you're not at a summit.

(a) By definition

$$\begin{aligned} \frac{\partial f}{\partial x}(0,0) &=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)-f(0,0)}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}\frac{\frac{(\Delta x^2)(0)}{\Delta x^2+0^2}-0}{\Delta x} \\ &=0 \end{aligned}$$

(b) By definition

$$\begin{aligned} \frac{\partial f}{\partial y}(0,0) &=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)-f(0,0)}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}\frac{\frac{(0^2)(\Delta y)}{0^2+\Delta y^2}-0}{\Delta y} \\ &=0 \end{aligned}$$

(c) By definition

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} f(t,t)\Big|_{t=0} &=\lim_{t\rightarrow 0}\frac{f(t,t)-f(0,0)}{t} \\ &=\lim_{t\rightarrow 0}\frac{\frac{(t^2)(t)}{t^2+t^2}-0}{t} \\ &=\lim_{t\rightarrow 0}\frac{t/2}{t} \\ &=\frac{1}{2} \end{aligned}$$

No such $f(x,y)$ exists, because if it were to exist, then we would have that $f_{xy}(x,y)=f_{yx}(x,y)$. But

$$\begin{aligned} f_{xy}(x,y)&=\frac{\partial }{\partial y}f_x(x,y)=\frac{\partial }{\partial y}e^y=e^y \\ f_{yx}(x,y)&=\frac{\partial }{\partial x}f_y(x,y)=\frac{\partial }{\partial x}e^x=e^x \end{aligned}$$

are not equal.

The problem is that $\frac{\partial w}{\partial x}$ is used to represent two completely different functions in the same equation. The careful way to write the equation is the following. Let $f(x,y,z)$ and $g(x,y)$ be continuously differentiable functions and define $w(x,y)=f\big(x,y,g(x,y)\big)$. By the chain rule,

$$\begin{aligned} \frac{\partial w}{\partial x}(x,y) &=\frac{\partial f}{\partial x}\big(x,y,g(x,y)\big) \frac{\partial x}{\partial x} +\frac{\partial f}{\partial y}\big(x,y,g(x,y)\big) \frac{\partial y}{\partial x} +\frac{\partial f}{\partial z}\big(x,y,g(x,y)\big) \frac{\partial g}{\partial x}(x,y)\\ &=\frac{\partial f}{\partial x}\big(x,y,g(x,y)\big) +\frac{\partial f}{\partial z}\big(x,y,g(x,y)\big) \frac{\partial g}{\partial x}(x,y) \end{aligned}$$

While $w(x,y)=f\big(x,y,g(x,y)\big)$, it is not true that $\frac{\partial w}{\partial x}(x,y) =\frac{\partial f}{\partial x}\big(x,y,g(x,y)\big)$. For example, take $f(x,y,z)=x-z$ and $g(x,y)=x$. Then $w(x,y)=f\big(x,y,g(x,y)\big)=x-g(x,y)=0$ for all $(x,y)$, so that $\frac{\partial w}{\partial x}(x,y)=0$ while $\frac{\partial f}{\partial x}(x,y,z)=1$ for all $(x,y,z)$.

Write $F(x,y,z) = x^2+y^2+(z-1)^2-1$ and $G(x,y,z) = x^2+y^2+(z+1)^2-1$. Let $S_1$ denote the surface $F(x,y,z)=0$ and $S_2$ denote the surface $G(x,y,z)=0$. First note that $F(0,0,0)=G(0,0,0)=0$ so that the point $(0,0,0)$ lies on both $S_1$ and $S_2$. The gradients of $F$ and $G$ are

$$\begin{aligned} { \boldsymbol{\nabla} } F(x,y,z) &=\left<\frac{\partial F}{\partial x}(x,y,z)\,,\, \frac{\partial F}{\partial y}(x,y,z)\,,\, \frac{\partial F}{\partial z}(x,y,z)\right> =\left< 2x\,,\,2y\,,\,2(z-1)\right> \\ { \boldsymbol{\nabla} } G(x,y,z) &=\left<\frac{\partial G}{\partial x}(x,y,z)\,,\, \frac{\partial G}{\partial y}(x,y,z)\,,\, \frac{\partial G}{\partial z}(x,y,z)\right> =\left< 2x\,,\,2y\,,\,2(z+1)\right> \end{aligned}$$

In particular,

$${ \boldsymbol{\nabla} } F(0,0,0)=\left< 0,0,-2\right>\qquad { \boldsymbol{\nabla} } G(0,0,0)=\left< 0,0,2\right>$$

so that the vector $\hat{\mathbf{k}}=-\frac{1}{2}{ \boldsymbol{\nabla} } F(0,0,0) =\frac{1}{2}{ \boldsymbol{\nabla} } G(0,0,0)$ is normal to both surfaces at $(0,0,0)$. So the tangent plane to both $S_1$ and $S_2$ at $(0,0,0)$ is

$$\hat{\mathbf{k}}\cdot\left< x-0,y-0,z-0\right>=0\qquad\text{or}\qquad z=0$$

Denote by $P$ the plane $z=0$. Thus $S_1$ is tangent to $P$ at $(0,0,0)$ and $P$ is tangent to $S_2$ at $(0,0,0)$. So it is reasonable to say that $S_1$ and $S_2$ are tangent at $(0,0,0)$.

The partial derivatives, at a general point $(x,y,z)$ and also at the point of interest $(0,1,1)$, are

$$\begin{aligned} f_x(x,y,z)&=yz e^{xyz}\qquad & f_x(0,1,1)&= 1 \\ f_y(x,y,z)&=xz e^{xyz}\qquad & f_y(0,1,1)&= 0 \\ f_z(x,y,z)&=xy e^{xyz}\qquad & f_z(0,1,1)&= 0 \end{aligned}$$

So ${ \boldsymbol{\nabla} } f(0,1,1) =\left< 1,0,0\right>$ and the specified directional derivative is

$$\begin{aligned} D_{\frac{\left< 0,1,1\right>}{\sqrt{2}}}f(0,1,1) =\left< 1,0,0\right>\cdot\frac{\left< 0,1,1\right>}{\sqrt{2}} =0 \end{aligned}$$

In two dimensions, write $g(x,y) = y^2+\sin(xy)$. Then

$$\begin{aligned} { \boldsymbol{\nabla} } g =\left< g_x\,,\,g_y\right> =\left< y\cos(xy)\,,\, 2y + x\cos(xy) \right> \end{aligned}$$

In three dimensions, write $g(x,y,z) = y^2+\sin(xy)$. Then

$$\begin{aligned} { \boldsymbol{\nabla} } g =\left< g_x\,,\,g_y\,,\,g_z\right> =\left< y\cos(xy)\,,\, 2y + x\cos(xy)\,,\, 0 \right> \end{aligned}$$

We start by evaluating $u_t(x,t)$ and $u_{xx}(x,t)+u(x,t)$ when $u(x,t)= e^{-t-x^2}$.

$$\begin{aligned} u(x,t)&= e^{-t-x^2} \\ u_t(x,t)&= -e^{-t-x^2} \\ u_x(x,t)&= -2x e^{-t-x^2} \\ u_{xx}(x,t)&= -2 e^{-t-x^2} +4x^2 e^{-t-x^2} \end{aligned}$$

So

$$\begin{aligned} u_{xx}(x,t)+u(x,t) &= \big[-2 e^{-t-x^2} +4x^2 e^{-t-x^2}\big] +e^{-t-x^2} \\ &= \big[4x^2-1]e^{-t-x^2} \end{aligned}$$

For this to equal $g(x)\, u_t(x,t) = -g(x)\, e^{-t-x^2}$, we need $g(x) = 1-4x^2$.

(a) Fix any $y_0$ and set $v(x)=u(x,y_0)$. Then

$$\begin{aligned} \frac{\mathrm{d}v}{\mathrm{d}x}(x) =\frac{\partial u}{\partial x}(x,y_0) = 0 \end{aligned}$$

So, for each fixed $y_0$, $v(x)=u(x,y_0)$, which is a function of $x$, has to be a constant. The constant may be different for each different choice of $y_0$. So $u(x,y_0) = C(y_0)$ with $C(y_0)$ depending only on $y_0$, not on $x$. Or, renaming $y_0$ back to $y$, $u(x,y)=C(y)$ with $C(y)$ being any function of the single variable $y$.