다변수 미분 전체 문제

(a) Since, in polar coordinates,

$$\frac{xy}{x^2+y^2}=\frac{r^2\cos\theta\sin\theta}{r^2} =\cos\theta\sin\theta$$

we have that the limit does not exist. For example,

• if we send $(x,y)$ to $(0,0)$ along the positive $x$-axis, so that $\theta=0$, we get the limit $\sin\theta\cos\theta\big|_{\theta=0}=0$,
• while if we send $(x,y)$ to $(0,0)$ along the line $y=x$ in the first quadrant, so that $\theta=\frac{\pi}{4}$, we get the limit $\sin\theta\cos\theta\big|_{\theta=\pi/4}=\frac{1}{2}$.

(b) This limit does not exist, since if it were to exist the limit

$$\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{x^2+y^2} =\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{\sin(xy)}\ \frac{\sin(xy)}{x^2+y^2} =\lim_{(x,y)\rightarrow(0,0)}\frac{xy}{\sin(xy)}\ \lim_{(x,y)\rightarrow(0,0)}\frac{\sin(xy)}{x^2+y^2}$$

would also exist. (Recall that $\displaystyle \lim_{t\rightarrow 0}\tfrac{\sin t}{t} =1$.)

(c) Since

$$\begin{aligned} \lim_{(x,y)\rightarrow(-1,1)}\big[x^2+2xy^2+y^4\big] &=(-1)^2+2(-1)(1)^2+(1)^4=0 \\ \lim_{(x,y)\rightarrow(-1,1)}\big[1+y^4\big] &=1+(1)^4=2 \end{aligned}$$

and the second limit is nonzero,

$$\lim_{(x,y)\rightarrow(-1,1)}\frac{x^2+2xy^2+y^4}{1+y^4}=\frac{0}{2}=0$$

(d) Since the limit along the positive $x$-axis

$$\lim_{\genfrac{}{}{0pt}{}{t\rightarrow 0}{t>0}}|y|^x\Big|_{(x,y)=(t,0)} =\lim_{\genfrac{}{}{0pt}{}{t\rightarrow 0}{t>0}}0^t =\lim_{\genfrac{}{}{0pt}{}{t\rightarrow 0}{t>0}}0 =0$$

and the limit along the $y$-axis

$$\lim_{t\rightarrow 0}|y|^x\Big|_{(x,y)=(0,t)} =\lim_{t\rightarrow 0}|t|^0 =\lim_{t\rightarrow 0}1 =1$$

are different, the limit as $(x,y)\rightarrow 0$ does not exist.

As $z(x,y)=f(x^2+y^2)$

$$\begin{aligned} \frac{\partial z}{\partial x}(x,y)&=2xf'(x^2+y^2) \\ \frac{\partial z}{\partial y}(x,y)&=2yf'(x^2+y^2) \end{aligned}$$

by the (ordinary single variable) chain rule. So

$$y\frac{\partial z}{\partial x}-x\frac{\partial z}{\partial y} =y(2x)f'(x^2+y^2)-x(2y)f'(x^2+y^2)=0$$

and the differential equation is always satisfied, assuming that $f$ is differentiable, so that the chain rule applies.

By definition

$$\begin{aligned} \frac{\partial f}{\partial x}(0,0) &=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)-f(0,0)}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}\frac{\frac{(\Delta x+2\times 0)^2}{\Delta x+0}-0}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}\frac{\Delta x}{\Delta x} \\ &=1 \end{aligned}$$

and

$$\begin{aligned} \frac{\partial f}{\partial y}(0,0) &=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)-f(0,0)}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}\frac{\frac{(0+2\Delta y)^2}{0+\Delta y}-0}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}\frac{4\Delta y}{\Delta y} \\ &=4 \end{aligned}$$

(b) $f(x,y)$ is not continuous at $(0,0)$, even though both partial derivatives exist there. To see this, make a change of coordinates from $(x,y)$ to $(X,y)$ with $X=x+y$ (the denominator). Of course, $(x,y)\rightarrow (0,0)$ if and only if $(X,y)\rightarrow (0,0)$. Now watch what happens when $(X,y)\rightarrow(0,0)$ with $X$ a lot smaller than $y$. For example, $X=ay^2$. Then

$$\begin{aligned} \frac{(x+2y)^2}{x+y}=\frac{(X+y)^2}{X}=\frac{(ay^2+y)^2}{ay^2} =\frac{(1+ay)^2}{a}\rightarrow\frac{1}{a} \end{aligned}$$

This depends on $a$. So approaching $(0,0)$ along different paths gives different limits. (You can see the same effect without changing coordinates by sending $(x,y)\rightarrow (0,0)$ with $x=-y+ay^2$.) Even more dramatically, watch what happens when $(X,y)\rightarrow(0,0)$ with $X=y^3$. Then

$$\frac{(x+2y)^2}{x+y}=\frac{(X+y)^2}{X}=\frac{(y^3+y)^2}{y^3} =\frac{{(1+y^2)}^2}{y}\rightarrow\pm\infty$$

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

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