다변수 미분 전체 문제

There is a saddle point at $(\pi ,\pi ),$ local maxima at $\left(\frac{\pi }{2},\frac{\pi }{2}\right)\,\text{and}\,\left(\frac{3\pi }{2},\frac{3\pi }{2}\right),$ and local minima at $\left(\frac{\pi }{2},\frac{3\pi }{2}\right)\,\text{and}\,\left(\frac{3\pi }{2},\frac{\pi }{2}\right).$

Answers may vary

Level curves are lines $y = (3/2)x-c/2$.

Level curves are hyperbolas $\frac{x^2}{c}-\frac{y^2}{c}=1$, except for $c=0$, where the level curve is the pair of lines $y=x$, $y=-x$.

Level curves are parabolas $x=y^2+c$.

Level curves are hyperbolas $(x-c)^2-(y-c)^2=1$, drawn in graph in different styles to differentiate the curves.

When $c\neq 0$, the level curves are circles, centered at $(1/c,-1/c)$ with radius $\sqrt{2/c^2-1}$. When $c=0$, the level curve is the line $y=x$.

Level curves are cubics of the form $y=x^3+cx+1$. Note how each curve passes through $(0,1)$ and that the function is not defined at $x=0$.

Level curves are ellipses of the form $\frac{x^2}{c^2}+\frac{y^2}{c^2/4}=1$, i.e., $a=c$ and $b=c/2$.

Level curves are ellipses of the form $\frac{x^2}{c}+\frac{y^2}{c/4}=1$, i.e., $a=\sqrt{c}$ and $b=\sqrt{c}/2$.

Solution 1

Let's start by finding an equation for this surface. Every level curve is a horizontal circle of radius one, so the equation should be of the form

$$(x-f_1)^2+(y-f_2)^2=1$$

where $f_1$ and $f_2$ are functions depending only on $z$. Since the centre of the circle at height $z$ is at position $x=0$, $y=z$, we see that the equation of our surface is

$$x^2+(y-z)^2=1$$

The height of the surface at the point $(x,y)$ is the $z(x,y)$ found by solving that equation. That is,

$$x^2+\big(y-z(x,y)\big)^2=1 \tag{$*$}$$

We differentiate this equation implicitly to find $z_x(x,y)$ and $z_y(x,y)$ at the desired point $(x,y)= (0,-1)$. First, differentiating $(*)$ with respect to $y$ gives

$$\begin{aligned} 0+2\big(y-z(x,y)\big)\big(1- z_y(x,y)\big)&=0 \\ 2(-1-0)\big(1-z_y(0,-1)\big)&=0& & \text{ at } (0,-1,0) \end{aligned}$$

so that the slope looking in the positive $y$ direction is $z_y(0,-1)=1$. Similarly, differentiating $(*)$ with respect to $x$ gives

$$\begin{aligned} 2x+2\big(y-z(x,y)\big)\cdot\big(0-z_x(x,y)\big)&=0 \\ 2x&=2\big(y-z(x,y)\big)\cdot z_x(x,y)\\ z_x(x,y)&=\frac{x}{y-z(x,y)}\\ z_x(0,-1)&=0 &\text{ at } (0,-1,0) \end{aligned}$$

The slope looking in the positive $x$ direction is $z_x(0,-1)=0$.

Solution 2

Standing at $(0,-1,0)$ and looking in the positive $y$ direction, the surface follows the straight line that

• passes through the point $(0,-1,0)$, and
• is parallel to the central line $z=y, x=0$ of the cylinder.

Shifting the central line one unit in the $y$-direction, we get the line $z=y+1$, $x=0$. (As a check, notice that $(0,-1,0)$ is indeed on $z=y+1$, $x=0$.) The slope of this line is 1.

Standing at $(0,-1,0)$ and looking in the positive $x$ direction, the surface follows the circle $x^2+y^2=1$, $z=0$, which is the intersection of the cylinder with the $xy$-plane. As we move along that circle our $z$ coordinate stays fixed at $0$. So the slope in that direction is 0.

(a) The specified graph is $z=f(x,y)=x^2-y^2$ or $F(x,y,z)=x^2-y^2-z=0$. Observe that $f(-2,1)=3$. The vector

$$\begin{aligned} { \boldsymbol{\nabla} } F(-2,1,3) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< 2x,-2y,-1\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< -4,-2,-1\right> \end{aligned}$$

is a normal vector to the graph at $(-2,1,3)$. So the tangent plane is

$$-4(x+2)-2(y-1)-(z-3)=0\text{ or } 4x+2y+z=-3$$

and the normal line is

$$\left< x,y,z\right>=\left< -2,1,3\right>+t\left< 4,2,1\right>$$

(b) The specified graph is $z=f(x,y)=e^{xy}$ or $F(x,y,z)=e^{xy}-z=0$. Observe that $f(2,0)=1$. The vector

$$\begin{aligned} { \boldsymbol{\nabla} } F(2,0,1) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< ye^{xy},xe^{xy},-1\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< 0,2,-1\right> \end{aligned}$$

is a normal vector to the graph at $(2,0,1)$. So the tangent plane is

$$0(x-2)+2(y-0)-(z-1)=0\text{ or } 2y-z=-1$$

and the normal line is

$$\left< x,y,z\right>=\left< 2,0,1\right>+t\left< 0,2,-1\right>$$

We are going to need the first order partial derivatives of $f(x,y)=ye^{-x}$ at $(x,y)=(1,e)$. Here they are.

$$\begin{aligned} f_x(x,y)&= -ye^{-x} & f_x(1,e)&=-e\,e^{-1}=-1 \\ f_y(x,y)&=e^{-x} & f_y(1,e)&= e^{-1} \end{aligned}$$

(a) The linear approximation to $f(x,y)$ at $(x,y)=(1,e)$ is

$$\begin{aligned} f(x,y) \approx f(1,e) + f_x(1,e)\,(x-1) +f_y(1,e)\,(y-e) = 1 -(x-1) +e^{-1}(y-e) \end{aligned}$$

The maximum error is then approximately

$$\begin{aligned} -1(-0.1) +e^{-1}(0.1) =\frac{1+e^{-1}}{10} \end{aligned}$$

(b) The equation of the graph is $g(x,y,z) = f(x,y) -z =0$. Any vector that is a nonzero constant times

$$\begin{aligned} { \boldsymbol{\nabla} } g(1,e,1) =\left< f_x(1,e)\,,\,f_y(1,e)\,,\,-1\right> =\left< -1\,,\,e^{-1}\,,\,-1\right> \end{aligned}$$

is perpendicular to $g=0$ at $(1,e,1)$.

Since

$$\begin{aligned} f_x(2,1) & = -2x e^{-x^2 +4y^2}\Big|_{(x,y)=(2,1)} = -4 \\ f_y(2,1) & = \phantom{-}8y e^{-x^2 +4y^2}\Big|_{(x,y)=(2,1)} = 8 \end{aligned}$$

The tangent plane to $z=f(x,y)$ at $(2,1)$ is

$$\begin{aligned} z &= f(2,1) +f_x(2,1)\,(x-2) +f_y(2,1)\,(y-1) = 1 -4 (x-2) +8(y-1) \\ &= 1-4x +8y \end{aligned}$$

and the tangent plane approximation to the value of $f(1.99, 1.01)$ is

$$\begin{aligned} f(1.99, 1.01) \approx 1 -4 (1.99-2) +8(1.01-1) = 1.12 \end{aligned}$$