다변수 미분 전체 문제

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.

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

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

(a) The linear approximation to $f(x,y)$ at $(a,b)$ is

$$\begin{aligned} f(x,y) &\approx f(a,b) + f_x(a,b)\,(x-a) + f_y(a,b)\,(y-b) \\ &= \ln(4a^2 + b^2) + \frac{8a}{4a^2+b^2}\, (x-a) + \frac{2b}{4a^2+b^2}\, (y-b) \end{aligned}$$

In particular, for $a=0$ and $b=1$,

$$\begin{aligned} f(x,y) &\approx 2\, (y-1) \end{aligned}$$

and, for $x=0.1$ and $y=1.2$,

$$\begin{aligned} f(0.1,1.2) &\approx 0.4 \end{aligned}$$

(b) The point $(a,b,c)$ is on the surface $z=f(x,y)$ if and only if

$$c = f(a,b) = \ln(4a^2 + b^2)$$

Note that this forces $4a^2 + b^2$ to be nonzero. The tangent plane to the surface $z = f (x, y)$ at the point $(a,b,c)$ is parallel to the plane $2x + 2y - z = 3$ if and only if $\left< 2\,,\,2\,,\,-1\right>$ is a normal vector for the tangent plane. That is, there is a nonzero number $t$ such that

$$\begin{aligned} \left< 2\,,\,2\,,\,-1\right> =t\left< f_x(a,b)\,,\,f_y(a,b)\,,\ -1\right> =t\left< \frac{8a}{4a^2+b^2} \,,\, \frac{2b}{4a^2+b^2} \,,\,-1\right> \end{aligned}$$

For the $z$–coordinates to be equal, $t$ must be $1$. Then, for the $x$– and $y$–coordinates to be equal, we need

$$\begin{aligned} \frac{8a}{4a^2+b^2} &= 2 \\ \frac{2b}{4a^2+b^2} &= 2 \end{aligned}$$

Note that these equations force both $a$ and $b$ to be nonzero. Dividing these equations gives $\frac{8a}{2b}=1$ and hence $b=4a$. Substituting $b=4a$ into either of the two equations gives

$$\begin{aligned} \frac{8a}{20a^2}=2 \implies a=\frac{1}{5} \end{aligned}$$

So $a=\frac{1}{5}$, $b=\frac{4}{5}$ and

$$\begin{aligned} c=\ln\left(\frac{4}{5^2}+\frac{4^2}{5^2}\right) =\ln\frac{4}{5} \end{aligned}$$

(a) Since

$$z=-\frac{100}{x^2+2x+4y^2+11}=-\frac{100}{(x+1)^2+4y^2+10}$$

the bottom of the crater is at $x=-1$, $y=0$ (where the denominator is a minimum) and the contours (level curves) are ellipses having equations $(x+1)^2+4y^2=C$. In the sketch below, the filled dot represents the bottom of the crater and the open dot represents the car park. The contours sketched are (from inside out) $z=-7.5, -5, -2.5, -1$. Note that the trail crosses the contour lines at right angles.

(b) The trail is to be parallel to $${ \boldsymbol{\nabla} } z = \frac{100}{{(x^2+2x+4y^2+11)}^2}(2x+2,8y)$$ At the car park ${ \boldsymbol{\nabla} } z(4,5)\parallel \left< 10, 40\right>\parallel \left< 1,4\right>$. To move towards the bottom of the crater, we should leave in the direction $-\left< 1,4\right>$.