편도함수

The sign is negative.

The partial derivative is zero at the origin.

Answers may vary

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.