방향도함수와 기울기벡터

The first order partial derivatives of $f$, both at a general point $(x,y,z)$ and at the point $(1, 0, \pi/2)$, are

$$\begin{aligned} f_x(x,y,z)&= 2x\qquad & f_x(1, 0, \pi/2)&= 2 \\ f_y(x,y,z)&= -z\sin(yz)\qquad & f_y(1, 0, \pi/2)&= 0 \\ f_z(x,y,z)&= -y\sin(yz)\qquad & f_z(1, 0, \pi/2)&= 0 \end{aligned}$$

(a) The rate of increase of $f$ is largest in the direction of ${ \boldsymbol{\nabla} } f(1, 0, \pi/2)=\left< 2,0,0\right>$. A unit vector in that direction is $\hat{\boldsymbol{\imath}}$.

(b) The gradient vector ${ \boldsymbol{\nabla} } f(1, 0, \pi/2)=\left< 2,0,0\right>$ is a normal vector to the surface $f=1$ at $(1, 0, \pi/2)$. So the specified tangent plane is

$$\begin{aligned} \left< 2,0,0\right> \cdot \left< x-1\,,\,y-0\,,\,z-\pi/2\right>=0\qquad\text{or}\qquad x=1 \end{aligned}$$

(c) The vector from the point $(0,1,0)$ to the point $(1,1,0)$, on $T$, is $\left< 1,0,0\right>$, which is perpendicular to $T$. So $(1,1,0)$ is the point on $T$ nearest $(0,1,0)$ and the distance from $(0,1,0)$ to $T$ is $|\left< 1,0,0\right>|=1$.

(d) The vector $\left< 1,0,1\right>$ is perpendicular to the plane $x+z=0$. So the angle between the planes $T$ and $x+z=0$ is the same as the angle $\theta$ between the vectors $\left< 1,0,0\right>$ and $\left< 1,0,1\right>$, which obeys

$$\begin{aligned} & |\left< 1,0,0\right>| \ |\left< 1,0,1\right>|\ \cos\theta =|\left< 1,0,0\right>\cdot\left< 1,0,1\right>| =1 \\ &\implies \cos\theta=\frac{1}{\sqrt{2}} \implies \theta=\frac{\pi}{4} \end{aligned}$$