방향도함수와 기울기벡터

The gradient of $f(x,y)$ is ${ \boldsymbol{\nabla} } f(x,y)=\left< y,x\right>$. In particular, the gradient of $f$ at the point $(2,0)$ is ${ \boldsymbol{\nabla} } f(2,0)=\left< 0,2 \right>$. So the rate of change in the direction that makes angle $\theta$ with respect to the $x$-axis, that is, in the direction $\left< \cos\theta,\sin\theta\right>$ is

$$\left< \cos\theta,\sin\theta\right> \cdot{ \boldsymbol{\nabla} } f(2,0) =\left< \cos\theta,\sin\theta\right> \cdot\left< 0, 2\right> =2\sin\theta$$

(a) To get a rate $-1$, we need

$$\sin\theta=-\frac{1}{2} \implies \theta=-30^\circ,\ -150^\circ$$

So the desired directions are

$$\left< \cos\theta,\sin\theta\right> =\left<\pm\frac{\sqrt{3}}{2} ,-\frac{1}{2} \right>$$

(b) To get a rate $-2$, we need

$$\sin\theta=-1 \implies \theta=-90^\circ$$

So the desired direction is

$$\left< \cos\theta,\sin\theta\right> =\left< 0 ,-1 \right>$$

(c) To get a rate $-3$, we need

$$\sin\theta=-\frac{3}{2}$$

No $\theta$ obeys this, since $-1\le\sin\theta\le 1$ for all $\theta$. So no direction works!