방향도함수와 기울기벡터

We are going to need, in both parts of this question, the gradient of $f(x,y)$ at $(x,y)=\left(1, -\frac{4}{3}\right)$. So we find it first.

$$\begin{aligned} f_x(x,y)&=4x+3y\qquad & f_x(1,-4/3)&=0 \\ f_y(x,y)&=3x+2y\qquad & f_y(1,-4/3)&=\frac{1}{3} \end{aligned}$$

so ${ \boldsymbol{\nabla} } f\left(1, -\frac{4}{3}\right) =\left< 0,\frac{1}{3}\right>$.

(a) The maximum rate of change of $f$ at $P$ is

$$\begin{aligned} \left|{ \boldsymbol{\nabla} } f\left(1, -\tfrac{4}{3}\right)\right| =\left|\left< 0,\tfrac{1}{3}\right>\right| =\tfrac{1}{3} \end{aligned}$$

(b) If $\left< a,b\right>$ is a unit vector, the directional derivative of $f$ at $P$ in the direction $\left< a,b\right>$ is

$$\begin{aligned} D_{\left< a,b\right>}f\left(1, -\tfrac{4}{3}\right) ={ \boldsymbol{\nabla} } f\left(1, -\tfrac{4}{3}\right)\cdot \left< a,b\right> =\left< 0,\tfrac{1}{3}\right>\cdot \left< a,b\right> =\tfrac{b}{3} \end{aligned}$$

So we need $\frac{b}{3}=\frac{1}{5}$ and hence $b=\frac{3}{5}$. For $\left< a,b\right>$ to be a unit vector, we also need

$$\begin{aligned} a^2+b^2=1 \iff a^2=1-b^2=1-\frac{3^2}{5^2}=\frac{16}{25} \iff a=\pm\frac{4}{5} \end{aligned}$$

So the allowed directions are $\left< \pm\frac{4}{5}\,,\,\frac{3}{5}\right>$.