방향도함수와 기울기벡터

(a) The expression $\lim_{t\rightarrow 0}\frac{f((1,1)+t\mathbf{u})-f(1,1)}{t}$ is the directional derivative of $f$ at $(1,1)$ in the direction $\mathbf{u}$, which is $D_\mathbf{u} f(1,1)={ \boldsymbol{\nabla} } f(1,1)\cdot\mathbf{u}$. This is mazimized when $\mathbf{u}$ is parallel to ${ \boldsymbol{\nabla} } f(1,1)$. Since

$$f_x(x,y)=2xy^2e^{-x-y}-x^2y^2e^{-x-y}\qquad f_y(x,y)=2x^2ye^{-x-y}-x^2y^2e^{-x-y}$$

we have

$${ \boldsymbol{\nabla} } f(1,1)=e^{-2}\left< 1,1\right>$$

so that the desired unit vector $\mathbf{u}$ is $\frac{1}{\sqrt{2}}\left< 1,1\right>$.

(b) In order to remain at elevation $e^{-2}$, the ant must move so that $D_\mathbf{u} f(1,1)=0$. This is the case if $\mathbf{u}\perp{ \boldsymbol{\nabla} } f(1,1)$. For example, we can take $\mathbf{u}=\left< 1,-1\right>$. When the ant moves in this direction, while remaining on the surface of the hill, its vertical component of velocity is zero. So $\mathbf{v}=c\left< 1,-1,0\right>$ for any nonzero constant $c$.

(c) In order to maximize its instantaneous rate of level increase, the ant must choose the $x$ and $y$ coordinates of its velocity vector in the same direction as ${ \boldsymbol{\nabla} } f(1,1)$. Namely $\mathbf{u}=c\left< 1,1\right>$ for any $c>0$. To make $\mathbf{u}$ a unit vector, we choose $c=\frac{1}{\sqrt{2}}$. The corresponding value of the $z$ coordinate of its velocity vector is the rate of change of $f$ per unit horizontal distance travelled, which is the directional derivative

$$D_\mathbf{u} f(1,1)={ \boldsymbol{\nabla} } f(1,1)\cdot\mathbf{u}=e^{-2}\left< 1,1\right> \cdot\left< c,c\right> =2ce^{-2}$$

So $\mathbf{v}=\frac{1}{\sqrt{2}}\left< 1,1,2e^{-2}\right>$. Any positive multiple of this vector is also a correct answer.