방향도함수와 기울기벡터

(a) The gradient of $f$ is ${ \boldsymbol{\nabla} } f(x,y)=\left< 3,-4\right>$. So the specified rate of change is

$$\left< 3,-4\right>\cdot\frac{\left< -2,0\right>}{|\left< -2,0\right>|}=-3$$

(b) The gradient of $f$ is ${ \boldsymbol{\nabla} } f(x,y,z)=\left< -x^{-2},-y^{-2},-z^{-2}\right>$. In particular, the gradient of $f$ at the point $(2,-3,4)$ is ${ \boldsymbol{\nabla} } f(2,-3,4)=\left< -\frac{1}{4},-\frac{1}{9},-\frac{1}{16}\right>$. So the specified rate of change is

$$\left< -\frac{1}{4},-\frac{1}{9},-\frac{1}{16}\right>\cdot \frac{\left< 1,1,1\right>}{\sqrt{3}}= -\frac{61}{144\sqrt{3}}\approx-0.2446$$

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!

Denote ${ \boldsymbol{\nabla} } f(a,b)=\left< \alpha ,\beta \right>$. We are told that

$$\begin{aligned} \left<\alpha ,\beta \right>\cdot\left<\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right>&=3\sqrt{2} & &\quad\text{or}\quad & \alpha +\beta &=6\\ \left<\alpha ,\beta \right>\cdot\left<\frac{3}{5},-\frac{4}{5}\right>&=5 & &\quad\text{or}\quad & 3\alpha -4\beta &=25 \end{aligned}$$

Adding 4 times the first equation to the second equation gives $7\alpha =49$. Substituting $\alpha =7$ into the first equation gives $\beta =-1$. So ${ \boldsymbol{\nabla} } f(a,b)=\left< 7,-1\right>$.

Use a coordinate system with the positive $y$–axis pointing north, with the positive $x$–axis pointing east and with our current location being $x=y=0$. Denote by $z(x,y)$ the elevation of the earth's surface at $(x,y)$. We are told that

$$\begin{aligned} { \boldsymbol{\nabla} } z(0,0)\cdot(-\hat{\boldsymbol{\jmath}}) &= 4 \\ { \boldsymbol{\nabla} } z(0,0)\cdot\left(\frac{\hat{\boldsymbol{\imath}}-\hat{\boldsymbol{\jmath}}}{\sqrt{2}}\right) &= \sqrt{2} \end{aligned}$$

The first equation implies that $z_y(0,0)=-4$ and the second equation implies that

$$\begin{aligned} \frac{z_x(0,0)-z_y(0,0)}{\sqrt{2}}=\sqrt{2} \implies z_x(0,0)=z_y(0,0)+2 = -2 \end{aligned}$$

So the slope in the eastern direction is

$$\begin{aligned} { \boldsymbol{\nabla} } z(0,0)\cdot\hat{\boldsymbol{\imath}} = z_x(0,0) = -2 \end{aligned}$$

(a) Use ${ \boldsymbol{\nabla} } f(P)$ to denote the gradient vector of $f$ at $P$. We are told that

• directional derivative of $f$ at $P$ is a maximum in the direction $2\hat{\boldsymbol{\imath}} - \hat{\boldsymbol{\jmath}} + \hat{\mathbf{k}}$, which implies that ${ \boldsymbol{\nabla} } f(P)$ is parallel to $2\hat{\boldsymbol{\imath}} - \hat{\boldsymbol{\jmath}} + \hat{\mathbf{k}}$, and
• the magnitude of the directional derivative in that direction is $3\sqrt{6}$, which implies that $|{ \boldsymbol{\nabla} } f(P)|=3\sqrt{6}$.

So

$$\begin{aligned} { \boldsymbol{\nabla} } f(P) = 3\sqrt{6} \frac{2\hat{\boldsymbol{\imath}} - \hat{\boldsymbol{\jmath}} + \hat{\mathbf{k}}}{|2\hat{\boldsymbol{\imath}} - \hat{\boldsymbol{\jmath}} + \hat{\mathbf{k}}|} = 6\hat{\boldsymbol{\imath}} - 3\hat{\boldsymbol{\jmath}} + 3\hat{\mathbf{k}} \end{aligned}$$

(b) The directional derivative of $f$ at $P$ in the direction $\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}$ is

$${ \boldsymbol{\nabla} } f(P) \cdot\frac{\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}}{|\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}|} =\frac{1}{\sqrt{2}}\big(6\hat{\boldsymbol{\imath}} - 3\hat{\boldsymbol{\jmath}} + 3\hat{\mathbf{k}}\big)\cdot\big(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}\big) =\frac{3}{\sqrt{2}}$$

(a) The gradient of $f$ at $(x,y)=(2,1)$ is

$$\begin{aligned} { \boldsymbol{\nabla} } f(2,1) = \left< -y^2\,,\,-2xy \right>\Big|_{(x,y)=(2,1)} = \left< -1\,,\, -4\right> \end{aligned}$$

So the path of steepest ascent is in the direction $-\frac{1}{\sqrt{17}}\left< 1\,,\, 4\right>$, which is a little west of south. The slope is

$$|{ \boldsymbol{\nabla} } f(2,1)| = |\left< -1\,,\, -4\right>| = \sqrt{17}$$

(b) The directional derivative in the north direction is

$$D_{\left< 0,1\right>}f(2,1) = { \boldsymbol{\nabla} } f(2,1)\cdot\left< 0,1\right> = \left< -1\,,\, -4\right> \cdot \left< 0,1\right> = -4$$

So the hiker descends with slope $|-4|=4$.

(c) To contour, i.e. remain at the same height, the hiker should walk in a direction perpendicular to ${ \boldsymbol{\nabla} } f(2,1)= \left< -1\,,\, -4\right>$. Two unit vectors perpendicular to $\left< -1\,,\, -4\right>$ are $\pm\frac{1}{\sqrt{17}}\left< 4, -1\right>$.

The gradient of $h(x,y)=1000-2x^2-3y^2$ is ${ \boldsymbol{\nabla} } h(x,y)=(-4x,-6y)$. This gradient (which points in the direction of steepest ascent) must be parallel to the tangent to $y=ax^b$ at all points on $y=ax^b$. A tangent to $y=ax^b$ is $\left< 1,\frac{\mathrm{d}y}{\mathrm{d}x}\right> =\left< 1, abx^{b-1}\right>$.

$$\left< -4x,-6y\right> \parallel \left< 1, abx^{b-1}\right> \implies \frac{abx^{b-1}}{1}=\frac{-6y}{-4x} \implies \frac{3}{2}y=abx^b$$

This is true at all points on $y=ax^b$ if and only if $b=\frac{3}{2}$. As $(1,1)$ must also be on $y=ax^b$, we need $1=a1^b$, which forces $a=1$, $b=\frac{3}{2}$. Here is a contour map showing the hiking trail.

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 $\left< \pm\frac{4}{5}\,,\,\frac{3}{5}\right>$.

Denote by $\left< a\,,\,b\,,\,c\right>$ the gradient of the function $f$ at $P$. We are told

$$\begin{aligned} \left< a\,,\,b\,,\,c\right> \cdot \left< 1\,,\,0\,,\,0\right> &= 2 \\ \left< a\,,\,b\,,\,c\right> \cdot \frac{1}{\sqrt{2}}\left< 1\,,\,1\,,\,0\right> &= -\sqrt{2} \\ \left< a\,,\,b\,,\,c\right> \cdot \frac{1}{\sqrt{3}}\left< 1\,,\,1\,,\,1\right> &= -\frac{5}{\sqrt{3}} \end{aligned}$$

Simplifying

$$\begin{aligned} a&=2 \\ a+b&=-2 \\ a+b+c&=-5 \end{aligned}$$

From these equations we read off, in order, $a=2$, $b=-4$ and $c=-3$. The function $f$ has maximum rate of change at $P$ in the direction if the gradient of $f$. The unit vector in that direction is

$$\frac{\left< 2\,,\,-4\,,\,-3\right>}{|\left< 2\,,\,-4\,,\,-3\right>|} =\frac{\left< 2\,,\,-4\,,\,-3\right>}{\sqrt{29}}$$

The maximum rate of change is the magnitude of the gradient, which is $\sqrt{29}$.

We are told that the direction of fastest increase for the function $f(x,y)$ at the origin is given by the vector $\left< 1, 2\right>$. This implies that ${ \boldsymbol{\nabla} } f(0,0)$ is parallel to $\left< 1, 2\right>$. This in turn implies that $\left< 1, 2\right>$ is normal to the level curve of $f(x,y)$ that passes through the origin. So $\left< 2, -1\right>$, being perpendicular to $\left< 1, 2\right>$, is tangent to the level curve of $f(x,y)$ that passes through the origin. The unit vectors that are parallel to $\left< 2, -1\right>$ are $\pm\frac{1}{\sqrt{5}}\left< 2, -1\right>$.