방향도함수와 기울기벡터

$\nabla F = \left\langle y^2-z^2, 2xy, -2xz\right\rangle$; at $P$, $\nabla F = \left\langle 0, 4, 4\right\rangle$

(a) $\ell_{\vec n}(t) = \left\{\begin{array}{l} x= 2 \\ y = 1+4t\\ z = -1+4t \end{array}\right.$ (b) $4(y-1) + 4(z+1) = 0$.

$\nabla F = \left\langle y\cos(xy), x\cos(xy)-z\sin(yz), -y\sin(yz)\right\rangle$; at $P$, $\nabla F = \left\langle \frac{\pi}{8\sqrt{3}}, -\sqrt{3},-\frac{\pi}{8\sqrt{3}}\right\rangle$

(a) $\ell_{\vec n}(t) = \left\{\begin{array}{l} x= 2 +\frac{\pi}{8\sqrt{3}}t\\ y = \frac{\pi}{12}-\sqrt{3}t\\ z = 4-\frac{\pi}{8\sqrt{3}}t \end{array}\right.$ (b) $\frac{\pi}{8\sqrt{3}}(x-2) -\sqrt{3}(y-\frac{\pi}{12}) -\frac{\pi}{8\sqrt{3}}(z-4) = 0$.

The partial derivatives, at a general point $(x,y,z)$ and also at the point of interest $(0,1,1)$, are

$$\begin{aligned} f_x(x,y,z)&=yz e^{xyz}\qquad & f_x(0,1,1)&= 1 \\ f_y(x,y,z)&=xz e^{xyz}\qquad & f_y(0,1,1)&= 0 \\ f_z(x,y,z)&=xy e^{xyz}\qquad & f_z(0,1,1)&= 0 \end{aligned}$$

So ${ \boldsymbol{\nabla} } f(0,1,1) =\left< 1,0,0\right>$ and the specified directional derivative is

$$\begin{aligned} D_{\frac{\left< 0,1,1\right>}{\sqrt{2}}}f(0,1,1) =\left< 1,0,0\right>\cdot\frac{\left< 0,1,1\right>}{\sqrt{2}} =0 \end{aligned}$$

In two dimensions, write $g(x,y) = y^2+\sin(xy)$. Then

$$\begin{aligned} { \boldsymbol{\nabla} } g =\left< g_x\,,\,g_y\right> =\left< y\cos(xy)\,,\, 2y + x\cos(xy) \right> \end{aligned}$$

In three dimensions, write $g(x,y,z) = y^2+\sin(xy)$. Then

$$\begin{aligned} { \boldsymbol{\nabla} } g =\left< g_x\,,\,g_y\,,\,g_z\right> =\left< y\cos(xy)\,,\, 2y + x\cos(xy)\,,\, 0 \right> \end{aligned}$$

(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.

(a) The temperature gradient at $(3,2,1)$ is

$$\begin{aligned} { \boldsymbol{\nabla} } T(3,2,1) = \left< 4x\,,\,2y\,,\,-2z\right>\Big|_{(x,y,z)=(3,2,1)} = \left< 12\,,\,4\,,\,-2\right> \end{aligned}$$

She wishes to fly in a direction that is perpendicular to ${ \boldsymbol{\nabla} } T(3,2,1)$. That is, she wishes to fly in a direction $\left< a\,,\,b\,,\,c\right>$ that obeys

$$0 = \left< 12\,,\,4\,,\,-2\right>\cdot \left< a\,,\,b\,,\,c\right> = 12a+4b-2c$$

Any nonzero $\left< a\,,\,b\,,\,c\right>$ that obeys $12a+4b-2c=0$ is an allowed direction. Four allowed unit vectors are $\pm\frac{\left< 0\,,\,1\,,\,2\right>}{\sqrt{5}}$ and $\pm\frac{\left< 1\,,\,-3\,,\,0\right>}{\sqrt{10}}$.

(b) No they need not be the same. Four different explicit directions were given in part (a).

(c) To cool down as quickly as possible, she should move in the direction opposite to the temperature gradient. A unit vector in that direction is $-\frac{\left< 6\,,\,2\,,\,-1\right>}{\sqrt{41}}$.