방향도함수와 기울기벡터

(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 $\pm\frac{1}{\sqrt{5}}\left< 2, -1\right>$.

Write $h(x,y) = 1000 -0.02\,x^2-0.01 y^2$ so that the hill is $z=h(x,y)$.

(a) The direction of steepest ascent at $(0,100,900)$ is the direction of maximum rate of increase of $h(x,y)$ at $(0,100)$ which is ${ \boldsymbol{\nabla} } h(0,100) = \left< 0\,,\, -0.01(2)(100)\right> = \left< 0\,,\,-2\right>$. In compass directions that is South.

(b) The slope of the hill there is

$$\begin{aligned} { \boldsymbol{\nabla} } h(0,100)\cdot\left< 0,-1\right> =-\frac{\partial h}{\partial y}(0,100) = 2 \end{aligned}$$

(c) Denote by $\big(x(t),y(t),z(t)\big)$ your position at time $t$ and suppose that you are at $(0,100,900)$ at time $t=0$. Then we know

• $z(t) = 1000 -0.02\,x(t)^2-0.01 y(t)^2$, so that $z'(t) = -0.04\,x(t)x'(t)-0.02 y(t)y'(t)$, since you are on the hill and
• $x'(0)=0$ and $y'(0)>0$ since you are going in the direction of steepest descent and
• $x'(0)^2+y'(0)^2+z'(0)^2=25$ since you are moving at speed $5$.

Since $x(0)$ and $y(0)=100$, we have $z'(0)= -0.02(100)y'(0) = -2y'(0)$. So

$$\begin{aligned} 25 = x'(0)^2+y'(0)^2+z'(0)^2 =5\ y'(0)^2 &\implies y'(0) = \sqrt{5} \\ &\implies \left< x'(0)\,,\,y'(0)\,,\,z'(0) \right> =\left< 0\,,\, \sqrt{5}\,,\, -2\sqrt{5}\right> \end{aligned}$$

and your rate of change of altitude is

$$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t} h\big(x(t)\,,\,y(t)\big)\Big|_{t=0} &={ \boldsymbol{\nabla} } h(0,100)\cdot \left< x'(0)\,,\,y'(0) \right> =\left< 0\,,\,-2\right> \cdot \left< 0\,,\, \sqrt{5}\right> =-2\sqrt{5} \end{aligned}$$

The first order partial derivatives of $f$, both at a general point $(x,y,z)$ and at the point $(1, 0, \pi/2)$, are

$$\begin{aligned} f_x(x,y,z)&= 2x\qquad & f_x(1, 0, \pi/2)&= 2 \\ f_y(x,y,z)&= -z\sin(yz)\qquad & f_y(1, 0, \pi/2)&= 0 \\ f_z(x,y,z)&= -y\sin(yz)\qquad & f_z(1, 0, \pi/2)&= 0 \end{aligned}$$

(a) The rate of increase of $f$ is largest in the direction of ${ \boldsymbol{\nabla} } f(1, 0, \pi/2)=\left< 2,0,0\right>$. A unit vector in that direction is $\hat{\boldsymbol{\imath}}$.

(b) The gradient vector ${ \boldsymbol{\nabla} } f(1, 0, \pi/2)=\left< 2,0,0\right>$ is a normal vector to the surface $f=1$ at $(1, 0, \pi/2)$. So the specified tangent plane is

$$\begin{aligned} \left< 2,0,0\right> \cdot \left< x-1\,,\,y-0\,,\,z-\pi/2\right>=0\qquad\text{or}\qquad x=1 \end{aligned}$$

(c) The vector from the point $(0,1,0)$ to the point $(1,1,0)$, on $T$, is $\left< 1,0,0\right>$, which is perpendicular to $T$. So $(1,1,0)$ is the point on $T$ nearest $(0,1,0)$ and the distance from $(0,1,0)$ to $T$ is $|\left< 1,0,0\right>|=1$.

(d) The vector $\left< 1,0,1\right>$ is perpendicular to the plane $x+z=0$. So the angle between the planes $T$ and $x+z=0$ is the same as the angle $\theta$ between the vectors $\left< 1,0,0\right>$ and $\left< 1,0,1\right>$, which obeys

$$\begin{aligned} & |\left< 1,0,0\right>| \ |\left< 1,0,1\right>|\ \cos\theta =|\left< 1,0,0\right>\cdot\left< 1,0,1\right>| =1 \\ &\implies \cos\theta=\frac{1}{\sqrt{2}} \implies \theta=\frac{\pi}{4} \end{aligned}$$

(a) We are being asked for the directional derivative of $T$ in the direction of the unit vector from $P=(2,1,1)$ to $Q=(3,2,2)$, which is $\frac{\left< 1,1,1\right>}{\sqrt{3}}$. That directional derivative is