방향도함수와 기울기벡터

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

$$\begin{aligned} { \boldsymbol{\nabla} } T(P)\cdot \frac{\left< 1,1,1\right>}{\sqrt{3}} =\left< 1,2,3\right> \cdot \frac{\left< 1,1,1\right>}{\sqrt{3}} =2\sqrt{3} \end{aligned}$$

(b) The linear approximation to $T$ at $P$ is

$$\begin{aligned} T(2+\Delta x\,,\,1+\Delta y\,,\,1+\Delta z) &\approx T(P) + T_x(P)\,\Delta x + T_y(P)\,\Delta y + T_z(P)\,\Delta z \\ &= 5 +\Delta x +2\,\Delta y +3\,\Delta z \end{aligned}$$

Applying this with $\Delta x = -0.1$, $\Delta y = 0$, $\Delta z=0.2$ gives

$$\begin{aligned} T(1.9\,,\,1\,,\,1.2) &\approx 5 +(-0.1) +2\,(0) +3\,(0.2) =5.5 \end{aligned}$$

(c) For the rate of change of $T$ to be zero, the direction of motion must be perpendicular to ${ \boldsymbol{\nabla} } T(P) = \left< 1,2,3\right>$. For the rate of change of $S$ to also be zero, the direction of motion must also be perpendicular to ${ \boldsymbol{\nabla} } S(P) = \left< 1,0,1\right>$. The vector

$$\begin{aligned} \left< 1,2,3\right> \times \left< 1,0,1\right> &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}}\\ 1 & 2 & 3 \\ 1 & 0 & 1\end{matrix}\right] =\left< 2,2,-2\right> \end{aligned}$$

is perpendicular to both ${ \boldsymbol{\nabla} } T(P)$ and ${ \boldsymbol{\nabla} } S(P)$. So the desired unit vectors are $\pm\frac{\left< 1,1,-1\right>}{\sqrt{3}}$.

We are going to need the gradients of both $F$ and $G$ at $(0,1,2)$. So we compute

$$\begin{aligned} \frac{\partial F}{\partial x}(x,y,z)&=y^2+z & \frac{\partial F}{\partial y}(x,y,z)&=2xy & \frac{\partial F}{\partial z}(x,y,z)&=3z^2+x \\ \frac{\partial G}{\partial x}(x,y,z)&=3 & \frac{\partial G}{\partial y}(x,y,z)&=-1 & \frac{\partial G}{\partial z}(x,y,z)&=4 \end{aligned}$$

and then

$${ \boldsymbol{\nabla} } F(0,1,2) = \left< 3,0,12 \right>\qquad { \boldsymbol{\nabla} } G(0,1,2) = \left< 3,-1,4 \right>$$

(a) The linear approximation to $F$ at $(0,1,2)$ is

$$\begin{aligned} F(x,y,z) &\approx F(0,1,2) + F_x(0,1,2)\,x + F_y(0,1,2)\,(y-1) + F_z(0,1,2)\,(z-2) \\ &=8+ 3 x + 12 (z-2) \end{aligned}$$

In particular

$$\begin{aligned} F(0.1\,,\,0.9\,,\,1.8) - F(0,1,2) &\approx 3(0.1) + 12(-0.2) =-2.1 \end{aligned}$$

(b) The direction along which $G$ increases most rapidly at $P$ is ${ \boldsymbol{\nabla} } G(0,1,2) = \left< 3,-1,4 \right>$. The directional derivative of $F$ in that direction is

$$\begin{aligned} D_{\frac{\left< 3,-1,4\right>}{\sqrt{26}}}F(0,1,2) = { \boldsymbol{\nabla} } F(0,1,2) \cdot \frac{\left< 3,-1,4\right>}{\sqrt{26}} = \left< 3,0,12 \right>\cdot \frac{\left< 3,-1,4\right>}{\sqrt{26}} > 0 \end{aligned}$$

So $F$ increases.

(c) For the rate of change of $F$ to be zero, $\left< a\,,\,b\,,\,c\right>$ must be perpendicular to ${ \boldsymbol{\nabla} } F(0,1,2) = \left< 3,0,12 \right>$.

For the rate of change of $G$ to be zero, $\left< a\,,\,b\,,\,c\right>$ must be perpendicular to ${ \boldsymbol{\nabla} } G(0,1,2) = \left< 3,-1,4 \right>$.

So any nonzero constant times

$$\begin{aligned} \det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 3 & 0 & 12 \\ 3 & -1 & 4 \end{matrix}\right] =\left< 12 \,,\, 24 \,,\, -3 \right> =3 \left< 4 \,,\, 8 \,,\, -1 \right> \end{aligned}$$

is an allowed direction.

(a) Since

$$z=-\frac{100}{x^2+2x+4y^2+11}=-\frac{100}{(x+1)^2+4y^2+10}$$

the bottom of the crater is at $x=-1$, $y=0$ (where the denominator is a minimum) and the contours (level curves) are ellipses having equations $(x+1)^2+4y^2=C$. In the sketch below, the filled dot represents the bottom of the crater and the open dot represents the car park. The contours sketched are (from inside out) $z=-7.5, -5, -2.5, -1$. Note that the trail crosses the contour lines at right angles.

(b) The trail is to be parallel to $${ \boldsymbol{\nabla} } z = \frac{100}{{(x^2+2x+4y^2+11)}^2}(2x+2,8y)$$ At the car park ${ \boldsymbol{\nabla} } z(4,5)\parallel \left< 10, 40\right>\parallel \left< 1,4\right>$. To move towards the bottom of the crater, we should leave in the direction $-\left< 1,4\right>$.

We have

$${ \boldsymbol{\nabla} } h(x,y)= -200 e^{-(x^2+2y^2)}\left< x,2y\right>\text{ and, in particular, } { \boldsymbol{\nabla} } h(3,2)= -200 e^{-17}\left< 3,4\right>$$

(a) At $(3,2)$ the dune slopes downward the most steeply in the direction opposite ${ \boldsymbol{\nabla} } h(3,2)$, which is (any positive multiple of) $\left< 3,4\right>$.

(b) The rate is $D_{\hat{\boldsymbol{\jmath}}} h(3,2)={ \boldsymbol{\nabla} } h(3,2)\cdot\hat{\boldsymbol{\jmath}}=-800 e^{-17}$.

(c) To remain at the same height, you should walk perpendicular to ${ \boldsymbol{\nabla} } h(3,2)$. So you should walk in one of the directions $\pm\big(\frac{4}{5},-\frac{3}{5}\big)$.

(d) Suppose that you are walking along a steepest descent curve. Then the direction from $(x,y)$ to $(x+\mathrm{d}x, y+\mathrm{d}y)$, with $(\mathrm{d}x,\mathrm{d}y)$ infinitesmal, must be opposite to ${ \boldsymbol{\nabla} } h(x,y)= -200 e^{-(x^2+2y^2)}(x,2y)$. Thus $(\mathrm{d}x,\mathrm{d}y)$ must be parallel to $(x,2y)$ so that the slope

$$\begin{aligned} \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{2y}{x} \implies \frac{\mathrm{d}y}{y}=2\frac{\mathrm{d}x}{x} \implies \ln y=2\ln x+C \end{aligned}$$

We must choose $C$ to obey $\ln 2=2\ln 3+C$ in order to pass through the point $(3,2)$. Thus $C=\ln\frac{2}{9}$ and the curve is $\ln y=2\ln x+\ln\frac{2}{9}$ or $y=\frac{2}{9}x^2$.