방향도함수와 기울기벡터

(a) The direction of motion at $s=1$ is given by the tangent vector

$$\mathbf{r}'(s)=\left< s^2,2,2s\right>\big|_{s=1}=\left< 1,2,2\right>$$

Since the length of the velocity vector must be $3$,

$$\text{velocity}=\mathbf{v}=3\frac{\left< 1,2,2\right>}{|\left< 1,2,2\right>|}=\left< 1,2,2\right>$$

(b) The rate of change of temperature per unit distance felt by the sparrow at $s=1$ is ${ \boldsymbol{\nabla} } T\big(\frac{1}{3},2,1\big)\cdot\frac{\mathbf{v}}{|\mathbf{v}|}$. The rate of change of temperature per unit time felt by the sparrow at $s=1$ is

$$\begin{aligned} { \boldsymbol{\nabla} } T\left(\frac{1}{3},2,1\right)\cdot\frac{\mathbf{v}}{|\mathbf{v}|}\ |\mathbf{v}| &={ \boldsymbol{\nabla} } T\left(\frac{1}{3},2,1\right)\cdot\mathbf{v} =\mathbf{v}\cdot\left< 6x,y,4z\right>\Big|_{({1\over3},2,1\big)} \\ &=\left< 1,2,2\right>\cdot\left< 2,2,4\right>=14^\circ/{\rm s} \end{aligned}$$

(c) The temperature decreases at maximum rate in the direction opposite the temperature gradient, which is (any positive constant times) $-\left< 2,2,4\right>$.

(d) The eagle is moving at right angles to the direction of motion of the sparrow, which is $\left< 1,2,2\right>$. As the eagle is also moving in a direction for which the temperature remains constant, it must be moving perpendicularly to the temperature gradient, $\left< 2,2,4\right>$. So the direction of the eagle must be (a posiitve constant times) one of

$$\begin{aligned} \pm\left< 1,2,2\right> \times\left< 2,2,4\right> =\pm\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 1 & 2 & 2 \\ 2 & 2 & 4 \end{matrix}\right] =\pm\left< 4,0,-2\right> \end{aligned}$$

or equivalently, any positive constant times $\pm\left< 2,0,-1\right>$.

(a) The moth is moving the direction of the temperature gradient at $(3,4,0)$, which is

$${ \boldsymbol{\nabla} } T(3,4,0) =-200\frac{2x\hat{\boldsymbol{\imath}}+2y\hat{\boldsymbol{\jmath}}+2z\hat{\mathbf{k}}}{{(1+x^2+y^2+z^2)}^2}\bigg|_{(3,4,0)} =-400\frac{3\hat{\boldsymbol{\imath}}+4\hat{\boldsymbol{\jmath}}}{26^2}$$

Since the speed of the moth is $1 {\rm m/s}$ its velocity vector is a vector of length one in direction $-\frac{400}{26^2}\left< 3,4,0\right>$ and hence is $\mathbf{v}=-\frac{\left< 3,4,0\right>}{|\left< 3,4,0\right>|} =-\left<\frac{3}{5},\frac{4}{5},0\right>$.

(b) The rate of change of temperature (per unit time) the moth feels at that time is

$${ \boldsymbol{\nabla} } T(3,4,0)\cdot\mathbf{v} =\frac{400}{26^2}\left< 3,4,0\right>\cdot \left<\frac{3}{5},\frac{4}{5},0\right> =\frac{400\times25}{26^2\times 5} =\frac{500}{169}\approx 2.96^\circ/{\rm s}$$

(a) We are told that $T(x,y,z)=\frac{k}{|\left< x,y,z\right>|} =\frac{k}{\sqrt{x^2+y^2+z^2}}$ for some constant $k$ and that

$$120=T(1,2,2)=\frac{k}{|\left< 1,2,2\right>|} \implies k=120\times\sqrt{1+2^2+2^2} = 360$$

(b) The (unit) direction from $(1,2,2)$ to $(2,1,3)$ is $\mathbf{d}=\frac{\left< 2,1,3\right>-\left< 1,2,2\right>}{|\left< 2,1,3\right>-\left< 1,2,2\right>|} =\frac{\left< 1,-1,1\right>}{|\left< 1,-1,1\right>|}=\frac{1}{\sqrt{3}}\left< 1,-1,1\right>$. The desired rate of change of temperature is

$$\begin{aligned} D_{\mathbf{d}} T(1,2,2) &={ \boldsymbol{\nabla} } T(1,2,2)\cdot\mathbf{d} =-360\frac{x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}+z\hat{\mathbf{k}}}{{(x^2+y^2+z^2)}^{3/2}}\Big|_{\left< 1,2,2\right>}\cdot\mathbf{d} \\ &=-360\frac{\left< 1,2,2\right>}{27}\cdot\frac{\left< 1,-1,1\right>}{\sqrt{3}} =-\frac{40}{3\sqrt{3}}\approx-7.70 \end{aligned}$$

degrees per unit distance.

(c) At $(x,y,z)$, the direction of greatest increase is in the direction of the temperature gradient at $(x,y,z)$, which is ${ \boldsymbol{\nabla} } T(x,y,z) =-360\frac{x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}+z\hat{\mathbf{k}}}{{(x^2+y^2+z^2)}^{3/2}}$ and which points opposite to the radius vector. That is, it points towards the origin. This argument only fails at $(x,y,z)=(0,0,0)$, where the gradient, and indeed $T(x,y,z)$, is not defined.

(a) The shoreline is $f(x,y)=0$ or $x^2+4x+4y^2=32$ or $(x+2)^2+4y^2=36$, which is an ellipse centred on $(-2,0)$ with semiaxes $6$ in the $x$-direction and $3$ in the $y$-direction.

(b,c) The gradient of $f$ at $(-1,1)$ is

$${ \boldsymbol{\nabla} } f(-1,1)=\big[(-2x-4)\,\hat{\boldsymbol{\imath}}-8y\,\hat{\boldsymbol{\jmath}}\big]_{(-1,1)}=-2\,\hat{\boldsymbol{\imath}}-8\,\hat{\boldsymbol{\jmath}}$$

To remain at constant depth, he should swim perpendicular to the depth gradient. So he should swim in direction $\pm\frac{1}{\sqrt{17}}\left< 4,-1\right>$. To increase his depth as rapidly as possible, he should swim in the direction of the depth gradient, which is $-\frac{1}{\sqrt{17}}\left< 1,4\right>$.