벡터 미적분 전체 문제

To parametrize $C$, we note the vector between the two points is $(2-1,4-2,5-3)=(1,2,2)$. So, the line is $(1,2,3)+t(1,2,2)$ for $0 \le t \le 1$. That is, $x(t)=1+t$, $y(t)=2+2t$, and $z(t)=3+2t$.

$$\begin{aligned} \int_C(xy+z)\mathrm{d}s&=\int_0^1 \left((1+t)(2+2t)+(3+2t)\right)\sqrt{1^1+2^2+2^2}\mathrm{d}t\\ &= \int_0^1 3\big(5 + 6t + 2t^2\big)\ \mathrm{d}t=26 \end{aligned}$$

(a) In this case $\mathbf{r}(t)=t\hat{\boldsymbol{\imath}}+t^2\hat{\boldsymbol{\jmath}}$, so that $\mathbf{v}(t)=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t)=\hat{\boldsymbol{\imath}}+2t\hat{\boldsymbol{\jmath}}$ and $\frac{\mathrm{d}s}{\mathrm{d}t}=\sqrt{1+4t^2}$. Hence

$$\begin{aligned} \int_\mathcal{C} f(x,y,z)\,\mathrm{d}s &=\int_0^1 x(t)\cos z(t) \frac{\mathrm{d}s}{\mathrm{d}t}\,\mathrm{d}t =\int_0^1 t(\cos 0) \sqrt{1+4t^2}\,\mathrm{d}t =\frac{1}{8}\frac{ {(1+4t^2)}^{3/2}}{3/2}\bigg|_0^1\\ &=\frac{5^{3/2}-1}{12} \end{aligned}$$

(b) In this case $\mathbf{r}(t)=\big(t,\frac{2}{3}t^{3/2},t\big)$, so that $\mathbf{v}(t)=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t)=\big(1,t^{1/2},1\big)$ and $\frac{\mathrm{d}s}{\mathrm{d}t}=\sqrt{2+t}$. Hence

$$\begin{aligned} \int_\mathcal{C} f(x,y,z)\,\mathrm{d}s &=\int_1^2 \frac{x(t)+y(t)}{y(t)+z(t)} \, \frac{\mathrm{d}s}{\mathrm{d}t}\,\mathrm{d}t =\int_1^2 \frac{t+{2\over 3}t^{3/2}}{{2\over3}t^{3/2}+t} \sqrt{2+t}\,\mathrm{d}t =\frac{(2+t)^{3/2}}{3/2}\bigg|_1^2 \\ &=\frac{8-3^{3/2}}{3/2} \end{aligned}$$

In the figure below, we construct a triangle with $\theta=\mathop{\mathrm{arcsec}} t$; the hypotenuse has length $t$, while the side adjacent to $\theta$ has length 1. By the Pythagorean Theorem, the remaining side has length $\sqrt{t^2-1}$, so $\sin\theta=\sin(\mathop{\mathrm{arcsec}} t)=\frac{\sqrt{t^2-1}}{t}$.

Remember $\frac{\mathrm{d}}{\mathrm{d}t}\{\ln t\} = \frac{1}{t}$ and $\frac{\mathrm{d}}{\mathrm{d}t}\{\mathop{\mathrm{arcsec}} t\} = \frac{1}{|t|\sqrt{t^2-1}}$. In our range, $1 \le t \le \sqrt 2$, we have $|t|=t$.

$$\begin{aligned} \int_C \sin x \,\mathrm{d}s&=\int_{1}^{\sqrt 2}\sin\left(\mathop{\mathrm{arcsec}} t \right)\sqrt{\left(\frac{1}{t\sqrt{t^2-1}} \right)^2+\left(\frac1t \right)^2}\,\mathrm{d}t\\ &=\int_1^{\sqrt{2}} \frac{\sqrt{t^2-1}}{t}\sqrt{\frac{1}{t^2(t^2-1)}+\frac{1}{t^2}} \,\mathrm{d}t\\ &=\int_1^{\sqrt 2}\frac1t\,\mathrm{d}t=\frac{1}{2}\ln 2 \end{aligned}$$

Set your face to be at the origin of our coordinate system, $(0,0,0)$. A bee at position $(x,y,z)$ is a distance of $\sqrt{x^2+y^2+z^2}$ from your face, heading in the direction $(-x,-y,-z)$. So, the unit vector indicating the direction of travel of one bee is $\frac{-1}{\sqrt{x^2+y^2+z^2}}(x,y,z)$. Now all we need to find is the length of the velocity vector, i.e. the speed of the bee.

The speed of the friendly bee is inversely proportional to $\sqrt{x^2+y^2+z^2}$, its distance from your face. (Bees that are closer to you are buzzing towards you more excitedly.) So, the speed is given by $\frac{\alpha}{\sqrt{x^2+y^2+z^2}}$ for some constant $\alpha$.

The bee velocity has the direction of the unit vector $\frac{-1}{\sqrt{x^2+y^2+z^2}}(x,y,z)$ with length $\frac{\alpha}{\sqrt{x^2+y^2+z^2}}$ for some positive constant $\alpha$. That is, $$\mathbf{v}(x,y,z)=-\frac{\alpha}{{x^2+y^2+z^2}}(x,y,z)$$

Beginning as in the text, we note

$\mathbf{v}(x,y)\cdot\hat{\boldsymbol{\imath}}=x^2 \begin{cases} >0 & x \neq 0\\ =0 & x=0 \end{cases}$ and $\mathbf{v}(x,y)\cdot\hat{\boldsymbol{\jmath}}=y \begin{cases} >0 & y > 0\\ =0 & y=0\\ <0 & y<0 \end{cases}$ .

That leads to the following picture:

This gives us a general idea to start with. Refining, we notice that when $x^2>|y|$, then the vector $\mathbf{v}(x,y)$ will be more horizontal than vertical. As we move away from the $y$-axis in a horizontal line, the difference between $x^2$ and $|y|$ grows, so the vectors get more and more horizontal. However, for a fixed value of $x$, vectors farther from the axis will be more vertical than vectors closer to it.

Although ultimately we'll sketch only unit-length vectors, we can still find the direction of $\mathbf{v}(x,y)$ by finding its $x$- and $y$ components.

Note $\mathbf{v}(x,y)\cdot \hat{\boldsymbol{\imath}}$ is the distance from $(x,y)$ to the origin, while $\mathbf{v}(x,y)\cdot\hat{\boldsymbol{\jmath}}$ is the distance from $(x,y)$ to the point $(1,1)$. Both these numbers are always nonnegative. This leads to the following sketch:

When $(x,y)$ is far from the origin, its distance from $(0,0)$ is almost the same as its distance from $(1,0)$. So, we expect $\mathbf{v}(x,y)$ to be approximately a scalar multiple of $(1,1)$.

At $(0,0)$, $v(0,0)\cdot \hat{\boldsymbol{\imath}}=0$, so our vector is horizontal; similarly, $v(1,1)\cdot\hat{\boldsymbol{\jmath}}=0$ so this vector is horizontal. Vectors very near to $(0,0)$ are nearly horizontal, while vectors near to $(1,1)$ are nearly vertical.

For the direction field, we normalize our vectors to have unit length.

The field $\mathbf{v}(x,y)$ is the sum, scaled by 1/3, of the unit vector pointing away from the origin and the unit vector pointing away from $(1,0)$. This tells us about a few regions:

• Along the $x$ axis between $(0,0)$ and $(1,0)$, the vectors away from these points are pointing in opposite directions (and have the same length), so they cancel each other out. That is, $v(x,0)=0$ for all $x \in (0,1)$.
• $v(0,0)$ and $v(1,0)$ are not defined.
• Along the $x$-axis outside of $[0,1]$, the vector pointing away from the point $(0,0)$ is the same as the vector pointing away from the point $(1,0)$. So, $v(x,0)=(-2/3,0)$ for $x<0$ and $v(x,0)=(2/3,0)$ for $x>1$.

• As the distance from $(x,y)$ to the origin grows, the vector pointing away from $(0,0)$ looks more and more like the vector pointing away from $(1,0)$. So, our vectors far away from the origin look like vectors of length about 2/3, pointing away from the origin.

A particle of unit mass at position $(x,y)$ has distance $D_1=\sqrt{x^2+y^2}$ from the 5kg mass, so that mass exerts a force of magnitude $\frac{G(5)}{x^2+y^2}$ on the particle. This force has direction $(-x,-y)$. So, the force exerted by the 5kg mass is $\mathbf{f}_1(x,y)=\frac{-5G}{(x^2+y^2)^{3/2}}(x,y)$.

Similarly, the 3 kg mass at $(2,3)$ exerts a force of $\mathbf{f}_2(x,y)=\frac{3G}{((x-2)^2+(y-3)^2)^{3/2}}(2-x,3-y)$; and the 7 kg mass at $(4,0)$ exerts a force of $\mathbf{f}_3(x,y)=\frac{7G}{((x-4)^2+y^2)^{3/2}}(4-x,-y)$.

The net force on a unit mass is therefore

$$\begin{aligned} \mathbf{f}(x,y)&=\mathbf{f}_1(x,y)+\mathbf{f}_2(x,y)+\mathbf{f}_3(x,y)\\ &=\frac{-5G(x,y)}{(x^2+y^2)^{3/2}}+\frac{3G(2-x,3-y)}{((x-2)^2+(y-3)^2)^{3/2}}+\frac{7G(4-x,-y)}{((x-4)^2+y^2)^{3/2}} \end{aligned}$$

The field lines obey

$$\frac{\mathrm{d}x}{2y}=\frac{\mathrm{d}y}{x/y^2}=\frac{\mathrm{d}z}{e^y} \qquad\text{ if $x,y\ne 0$}$$

In particular

$$\begin{aligned} \frac{\mathrm{d}x}{2y}=\frac{y^2\,\mathrm{d}y}{x} \implies x\,\mathrm{d}x=2y^3\,\mathrm{d}y \implies \frac{1}{2}x^2=\frac{1}{2}y^4+C \end{aligned}$$

Since $y=1$ when $x=1$, $C=0$. So $x=y^2$ and

$$\frac{\mathrm{d}y}{x/y^2}=\frac{\mathrm{d}z}{e^y} \implies e^ydy=dz \implies z=e^y+D$$

Since $z=e$ when $y=1$, $D=0$. So the field line is

$$x=y^2 \qquad z=e^y$$

The field lines obey

$$\begin{aligned} &\frac{\mathrm{d}x}{x}=\frac{\mathrm{d}y}{3y}\qquad\text{ if $x,y\ne 0$}\\ &\implies 3\ln |x|=\ln |y|+C\\ &\implies |x|^3=e^C|y| \\ &\implies y=\pm e^{-C}x^3 \\ &\implies y=C'x^3 \end{aligned}$$

with $C'$ a nonzero constant. $x=0$ and $y=0$ are also field lines, since on the $y$-axis $\mathbf{F}\parallel\hat{\boldsymbol{\jmath}}$ and on the $x$-axis $\mathbf{F}\parallel\hat{\boldsymbol{\imath}}$.

If $\varphi$ is a potential for $\mathbf{F}$, then:

• $\frac{\partial \varphi}{\partial x}=x+y$, so $\varphi = \frac{1}{2}x^2+xy+\psi_1(y)$
• $\frac{\partial \varphi}{\partial y}=x-y$, so $\varphi = xy-\frac{1}{2}y^2+\psi_2(x)$

So, for instance, $\varphi = \frac{1}{2}x^2+xy-\frac{1}{2}y^2$ is a potential for $\mathbf{F}$.

If $\varphi$ is a potential for $\mathbf{F}$, then:

• $\frac{\partial \varphi}{\partial x}=\frac1x-\frac1y$, so $\varphi = \ln|x|-\frac{x}{y}+\psi_1(y)$
• $\frac{\partial \varphi}{\partial y}=\frac{x}{y^2}$, so $\varphi = -\frac{x}{y}+\psi_2(x)$

So, for instance, $\varphi=\ln |x| - \frac{x}{y}$ is a potential for $\mathbf{F}$.