벡터 미적분 전체 문제

$||E||=\frac{c}{{x}^{2}+{y}^{2}}\sqrt{{x}^{2}+{y}^{2}}=\frac{c}{\sqrt{{x}^{2}+{y}^{2}}}=\frac{c}{r}$

$\text{c}'(t)=(\cos \,t,-\sin \,t,{e}^{-t})=\text{F}(\text{c}(t))$

${\int }_{C}\text{F}\cdot d\text{r}=-1$

The net flux is zero.

(a) The curve is $\mathbf{r}(\theta) = x(\theta)\,\hat{\boldsymbol{\imath}}+ y(\theta)\,\hat{\boldsymbol{\jmath}}$ with $x(\theta)=r(\theta)\cos\theta$, $y(\theta)=r(\theta)\sin\theta$ and $\theta_1\le \theta\le \theta_2$. On this curve

$$\begin{aligned} \mathbf{v}(\theta) =\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}\theta}(\theta)&= x'(\theta)\hat{\boldsymbol{\imath}}+ y'(\theta)\hat{\boldsymbol{\jmath}} =\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]\hat{\boldsymbol{\imath}}+ \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]\hat{\boldsymbol{\jmath}}\\ \implies \frac{\mathrm{d}s}{\mathrm{d}\theta}(\theta) &=\sqrt{\big[r'(\theta)\cos\theta-r(\theta)\sin\theta\big]^2 + \big[r'(\theta)\sin\theta+r(\theta)\cos\theta\big]^2}\\ &=\sqrt{r'(\theta)^2+r(\theta)^2} \end{aligned}$$

Hence

$$\begin{aligned} \int_\mathcal{C} f(x,y)\,ds &= \int_{\theta_1}^{\theta_2}\!\!\!f\big(x(\theta), y(\theta)\big) \frac{\mathrm{d}s}{\mathrm{d}\theta}\,\mathrm{d}\theta \\ &=\int_{\theta_1}^{\theta_2}\!\!\!f\big(r(\theta)\cos\theta, r(\theta)\sin\theta\big) \sqrt{r(\theta)^2 +\left(\frac{\mathrm{d}r}{\mathrm{d}\theta}(\theta)\right)^2}\, \mathrm{d}\theta \end{aligned}$$

(b) In this case $f(x,y)=1$, $r(\theta)=1+\cos\theta$, $\theta_1=0$ and $\theta_2=2\pi$,

$$\begin{aligned} \int_C \,ds &=\int_0^{2\pi} \sqrt{[1+\cos\theta]^2+[-\sin\theta]^2}\,\mathrm{d}\theta =\int_0^{2\pi} \sqrt{2(1+\cos\theta)}\,\mathrm{d}\theta\\ &=\int_0^{2\pi} \sqrt{4\cos^2\frac{\theta}{2}}\,\mathrm{d}\theta =2\int_0^{2\pi} \left|\cos\frac{\theta}{2}\right|\,\mathrm{d}\theta =4\int_0^{\pi} \cos\frac{\theta}{2}\ \mathrm{d}\theta =8 \sin\frac{\theta}{2}\bigg|_0^{\pi} =8 \end{aligned}$$

a. If $\mathbf{F}+\mathbf{G}$ is conservative for any particular $\mathbf{F}$ and $\mathbf{G}$, then by definition, there exists a potential $\varphi$ with $\mathbf{F}+\mathbf{G} = { \boldsymbol{\nabla} } \varphi$.

Since $\mathbf{F}$ is conservative, there also exists a potential $\psi$ with $\mathbf{F} = { \boldsymbol{\nabla} } \psi$.

But now $\mathbf{G} = (\mathbf{F}+\mathbf{G})-\mathbf{F}={ \boldsymbol{\nabla} } \varphi - { \boldsymbol{\nabla} } \psi = { \boldsymbol{\nabla} }(\varphi-\psi)$. That means the function $(\varphi-\psi)$ is a potential for $G$. However, this is impossible: since $\mathbf{G}$ is non-conservative, no function with this property exists.

So it is not possible that $\mathbf{F}+\mathbf{G}$ is conservative. It must be non-conservative. b. Counterexample: if $\mathbf{F} = -\mathbf{G}$, then $\mathbf{F}+\mathbf{G} = \mathbf 0 = { \boldsymbol{\nabla} } c$ for any constant $c$. c. Since both fields are conservative, they both have potentials, say $\mathbf{F}={ \boldsymbol{\nabla} } \varphi$ and $\mathbf{G} = { \boldsymbol{\nabla} } \psi$. Then $\mathbf{F}+\mathbf{G} = { \boldsymbol{\nabla} }\varphi+{ \boldsymbol{\nabla} }\psi={ \boldsymbol{\nabla} }(\varphi+\psi)$. That is, $(\varphi+\psi)$ is a potential for $\mathbf{F}+\mathbf{G}$, so $\mathbf{F}+\mathbf{G}$ is conservative.

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 $\mathbf{F}$.

If $\varphi(x,y,z)$ is a potential for $\mathbf{F}(x,y,z)$, then:

• $\frac{\partial \varphi}{\partial x}(x,y,z)=\frac{x}{x^2+y^2+z^2}$, so $\varphi(x,y,z) = \frac12\ln(x^2+y^2+z^2)+\psi_1(y,z)$
• $\frac{\partial \varphi}{\partial y}(x,y,z)=\frac{y}{x^2+y^2+z^2}$, so $\varphi(x,y,z) = \frac12\ln(x^2+y^2+z^2)+\psi_2(x,z)$
• $\frac{\partial \varphi}{\partial z}(x,y,z)=\frac{z}{x^2+y^2+z^2}$, so $\varphi(x,y,z) = \frac12\ln(x^2+y^2+z^2)+\psi_2(x,y)$

So, for instance, $\varphi(x,y,z)= \frac{1}{2}\ln(x^2+y^2+z^2)$ is a potential for $\mathbf{F}(x,y,z)$.

The line integral is $\int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r}$ with $\mathbf{F} = (ye^x + \sin y)\,\hat{\boldsymbol{\imath}} + (e^x + \sin y + x \cos y)\,\hat{\boldsymbol{\jmath}}$. We are to show that it is independent of path. That is the case if and only if $\mathbf{F}$ is conservative. So let's look for a potential $\varphi$ for $\mathbf{F}$. That is, let's look for a function $\varphi$ that obeys

$$\begin{aligned} \frac{\partial \varphi}{\partial x}(x,y) &= ye^x + \sin y \\ \frac{\partial \varphi}{\partial y}(x,y) &= e^x + \sin y + x \cos y \end{aligned}$$

Integrating the first of these equations gives

$$\varphi(x,y) = ye^x + x\sin y + f(y)$$

Substituting this into the second equation gives

$$e^x + x\cos y + f'(y) = e^x + \sin y + x \cos y\qquad\text{or}\quad f'(y) = \sin y$$

which integrates to

$$f(y) = -\cos y + C$$

So $\mathbf{F}$ is indeed conservative with one potential being $\varphi(x,y) = ye^x + x\sin y -\cos y$ and the line integral is

$$\begin{aligned} \int_C (ye^x + \sin y)\,\mathrm{d}x + (e^x + \sin y + x \cos y)\,\mathrm{d}y &=\int_C \mathbf{F}\cdot\mathrm{d}\mathbf{r} =\varphi(x,y)\Big|^{(0,\pi/2)}_{(1,0)} \\ &=\Big[ye^x + x\sin y -\cos y\Big]^{(0,\pi/2)}_{(1,0)} \\ &=1+\frac{\pi}{2} \end{aligned}$$

(a) In the sketch below, the point $(x,y,z)$ deviates from the centre $(2,2,4)$ by $\sin\theta$ units in the $\hat{\mathbf{k}}$ direction, and by $\cos\theta$ units in the $\sqrt \frac1{\sqrt 2}(\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}})$ direction. So, $(x,y,z)=(2+\frac1{\sqrt 2}\cos\theta~,~2+\frac1{\sqrt 2}\cos\theta~,~4+\sin\theta)$.

So, we can parametrize the circle as $(x,y,z)=(2+\frac1{\sqrt 2}\cos\theta~,~2+\frac1{\sqrt 2}\cos\theta~,~4+\sin\theta)$, with $0 \le \theta \le 2\pi$.

Remark: it's easy to check that this equation satisfies the two properties we desire. Since the $x$- and $y$ coordinates match, it's in the plane $x=y$. To check that it's a circle centred at $(2,2,4)$, we note the distance from $(x,y,z)$ to $(2,2,4)$ is:

$$\begin{aligned} d&=\sqrt{(x-2)^2+(y-2)^2+(z-4)^2}=\sqrt{\left(\frac1{\sqrt 2}\cos\theta\right)^2+\left(\frac1{\sqrt 2}\cos\theta\right)^2+\left(\sin\theta\right)^2}\\ &=\sqrt{\frac12\cos^2\theta+\frac12\cos^2\theta+\sin^2\theta}=\sqrt{\cos^2\theta+\sin^2\theta}=1 \end{aligned}$$

So, our points all have distance one from the same point — that is, they lie on a circle of radius 1.

(b) Consider a point $(x,y,z)=(2+\frac1{\sqrt 2}\cos\theta~,~2+\frac1{\sqrt 2}\cos\theta~,~4+\sin\theta)$, rotating $\phi$ radians about the line $x=y=4$.

The new position of the point has the same height, $z=4+\sin\theta$. Its distance from the line $x=y=4$ is also preserved: $R=\sqrt{(x-4)^2+(y-4)^2+(z-z)^2}=\sqrt{(\frac1{\sqrt 2}\cos\theta-2)^2+(\frac1{\sqrt 2}\cos\theta-2)^2+0)}=\cos\theta-2\sqrt2$.