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 $|t|=t$.

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

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

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

(a) We shall show that $\mathbf{F}(x,y,z)$ is conservative by finding a potential for it. $\varphi(x,y,z)$ is a potential for this $\mathbf{F}$ if and only if

$$\begin{aligned} \frac{\partial \varphi}{\partial x}(x,y,z) &= x \\ \frac{\partial \varphi}{\partial y}(x,y,z) &= -2y \\ \frac{\partial \varphi}{\partial z}(x,y,z) &= 3z \end{aligned}$$

Integrating the first of these equations gives

$$\varphi(x,y,z) = \frac{x^2}{2} + f(y,z)$$

Substituting this into the second equation gives

$$\frac{\partial f}{\partial y}(y,z) = -2y$$

which integrates to

$$f(y,z) = -y^2+ g(z)$$

Finally, substituting $\varphi(x,y,z) = \frac{x^2}{2} -y^2 + g(z)$ into the last equation gives

$$g'(z) = 3z$$

which integrates to

$$g(z) = \frac{3}{2} z^2 +C$$

with $C$ being an arbitrary constant. So, $\mathbf{F}(x,y,z)$ is conservative and $\varphi(x,y,z)=\tfrac{1}{2} x^2-y^2+\frac{3}{2}z^2$ is one allowed potential.

(b) The field $\mathbf{F}= F_1\,\hat{\boldsymbol{\imath}} + F_2\,\hat{\boldsymbol{\jmath}}$ can be conservative only if it passes the screening test

$$\frac{\partial F_1}{\partial y}=\frac{\partial F_2}{\partial x}$$

In this case

$$\frac{\partial F_1}{\partial y} =\frac{\partial}{\partial y}\Big(\frac{x}{x^2+y^2}\Big) =-\frac{2xy}{{(x^2+y^2)}^2}$$

is different from

$$\frac{\partial F_2}{\partial x} =\frac{\partial}{\partial x}\Big(\frac{-y}{x^2+y^2}\Big) =\frac{2xy}{{(x^2+y^2)}^2}$$

for all $(x,y)$ with $x$ and $y$ both nonzero. So $\mathbf{F}$ is not conservative.

Note that:

• Along the line segment from $(0, 0)$ to $(1,0)$, $x$ increases from $0$ to $1$, while $y$ is held fixed at $y=0$. So we may parametrize this segment by $\mathbf{r}(x) = x\,\hat{\boldsymbol{\imath}}$, $0\le x\le 1$.
• Along the line segment from $(1, 0)$ to $(1,\pi)$, $y$ increases from $0$ to $\pi$, while $x$ is held fixed at $x=1$. So we may parametrize this segment by $\mathbf{r}(x) = \hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}}$, $0\le y\le\pi$.
• Along the line segment from $(1, \pi)$ to $(0,\pi)$, $x$ decreases from $1$ to $0$, while $y$ is held fixed at $y=\pi$. So we may parametrize this segment by $\mathbf{r}(x) = x\,\hat{\boldsymbol{\imath}}+\pi\,\hat{\boldsymbol{\jmath}}$ with $x$ running from $1$ to $0$.

Hence

$$\begin{aligned} \int_C \mathbf{V}\cdot \mathrm{d}\mathbf{r} &=\int_0^1 \mathbf{V}(x,0)\cdot\hat{\boldsymbol{\imath}}\ \mathrm{d}x+\int_0^\pi \mathbf{V}(1,y)\cdot\hat{\boldsymbol{\jmath}}\ \mathrm{d}y +\int_1^0 \mathbf{V}(x,\pi)\cdot\hat{\boldsymbol{\imath}}\ \mathrm{d}x\\ &=\int_0^1 (e^x+x^2)\ \mathrm{d}x +\int_0^\pi (y+3)\ \mathrm{d}y+\int_1^0 (-e^x+x^2)\ \mathrm{d}x\\ &=2\int_0^1 e^x\ \mathrm{d}x+\int_0^\pi (y+3)\ \mathrm{d}y\\ &=2(e-1)+\frac{\pi^2}{2}+3\pi \end{aligned}$$

(a) We may parametrize the curve by $\mathbf{r}(t)=t\,\hat{\boldsymbol{\imath}}+t^2\,\hat{\boldsymbol{\jmath}}$ with $0\le t\le 1$. Then $\mathbf{v}(t)=\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t)=\hat{\boldsymbol{\imath}}+2t\,\hat{\boldsymbol{\jmath}}$ and $\mathbf{F}\big(x(t),y(t)\big)=t^3\,\hat{\boldsymbol{\imath}}-t^2\,\hat{\boldsymbol{\jmath}}$ so

$$\begin{aligned} \int_\mathcal{C}\mathbf{F}\cdot \mathrm{d}\mathbf{r} &=\int_0^1 \mathbf{F}\big(x(t),y(t)\big)\cdot \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t)\ \mathrm{d}t =\int_0^1\big[t^3\,\hat{\boldsymbol{\imath}}-t^2\,\hat{\boldsymbol{\jmath}}\big] \cdot\big[\hat{\boldsymbol{\imath}}+2t\,\hat{\boldsymbol{\jmath}}\big]\,\mathrm{d}t =\int_0^1\big[-t^3\big]\,\mathrm{d}t \\ &=-\frac{1}{4} \end{aligned}$$

(b) The path is the union of three line segments.

• On the first segment of the path $y=z=0$ so $\mathbf{F}$ simplifies to $x\,\hat{\boldsymbol{\imath}}-x\,\hat{\mathbf{k}}$ and $\mathrm{d}\mathbf{r}=\hat{\boldsymbol{\imath}}\ \mathrm{d}x$ (i.e. we can parametrize the first segment of the path by $\mathbf{r}(x)=x\,\hat{\boldsymbol{\imath}}$ with $0\le x\le 1$), so $\mathbf{F}\cdot \mathrm{d}\mathbf{r}=x\,\mathrm{d}x$.
• On the second segment of the path $x=1$, $z=0$ so $\mathbf{F}$ simplifies to $\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}-(1+y)\hat{\mathbf{k}}$ and $\mathrm{d}\mathbf{r}=\hat{\boldsymbol{\jmath}}\, \mathrm{d}y$ (parametrize the second segment of the path by $\mathbf{r}(y)=\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}$ with $0\le y\le 1$), so $\mathbf{F}\cdot \mathrm{d}\mathbf{r}=y\,\mathrm{d}y$.
• On the final segment of the path $x=y=1$ so $\mathbf{F}$ simplifies to $(1-z)\hat{\boldsymbol{\imath}}+(1-z)\hat{\boldsymbol{\jmath}}-2\hat{\mathbf{k}}$ and $\mathrm{d}\mathbf{r}=\hat{\mathbf{k}}\, \mathrm{d}z$ (parametrize the third segment of the path by $\mathbf{r}(z)=\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}$ with $0\le z\le 1$), so $\mathbf{F}\cdot \mathrm{d}\mathbf{r}=-2\,\mathrm{d}z$.

So

$$\int_\mathcal{C}\mathbf{F}\cdot \mathrm{d}\mathbf{r}=\int_0^1 x\,\mathrm{d}x +\int_0^1 y\,\mathrm{d}y+\int_0^1(-2)\,\mathrm{d}z =\frac{1}{2}+\frac{1}{2}-2 =-1$$

Parametrize the curve using $y$ as a parameter. Then $y=t$, $x=2y=2t$ and $z=\frac{8}{xy}=\frac{8}{2t^2}$ so that:

$$\begin{aligned} \mathbf{r}(t)&=2t\,\hat{\boldsymbol{\imath}}+t\,\hat{\boldsymbol{\jmath}}+\frac{4}{t^2}\,\hat{\mathbf{k}},\qquad 1\le t\le 2\\ \mathbf{r}'(t)&=2\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}-\frac{8}{t^3}\,\hat{\mathbf{k}}\\ \mathbf{F}(\mathbf{r}(t))&= 4t^2\,\hat{\boldsymbol{\imath}}+4t^3\,\hat{\mathbf{k}}\\ \mathbf{F}(\mathbf{r}(t))\cdot \mathbf{r}'(t)&=8t^2-32 \end{aligned}$$