발산과 회전 연습문제 | 빌드라인 스쿨

1.CLP-4 · 4.1절 · 13번대학 수준우리말 번역

벡터장 $\mathbf{F}$ 가 $\mathbb{R}^3$ 전체에서 ${ \boldsymbol{\nabla} }\cdot \mathbf{F}=0$ 을 만족한다고 가정하세요.

$\mathbf{r}(t)=tx\,\hat{\boldsymbol{\imath}}+ty\,\hat{\boldsymbol{\jmath}}+tz\,\hat{\mathbf{k}},\qquad 0\le t\le 1$

을 원점에서 $(x,y,z)$ 까지의 선분의 매개변수 표현이라고 하고,

$\mathbf{G}(x,y,z)=\int_0^1 t\,\mathbf{F}\big(\mathbf{r}(t)\big)\times\frac{d\mathbf{r}}{dt}(t)\,dt$

로 정의합니다. $\mathbb{R}^3$ 전체에서 ${ \boldsymbol{\nabla} }\times \mathbf{G}=\mathbf{F}$ 임을 보이세요.

영어 원문 보기

Suppose that the vector field $\mathbf{F}$ obeys ${ \boldsymbol{\nabla} }\cdot \mathbf{F}=0$ in all of $\mathbb{R}^3$ . Let

$\mathbf{r}(t)=tx\,\hat{\boldsymbol{\imath}}+ty\,\hat{\boldsymbol{\jmath}}+tz\,\hat{\mathbf{k}},\qquad 0\le t\le 1$

be a parametrization of the line segment from the origin to $(x,y,z)$ . Define

$\mathbf{G}(x,y,z)=\int_0^1 t\,\mathbf{F}\big(\mathbf{r}(t)\big)\times\frac{d\mathbf{r}}{dt}(t)\,dt$

Show that ${ \boldsymbol{\nabla} }\times \mathbf{G}=\mathbf{F}$ throughout $\mathbb{R}^3$ .

정답·풀이 보기

$\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}=F_1$ 임을 보이겠다. 나머지 성분도 이와 비슷하다. 먼저

$\begin{aligned} t\,\mathbf{F}\big(\mathbf{r}(t)\big)\times\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t) &=t\,\mathbf{F}\big(tx,ty,tz\big)\times\big(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}\big) =t\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ F_1 & F_2 & F_3\\ x & y & z\end{matrix}\right] \end{aligned}$

이다. 행렬식의 $\hat{\mathbf{k}}$ 성분과 $\hat{\boldsymbol{\jmath}}$ 성분을 읽으면

$\begin{aligned} G_3(x,y,z)&=\int_0^1 t\big[F_1\big(tx,ty,tz\big)\,y-F_2\big(tx,ty,tz\big)\,x\big]\,dt\\ G_2(x,y,z)&=\int_0^1 t\big[F_3\big(tx,ty,tz\big)\,x-F_1\big(tx,ty,tz\big)\,z\big]\,dt \end{aligned}$

이다. 따라서

$\begin{aligned} \frac{\partial G_3}{\partial y} &=\int_0^1 t\Big[F_1\big(tx,ty,tz\big)+ \frac{\partial F_1}{\partial y}\big(tx,ty,tz\big)\,ty -\frac{\partial F_2}{\partial y}\big(tx,ty,tz\big)\,tx\Big]\,dt\\ \frac{\partial G_2}{\partial z}&=\int_0^1 t\Big[ \frac{\partial F_3}{\partial z}\big(tx,ty,tz\big)\,tx -\frac{\partial F_1}{\partial z}\big(tx,ty,tz\big)\,tz -F_1\big(tx,ty,tz\big)\Big]\,dt\\ \Rightarrow \frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z} &=\int_0^1 \Big[2t\,F_1\big(tx,ty,tz\big)+ t^2y\,\frac{\partial F_1}{\partial y}\big(tx,ty,tz\big) +t^2z\frac{\partial F_1}{\partial z}\big(tx,ty,tz\big)\\ & -t^2x\,\frac{\partial F_2}{\partial y}\big(tx,ty,tz\big) -t^2x\,\frac{\partial F_3}{\partial z}\big(tx,ty,tz\big)\Big]\,dt \end{aligned}$

이다. 가정에 의해 ${ \boldsymbol{\nabla} }\cdot\mathbf{F}=\frac{\partial F_1}{\partial x} +\frac{\partial F_2}{\partial y} +\frac{\partial F_3}{\partial z}=0$ 이므로 마지막 두 항은

$\begin{aligned} -t^2x\Big\{\frac{\partial F_2}{\partial y}\big(tx,ty,tz\big) +\frac{\partial F_3}{\partial z}\big(tx,ty,tz\big)\Big\} =-t^2x\Big\{-\frac{\partial F_1}{\partial x}\big(tx,ty,tz\big)\Big\} \end{aligned}$

이고, 따라서

$\begin{aligned} &\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z} \\ &=\int_0^1 \Big[2tF_1\big(tx,ty,tz\big)+ t^2x\frac{\partial F_1}{\partial x}\big(tx,ty,tz\big) +t^2y\frac{\partial F_1}{\partial y}\big(tx,ty,tz\big) +t^2z\frac{\partial F_1}{\partial z}\big(tx,ty,tz\big)\Big]\,dt\\ &=\int_0^1 \frac{d}{dt}\Big[ t^2 F_1(tx,ty,tz)\Big]\,dt =\Big[ t^2 F_1(tx,ty,tz)\Big]^{t=1}_{t=0} = F_1(x,y,z) \end{aligned}$

이다.

원본 교재 풀이(영어)

We shall show that $\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}=F_1$ . The other components are similar. First we have

Reading off the $\hat{\mathbf{k}}$ and $\hat{\boldsymbol{\jmath}}$ components of the determinant gives

Since, by hypothesis, ${ \boldsymbol{\nabla} }\cdot\mathbf{F}=\frac{\partial F_1}{\partial x} +\frac{\partial F_2}{\partial y} +\frac{\partial F_3}{\partial z}=0$ , the last two terms

so that

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 4번

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