발산과 회전 연습문제 | 빌드라인 스쿨

41.OpenStax 3권 · 6.5절 · 46번대학 수준우리말 번역

다음 연습문제에서, 주어진 점에서 F 의 회전을 구하세요.

$\text{F}(x,y,z)=xyz\text{i}+y\text{j}+x\text{k}$ 의 점 $(1,2,3)$ 에서

영어 원문 보기

For the following exercises, find the curl of F at the given point.

$\text{F}(x,y,z)=xyz\text{i}+y\text{j}+x\text{k}$ at $(1,2,3)$

정답·풀이 보기

$\text{curl}\,\text{F}=\text{j}-3\text{k}$

원본 교재 풀이(영어)

$\text{curl}\,\text{F}=\text{j}-3\text{k}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 36번

# 발산과 회전계산자동 분류

42.OpenStax 3권 · 6.5절 · 47번대학 수준우리말 번역

다음 연습문제에서, 주어진 점에서 F 의 회전(curl)을 구하세요.

$\text{F}(x,y,z)={e}^{-xy}\text{i}+{e}^{xz}\text{j}+{e}^{yz}\text{k}$ 에서 점 (3, 2, 0)

영어 원문 보기

For the following exercises, find the curl of F at the given point.

$\text{F}(x,y,z)={e}^{-xy}\text{i}+{e}^{xz}\text{j}+{e}^{yz}\text{k}$ at (3, 2, 0)

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 50번

# 발산과 회전계산자동 분류

43.OpenStax 3권 · 6.5절 · 48번대학 수준우리말 번역

다음 연습문제에서, 주어진 점에서 F 의 회전(curl)을 구하세요.

$\text{F}(x,y,z)=xyz\text{i}+y\text{j}+z\text{k}$ 에서 점 (1, 2, 1)

영어 원문 보기

For the following exercises, find the curl of F at the given point.

$\text{F}(x,y,z)=xyz\text{i}+y\text{j}+z\text{k}$ at (1, 2, 1)

정답·풀이 보기

$\text{curl}\,\text{F}=2\text{j}-\text{k}$

원본 교재 풀이(영어)

$\text{curl}\,\text{F}=2\text{j}-\text{k}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 62번

# 발산과 회전계산자동 분류

44.OpenStax 3권 · 6.5절 · 49번대학 수준우리말 번역

다음 연습문제에서, 주어진 점에서 F 의 회전(curl)을 구하세요.

$\text{F}(x,y,z)={e}^{x}\sin y\text{i}-{e}^{x}\cos y\text{j}$ 에서, 점 (0, 0, 3)에서의 회전을 구하세요

영어 원문 보기

For the following exercises, find the curl of F at the given point.

$\text{F}(x,y,z)={e}^{x}\sin y\text{i}-{e}^{x}\cos y\text{j}$ at (0, 0, 3)

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 22번

# 발산과 회전계산자동 분류

45.OpenStax 3권 · 6.5절 · 50번대학 수준우리말 번역

다음 연습문제에서, 주어진 점에서 F 의 회전(curl)을 구하세요.

$\text{F}(x,y,z)=(3{x}^{2}y+az)\text{i}+{x}^{3}\text{j}+(3x+3{z}^{2})\text{k}$ 라고 하자. a 의 값이 얼마일 때 F 가 보존적 벡터장인가?

영어 원문 보기

For the following exercises, find the curl of F at the given point.

Let $\text{F}(x,y,z)=(3{x}^{2}y+az)\text{i}+{x}^{3}\text{j}+(3x+3{z}^{2})\text{k}.$ For what value of a is F conservative?

답 쓰고 채점

정답·풀이 보기

$a=3$

원본 교재 풀이(영어)

$a=3$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 48번

# 발산과 회전계산자동 분류

46.OpenStax 3권 · 6.5절 · 51번대학 수준우리말 번역

다음 연습문제에서, 주어진 점에서 F 의 회전(curl)을 구하세요.

정의역 $D={\mathbb{R}}^{2}-\{(0,0)\}=\{(x,y)\in {\mathbb{R}}^{2}|(x,y)\ne (0,0)\}$ 에서 주어진 벡터장 $\text{F}(x,y)=\frac{1}{{x}^{2}+{y}^{2}}<-y,x>$ 에 대해, F 는 보존적인가요?

영어 원문 보기

For the following exercises, find the curl of F at the given point.

Given vector field $\text{F}(x,y)=\frac{1}{{x}^{2}+{y}^{2}}<-y,x>$ on domain $D={\mathbb{R}}^{2}-\{(0,0)\}=\{(x,y)\in {\mathbb{R}}^{2}|(x,y)\ne (0,0)\},$ is F conservative?

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 36번

# 발산과 회전계산자동 분류

47.OpenStax 3권 · 6.5절 · 52번대학 수준우리말 번역

다음 연습문제에서 주어진 점에서 F 의 회전(curl)을 구하세요.

정의역 $D={\mathbb{R}}^{2}-\{(0,0)\}=\{(x,y)\in {\mathbb{R}}^{2}|(x,y)\ne (0,0)\}$ , 위에서 벡터장 $\text{F}(x,y)=\frac{1}{{x}^{2}+{y}^{2}}<x,y>$ 가 주어졌을 때, F 는 보존적인지 판단하세요.

영어 원문 보기

For the following exercises, find the curl of F at the given point.

Given vector field $\text{F}(x,y)=\frac{1}{{x}^{2}+{y}^{2}}<x,y>$ on domain $D={\mathbb{R}}^{2}-\{(0,0)\}=\{(x,y)\in {\mathbb{R}}^{2}|(x,y)\ne (0,0)\},$ is F conservative?

정답·풀이 보기

F 는 보존적이다.

원본 교재 풀이(영어)

F is conservative.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 50번

# 발산과 회전계산자동 분류

48.OpenStax 3권 · 6.5절 · 53번대학 수준우리말 번역

힘의 장 $\text{F}(x,y)={e}^{-y}\text{i}-x{e}^{-y}\text{j}$ 가 물체를 P(0, 1)에서 Q(2, 0)까지 움직일 때 한 일을 구하세요. 이 힘의 장은 보존적인지 판단하세요.

영어 원문 보기

Find the work done by force field $\text{F}(x,y)={e}^{-y}\text{i}-x{e}^{-y}\text{j}$ in moving an object from P(0, 1) to Q(2, 0). Is the force field conservative?

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 16번

# 발산과 회전계산자동 분류

49.OpenStax 3권 · 6.5절 · 54번대학 수준우리말 번역

다음 문제에서, 주어진 점에서 F의 회전을 구하세요.

$\text{F}=(\sinh \,x)\text{i}+(\cosh \,y)\text{j}-xyz\text{k}$ 의 발산을 구하세요.

영어 원문 보기

For the following exercises, find the curl of F at the given point.

Compute the divergence of $\text{F}=(\sinh \,x)\text{i}+(\cosh \,y)\text{j}-xyz\text{k}.$

정답·풀이 보기

$\text{div}\,\text{F}=\cosh \,x+\sinh \,y-xy$

원본 교재 풀이(영어)

$\text{div}\,\text{F}=\cosh \,x+\sinh \,y-xy$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 52번

# 발산과 회전계산자동 분류

50.OpenStax 3권 · 6.5절 · 55번대학 수준우리말 번역

다음 연습문제에서 주어진 점에서 F 의 회전(curl)을 구하세요.

$\text{F}=(\sinh \,x)\text{i}+(\cosh \,y)\text{j}-xyz\text{k}$ 의 회전(curl)을 구하세요.

영어 원문 보기

For the following exercises, find the curl of F at the given point.

Compute the curl of $\text{F}=(\sinh \,x)\text{i}+(\cosh \,y)\text{j}-xyz\text{k}.$

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 14번

# 발산과 회전계산자동 분류

51.OpenStax 3권 · 6.5절 · 59번대학 수준우리말 번역

다음 연습문제에서, $\nabla \cdot \text{F}=0$ 이고 $\nabla \cdot \text{G}=0$ 이라고 하세요.

$\text{F}+\text{G}$ 는 반드시 발산이 0인가요?

영어 원문 보기

In the following exercises, suppose that $\nabla \cdot \text{F}=0$ and $\nabla \cdot \text{G}=0.$

Does $\text{F}+\text{G}$ necessarily have zero divergence?

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 5번

# 발산과 회전계산자동 분류

52.OpenStax 3권 · 6.5절 · 60번대학 수준우리말 번역

다음 연습문제에서, $\nabla \cdot \text{F}=0$ 이고 $\nabla \cdot \text{G}=0$ 이라고 하세요.

$\text{F}\,\times \,\text{G}$ 는 반드시 발산이 0인가요?

영어 원문 보기

In the following exercises, suppose that $\nabla \cdot \text{F}=0$ and $\nabla \cdot \text{G}=0.$

Does $\text{F}\,\times \,\text{G}$ necessarily have zero divergence?

정답·풀이 보기

$\text{F}\,\times \,\text{G}$ 는 발산이 0이 아니다.

원본 교재 풀이(영어)

$\text{F}\,\times \,\text{G}$ does not have zero divergence.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 58번

# 발산과 회전계산자동 분류

53.CLP-4 · 4.1절 · 2번대학 수준우리말 번역

${ \boldsymbol{\nabla} }\times \mathbf{F}$ 는 반드시 $\mathbf{F}$ 에 수직인가요?

영어 원문 보기

Does ${ \boldsymbol{\nabla} }\times \mathbf{F}$ have to be perpendicular to $\mathbf{F}$ ?

답 쓰고 채점

정답·풀이 보기

아니요. 벡터장 $\mathbf{F}(x,y,z) = \hat{\boldsymbol{\imath}} +y\,\hat{\mathbf{k}}$ 에 대해

$\begin{aligned} { \boldsymbol{\nabla} }\times \mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ 1 & 0 & y \end{matrix}\right] \\ &=\hat{\boldsymbol{\imath}} \end{aligned}$

이고, 이 벡터는 모든 $x$ , $y$ , $z$ 에 대해 $\mathbf{F}(x,y,z)$ 와의 내적이 $1$ 이므로 이 벡터장과 수직이 아니다.

원본 교재 풀이(영어)

No. The vector field $\mathbf{F}(x,y,z) = \hat{\boldsymbol{\imath}} +y\,\hat{\mathbf{k}}$ has

$\begin{aligned} { \boldsymbol{\nabla} }\times \mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ 1 & 0 & y \end{matrix}\right] \\ &=\hat{\boldsymbol{\imath}} \end{aligned}$

has dot product $1$ with $\mathbf{F}(x,y,z)$ (for all $x$ , $y$ , $z$ ) and so is not perpendicular to it.

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 3번

# 발산과 회전계산기본자동 분류

54.CLP-4 · 4.1절 · 3번대학 수준우리말 번역

다음 벡터 항등식을 확인하세요.

(a) ${ \boldsymbol{\nabla} }\cdot(f\mathbf{F})=f{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f$ (b) ${ \boldsymbol{\nabla} }\cdot(\mathbf{F}\times\mathbf{G}) =\mathbf{G}\cdot({ \boldsymbol{\nabla} }\times\mathbf{F})- \mathbf{F}\cdot({ \boldsymbol{\nabla} }\times\mathbf{G})$ (c) ${ \boldsymbol{\nabla} }^2(fg)=f\,{ \boldsymbol{\nabla} }^2 g+2{ \boldsymbol{\nabla} } f\cdot{ \boldsymbol{\nabla} } g+g\,{ \boldsymbol{\nabla} }^2 f$

영어 원문 보기

Verify the vector identities

(a) ${ \boldsymbol{\nabla} }\cdot(f\mathbf{F})=f{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f$ (b) ${ \boldsymbol{\nabla} }\cdot(\mathbf{F}\times\mathbf{G}) =\mathbf{G}\cdot({ \boldsymbol{\nabla} }\times\mathbf{F})- \mathbf{F}\cdot({ \boldsymbol{\nabla} }\times\mathbf{G})$ (c) ${ \boldsymbol{\nabla} }^2(fg)=f\,{ \boldsymbol{\nabla} }^2 g+2{ \boldsymbol{\nabla} } f\cdot{ \boldsymbol{\nabla} } g+g\,{ \boldsymbol{\nabla} }^2 f$

정답·풀이 보기

(a) 곱의 미분법에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(f\mathbf{F}) &=\frac{\partial }{\partial x}(fF_1) +\frac{\partial }{\partial y}(fF_2) +\frac{\partial }{\partial z}(fF_3) \\ &=\phantom{+}f\frac{\partial F_1}{\partial x} +f\frac{\partial F_2}{\partial y} +f\frac{\partial F_3}{\partial z} \\ &\phantom{=}+F_1\frac{\partial f}{\partial x} +F_2\frac{\partial f}{\partial y} +F_3\frac{\partial f}{\partial z} \\ &=f\,{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f \end{aligned}$

(b) 다시 곱의 미분법에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(\mathbf{F}\times\mathbf{G}) &=\frac{\partial }{\partial x}(F_2G_3-F_3G_2) +\frac{\partial }{\partial y}(F_3G_1-F_1G_3) +\frac{\partial }{\partial z}(F_1G_2-F_2G_1) \\ &=\phantom{+}\frac{\partial F_2}{\partial x}G_3 -\frac{\partial F_3}{\partial x}G_2 +\frac{\partial F_3}{\partial y}G_1 - \frac{\partial F_1}{\partial y}G_3 +\frac{\partial F_1}{\partial z}G_2 - \frac{\partial F_2}{\partial z}G_1 \\ &\phantom{=}+F_2\frac{\partial G_3}{\partial x} -F_3\frac{\partial G_2}{\partial x} +F_3\frac{\partial G_1}{\partial y} -F_1\frac{\partial G_3}{\partial y} +F_1\frac{\partial G_2}{\partial z} - F_2\frac{\partial G_1}{\partial z} \\ &=\phantom{+} \left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)G_1 +\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)G_2 +\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)G_3 \\ &\phantom{=} -F_1\left(\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}\right) -F_2\left(\frac{\partial G_1}{\partial z}-\frac{\partial G_3}{\partial x}\right) -F_3\left(\frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y}\right) \\ &=\mathbf{G}\cdot({ \boldsymbol{\nabla} }\times\mathbf{F})- \mathbf{F}\cdot({ \boldsymbol{\nabla} }\times\mathbf{G}) \end{aligned}$

(c) ${ \boldsymbol{\nabla} }^2(fg) = { \boldsymbol{\nabla} }\cdot\big[{ \boldsymbol{\nabla} }(fg)\big]$ 임을 떠올리자. 먼저

$\begin{aligned} { \boldsymbol{\nabla} }(fg) &=\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x}(fg) +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y}(fg) +\hat{\mathbf{k}}\frac{\partial }{\partial z}(fg) \\ &=\phantom{+}\hat{\boldsymbol{\imath}} g\frac{\partial f}{\partial x} +\hat{\boldsymbol{\jmath}} g\frac{\partial f}{\partial y} +\hat{\mathbf{k}} g\frac{\partial f}{\partial z} \\ &\phantom{=}+\hat{\boldsymbol{\imath}} f\frac{\partial g}{\partial x} +\hat{\boldsymbol{\jmath}} f\frac{\partial g}{\partial y} +\hat{\mathbf{k}} f\frac{\partial g}{\partial z} \\ &=g{ \boldsymbol{\nabla} } f+f{ \boldsymbol{\nabla} } g \end{aligned}$

따라서 (a)를 두 번 이용하면

$\begin{aligned} { \boldsymbol{\nabla} }^2(fg) & = { \boldsymbol{\nabla} }\cdot\big(g{ \boldsymbol{\nabla} } f\big) +{ \boldsymbol{\nabla} }\cdot\big(f{ \boldsymbol{\nabla} } g\big) \\ &= g\big({ \boldsymbol{\nabla} }\cdot { \boldsymbol{\nabla} } f\big) + \big({ \boldsymbol{\nabla} } g\big)\cdot\big({ \boldsymbol{\nabla} } f\big) +f\big({ \boldsymbol{\nabla} }\cdot{ \boldsymbol{\nabla} } g\big) + \big({ \boldsymbol{\nabla} } f\big)\cdot\big({ \boldsymbol{\nabla} } g\big) \\ &=f\,{ \boldsymbol{\nabla} }^2 g+2{ \boldsymbol{\nabla} } f\cdot{ \boldsymbol{\nabla} } g+g\,{ \boldsymbol{\nabla} }^2 f \end{aligned}$

원본 교재 풀이(영어)

(a) By the product rule

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(f\mathbf{F}) &=\frac{\partial }{\partial x}(fF_1) +\frac{\partial }{\partial y}(fF_2) +\frac{\partial }{\partial z}(fF_3) \\ &=\phantom{+}f\frac{\partial F_1}{\partial x} +f\frac{\partial F_2}{\partial y} +f\frac{\partial F_3}{\partial z} \\ &\phantom{=}+F_1\frac{\partial f}{\partial x} +F_2\frac{\partial f}{\partial y} +F_3\frac{\partial f}{\partial z} \\ &=f\,{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f \end{aligned}$

(b) Again by the product rule

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(\mathbf{F}\times\mathbf{G}) &=\frac{\partial }{\partial x}(F_2G_3-F_3G_2) +\frac{\partial }{\partial y}(F_3G_1-F_1G_3) +\frac{\partial }{\partial z}(F_1G_2-F_2G_1) \\ &=\phantom{+}\frac{\partial F_2}{\partial x}G_3 -\frac{\partial F_3}{\partial x}G_2 +\frac{\partial F_3}{\partial y}G_1 - \frac{\partial F_1}{\partial y}G_3 +\frac{\partial F_1}{\partial z}G_2 - \frac{\partial F_2}{\partial z}G_1 \\ &\phantom{=}+F_2\frac{\partial G_3}{\partial x} -F_3\frac{\partial G_2}{\partial x} +F_3\frac{\partial G_1}{\partial y} -F_1\frac{\partial G_3}{\partial y} +F_1\frac{\partial G_2}{\partial z} - F_2\frac{\partial G_1}{\partial z} \\ &=\phantom{+} \left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)G_1 +\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)G_2 +\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)G_3 \\ &\phantom{=} -F_1\left(\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}\right) -F_2\left(\frac{\partial G_1}{\partial z}-\frac{\partial G_3}{\partial x}\right) -F_3\left(\frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y}\right) \\ &=\mathbf{G}\cdot({ \boldsymbol{\nabla} }\times\mathbf{F})- \mathbf{F}\cdot({ \boldsymbol{\nabla} }\times\mathbf{G}) \end{aligned}$

(c) Recall that ${ \boldsymbol{\nabla} }^2(fg) = { \boldsymbol{\nabla} }\cdot\big[{ \boldsymbol{\nabla} }(fg)\big]$ . First

$\begin{aligned} { \boldsymbol{\nabla} }(fg) &=\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x}(fg) +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y}(fg) +\hat{\mathbf{k}}\frac{\partial }{\partial z}(fg) \\ &=\phantom{+}\hat{\boldsymbol{\imath}} g\frac{\partial f}{\partial x} +\hat{\boldsymbol{\jmath}} g\frac{\partial f}{\partial y} +\hat{\mathbf{k}} g\frac{\partial f}{\partial z} \\ &\phantom{=}+\hat{\boldsymbol{\imath}} f\frac{\partial g}{\partial x} +\hat{\boldsymbol{\jmath}} f\frac{\partial g}{\partial y} +\hat{\mathbf{k}} f\frac{\partial g}{\partial z} \\ &=g{ \boldsymbol{\nabla} } f+f{ \boldsymbol{\nabla} } g \end{aligned}$

So by part (a), twice,

$\begin{aligned} { \boldsymbol{\nabla} }^2(fg) & = { \boldsymbol{\nabla} }\cdot\big(g{ \boldsymbol{\nabla} } f\big) +{ \boldsymbol{\nabla} }\cdot\big(f{ \boldsymbol{\nabla} } g\big) \\ &= g\big({ \boldsymbol{\nabla} }\cdot { \boldsymbol{\nabla} } f\big) + \big({ \boldsymbol{\nabla} } g\big)\cdot\big({ \boldsymbol{\nabla} } f\big) +f\big({ \boldsymbol{\nabla} }\cdot{ \boldsymbol{\nabla} } g\big) + \big({ \boldsymbol{\nabla} } f\big)\cdot\big({ \boldsymbol{\nabla} } g\big) \\ &=f\,{ \boldsymbol{\nabla} }^2 g+2{ \boldsymbol{\nabla} } f\cdot{ \boldsymbol{\nabla} } g+g\,{ \boldsymbol{\nabla} }^2 f \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 13번

# 발산과 회전계산기본자동 분류

55.CLP-4 · 4.1절 · 4번대학 수준우리말 번역

다음 각 벡터장에 대해 ${ \boldsymbol{\nabla} }\cdot\mathbf{F}$ 와 ${ \boldsymbol{\nabla} }\times\mathbf{F}$ 의 값을 구하세요.

(a) $\mathbf{F}=x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}$ (b) $\mathbf{F}=xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}$ (c) $\mathbf{F}=\frac{x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}}$ (2차원에서의 극좌표 기저 벡터 $\hat{\bf r}$ ) (d) $\mathbf{F}=\frac{-y\hat{\boldsymbol{\imath}}+x\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}}$ (2차원에서의 극좌표 기저 벡터 $\hat{\boldsymbol{\theta}}$ )

영어 원문 보기

Evaluate ${ \boldsymbol{\nabla} }\cdot\mathbf{F}$ and ${ \boldsymbol{\nabla} }\times\mathbf{F}$ for each of the following vector fields.

(a) $\mathbf{F}=x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}$ (b) $\mathbf{F}=xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}$ (c) $\mathbf{F}=\frac{x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}}$ (the polar basis vector $\hat{\bf r}$ in 2d) (d) $\mathbf{F}=\frac{-y\hat{\boldsymbol{\imath}}+x\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}}$ (the polar basis vector $\hat{\boldsymbol{\theta}}$ in 2d)

정답·풀이 보기

(a) 정의에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}) &=\frac{\partial }{\partial x}\big(x\big) +\frac{\partial }{\partial y}\big(y\big) +\frac{\partial }{\partial z}\big(z\big) = 3 \\ { \boldsymbol{\nabla} }\times(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] x & y & z\end{matrix}\right] = \mathbf{0} \end{aligned}$

(b) 정의에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}) &=\frac{\partial }{\partial x}\big(xy^2\big) +\frac{\partial }{\partial y}\big(-yz^2\big) +\frac{\partial }{\partial z}\big(zx^2\big) =y^2-z^2+x^2 \\ { \boldsymbol{\nabla} }\times(xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] xy^2 & -yz^2 & zx^2\end{matrix}\right] = 2yz\,\hat{\boldsymbol{\imath}}-2xz\,\hat{\boldsymbol{\jmath}}-2xy\,\hat{\mathbf{k}} \end{aligned}$

(c) 정의에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\left(\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\frac{\partial }{\partial x}\left(\frac{x}{\sqrt{x^2+y^2}}\right) +\frac{\partial }{\partial y}\left(\frac{y}{\sqrt{x^2+y^2}}\right) \\ &= \frac{1}{\sqrt{x^2+y^2}} - \frac{x^2}{{[x^2+y^2]}^{3/2}} +\frac{1}{\sqrt{x^2+y^2}} - \frac{y^2}{{[x^2+y^2]}^{3/2}} \\ &=\frac{x^2+y^2\ -x^2\ +x^2+y^2\ -\ y^2}{{[x^2+y^2]}^{3/2}} \\ &=\frac{1}{\sqrt{x^2+y^2}} \\ { \boldsymbol{\nabla} }\times\left(\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} & 0\end{matrix}\right] \\[0.05in] &=\left(- \frac{xy}{{[x^2+y^2]}^{3/2}}+ \frac{xy}{{[x^2+y^2]}^{3/2}}\right)\hat{\mathbf{k}} = \mathbf{0} \end{aligned}$

(d) 정의에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\left(-\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\frac{\partial }{\partial x}\left(-\frac{y}{\sqrt{x^2+y^2}}\right) +\frac{\partial }{\partial y}\left(\frac{x}{\sqrt{x^2+y^2}}\right) \\ &= \frac{xy}{{[x^2+y^2]}^{3/2}} - \frac{xy}{{[x^2+y^2]}^{3/2}} =0 \\ { \boldsymbol{\nabla} }\times\left(-\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}} +\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] -\frac{y}{\sqrt{x^2+y^2}} & \frac{x}{\sqrt{x^2+y^2}} & 0\end{matrix}\right] \\[0.05in] &=\!\left(\!\frac{1}{\sqrt{x^2+y^2}} - \frac{x^2}{{[x^2+y^2]}^{3/2}} +\frac{1}{\sqrt{x^2+y^2}} - \frac{y^2}{{[x^2+y^2]}^{3/2}}\!\right)\!\hat{\mathbf{k}} \\ &= \frac{x^2+y^2\ -x^2\ +x^2+y^2\ -\ y^2}{{[x^2+y^2]}^{3/2}}\ \hat{\mathbf{k}} = \frac{\hat{\mathbf{k}}}{\sqrt{x^2+y^2}} \end{aligned}$

원본 교재 풀이(영어)

(a) By definition

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}) &=\frac{\partial }{\partial x}\big(x\big) +\frac{\partial }{\partial y}\big(y\big) +\frac{\partial }{\partial z}\big(z\big) = 3 \\ { \boldsymbol{\nabla} }\times(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] x & y & z\end{matrix}\right] = \mathbf{0} \end{aligned}$

(b) By definition

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}) &=\frac{\partial }{\partial x}\big(xy^2\big) +\frac{\partial }{\partial y}\big(-yz^2\big) +\frac{\partial }{\partial z}\big(zx^2\big) =y^2-z^2+x^2 \\ { \boldsymbol{\nabla} }\times(xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] xy^2 & -yz^2 & zx^2\end{matrix}\right] = 2yz\,\hat{\boldsymbol{\imath}}-2xz\,\hat{\boldsymbol{\jmath}}-2xy\,\hat{\mathbf{k}} \end{aligned}$

(c) By definition

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\left(\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\frac{\partial }{\partial x}\left(\frac{x}{\sqrt{x^2+y^2}}\right) +\frac{\partial }{\partial y}\left(\frac{y}{\sqrt{x^2+y^2}}\right) \\ &= \frac{1}{\sqrt{x^2+y^2}} - \frac{x^2}{{[x^2+y^2]}^{3/2}} +\frac{1}{\sqrt{x^2+y^2}} - \frac{y^2}{{[x^2+y^2]}^{3/2}} \\ &=\frac{x^2+y^2\ -x^2\ +x^2+y^2\ -\ y^2}{{[x^2+y^2]}^{3/2}} \\ &=\frac{1}{\sqrt{x^2+y^2}} \\ { \boldsymbol{\nabla} }\times\left(\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} & 0\end{matrix}\right] \\[0.05in] &=\left(- \frac{xy}{{[x^2+y^2]}^{3/2}}+ \frac{xy}{{[x^2+y^2]}^{3/2}}\right)\hat{\mathbf{k}} = \mathbf{0} \end{aligned}$

(d) By definition

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\left(-\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\frac{\partial }{\partial x}\left(-\frac{y}{\sqrt{x^2+y^2}}\right) +\frac{\partial }{\partial y}\left(\frac{x}{\sqrt{x^2+y^2}}\right) \\ &= \frac{xy}{{[x^2+y^2]}^{3/2}} - \frac{xy}{{[x^2+y^2]}^{3/2}} =0 \\ { \boldsymbol{\nabla} }\times\left(-\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}} +\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] -\frac{y}{\sqrt{x^2+y^2}} & \frac{x}{\sqrt{x^2+y^2}} & 0\end{matrix}\right] \\[0.05in] &=\!\left(\!\frac{1}{\sqrt{x^2+y^2}} - \frac{x^2}{{[x^2+y^2]}^{3/2}} +\frac{1}{\sqrt{x^2+y^2}} - \frac{y^2}{{[x^2+y^2]}^{3/2}}\!\right)\!\hat{\mathbf{k}} \\ &= \frac{x^2+y^2\ -x^2\ +x^2+y^2\ -\ y^2}{{[x^2+y^2]}^{3/2}}\ \hat{\mathbf{k}} = \frac{\hat{\mathbf{k}}}{\sqrt{x^2+y^2}} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 5번

# 발산과 회전계산보통자동 분류

56.CLP-4 · 4.1절 · 5번대학 수준우리말 번역

(a) $\mathbf{r}=(x,y,z)$ , $r=|(x,y,z)|$ 일 때 ${ \boldsymbol{\nabla} }\cdot\big(\frac{\mathbf{r}}{r}\big)$ 를 계산하고 간단히 하세요. 답을 $r$ 에 관한 식으로 나타내세요. (b) ${ \boldsymbol{\nabla} }\times\big(yz\,\hat{\boldsymbol{\imath}} + 2xz\,\hat{\boldsymbol{\jmath}} + e^{xy}\,\hat{\mathbf{k}}\big)$ 를 계산하세요.

영어 원문 보기

(a) Compute and simplify ${ \boldsymbol{\nabla} }\cdot\big(\frac{\mathbf{r}}{r}\big)$ for $\mathbf{r}=(x,y,z)$ and $r=|(x,y,z)|$ . Express your answer in terms of $r$ . (b) Compute ${ \boldsymbol{\nabla} }\times\big(yz\,\hat{\boldsymbol{\imath}} + 2xz\,\hat{\boldsymbol{\jmath}} + e^{xy}\,\hat{\mathbf{k}}\big)$ .

정답·풀이 보기

(a) $\frac{\mathbf{r}}{r} =\frac{x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}}{[x^2+y^2+z^2]^{1/2}}$ 의 발산을 계산하면 된다.

$\begin{aligned} \frac{\partial }{\partial x}\frac{x}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{x(2x)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{y^2+z^2}{{[x^2+y^2+z^2]}^{3/2}} \\ \frac{\partial }{\partial y}\frac{y}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{y(2y)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{x^2+z^2}{{[x^2+y^2+z^2]}^{3/2}} \\ \frac{\partial }{\partial z}\frac{z}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{z(2z)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{x^2+y^2}{{[x^2+y^2+z^2]}^{3/2}} \end{aligned}$

이므로 구하는 발산은 다음과 같다.

$\begin{aligned} { \boldsymbol{\nabla} }\left(\frac{\mathbf{r}}{r}\right) &= \frac{2x^2+2y^2+2z^2}{{[x^2+y^2+z^2]}^{3/2}} =\frac{2r^2}{r^3} =\frac{2}{r} \end{aligned}$

(b)

$\begin{aligned} { \boldsymbol{\nabla} }\times\big(yz\,\hat{\boldsymbol{\imath}} + 2xz\,\hat{\boldsymbol{\jmath}} + e^{xy}\,\hat{\mathbf{k}}\big) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in] \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\[0.03in] yz&2xz&e^{xy}\end{matrix}\right] =\big(xe^{xy}-2x\big)\,\hat{\boldsymbol{\imath}}-\big(ye^{xy}-y\big)\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}} \end{aligned}$

원본 교재 풀이(영어)

(a) We are to compute the divergence of $\frac{\mathbf{r}}{r} =\frac{x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}}{[x^2+y^2+z^2]^{1/2}}$ . Since

$\begin{aligned} \frac{\partial }{\partial x}\frac{x}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{x(2x)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{y^2+z^2}{{[x^2+y^2+z^2]}^{3/2}} \\ \frac{\partial }{\partial y}\frac{y}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{y(2y)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{x^2+z^2}{{[x^2+y^2+z^2]}^{3/2}} \\ \frac{\partial }{\partial z}\frac{z}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{z(2z)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{x^2+y^2}{{[x^2+y^2+z^2]}^{3/2}} \end{aligned}$

the specified divergence is

$\begin{aligned} { \boldsymbol{\nabla} }\left(\frac{\mathbf{r}}{r}\right) &= \frac{2x^2+2y^2+2z^2}{{[x^2+y^2+z^2]}^{3/2}} =\frac{2r^2}{r^3} =\frac{2}{r} \end{aligned}$

(b)

$\begin{aligned} { \boldsymbol{\nabla} }\times\big(yz\,\hat{\boldsymbol{\imath}} + 2xz\,\hat{\boldsymbol{\jmath}} + e^{xy}\,\hat{\mathbf{k}}\big) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in] \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\[0.03in] yz&2xz&e^{xy}\end{matrix}\right] =\big(xe^{xy}-2x\big)\,\hat{\boldsymbol{\imath}}-\big(ye^{xy}-y\big)\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 6번

# 발산과 회전계산보통자동 분류

57.CLP-4 · 4.1절 · 6번대학 수준우리말 번역

다음에서 기호 $\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}}$ , $r = |\mathbf{r}|$ 를 사용하고, $k$ 는 $k = 0, 1, -1, 2, -2, \dots$ 중 어떤 수이다.

(a) 다음을 만족하는 $k$ 의 값을 구하세요.

${ \boldsymbol{\nabla} } (r^k) = -3\frac{\mathbf{r}}{r^5}$ (b) 다음을 만족하는 $k$ 의 값을 구하세요.

${ \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) = 5r^2$ (c) 다음을 만족하는 $k$ 의 값을 구하세요.

${ \boldsymbol{\nabla} }^2 (r^k) = \frac{2}{r^4}$

영어 원문 보기

In the following, we use the notation $\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}}$ , $r = |\mathbf{r}|$ , and $k$ is some number $k = 0, 1, -1, 2, -2, \dots$ .

(a) Find the value $k$ for which

${ \boldsymbol{\nabla} } (r^k) = -3\frac{\mathbf{r}}{r^5}$ (b) Find the value $k$ for which

${ \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) = 5r^2$ (c) Find the value $k$ for which

${ \boldsymbol{\nabla} }^2 (r^k) = \frac{2}{r^4}$

답 쓰고 채점

(a)

(b)

(c)

정답·풀이 보기

(a) $r^k=\big(x^2+y^2+z^2\big)^{k/2}$ 이므로,

$\begin{aligned} \frac{\partial }{\partial x} r^k & = 2x\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\boldsymbol{\imath}})\ r^{k-2} \\ \frac{\partial }{\partial y} r^k & = 2y\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\boldsymbol{\jmath}})\ r^{k-2} \\ \frac{\partial }{\partial z} r^k & = 2z\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\mathbf{k}})\ r^{k-2} \end{aligned}$

구하는 값은 $k=-3$ 이다.

(b) (a)의 계산을 이용하면

$\begin{aligned} { \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) &= \frac{\partial }{\partial x} (x r^k) + \frac{\partial }{\partial y} (y r^k) + \frac{\partial }{\partial z} (z r^k) \\ &= 3r^k + x \frac{\partial }{\partial x} r^k + y \frac{\partial }{\partial y} r^k + z \frac{\partial }{\partial z} r^k \\ &= 3r^k + x \big(kx\ r^{k-2}\big) + y \big(ky\ r^{k-2}\big) + z \big(kz\ r^{k-2}\big) \\ &= \big(3+k\big)r^k \end{aligned}$

구하는 값은 $k=2$ 이다.

(c) ${ \boldsymbol{\nabla} }^2 = { \boldsymbol{\nabla} }\cdot{ \boldsymbol{\nabla} }$ 임을 떠올리면,

$\begin{aligned} { \boldsymbol{\nabla} }^2 (r^k ) &={ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^k)\big) \\ &={ \boldsymbol{\nabla} }\cdot(k r^{k-2}\,\mathbf{r}) &\text{(a)에 의해} \\ &=k(3+k-2)r^{k-2} &\text{(b)에 의해, 단 $k$ 를 $k-2$ 로 바꾼다} \end{aligned}$

구하는 값은 $k=-2$ 이다.

원본 교재 풀이(영어)

(a) Since $r^k=\big(x^2+y^2+z^2\big)^{k/2}$ ,

$\begin{aligned} \frac{\partial }{\partial x} r^k & = 2x\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\boldsymbol{\imath}})\ r^{k-2} \\ \frac{\partial }{\partial y} r^k & = 2y\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\boldsymbol{\jmath}})\ r^{k-2} \\ \frac{\partial }{\partial z} r^k & = 2z\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\mathbf{k}})\ r^{k-2} \end{aligned}$

We want $k=-3$ .

(b) Using the computation in part (a)

$\begin{aligned} { \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) &= \frac{\partial }{\partial x} (x r^k) + \frac{\partial }{\partial y} (y r^k) + \frac{\partial }{\partial z} (z r^k) \\ &= 3r^k + x \frac{\partial }{\partial x} r^k + y \frac{\partial }{\partial y} r^k + z \frac{\partial }{\partial z} r^k \\ &= 3r^k + x \big(kx\ r^{k-2}\big) + y \big(ky\ r^{k-2}\big) + z \big(kz\ r^{k-2}\big) \\ &= \big(3+k\big)r^k \end{aligned}$

We want $k=2$ .

(c) Recalling that ${ \boldsymbol{\nabla} }^2 = { \boldsymbol{\nabla} }\cdot{ \boldsymbol{\nabla} }$ ,

$\begin{aligned} { \boldsymbol{\nabla} }^2 (r^k ) &={ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^k)\big) \\ &={ \boldsymbol{\nabla} }\cdot(k r^{k-2}\,\mathbf{r}) &\text{by part (a)} \\ &=k(3+k-2)r^{k-2} &\text{by part (b), but with $k$ replaced by $k-2$} \end{aligned}$

We want $k=-2$ .

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 7번

# 발산과 회전계산보통자동 분류

58.CLP-4 · 4.1절 · 7번대학 수준우리말 번역

$\mathbf{r}$ 를 벡터장 $\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}}$ 라 하고, $r$ 를 함수 $r = |\mathbf{r}|$ 라 하자. $\mathbf{a}$ 를 상수 벡터 $\mathbf{a} = a_1\,\hat{\boldsymbol{\imath}} + a_2\,\hat{\boldsymbol{\jmath}} + a_3\,\hat{\mathbf{k}}$ 라 하자. 다음 양을 계산하여 간단히 하세요. 답은 $\mathbf{a}$ , $\mathbf{r}$ , $r$ 로 나타내야 합니다. 답에 $x$ , $y$ , $z$ 가 들어 있으면 안 됩니다.

(a) ${ \boldsymbol{\nabla} }\cdot\mathbf{r}$ (b) ${ \boldsymbol{\nabla} }(r^2)$ (c) ${ \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a})$ (d) ${ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big)$

영어 원문 보기

Let $\mathbf{r}$ be the vector field $\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}}$ and let $r$ be the function $r = |\mathbf{r}|$ . Let $\mathbf{a}$ be the constant vector $\mathbf{a} = a_1\,\hat{\boldsymbol{\imath}} + a_2\,\hat{\boldsymbol{\jmath}} + a_3\,\hat{\mathbf{k}}$ . Compute and simplify the following quantities. Answers must be expressed in terms of $\mathbf{a}$ , $\mathbf{r}$ , and $r$ . There should be no $x$ 's, $y$ 's, or $z$ 's in your answers.

(a) ${ \boldsymbol{\nabla} }\cdot\mathbf{r}$ (b) ${ \boldsymbol{\nabla} }(r^2)$ (c) ${ \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a})$ (d) ${ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big)$

정답·풀이 보기

(a)

${ \boldsymbol{\nabla} }\cdot\mathbf{r} =\frac{\partial x}{\partial x} +\frac{\partial y}{\partial y} +\frac{\partial z}{\partial z} =3$

(b)

${ \boldsymbol{\nabla} }(r^2) =\left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big) =2x\,\hat{\boldsymbol{\imath}} +2y\,\hat{\boldsymbol{\jmath}} +2z\,\hat{\mathbf{k}} =2\mathbf{r}$

(c) 다음과 같으므로

$\mathbf{r}\times\mathbf{a} = \det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ x & y & z \\ a_1 & a_2 & a_3 \end{matrix}\right] =\hat{\boldsymbol{\imath}}\big(a_3y-a_2z\big) +\hat{\boldsymbol{\jmath}}\big(a_1z-a_3x\big) +\hat{\mathbf{k}}\big(a_2x-a_1y\big)$

다음을 얻는다.

$\begin{aligned} { \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a}) &= \det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in] \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\[0.03in] a_3y-a_2z& a_1z-a_3x & a_2x-a_1y\end{matrix}\right] =-2a_1\,\hat{\boldsymbol{\imath}} -2a_2\,\hat{\boldsymbol{\jmath}} -2a_3\,\hat{\mathbf{k}} \\ &=-2\mathbf{a} \end{aligned}$

(d) 다음과 같으므로

$\begin{aligned} { \boldsymbol{\nabla} }(r) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{1/2} \\ &=\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} +\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{1/2}} +\hat{\mathbf{k}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} \end{aligned}$

다음을 얻는다.

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big) &= \frac{\partial }{\partial x} \frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} +\frac{\partial }{\partial y} \frac{y}{\big(x^2+y^2+z^2\big)^{1/2}} +\frac{\partial }{\partial z} \frac{z}{\big(x^2+y^2+z^2\big)^{1/2}} \\ &= \frac{3}{\big(x^2+y^2+z^2\big)^{1/2}} -\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{3/2}} =\frac{2}{\big(x^2+y^2+z^2\big)^{1/2}} =\frac{2}{r} \end{aligned}$

원본 교재 풀이(영어)

(a)

${ \boldsymbol{\nabla} }\cdot\mathbf{r} =\frac{\partial x}{\partial x} +\frac{\partial y}{\partial y} +\frac{\partial z}{\partial z} =3$

(b)

${ \boldsymbol{\nabla} }(r^2) =\left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big) =2x\,\hat{\boldsymbol{\imath}} +2y\,\hat{\boldsymbol{\jmath}} +2z\,\hat{\mathbf{k}} =2\mathbf{r}$

(c) Since

$\mathbf{r}\times\mathbf{a} = \det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ x & y & z \\ a_1 & a_2 & a_3 \end{matrix}\right] =\hat{\boldsymbol{\imath}}\big(a_3y-a_2z\big) +\hat{\boldsymbol{\jmath}}\big(a_1z-a_3x\big) +\hat{\mathbf{k}}\big(a_2x-a_1y\big)$

we have

$\begin{aligned} { \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a}) &= \det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in] \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\[0.03in] a_3y-a_2z& a_1z-a_3x & a_2x-a_1y\end{matrix}\right] =-2a_1\,\hat{\boldsymbol{\imath}} -2a_2\,\hat{\boldsymbol{\jmath}} -2a_3\,\hat{\mathbf{k}} \\ &=-2\mathbf{a} \end{aligned}$

(d) Since

$\begin{aligned} { \boldsymbol{\nabla} }(r) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{1/2} \\ &=\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} +\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{1/2}} +\hat{\mathbf{k}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} \end{aligned}$

we have

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big) &= \frac{\partial }{\partial x} \frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} +\frac{\partial }{\partial y} \frac{y}{\big(x^2+y^2+z^2\big)^{1/2}} +\frac{\partial }{\partial z} \frac{z}{\big(x^2+y^2+z^2\big)^{1/2}} \\ &= \frac{3}{\big(x^2+y^2+z^2\big)^{1/2}} -\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{3/2}} =\frac{2}{\big(x^2+y^2+z^2\big)^{1/2}} =\frac{2}{r} \end{aligned}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 30번

# 발산과 회전계산보통자동 분류

59.CLP-4 · 4.1절 · 8번대학 수준우리말 번역

다음과 같이 두자.

$\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}},\qquad r = |\mathbf{r}|$

(a) ${ \boldsymbol{\nabla} }\big(\frac{1}{r}\big) =- r^a\,\mathbf{r}$ 를 만족하는 $a$ 의 값을 구하세요. (b) ${ \boldsymbol{\nabla} }\cdot\big(r\,\mathbf{r}\big) = ar$ 를 만족하는 $a$ 의 값을 구하세요. (c) ${ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^3)\big) = ar$ 를 만족하는 $a$ 의 값을 구하세요.

영어 원문 보기

Let

$\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}},\qquad r = |\mathbf{r}|$

(a) Compute $a$ where ${ \boldsymbol{\nabla} }\big(\frac{1}{r}\big) =- r^a\,\mathbf{r}$ . (b) Compute $a$ where ${ \boldsymbol{\nabla} }\cdot\big(r\,\mathbf{r}\big) = ar$ . (c) Compute $a$ where ${ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^3)\big) = ar$ .

답 쓰고 채점

(a)

(b)

(c)

정답·풀이 보기

(a) 다음과 같으므로

$\begin{aligned} { \boldsymbol{\nabla} }\left(\frac{1}{r}\right) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{-1/2} \\ &=-\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{3/2}} -\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{3/2}} -\hat{\mathbf{k}}\frac{z}{\big(x^2+y^2+z^2\big)^{3/2}} \\ &=-\hat{\boldsymbol{\imath}}\frac{x}{r^3} -\hat{\boldsymbol{\jmath}}\frac{y}{r^3} -\hat{\mathbf{k}}\frac{x}{r^3} \end{aligned}$

$a=-3$ 이다.

(b) 다음과 같으므로

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big(r\,\mathbf{r}\big) &= \frac{\partial }{\partial x} \Big[\big(x^2+y^2+z^2\big)^{1/2}x\Big] +\frac{\partial }{\partial y}\Big[\big(x^2+y^2+z^2\big)^{1/2}y\Big] +\frac{\partial }{\partial z} \Big[\big(x^2+y^2+z^2\big)^{1/2}z\Big] \\ &= 3\big(x^2+y^2+z^2\big)^{1/2} +\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{1/2}} =4\big(x^2+y^2+z^2\big)^{1/2} =4r \end{aligned}$

$a=4$ 이다.

(c) 다음과 같으므로

$\begin{aligned} { \boldsymbol{\nabla} }(r^3) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{3/2} \\ &=\hat{\boldsymbol{\imath}}\,3x\big(x^2+y^2+z^2\big)^{1/2} +\hat{\boldsymbol{\jmath}}\,3y\big(x^2+y^2+z^2\big)^{1/2} +\hat{\mathbf{k}}\,3z\big(x^2+y^2+z^2\big)^{1/2} \\ &=3r\mathbf{r} \end{aligned}$

다음을 얻는다.

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^3)\big) &= { \boldsymbol{\nabla} }\cdot\big(3r\mathbf{r}\big) = 3\,{ \boldsymbol{\nabla} }\cdot\big(r\mathbf{r}\big) = 3\,\big(4r\big) \qquad\text{(b)에 의해} \\ &= 12 r \end{aligned}$

따라서 $a=12$ 이다.

원본 교재 풀이(영어)

(a) Since

$\begin{aligned} { \boldsymbol{\nabla} }\left(\frac{1}{r}\right) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{-1/2} \\ &=-\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{3/2}} -\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{3/2}} -\hat{\mathbf{k}}\frac{z}{\big(x^2+y^2+z^2\big)^{3/2}} \\ &=-\hat{\boldsymbol{\imath}}\frac{x}{r^3} -\hat{\boldsymbol{\jmath}}\frac{y}{r^3} -\hat{\mathbf{k}}\frac{x}{r^3} \end{aligned}$

we have $a=-3$ .

(b) Since

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big(r\,\mathbf{r}\big) &= \frac{\partial }{\partial x} \Big[\big(x^2+y^2+z^2\big)^{1/2}x\Big] +\frac{\partial }{\partial y}\Big[\big(x^2+y^2+z^2\big)^{1/2}y\Big] +\frac{\partial }{\partial z} \Big[\big(x^2+y^2+z^2\big)^{1/2}z\Big] \\ &= 3\big(x^2+y^2+z^2\big)^{1/2} +\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{1/2}} =4\big(x^2+y^2+z^2\big)^{1/2} =4r \end{aligned}$

we have $a=4$ .

(c) Since

$\begin{aligned} { \boldsymbol{\nabla} }(r^3) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{3/2} \\ &=\hat{\boldsymbol{\imath}}\,3x\big(x^2+y^2+z^2\big)^{1/2} +\hat{\boldsymbol{\jmath}}\,3y\big(x^2+y^2+z^2\big)^{1/2} +\hat{\mathbf{k}}\,3z\big(x^2+y^2+z^2\big)^{1/2} \\ &=3r\mathbf{r} \end{aligned}$

we have

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^3)\big) &= { \boldsymbol{\nabla} }\cdot\big(3r\mathbf{r}\big) = 3\,{ \boldsymbol{\nabla} }\cdot\big(r\mathbf{r}\big) = 3\,\big(4r\big) \qquad\text{by part (b)} \\ &= 12 r \end{aligned}$

so that $a=12$ .

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 9번

# 발산과 회전계산보통자동 분류

60.CLP-4 · 4.1절 · 9번대학 수준우리말 번역

$\hat{\mathbf{k}}$ 성분이 $A_3=0$ 이고 다음의 벡터 퍼텐셜이 되는 벡터장 $\mathbf{A}$ 를 가능하면 구하세요.

(a) $\mathbf{F}=(1+yz)\hat{\boldsymbol{\imath}}+(2y+zx)\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}}$ (b) $\mathbf{G}= yz\hat{\boldsymbol{\imath}}+zx\hat{\boldsymbol{\jmath}}+xy\hat{\mathbf{k}}$

영어 원문 보기

Find, if possible, a vector field $\mathbf{A}$ that has $\hat{\mathbf{k}}$ component $A_3=0$ and that is a vector potential for

(a) $\mathbf{F}=(1+yz)\hat{\boldsymbol{\imath}}+(2y+zx)\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}}$ (b) $\mathbf{G}= yz\hat{\boldsymbol{\imath}}+zx\hat{\boldsymbol{\jmath}}+xy\hat{\mathbf{k}}$

정답·풀이 보기

(a) ${ \boldsymbol{\nabla} }\cdot\mathbf{F} =\frac{\partial }{\partial x}(1+yz)+\frac{\partial }{\partial y}(2y+zx)+\frac{\partial }{\partial z}(3z^2+xy)=2+6z\ne 0$ 이므로 $\mathbf{F}$ 는 판별 조건을 만족하지 못하여 벡터 퍼텐셜을 가질 수 없다.

(b) 벡터장 $\mathbf{A}=A_1\hat{\boldsymbol{\imath}}+A_2\hat{\boldsymbol{\jmath}}$ 가 $\mathbf{G}$ 의 벡터 퍼텐셜이 될 필요충분조건은 $\mathbf{G}={ \boldsymbol{\nabla} }\times\mathbf{A}$ 이고, 이는 다음이 성립할 필요충분조건과 같다.

$\begin{aligned} -\frac{\partial A_2}{\partial z}&= yz\quad & \iff&\quad& &A_2=-\frac{1}{2} yz^2+B_2(x,y)\\ \frac{\partial A_1}{\partial z}&=zx & \iff&\quad & &A_1=\phantom{-}\frac{1}{2} xz^2+B_1(x,y)\\ \frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y} &=xy & \iff& & & \frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy \end{aligned}$

$\frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy$ 의 해는 무수히 많다. 실제로 $B_2$ 는 완전히 임의로 정할 수 있다. $B_2=0$ 으로 택하면 $B_1=-\frac{1}{2} xy^2$ 이 조건을 만족한다. $B_1=0$ 으로 택하면 $B_2=\frac{1}{2} x^2y$ 가 조건을 만족한다. 따라서 두 가지 해는 $\mathbf{A}=\frac{1}{2}(z^2-y^2)x\hat{\boldsymbol{\imath}}-\frac{1}{2} yz^2\hat{\boldsymbol{\jmath}}$ 와 $\mathbf{A}=\frac{1}{2} xz^2\hat{\boldsymbol{\imath}}+\frac{1}{2}(x^2-z^2)y\hat{\boldsymbol{\jmath}}$ 이다.

원본 교재 풀이(영어)

(a) Since ${ \boldsymbol{\nabla} }\cdot\mathbf{F} =\frac{\partial }{\partial x}(1+yz)+\frac{\partial }{\partial y}(2y+zx)+\frac{\partial }{\partial z}(3z^2+xy)=2+6z\ne 0$ , $\mathbf{F}$ fails the screening test and cannot have a vector potential.

(b) The vector field $\mathbf{A}=A_1\hat{\boldsymbol{\imath}}+A_2\hat{\boldsymbol{\jmath}}$ is a vector potential for $\mathbf{G}$ if and only if $\mathbf{G}={ \boldsymbol{\nabla} }\times\mathbf{A}$ , which is the case if and only if

$\begin{aligned} -\frac{\partial A_2}{\partial z}&= yz\quad & \iff&\quad& &A_2=-\frac{1}{2} yz^2+B_2(x,y)\\ \frac{\partial A_1}{\partial z}&=zx & \iff&\quad & &A_1=\phantom{-}\frac{1}{2} xz^2+B_1(x,y)\\ \frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y} &=xy & \iff& & & \frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy \end{aligned}$

There are infinitely many solutions to $\frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy$ . In fact $B_2$ is completely arbitrary. If one chooses $B_2=0$ , then $B_1=-\frac{1}{2} xy^2$ does the job. If one chooses $B_1=0$ , then $B_2=\frac{1}{2} x^2y$ does the job. Thus two solutions are $\mathbf{A}=\frac{1}{2}(z^2-y^2)x\hat{\boldsymbol{\imath}}-\frac{1}{2} yz^2\hat{\boldsymbol{\jmath}}$ and $\mathbf{A}=\frac{1}{2} xz^2\hat{\boldsymbol{\imath}}+\frac{1}{2}(x^2-z^2)y\hat{\boldsymbol{\jmath}}$ .

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 2번

# 발산과 회전계산보통자동 분류