빌드라인 스쿨

문제 은행 · 미적분 · 대학 미적분 · 벡터 미적분

발산과 회전

벡터장의 발산(div)과 회전(curl)을 계산하고 뜻을 해석한다

유형전체개념 확인계산그래프·표 해석활용증명·논술
난이도전체기본보통도전
풀이전체우리말 풀이 있는 문제만

유형·난이도는 우리말 풀이가 붙은 문제는 선생님이 정했고, 나머지는 자동 분류라 다를 수 있어요. 난이도가 정해지지 않은 문제는 난이도로 거를 때 빠집니다.

72문제 · 4 / 4쪽

61.OpenStax 3권 · 6.5절 · 61번대학 수준우리말 번역

다음 연습문제에서, R3{\mathbb{R}}^{3} 안의 고체가 온도 분포 T(x,y,z)=100e−x2+y2+z2T(x,y,z)=100{e}^{-{x}^{2}+{y}^{2}+{z}^{2}} 을 갖는다고 하세요. 이 물체 안의 열류 벡터장은 F=−k∇T\text{F}=-k\nabla T 이며, 여기서 k>0k>0 은 물질의 성질입니다. 열류 벡터는 기울기의 반대 방향, 즉 온도가 가장 크게 감소하는 방향을 가리킵니다. 열류 벡터의 발산은 ∇⋅F=−k∇⋅∇T=−k∇2T\nabla \cdot \text{F}=-k\nabla \cdot \nabla T=-k{\nabla }^{2}T 입니다.

열류 벡터장을 계산하세요.

영어 원문 보기

In the following exercises, suppose a solid object in R3{\mathbb{R}}^{3} has a temperature distribution given by T(x,y,z)=100e−x2+y2+z2T(x,y,z)=100{e}^{-{x}^{2}+{y}^{2}+{z}^{2}}. The heat flow vector field in the object is F=−k∇T,\text{F}=-k\nabla T, where k>0k>0 is a property of the material. The heat flow vector points in the direction opposite to that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is ∇⋅F=−k∇⋅∇T=−k∇2T.\nabla \cdot \text{F}=-k\nabla \cdot \nabla T=-k{\nabla }^{2}T.

Compute the heat flow vector field.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 50번
# 발산과 회전활용자동 분류
OpenStax Calculus Volume 3 CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
62.OpenStax 3권 · 6.5절 · 62번대학 수준우리말 번역

다음 연습문제에서, R3{\mathbb{R}}^{3} 안의 입체가 T(x,y,z)=100e−x2+y2+z2T(x,y,z)=100{e}^{-{x}^{2}+{y}^{2}+{z}^{2}} 으로 주어지는 온도 분포를 가진다고 하자. 이 물체 안의 열 흐름 벡터장은 F=−k∇T\text{F}=-k\nabla T 이며, 여기서 k>0k>0 은 물질의 성질이다. 열 흐름 벡터는 기울기(그래디언트)와 반대 방향, 즉 온도가 가장 크게 감소하는 방향을 가리킨다. 열 흐름 벡터의 발산은 ∇⋅F=−k∇⋅∇T=−k∇2T\nabla \cdot \text{F}=-k\nabla \cdot \nabla T=-k{\nabla }^{2}T 이다.

발산을 계산하세요.

영어 원문 보기

In the following exercises, suppose a solid object in R3{\mathbb{R}}^{3} has a temperature distribution given by T(x,y,z)=100e−x2+y2+z2T(x,y,z)=100{e}^{-{x}^{2}+{y}^{2}+{z}^{2}}. The heat flow vector field in the object is F=−k∇T,\text{F}=-k\nabla T, where k>0k>0 is a property of the material. The heat flow vector points in the direction opposite to that of the gradient, which is the direction of greatest temperature decrease. The divergence of the heat flow vector is ∇⋅F=−k∇⋅∇T=−k∇2T.\nabla \cdot \text{F}=-k\nabla \cdot \nabla T=-k{\nabla }^{2}T.

Compute the divergence.

정답·풀이 보기

∇⋅F=−200k[1+2(x2+y2+z2)]e−x2+y2+z2\nabla \cdot \text{F}=-200k[1+2({x}^{2}+{y}^{2}+{z}^{2})]{e}^{-{x}^{2}+{y}^{2}+{z}^{2}}

원본 교재 풀이(영어)

∇⋅F=−200k[1+2(x2+y2+z2)]e−x2+y2+z2\nabla \cdot \text{F}=-200k[1+2({x}^{2}+{y}^{2}+{z}^{2})]{e}^{-{x}^{2}+{y}^{2}+{z}^{2}}

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 60번
# 발산과 회전활용자동 분류
OpenStax Calculus Volume 3 CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
63.OpenStax 3권 · 6.5절 · 63번대학 수준우리말 번역

[계산기 사용] 회전 속도장 v=⟨0,10z,−10y⟩\text{v}=\langle 0,10z,-10y\rangle 을 생각하자. 물레바퀴를 평면 x+y+z=1x+y+z=1 위에 그 축이 이 평면에 수직이 되도록 놓을 때, 컴퓨터 대수 시스템을 사용하여 물레바퀴가 단위 시간당 몇 바퀴 도는지 계산하세요.

회전 속도장을 나타낸 3차원 그림이다. 화살표들은 시계 방향의 회전을 보여 준다. 평면 x + y + z = 1 위에 물레바퀴가 놓여 있고, n이 평면에 수직으로 뻗어 나와 있다.

영어 원문 보기

[T] Consider rotational velocity field v=⟨0,10z,−10y⟩.\text{v}=\langle 0,10z,-10y\rangle . If a paddlewheel is placed in plane x+y+z=1x+y+z=1 with its axis normal to this plane, using a computer algebra system, calculate how fast the paddlewheel spins in revolutions per unit of time.

A three dimensional diagram of a rotational velocity field. The arrows are showing a rotation in a clockwise manner. A paddlewheel is shown in plan x + y + z = 1 with n extended out perpendicular to the plane.

원본 교재에 이 문제의 정답은 실려 있지 않습니다.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 62번
# 발산과 회전그래프·표 해석자동 분류
OpenStax Calculus Volume 3 CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
64.CLP-4 · 4.1절 · 2번대학 수준우리말 번역

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

영어 원문 보기

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

답 쓰고 채점
정답·풀이 보기

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

∇×F=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂z10y]=ı^\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 xx, yy, zz 에 대해 F(x,y,z)\mathbf{F}(x,y,z) 와의 내적이 11 이므로 이 벡터장과 수직이 아니다.

원본 교재 풀이(영어)

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

∇×F=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂z10y]=ı^\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 11 with F(x,y,z)\mathbf{F}(x,y,z) (for all xx, yy, zz) and so is not perpendicular to it.

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 3번
# 발산과 회전계산기본자동 분류
CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
65.CLP-4 · 4.1절 · 3번대학 수준우리말 번역

다음 벡터 항등식을 확인하세요.

(a) ∇⋅(fF)=f∇⋅F+F⋅∇f{ \boldsymbol{\nabla} }\cdot(f\mathbf{F})=f{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f (b) ∇⋅(F×G)=G⋅(∇×F)−F⋅(∇×G){ \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) ∇2(fg)=f ∇2g+2∇f⋅∇g+g ∇2f{ \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) ∇⋅(fF)=f∇⋅F+F⋅∇f{ \boldsymbol{\nabla} }\cdot(f\mathbf{F})=f{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f (b) ∇⋅(F×G)=G⋅(∇×F)−F⋅(∇×G){ \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) ∇2(fg)=f ∇2g+2∇f⋅∇g+g ∇2f{ \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

∇⋅(fF)=∂∂x(fF1)+∂∂y(fF2)+∂∂z(fF3)=+f∂F1∂x+f∂F2∂y+f∂F3∂z=+F1∂f∂x+F2∂f∂y+F3∂f∂z=f ∇⋅F+F⋅∇f\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

∇⋅(F×G)=∂∂x(F2G3−F3G2)+∂∂y(F3G1−F1G3)+∂∂z(F1G2−F2G1)=+∂F2∂xG3−∂F3∂xG2+∂F3∂yG1−∂F1∂yG3+∂F1∂zG2−∂F2∂zG1=+F2∂G3∂x−F3∂G2∂x+F3∂G1∂y−F1∂G3∂y+F1∂G2∂z−F2∂G1∂z=+(∂F3∂y−∂F2∂z)G1+(∂F1∂z−∂F3∂x)G2+(∂F2∂x−∂F1∂y)G3=−F1(∂G3∂y−∂G2∂z)−F2(∂G1∂z−∂G3∂x)−F3(∂G2∂x−∂G1∂y)=G⋅(∇×F)−F⋅(∇×G)\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) ∇2(fg)=∇⋅[∇(fg)]{ \boldsymbol{\nabla} }^2(fg) = { \boldsymbol{\nabla} }\cdot\big[{ \boldsymbol{\nabla} }(fg)\big] 임을 떠올리자. 먼저

∇(fg)=ı^∂∂x(fg)+ȷ^∂∂y(fg)+k^∂∂z(fg)=+ı^g∂f∂x+ȷ^g∂f∂y+k^g∂f∂z=+ı^f∂g∂x+ȷ^f∂g∂y+k^f∂g∂z=g∇f+f∇g\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

∇2(fg)=∇⋅(g∇f)+∇⋅(f∇g)=g(∇⋅∇f)+(∇g)⋅(∇f)+f(∇⋅∇g)+(∇f)⋅(∇g)=f ∇2g+2∇f⋅∇g+g ∇2f\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

∇⋅(fF)=∂∂x(fF1)+∂∂y(fF2)+∂∂z(fF3)=+f∂F1∂x+f∂F2∂y+f∂F3∂z=+F1∂f∂x+F2∂f∂y+F3∂f∂z=f ∇⋅F+F⋅∇f\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

∇⋅(F×G)=∂∂x(F2G3−F3G2)+∂∂y(F3G1−F1G3)+∂∂z(F1G2−F2G1)=+∂F2∂xG3−∂F3∂xG2+∂F3∂yG1−∂F1∂yG3+∂F1∂zG2−∂F2∂zG1=+F2∂G3∂x−F3∂G2∂x+F3∂G1∂y−F1∂G3∂y+F1∂G2∂z−F2∂G1∂z=+(∂F3∂y−∂F2∂z)G1+(∂F1∂z−∂F3∂x)G2+(∂F2∂x−∂F1∂y)G3=−F1(∂G3∂y−∂G2∂z)−F2(∂G1∂z−∂G3∂x)−F3(∂G2∂x−∂G1∂y)=G⋅(∇×F)−F⋅(∇×G)\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 ∇2(fg)=∇⋅[∇(fg)]{ \boldsymbol{\nabla} }^2(fg) = { \boldsymbol{\nabla} }\cdot\big[{ \boldsymbol{\nabla} }(fg)\big]. First

∇(fg)=ı^∂∂x(fg)+ȷ^∂∂y(fg)+k^∂∂z(fg)=+ı^g∂f∂x+ȷ^g∂f∂y+k^g∂f∂z=+ı^f∂g∂x+ȷ^f∂g∂y+k^f∂g∂z=g∇f+f∇g\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,

∇2(fg)=∇⋅(g∇f)+∇⋅(f∇g)=g(∇⋅∇f)+(∇g)⋅(∇f)+f(∇⋅∇g)+(∇f)⋅(∇g)=f ∇2g+2∇f⋅∇g+g ∇2f\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
CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
66.CLP-4 · 4.1절 · 4번대학 수준우리말 번역

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

(a) F=x ı^+y ȷ^+z k^\mathbf{F}=x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}} (b) F=xy2ı^−yz2ȷ^+zx2k^\mathbf{F}=xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}} (c) F=xı^+yȷ^x2+y2\mathbf{F}=\frac{x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}} (2차원에서의 극좌표 기저 벡터 r^\hat{\bf r}) (d) F=−yı^+xȷ^x2+y2\mathbf{F}=\frac{-y\hat{\boldsymbol{\imath}}+x\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}} (2차원에서의 극좌표 기저 벡터 θ^\hat{\boldsymbol{\theta}})

영어 원문 보기

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

(a) F=x ı^+y ȷ^+z k^\mathbf{F}=x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}} (b) F=xy2ı^−yz2ȷ^+zx2k^\mathbf{F}=xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}} (c) F=xı^+yȷ^x2+y2\mathbf{F}=\frac{x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}} (the polar basis vector r^\hat{\bf r} in 2d) (d) F=−yı^+xȷ^x2+y2\mathbf{F}=\frac{-y\hat{\boldsymbol{\imath}}+x\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}} (the polar basis vector θ^\hat{\boldsymbol{\theta}} in 2d)

정답·풀이 보기

(a) 정의에 의해

∇⋅(x ı^+y ȷ^+z k^)=∂∂x(x)+∂∂y(y)+∂∂z(z)=3∇×(x ı^+y ȷ^+z k^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂zxyz]=0\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

∇⋅(xy2ı^−yz2ȷ^+zx2k^)=∂∂x(xy2)+∂∂y(−yz2)+∂∂z(zx2)=y2−z2+x2∇×(xy2ı^−yz2ȷ^+zx2k^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂zxy2−yz2zx2]=2yz ı^−2xz ȷ^−2xy k^\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

∇⋅(xx2+y2ı^+yx2+y2ȷ^)=∂∂x(xx2+y2)+∂∂y(yx2+y2)=1x2+y2−x2[x2+y2]3/2+1x2+y2−y2[x2+y2]3/2=x2+y2 −x2 +x2+y2 − y2[x2+y2]3/2=1x2+y2∇×(xx2+y2ı^+yx2+y2ȷ^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂zxx2+y2yx2+y20]=(−xy[x2+y2]3/2+xy[x2+y2]3/2)k^=0\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

∇⋅(−yx2+y2ı^+xx2+y2ȷ^)=∂∂x(−yx2+y2)+∂∂y(xx2+y2)=xy[x2+y2]3/2−xy[x2+y2]3/2=0∇×(−yx2+y2ı^+xx2+y2ȷ^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂z−yx2+y2xx2+y20]= ⁣( ⁣1x2+y2−x2[x2+y2]3/2+1x2+y2−y2[x2+y2]3/2 ⁣) ⁣k^=x2+y2 −x2 +x2+y2 − y2[x2+y2]3/2 k^=k^x2+y2\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

∇⋅(x ı^+y ȷ^+z k^)=∂∂x(x)+∂∂y(y)+∂∂z(z)=3∇×(x ı^+y ȷ^+z k^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂zxyz]=0\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

∇⋅(xy2ı^−yz2ȷ^+zx2k^)=∂∂x(xy2)+∂∂y(−yz2)+∂∂z(zx2)=y2−z2+x2∇×(xy2ı^−yz2ȷ^+zx2k^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂zxy2−yz2zx2]=2yz ı^−2xz ȷ^−2xy k^\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

∇⋅(xx2+y2ı^+yx2+y2ȷ^)=∂∂x(xx2+y2)+∂∂y(yx2+y2)=1x2+y2−x2[x2+y2]3/2+1x2+y2−y2[x2+y2]3/2=x2+y2 −x2 +x2+y2 − y2[x2+y2]3/2=1x2+y2∇×(xx2+y2ı^+yx2+y2ȷ^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂zxx2+y2yx2+y20]=(−xy[x2+y2]3/2+xy[x2+y2]3/2)k^=0\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

∇⋅(−yx2+y2ı^+xx2+y2ȷ^)=∂∂x(−yx2+y2)+∂∂y(xx2+y2)=xy[x2+y2]3/2−xy[x2+y2]3/2=0∇×(−yx2+y2ı^+xx2+y2ȷ^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂z−yx2+y2xx2+y20]= ⁣( ⁣1x2+y2−x2[x2+y2]3/2+1x2+y2−y2[x2+y2]3/2 ⁣) ⁣k^=x2+y2 −x2 +x2+y2 − y2[x2+y2]3/2 k^=k^x2+y2\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
CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
67.CLP-4 · 4.1절 · 5번대학 수준우리말 번역

(a) r=(x,y,z)\mathbf{r}=(x,y,z), r=∣(x,y,z)∣r=|(x,y,z)| 일 때 ∇⋅(rr){ \boldsymbol{\nabla} }\cdot\big(\frac{\mathbf{r}}{r}\big) 를 계산하고 간단히 하세요. 답을 rr 에 관한 식으로 나타내세요. (b) ∇×(yz ı^+2xz ȷ^+exy k^){ \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 ∇⋅(rr){ \boldsymbol{\nabla} }\cdot\big(\frac{\mathbf{r}}{r}\big) for r=(x,y,z)\mathbf{r}=(x,y,z) and r=∣(x,y,z)∣r=|(x,y,z)|. Express your answer in terms of rr. (b) Compute ∇×(yz ı^+2xz ȷ^+exy k^){ \boldsymbol{\nabla} }\times\big(yz\,\hat{\boldsymbol{\imath}} + 2xz\,\hat{\boldsymbol{\jmath}} + e^{xy}\,\hat{\mathbf{k}}\big).

정답·풀이 보기

(a) rr=x ı^+y ȷ^+z k^[x2+y2+z2]1/2\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.

∂∂xx[x2+y2+z2]1/2=1[x2+y2+z2]1/2−12x(2x)[x2+y2+z2]3/2 = y2+z2[x2+y2+z2]3/2∂∂yy[x2+y2+z2]1/2=1[x2+y2+z2]1/2−12y(2y)[x2+y2+z2]3/2 = x2+z2[x2+y2+z2]3/2∂∂zz[x2+y2+z2]1/2=1[x2+y2+z2]1/2−12z(2z)[x2+y2+z2]3/2 = x2+y2[x2+y2+z2]3/2\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.

∇(rr)=2x2+2y2+2z2[x2+y2+z2]3/2=2r2r3=2r\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)

∇×(yz ı^+2xz ȷ^+exy k^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂zyz2xzexy]=(xexy−2x) ı^−(yexy−y) ȷ^+z k^\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 rr=x ı^+y ȷ^+z k^[x2+y2+z2]1/2\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

∂∂xx[x2+y2+z2]1/2=1[x2+y2+z2]1/2−12x(2x)[x2+y2+z2]3/2 = y2+z2[x2+y2+z2]3/2∂∂yy[x2+y2+z2]1/2=1[x2+y2+z2]1/2−12y(2y)[x2+y2+z2]3/2 = x2+z2[x2+y2+z2]3/2∂∂zz[x2+y2+z2]1/2=1[x2+y2+z2]1/2−12z(2z)[x2+y2+z2]3/2 = x2+y2[x2+y2+z2]3/2\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

∇(rr)=2x2+2y2+2z2[x2+y2+z2]3/2=2r2r3=2r\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)

∇×(yz ı^+2xz ȷ^+exy k^)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂zyz2xzexy]=(xexy−2x) ı^−(yexy−y) ȷ^+z k^\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
CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
68.CLP-4 · 4.1절 · 6번대학 수준우리말 번역

다음에서 기호 r=x ı^+y ȷ^+z k^\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}}, r=∣r∣r = |\mathbf{r}| 를 사용하고, kk 는 k=0,1,−1,2,−2,…k = 0, 1, -1, 2, -2, \dots 중 어떤 수이다.

(a) 다음을 만족하는 kk 의 값을 구하세요.

∇(rk)=−3rr5{ \boldsymbol{\nabla} } (r^k) = -3\frac{\mathbf{r}}{r^5} (b) 다음을 만족하는 kk 의 값을 구하세요.

∇⋅(rkr)=5r2{ \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) = 5r^2 (c) 다음을 만족하는 kk 의 값을 구하세요.

∇2(rk)=2r4{ \boldsymbol{\nabla} }^2 (r^k) = \frac{2}{r^4}

영어 원문 보기

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

(a) Find the value kk for which

∇(rk)=−3rr5{ \boldsymbol{\nabla} } (r^k) = -3\frac{\mathbf{r}}{r^5} (b) Find the value kk for which

∇⋅(rkr)=5r2{ \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) = 5r^2 (c) Find the value kk for which

∇2(rk)=2r4{ \boldsymbol{\nabla} }^2 (r^k) = \frac{2}{r^4}

답 쓰고 채점
(a)
(b)
(c)
정답·풀이 보기

(a) rk=(x2+y2+z2)k/2r^k=\big(x^2+y^2+z^2\big)^{k/2} 이므로,

∂∂xrk=2x k2(x2+y2+z2)k2−1=k (r⋅ı^) rk−2∂∂yrk=2y k2(x2+y2+z2)k2−1=k (r⋅ȷ^) rk−2∂∂zrk=2z k2(x2+y2+z2)k2−1=k (r⋅k^) rk−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}

구하는 값은 k=−3k=-3 이다.

(b) (a)의 계산을 이용하면

∇⋅(rkr)=∂∂x(xrk)+∂∂y(yrk)+∂∂z(zrk)=3rk+x∂∂xrk+y∂∂yrk+z∂∂zrk=3rk+x(kx rk−2)+y(ky rk−2)+z(kz rk−2)=(3+k)rk\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=2k=2 이다.

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

∇2(rk)=∇⋅(∇(rk))=∇⋅(krk−2 r)(a)에 의해=k(3+k−2)rk−2(b)에 의해, 단 k 를 k−2 로 바꾼다\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=−2k=-2 이다.

원본 교재 풀이(영어)

(a) Since rk=(x2+y2+z2)k/2r^k=\big(x^2+y^2+z^2\big)^{k/2},

∂∂xrk=2x k2(x2+y2+z2)k2−1=k (r⋅ı^) rk−2∂∂yrk=2y k2(x2+y2+z2)k2−1=k (r⋅ȷ^) rk−2∂∂zrk=2z k2(x2+y2+z2)k2−1=k (r⋅k^) rk−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=−3k=-3.

(b) Using the computation in part (a)

∇⋅(rkr)=∂∂x(xrk)+∂∂y(yrk)+∂∂z(zrk)=3rk+x∂∂xrk+y∂∂yrk+z∂∂zrk=3rk+x(kx rk−2)+y(ky rk−2)+z(kz rk−2)=(3+k)rk\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=2k=2.

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

∇2(rk)=∇⋅(∇(rk))=∇⋅(krk−2 r)by part (a)=k(3+k−2)rk−2by part (b), but with k replaced by k−2\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=−2k=-2.

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 7번
# 발산과 회전계산보통자동 분류
CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
69.CLP-4 · 4.1절 · 7번대학 수준우리말 번역

r\mathbf{r} 를 벡터장 r=x ı^+y ȷ^+z k^\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}} 라 하고, rr 를 함수 r=∣r∣r = |\mathbf{r}| 라 하자. a\mathbf{a} 를 상수 벡터 a=a1 ı^+a2 ȷ^+a3 k^\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 a\mathbf{a}, r\mathbf{r}, rr 로 나타내야 합니다. 답에 xx, yy, zz 가 들어 있으면 안 됩니다.

(a) ∇⋅r{ \boldsymbol{\nabla} }\cdot\mathbf{r} (b) ∇(r2){ \boldsymbol{\nabla} }(r^2) (c) ∇×(r×a){ \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a}) (d) ∇⋅(∇(r)){ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big)

영어 원문 보기

Let r\mathbf{r} be the vector field r=x ı^+y ȷ^+z k^\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}} and let rr be the function r=∣r∣r = |\mathbf{r}|. Let a\mathbf{a} be the constant vector a=a1 ı^+a2 ȷ^+a3 k^\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 a\mathbf{a}, r\mathbf{r}, and rr. There should be no xx's, yy's, or zz's in your answers.

(a) ∇⋅r{ \boldsymbol{\nabla} }\cdot\mathbf{r} (b) ∇(r2){ \boldsymbol{\nabla} }(r^2) (c) ∇×(r×a){ \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a}) (d) ∇⋅(∇(r)){ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big)

정답·풀이 보기

(a)

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

(b)

∇(r2)=(ı^∂∂x+ȷ^∂∂y+k^∂∂z)(x2+y2+z2)=2x ı^+2y ȷ^+2z k^=2r{ \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

r×a=det⁡[ı^ȷ^k^xyza1a2a3]=ı^(a3y−a2z)+ȷ^(a1z−a3x)+k^(a2x−a1y)\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.

∇×(r×a)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂za3y−a2za1z−a3xa2x−a1y]=−2a1 ı^−2a2 ȷ^−2a3 k^=−2a\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

∇(r)=(ı^∂∂x+ȷ^∂∂y+k^∂∂z)(x2+y2+z2)1/2=ı^x(x2+y2+z2)1/2+ȷ^y(x2+y2+z2)1/2+k^x(x2+y2+z2)1/2\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.

∇⋅(∇(r))=∂∂xx(x2+y2+z2)1/2+∂∂yy(x2+y2+z2)1/2+∂∂zz(x2+y2+z2)1/2=3(x2+y2+z2)1/2−122x2+2y2+2z2(x2+y2+z2)3/2=2(x2+y2+z2)1/2=2r\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)

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

(b)

∇(r2)=(ı^∂∂x+ȷ^∂∂y+k^∂∂z)(x2+y2+z2)=2x ı^+2y ȷ^+2z k^=2r{ \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

r×a=det⁡[ı^ȷ^k^xyza1a2a3]=ı^(a3y−a2z)+ȷ^(a1z−a3x)+k^(a2x−a1y)\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

∇×(r×a)=det⁡[ı^ȷ^k^∂∂x∂∂y∂∂za3y−a2za1z−a3xa2x−a1y]=−2a1 ı^−2a2 ȷ^−2a3 k^=−2a\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

∇(r)=(ı^∂∂x+ȷ^∂∂y+k^∂∂z)(x2+y2+z2)1/2=ı^x(x2+y2+z2)1/2+ȷ^y(x2+y2+z2)1/2+k^x(x2+y2+z2)1/2\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

∇⋅(∇(r))=∂∂xx(x2+y2+z2)1/2+∂∂yy(x2+y2+z2)1/2+∂∂zz(x2+y2+z2)1/2=3(x2+y2+z2)1/2−122x2+2y2+2z2(x2+y2+z2)3/2=2(x2+y2+z2)1/2=2r\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
CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
70.CLP-4 · 4.1절 · 8번대학 수준우리말 번역

다음과 같이 두자.

r=x ı^+y ȷ^+z k^,r=∣r∣\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}},\qquad r = |\mathbf{r}|

(a) ∇(1r)=−ra r{ \boldsymbol{\nabla} }\big(\frac{1}{r}\big) =- r^a\,\mathbf{r} 를 만족하는 aa 의 값을 구하세요. (b) ∇⋅(r r)=ar{ \boldsymbol{\nabla} }\cdot\big(r\,\mathbf{r}\big) = ar 를 만족하는 aa 의 값을 구하세요. (c) ∇⋅(∇(r3))=ar{ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^3)\big) = ar 를 만족하는 aa 의 값을 구하세요.

영어 원문 보기

Let

r=x ı^+y ȷ^+z k^,r=∣r∣\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}},\qquad r = |\mathbf{r}|

(a) Compute aa where ∇(1r)=−ra r{ \boldsymbol{\nabla} }\big(\frac{1}{r}\big) =- r^a\,\mathbf{r}. (b) Compute aa where ∇⋅(r r)=ar{ \boldsymbol{\nabla} }\cdot\big(r\,\mathbf{r}\big) = ar. (c) Compute aa where ∇⋅(∇(r3))=ar{ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^3)\big) = ar.

답 쓰고 채점
(a)
(b)
(c)
정답·풀이 보기

(a) 다음과 같으므로

∇(1r)=(ı^∂∂x+ȷ^∂∂y+k^∂∂z)(x2+y2+z2)−1/2=−ı^x(x2+y2+z2)3/2−ȷ^y(x2+y2+z2)3/2−k^z(x2+y2+z2)3/2=−ı^xr3−ȷ^yr3−k^xr3\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=−3a=-3 이다.

(b) 다음과 같으므로

∇⋅(r r)=∂∂x[(x2+y2+z2)1/2x]+∂∂y[(x2+y2+z2)1/2y]+∂∂z[(x2+y2+z2)1/2z]=3(x2+y2+z2)1/2+122x2+2y2+2z2(x2+y2+z2)1/2=4(x2+y2+z2)1/2=4r\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=4a=4 이다.

(c) 다음과 같으므로

∇(r3)=(ı^∂∂x+ȷ^∂∂y+k^∂∂z)(x2+y2+z2)3/2=ı^ 3x(x2+y2+z2)1/2+ȷ^ 3y(x2+y2+z2)1/2+k^ 3z(x2+y2+z2)1/2=3rr\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.

∇⋅(∇(r3))=∇⋅(3rr)=3 ∇⋅(rr)=3 (4r)(b)에 의해=12r\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=12a=12 이다.

원본 교재 풀이(영어)

(a) Since

∇(1r)=(ı^∂∂x+ȷ^∂∂y+k^∂∂z)(x2+y2+z2)−1/2=−ı^x(x2+y2+z2)3/2−ȷ^y(x2+y2+z2)3/2−k^z(x2+y2+z2)3/2=−ı^xr3−ȷ^yr3−k^xr3\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=−3a=-3.

(b) Since

∇⋅(r r)=∂∂x[(x2+y2+z2)1/2x]+∂∂y[(x2+y2+z2)1/2y]+∂∂z[(x2+y2+z2)1/2z]=3(x2+y2+z2)1/2+122x2+2y2+2z2(x2+y2+z2)1/2=4(x2+y2+z2)1/2=4r\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=4a=4.

(c) Since

∇(r3)=(ı^∂∂x+ȷ^∂∂y+k^∂∂z)(x2+y2+z2)3/2=ı^ 3x(x2+y2+z2)1/2+ȷ^ 3y(x2+y2+z2)1/2+k^ 3z(x2+y2+z2)1/2=3rr\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

∇⋅(∇(r3))=∇⋅(3rr)=3 ∇⋅(rr)=3 (4r)by part (b)=12r\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=12a=12.

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 9번
# 발산과 회전계산보통자동 분류
CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
71.CLP-4 · 4.1절 · 9번대학 수준우리말 번역

k^\hat{\mathbf{k}} 성분이 A3=0A_3=0 이고 다음의 벡터 퍼텐셜이 되는 벡터장 A\mathbf{A} 를 가능하면 구하세요.

(a) F=(1+yz)ı^+(2y+zx)ȷ^+(3z2+xy)k^\mathbf{F}=(1+yz)\hat{\boldsymbol{\imath}}+(2y+zx)\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}} (b) G=yzı^+zxȷ^+xyk^\mathbf{G}= yz\hat{\boldsymbol{\imath}}+zx\hat{\boldsymbol{\jmath}}+xy\hat{\mathbf{k}}

영어 원문 보기

Find, if possible, a vector field A\mathbf{A} that has k^\hat{\mathbf{k}} component A3=0A_3=0 and that is a vector potential for

(a) F=(1+yz)ı^+(2y+zx)ȷ^+(3z2+xy)k^\mathbf{F}=(1+yz)\hat{\boldsymbol{\imath}}+(2y+zx)\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}} (b) G=yzı^+zxȷ^+xyk^\mathbf{G}= yz\hat{\boldsymbol{\imath}}+zx\hat{\boldsymbol{\jmath}}+xy\hat{\mathbf{k}}

정답·풀이 보기

(a) ∇⋅F=∂∂x(1+yz)+∂∂y(2y+zx)+∂∂z(3z2+xy)=2+6z≠0{ \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 F\mathbf{F} 는 판별 조건을 만족하지 못하여 벡터 퍼텐셜을 가질 수 없다.

(b) 벡터장 A=A1ı^+A2ȷ^\mathbf{A}=A_1\hat{\boldsymbol{\imath}}+A_2\hat{\boldsymbol{\jmath}} 가 G\mathbf{G} 의 벡터 퍼텐셜이 될 필요충분조건은 G=∇×A\mathbf{G}={ \boldsymbol{\nabla} }\times\mathbf{A} 이고, 이는 다음이 성립할 필요충분조건과 같다.

−∂A2∂z=yz  ⟺  A2=−12yz2+B2(x,y)∂A1∂z=zx  ⟺  A1=−12xz2+B1(x,y)∂A2∂x−∂A1∂y=xy  ⟺  ∂B2∂x−∂B1∂y=xy\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}

∂B2∂x−∂B1∂y=xy\frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy 의 해는 무수히 많다. 실제로 B2B_2 는 완전히 임의로 정할 수 있다. B2=0B_2=0 으로 택하면 B1=−12xy2B_1=-\frac{1}{2} xy^2 이 조건을 만족한다. B1=0B_1=0 으로 택하면 B2=12x2yB_2=\frac{1}{2} x^2y 가 조건을 만족한다. 따라서 두 가지 해는 A=12(z2−y2)xı^−12yz2ȷ^\mathbf{A}=\frac{1}{2}(z^2-y^2)x\hat{\boldsymbol{\imath}}-\frac{1}{2} yz^2\hat{\boldsymbol{\jmath}} 와 A=12xz2ı^+12(x2−z2)yȷ^\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 ∇⋅F=∂∂x(1+yz)+∂∂y(2y+zx)+∂∂z(3z2+xy)=2+6z≠0{ \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, F\mathbf{F} fails the screening test and cannot have a vector potential.

(b) The vector field A=A1ı^+A2ȷ^\mathbf{A}=A_1\hat{\boldsymbol{\imath}}+A_2\hat{\boldsymbol{\jmath}} is a vector potential for G\mathbf{G} if and only if G=∇×A\mathbf{G}={ \boldsymbol{\nabla} }\times\mathbf{A}, which is the case if and only if

−∂A2∂z=yz  ⟺  A2=−12yz2+B2(x,y)∂A1∂z=zx  ⟺  A1=−12xz2+B1(x,y)∂A2∂x−∂A1∂y=xy  ⟺  ∂B2∂x−∂B1∂y=xy\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 ∂B2∂x−∂B1∂y=xy\frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy. In fact B2B_2 is completely arbitrary. If one chooses B2=0B_2=0, then B1=−12xy2B_1=-\frac{1}{2} xy^2 does the job. If one chooses B1=0B_1=0, then B2=12x2yB_2=\frac{1}{2} x^2y does the job. Thus two solutions are A=12(z2−y2)xı^−12yz2ȷ^\mathbf{A}=\frac{1}{2}(z^2-y^2)x\hat{\boldsymbol{\imath}}-\frac{1}{2} yz^2\hat{\boldsymbol{\jmath}} and A=12xz2ı^+12(x2−z2)yȷ^\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
CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)
72.CLP-4 · 4.1절 · 13번대학 수준우리말 번역

벡터장 F\mathbf{F} 가 R3\mathbb{R}^3 전체에서 ∇⋅F=0{ \boldsymbol{\nabla} }\cdot \mathbf{F}=0 을 만족한다고 가정하세요.

r(t)=tx ı^+ty ȷ^+tz k^,0≤t≤1\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)(x,y,z) 까지의 선분의 매개변수 표현이라고 하고,

G(x,y,z)=∫01t F(r(t))×drdt(t) dt\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. R3\mathbb{R}^3 전체에서 ∇×G=F{ \boldsymbol{\nabla} }\times \mathbf{G}=\mathbf{F} 임을 보이세요.

영어 원문 보기

Suppose that the vector field F\mathbf{F} obeys ∇⋅F=0{ \boldsymbol{\nabla} }\cdot \mathbf{F}=0 in all of R3\mathbb{R}^3. Let

r(t)=tx ı^+ty ȷ^+tz k^,0≤t≤1\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)(x,y,z). Define

G(x,y,z)=∫01t F(r(t))×drdt(t) dt\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 ∇×G=F{ \boldsymbol{\nabla} }\times \mathbf{G}=\mathbf{F} throughout R3\mathbb{R}^3.

정답·풀이 보기

∂G3∂y−∂G2∂z=F1\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}=F_1 임을 보이겠다. 나머지 성분도 이와 비슷하다. 먼저

t F(r(t))×drdt(t)=t F(tx,ty,tz)×(x ı^+y ȷ^+z k^)=tdet⁡[ı^ȷ^k^F1F2F3xyz]\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 k^\hat{\mathbf{k}} 성분과 ȷ^\hat{\boldsymbol{\jmath}} 성분을 읽으면

G3(x,y,z)=∫01t[F1(tx,ty,tz) y−F2(tx,ty,tz) x] dtG2(x,y,z)=∫01t[F3(tx,ty,tz) x−F1(tx,ty,tz) z] dt\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

∂G3∂y=∫01t[F1(tx,ty,tz)+∂F1∂y(tx,ty,tz) ty−∂F2∂y(tx,ty,tz) tx] dt∂G2∂z=∫01t[∂F3∂z(tx,ty,tz) tx−∂F1∂z(tx,ty,tz) tz−F1(tx,ty,tz)] dt⇒∂G3∂y−∂G2∂z=∫01[2t F1(tx,ty,tz)+t2y ∂F1∂y(tx,ty,tz)+t2z∂F1∂z(tx,ty,tz)−t2x ∂F2∂y(tx,ty,tz)−t2x ∂F3∂z(tx,ty,tz)] dt\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 ∇⋅F=∂F1∂x+∂F2∂y+∂F3∂z=0{ \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

−t2x{∂F2∂y(tx,ty,tz)+∂F3∂z(tx,ty,tz)}=−t2x{−∂F1∂x(tx,ty,tz)}\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

∂G3∂y−∂G2∂z=∫01[2tF1(tx,ty,tz)+t2x∂F1∂x(tx,ty,tz)+t2y∂F1∂y(tx,ty,tz)+t2z∂F1∂z(tx,ty,tz)] dt=∫01ddt[t2F1(tx,ty,tz)] dt=[t2F1(tx,ty,tz)]t=0t=1=F1(x,y,z)\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 ∂G3∂y−∂G2∂z=F1\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}=F_1. The other components are similar. First we have

t F(r(t))×drdt(t)=t F(tx,ty,tz)×(x ı^+y ȷ^+z k^)=tdet⁡[ı^ȷ^k^F1F2F3xyz]\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}

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

G3(x,y,z)=∫01t[F1(tx,ty,tz) y−F2(tx,ty,tz) x] dtG2(x,y,z)=∫01t[F3(tx,ty,tz) x−F1(tx,ty,tz) z] dt\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}

So

∂G3∂y=∫01t[F1(tx,ty,tz)+∂F1∂y(tx,ty,tz) ty−∂F2∂y(tx,ty,tz) tx] dt∂G2∂z=∫01t[∂F3∂z(tx,ty,tz) tx−∂F1∂z(tx,ty,tz) tz−F1(tx,ty,tz)] dt⇒∂G3∂y−∂G2∂z=∫01[2t F1(tx,ty,tz)+t2y ∂F1∂y(tx,ty,tz)+t2z∂F1∂z(tx,ty,tz)−t2x ∂F2∂y(tx,ty,tz)−t2x ∂F3∂z(tx,ty,tz)] dt\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}

Since, by hypothesis, ∇⋅F=∂F1∂x+∂F2∂y+∂F3∂z=0{ \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

−t2x{∂F2∂y(tx,ty,tz)+∂F3∂z(tx,ty,tz)}=−t2x{−∂F1∂x(tx,ty,tz)}\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}

so that

∂G3∂y−∂G2∂z=∫01[2tF1(tx,ty,tz)+t2x∂F1∂x(tx,ty,tz)+t2y∂F1∂y(tx,ty,tz)+t2z∂F1∂z(tx,ty,tz)] dt=∫01ddt[t2F1(tx,ty,tz)] dt=[t2F1(tx,ty,tz)]t=0t=1=F1(x,y,z)\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}

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