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.

2문제 · 1 / 1쪽

1.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)
2.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)