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문제 은행 · 미적분 · 대학 미적분 · 벡터 미적분

발산과 회전

먼저 혼자 풀어 보고, 풀이로 확인한 다음, 아래 「한 번 더」 문제로 이어서 연습하세요.

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 문제CLP-4 · 4.1절 · 13번우리말 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.

다 풀었으면 정답 확인

∂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.

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CLP-4 Vector Calculus CC BY-NC-SA 4.0우리말 번역: 빌드라인 스쿨 — AI 번역·자동 검토(원문과 같은 조건)

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