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발산과 회전

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

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 문제CLP-4 · 4.1절 · 5번우리말 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.

다 풀었으면 정답 확인

(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}

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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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