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문제 은행 · 미적분 · 대학 미적분 · 벡터 미적분

발산과 회전

벡터장의 발산(div)과 회전(curl)을 계산하고 뜻을 해석한다

유형전체개념 확인계산그래프·표 해석활용증명·논술
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6문제 · 1 / 1쪽

1.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)
2.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)
3.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)
4.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)
5.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)
6.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)