벡터 미적분 전체 연습문제

641.CLP-4 · 3.2절 · 2번대학 수준우리말 번역

점 $\mathbf{r}_0= (x_0,y_0,z_0)$ 가 곡면 $G(x,y,z)=0$ 위에 있다고 하자. ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)\ne\mathbf{0}$ 이라 가정한다. 매개변수로 나타낸 곡선 $\mathbf{r}(t)=\big(x(t),y(t),z(t)\big)$ 가 이 곡면에 포함되고 $\mathbf{r}(t_0)=\mathbf{r}_0$ 이라 하자. 점 $\mathbf{r}_0$ 에서 이 곡선의 접선이 점 $\mathbf{r}_0$ 에서 $G=0$ 의 접평면 위에 있음을 보이세요.

영어 원문 보기

Let the point $\mathbf{r}_0= (x_0,y_0,z_0)$ lie on the surface $G(x,y,z)=0$ . Assume that ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)\ne\mathbf{0}$ . Suppose that the parametrized curve $\mathbf{r}(t)=\big(x(t),y(t),z(t)\big)$ is contained in the surface and that $\mathbf{r}(t_0)=\mathbf{r}_0$ . Show that the tangent line to the curve at $\mathbf{r}_0$ lies in the tangent plane to $G=0$ at $\mathbf{r}_0$ .

정답·풀이 보기

곡면 $G(x,y,z)=0$ 을 $S$ , 매개변수로 나타낸 곡선 $\mathbf{r}(t)=\big(x(t),y(t),z(t)\big)$ 를 $C$ 라 하자. 먼저 $\mathbf{r}_0$ 에서 $S$ 의 접평면과 $\mathbf{r}_0$ 에서 $C$ 의 접선을 구한다.

$\mathbf{r}_0$ 에서 $C$ 의 접선벡터는 $\left( x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right)$ 이므로, $\mathbf{r}_0$ 에서 $C$ 의 접선의 매개변수방정식은

$x-x_0 = t x'(t_0)\qquad y-y_0 = t x'(t_0)\qquad z-z_0 = t x'(t_0) \tag{$E_1$}$

그레이디언트 $\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right)$ 는 $(x_0,y_0,z_0)$ 에서 곡면 $S$ 의 법선벡터이다. 따라서 $(x_0,y_0,z_0)$ 에서 곡면 $S$ 의 접평면은

$\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right) \cdot \left( x-x_0\,,\, y-y_0\,,\,z-z_0\right) = 0$

즉

$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ (x-x_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ (y-y_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ (z-z_0) = 0 \tag{$E_2$}$

이다.

다음으로, $\mathbf{r}_0$ 에서 $C$ 의 접선벡터 $\left( x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right)$ 와 $\mathbf{r}_0$ 에서 $S$ 의 법선벡터 $\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right)$ 가 서로 수직임을 보인다. 이를 위해, 모든 $t$ 에 대해 점 $\big(x(t),y(t),z(t)\big)$ 가 곡면 $G(x,y,z)=0$ 위에 있으므로 다음이 성립함을 이용한다.

$\begin{aligned} G\big(x(t),y(t),z(t)\big) =0 \end{aligned}$

이 등식의 양변을 $t$ 에 대해 미분하면 연쇄 법칙에 의해

$\begin{aligned} 0&= \frac{\mathrm{d}}{\mathrm{d}t}G\big(x(t),y(t),z(t)\big) \\ &=\frac{\partial G}{\partial x}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ x'(t) +\frac{\partial G}{\partial y}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ y'(t) +\frac{\partial G}{\partial z}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ z'(t) \end{aligned}$

을 얻는다. 여기에 $t=t_0$ 을 대입하면

$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) = 0 \tag{$E_3$}$

이다.

마지막으로, $(x,y,z)$ 가 $\mathbf{r}_0$ 에서 $C$ 의 접선 위의 임의의 점이면 $(x,y,z)$ 는 $\mathbf{r}_0$ 에서 $S$ 의 접평면 위에도 있음을 보인다. $(x,y,z)$ 가 $\mathbf{r}_0$ 에서 $C$ 의 접선 위에 있으므로, $(E_1)$ 에 의해 다음을 만족하는 $t$ 가 존재한다.

$\begin{aligned} &\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{x-x_0\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{y-y_0\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{z-z_0\} \\ &=\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,x'(t_0)\big\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,y'(t_0)\big\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,z'(t_0)\big\}\\ &= t\left[\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) \right] =0 \end{aligned}$

이며, 마지막 등호는 $(E_3)$ 에 의한 것이다. 즉 $(x,y,z)$ 는 $\mathbf{r}_0$ 에서 $S$ 의 접평면의 방정식 $(E_2)$ 를 만족하므로 그 접평면 위에 있다. 따라서 $\mathbf{r}_0$ 에서 $C$ 의 접선은 $\mathbf{r}_0$ 에서 $S$ 의 접평면에 포함된다.

원본 교재 풀이(영어)

Denote by $S$ the surface $G(x,y,z)=0$ and by $C$ the parametrized curve $\mathbf{r}(t)=\big(x(t),y(t),z(t)\big)$ . To start, we'll find the tangent plane to $S$ at $\mathbf{r}_0$ and the tangent line to $C$ at $\mathbf{r}_0$ .

The tangent vector to $C$ at $\mathbf{r}_0$ is $\left( x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right)$ , so the parametric equations for the tangent line to $C$ at $\mathbf{r}_0$ are

$x-x_0 = t x'(t_0)\qquad y-y_0 = t x'(t_0)\qquad z-z_0 = t x'(t_0) \tag{$E_1$}$

The gradient $\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right)$ is a normal vector to the surface $S$ at $(x_0,y_0,z_0)$ . So the tangent plane to the surface $S$ at $(x_0,y_0,z_0)$ is

$\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right) \cdot \left( x-x_0\,,\, y-y_0\,,\,z-z_0\right) = 0$

or

$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ (x-x_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ (y-y_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ (z-z_0) = 0 \tag{$E_2$}$

Next, we'll show that the tangent vector $\left( x'(t_0)\,,\,y'(t_0)\,,\,z'(t_0) \right)$ to $C$ at $\mathbf{r}_0$ and the normal vector $\left(\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\,,\, \frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\right)$ to $S$ at $\mathbf{r}_0$ are perpendicular to each other. To do so, we observe that, for every $t$ , the point $\big(x(t),y(t),z(t)\big)$ lies on the surface $G(x,y,z)=0$ and so obeys

$\begin{aligned} G\big(x(t),y(t),z(t)\big) =0 \end{aligned}$

Differentiating this equation with respect to $t$ gives, by the chain rule,

$\begin{aligned} 0&= \frac{\mathrm{d}}{\mathrm{d}t}G\big(x(t),y(t),z(t)\big) \\ &=\frac{\partial G}{\partial x}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ x'(t) +\frac{\partial G}{\partial y}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ y'(t) +\frac{\partial G}{\partial z}\big( x(t)\,,\,y(t)\,,\,z(t)\big)\ z'(t) \end{aligned}$

Then setting $t=t_0$ gives

$\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) = 0 \tag{$E_3$}$

Finally, we are in a position to show that if $(x,y,z)$ is any point on the tangent line to $C$ at $\mathbf{r}_0$ , then $(x,y,z)$ is also on the tangent plane to $S$ at $\mathbf{r}_0$ . As $(x,y,z)$ is on the tangent line to $C$ at $\mathbf{r}_0$ then there is a $t$ such that, by $(E_1)$ ,

$\begin{aligned} &\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{x-x_0\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{y-y_0\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \{z-z_0\} \\ &=\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,x'(t_0)\big\} +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,y'(t_0)\big\} +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ \big\{ t\,z'(t_0)\big\}\\ &= t\left[\frac{\partial G}{\partial x}\big( x_0\,,\,y_0\,,\,z_0\big)\ x'(t_0) +\frac{\partial G}{\partial y}\big( x_0\,,\,y_0\,,\,z_0\big)\ y'(t_0) +\frac{\partial G}{\partial z}\big( x_0\,,\,y_0\,,\,z_0\big)\ z'(t_0) \right] =0 \end{aligned}$

by $(E_3)$ . That is, $(x,y,z)$ obeys the equation, $(E_2)$ , of the tangent plane to $S$ at $\mathbf{r}_0$ and so is on that tangent plane. So the tangent line to $C$ at $\mathbf{r}_0$ is contained in the tangent plane to $S$ at $\mathbf{r}_0$ .

한 번 더 — 비슷한 문제 풀기CLP-4 · 3.2절 · 13번

# 매개변수 곡면증명·논술기본자동 분류

642.CLP-4 · 3.2절 · 13번대학 수준우리말 번역

곡면 $x^2 + 9y^2 + 4z^2 = 17$ 위의 점 중에서 접평면이 평면 $x - 8z = 0$ 과 평행인 점을 모두 구하세요.

영어 원문 보기

Find all the points on the surface $x^2 + 9y^2 + 4z^2 = 17$ where the tangent plane is parallel to the plane $x - 8z = 0$ .

정답·풀이 보기

곡면 $x^2 + 9y^2 + 4z^2 = 17$ 의 점 $(x,y,z)$ 에서의 법선벡터는 $( 2x\,,\, 18y\,,\,8z)$ 이다. 평면 $x - 8z = 0$ 의 법선벡터는 $( 1\,,\,0\,,\,-8)$ 이다. 따라서 $( 2x\,,\, 18y\,,\,8z)$ 가 $( 1\,,\,0\,,\,-8)$ 과 평행, 즉 $( 1\,,\,0\,,\,-8)$ 의 0이 아닌 상수배가 되어야 한다. 이는 $x\ne 0$ 이면서 $y=0$ 이고 $z=-2x$ 일 때 성립한다. 또한 $(x,y,z)$ 는 곡면 $x^2 + 9y^2 + 4z^2 = 17$ 위에 있어야 한다. 따라서 $y=0$ , $z=-2x$ 이고

$\begin{aligned} 17= x^2 + 9y^2 + 4z^2 =x^2 +4(-2x)^2=17x^2 \implies x=\pm 1 \end{aligned}$

이다. 그러므로 구하는 점은 $\pm(1,0,-2)$ 이다.

원본 교재 풀이(영어)

A normal vector to the surface $x^2 + 9y^2 + 4z^2 = 17$ at the point $(x,y,z)$ is $( 2x\,,\, 18y\,,\,8z)$ . A normal vector to the plane $x - 8z = 0$ is $( 1\,,\,0\,,\,-8)$ . So we want $( 2x\,,\, 18y\,,\,8z)$ to be parallel to $( 1\,,\,0\,,\,-8)$ , i.e. to be a nonzero constant times $( 1\,,\,0\,,\,-8)$ . This is the case whenever $y=0$ and $z=-2x$ with $x\ne 0$ . In addition, we want $(x,y,z)$ to lie on the surface $x^2 + 9y^2 + 4z^2 = 17$ . So we want $y=0$ , $z=-2x$ and

$\begin{aligned} 17= x^2 + 9y^2 + 4z^2 =x^2 +4(-2x)^2=17x^2 \implies x=\pm 1 \end{aligned}$

So the allowed points are $\pm(1,0,-2)$ .

한 번 더 — 비슷한 문제 풀기CLP-4 · 3.2절 · 17번

# 매개변수 곡면계산보통자동 분류

643.CLP-4 · 3.2절 · 15번대학 수준우리말 번역

쌍곡면 $z^2=4x^2+y^2-1$ 위의 점 중에서 접평면이 평면 $2x-y+z=0$ 과 평행인 점을 모두 구하세요.

영어 원문 보기

Find all points on the hyperboloid $z^2=4x^2+y^2-1$ where the tangent plane is parallel to the plane $2x-y+z=0$ .

정답·풀이 보기

$(x_0,y_0,z_0)$ 을 쌍곡면 $z^2=4x^2+y^2-1$ 위의 점 중에서 접평면이 평면 $2x-y+z=0$ 과 평행인 점이라 하자. 평면 $2x-y+z=0$ 의 법선벡터는 $( 2,-1,1)$ 이다. 쌍곡면은 $G(x,y,z)=4x^2+y^2-z^2-1$ 이고 ${ \boldsymbol{\nabla} } G(x,y,z) = ( 8x,2y,-2z)$ 이므로, $(x_0,y_0,z_0)$ 에서 쌍곡면의 법선벡터는 ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)=( 8x_0,2y_0,-2z_0)$ 이다. 따라서 $(x_0,y_0,z_0)$ 이 주어진 조건을 만족시키는 것은 다음을 만족시키는 0이 아닌 $t$ 가 존재하는 것과 같다.

$\begin{aligned} &( 8x_0,2y_0,-2z_0) =t( 2,-1,1) \text{ 그리고 } z_0^2=4x_0^2+y_0^2-1\\ &\iff x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\text{ 그리고 } z_0^2=4x_0^2+y_0^2-1\\ &\iff \frac{t^2}{4}= \frac{t^2}{4}+ \frac{t^2}{4}-1\text{ 그리고 } x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\\ & \iff t=\pm 2\qquad (x_0,y_0,z_0)=\pm \big(\tfrac{1}{2},-1,-1\big) \end{aligned}$

원본 교재 풀이(영어)

Let $(x_0,y_0,z_0)$ be a point on the hyperboloid $z^2=4x^2+y^2-1$ where the tangent plane is parallel to the plane $2x-y+z=0$ . A normal vector to the plane $2x-y+z=0$ is $( 2,-1,1)$ . Because the hyperboloid is $G(x,y,z)=4x^2+y^2-z^2-1$ and ${ \boldsymbol{\nabla} } G(x,y,z) = ( 8x,2y,-2z)$ , a normal vector to the hyperboloid at $(x_0,y_0,z_0)$ is ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)=( 8x_0,2y_0,-2z_0)$ . So $(x_0,y_0,z_0)$ satisfies the required conditions if and only if there is a nonzero $t$ obeying

$\begin{aligned} &( 8x_0,2y_0,-2z_0) =t( 2,-1,1) \text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\text{ and } z_0^2=4x_0^2+y_0^2-1\\ &\iff \frac{t^2}{4}= \frac{t^2}{4}+ \frac{t^2}{4}-1\text{ and } x_0=\frac{t}{4},\ y_0=z_0=-\frac{t}{2}\\ & \iff t=\pm 2\qquad (x_0,y_0,z_0)=\pm \big(\tfrac{1}{2},-1,-1\big) \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 3.2절 · 18번

# 매개변수 곡면계산보통자동 분류

644.CLP-4 · 3.2절 · 16번대학 수준우리말 번역

(a) 방정식이 $x^2+z^2=10$ 인 곡면에 점 $(1,1,3)$ 에서 수직인 벡터를 구하세요. (b) (a)의 곡면과 곡면 $y^2+z^2=10$ 이 만나서 생기는 곡선에 같은 점에서 접하는 벡터를 구하세요. (c) 그 점에서 그 곡선에 접하는 접선의 매개변수방정식을 구하세요.

영어 원문 보기

(a) Find a vector perpendicular at the point $(1,1,3)$ to the surface with equation $x^2+z^2=10$ . (b) Find a vector tangent at the same point to the curve of intersection of the surface in part (a) with surface $y^2+z^2=10$ . (c) Find parametric equations for the line tangent to that curve at that point.

정답·풀이 보기

(a) 점 $(1,1,3)$ 에서 $x^2+z^2=10$ 에 수직인 벡터는

${ \boldsymbol{\nabla} }(x^2+z^2)\big|_{(1,1,3)} =(2x\hat{\boldsymbol{\imath}}+2z\hat{\mathbf{k}})\big|_{(1,1,3)} =2\hat{\boldsymbol{\imath}}+6\hat{\mathbf{k}}\text{ 또는 } \frac{1}{2} ( 2,0,6)=( 1,0,3)$

(b) 점 $(1,1,3)$ 에서 $y^2+z^2=10$ 에 수직인 벡터는

${ \boldsymbol{\nabla} }(y^2+z^2)\big|_{(1,1,3)} =(2y\hat{\boldsymbol{\jmath}}+2z\hat{\mathbf{k}})\big|_{(1,1,3)} =2\hat{\boldsymbol{\jmath}}+6\hat{\mathbf{k}}\text{ 또는 }\frac{1}{2} ( 0,2,6)=( 0,1,3)$

어떤 벡터가 주어진 점에서 주어진 곡선에 접할 필요충분조건은 그 벡터가 $(1,0,3)$ 과 $(0,1,3)$ 에 모두 수직인 것이다. 이러한 벡터 중 하나는

$( 0,1,3)\times(1,0,3) =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 0 & 1 & 3 \\ 1 & 0 & 3 \end{matrix}\right] =( 3,3,-1)$

(c) 구하는 접선은 점 $(1,1,3)$ 을 지나고 방향벡터가 $( 1,1,3)$ 이므로 벡터 매개변수방정식은 $\mathbf{r}(t)=( 1,1,3)+t( 3,3,-1)$

원본 교재 풀이(영어)

(a) A vector perpendicular to $x^2+z^2=10$ at $(1,1,3)$ is

${ \boldsymbol{\nabla} }(x^2+z^2)\big|_{(1,1,3)} =(2x\hat{\boldsymbol{\imath}}+2z\hat{\mathbf{k}})\big|_{(1,1,3)} =2\hat{\boldsymbol{\imath}}+6\hat{\mathbf{k}}\text{ or } \frac{1}{2} ( 2,0,6)=( 1,0,3)$

(b) A vector perpendicular to $y^2+z^2=10$ at $(1,1,3)$ is

${ \boldsymbol{\nabla} }(y^2+z^2)\big|_{(1,1,3)} =(2y\hat{\boldsymbol{\jmath}}+2z\hat{\mathbf{k}})\big|_{(1,1,3)} =2\hat{\boldsymbol{\jmath}}+6\hat{\mathbf{k}}\text{ or }\frac{1}{2} ( 0,2,6)=( 0,1,3)$

A vector is tangent to the specified curve at the specified point if and only if it perpendicular to both $(1,0,3)$ and $(0,1,3)$ . One such vector is

$( 0,1,3)\times(1,0,3) =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 0 & 1 & 3 \\ 1 & 0 & 3 \end{matrix}\right] =( 3,3,-1)$

(c) The specified tangent line passes through $(1,1,3)$ and has direction vector $( 1,1,3)$ and so has vector parametric equation $\mathbf{r}(t)=( 1,1,3)+t( 3,3,-1)$

한 번 더 — 비슷한 문제 풀기CLP-4 · 3.2절 · 15번

# 매개변수 곡면계산도전자동 분류

645.CLP-4 · 3.2절 · 17번대학 수준우리말 번역

곡선

$\mathbf{r}(t) = t^3\,\hat{\boldsymbol{\imath}} + t\,\hat{\boldsymbol{\jmath}} + t^2\,\hat{\mathbf{k}},\qquad (0 \le t <\infty)$

이 곡면

$z^3 + xyz -2 = 0$

과 만나는 점을 $P$ 라 하자.

$P$ 에서 곡선과 곡면 사이의 (예각) 각을 구하세요.

영어 원문 보기

Let $P$ be the point where the curve

$\mathbf{r}(t) = t^3\,\hat{\boldsymbol{\imath}} + t\,\hat{\boldsymbol{\jmath}} + t^2\,\hat{\mathbf{k}},\qquad (0 \le t <\infty)$

intersects the surface

$z^3 + xyz -2 = 0$

Find the (acute) angle between the curve and the surface at $P$ .

정답·풀이 보기

$\mathbf{r}(t)=( x(t)\,,\,y(t)\,,\,z(t))$ 가 $z^3 + xyz -2 = 0$ 과 만나는 것은 다음과 같을 때이다.

$z(t)^3+x(t)\,y(t)\,z(t)-2=0\iff \big(t^2\big)^3 + \big(t^3)(t)\big(t^2\big)-2=0 \iff 2t^6=2\iff t=1$

$t$ 는 양수여야 하기 때문이다. $t=1$ 에서 곡선의 방향벡터는

$\mathbf{r}'(1)=3\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+2\,\hat{\mathbf{k}}$

$\mathbf{r}(1)=( 1,1,1)$ 에서 곡면의 법선벡터는

${ \boldsymbol{\nabla} }(z^3+xyz)\big|_{(1,1,1)}=[yz\hat{\boldsymbol{\imath}}+xz\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}}]_{(1,1,1)} =\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+4\hat{\mathbf{k}}$

곡선과 곡면의 법선벡터 사이의 각 $\theta$ 는 다음으로 정해진다.

$\begin{aligned} \big|( 3,1,2)\big|\,\big|( 1,1,4)\big|\cos\theta =( 3,1,2) \cdot( 1,1,4) &\iff \sqrt{14}\sqrt{18}\cos\theta=12 \\ &\iff \sqrt{7\times 36}\cos\theta=12 \\ &\iff \cos\theta=\frac{2}{\sqrt{7}} \\ &\iff \theta=40.89^\circ \end{aligned}$

곡선과 곡면 사이의 각은 $90-40.89=49.11^\circ$ (소수 둘째 자리까지)이다.

원본 교재 풀이(영어)

$\mathbf{r}(t)=( x(t)\,,\,y(t)\,,\,z(t))$ intersects $z^3 + xyz -2 = 0$ when

$z(t)^3+x(t)\,y(t)\,z(t)-2=0\iff \big(t^2\big)^3 + \big(t^3)(t)\big(t^2\big)-2=0 \iff 2t^6=2\iff t=1$

since $t$ is required to be positive. The direction vector for the curve at $t=1$ is

$\mathbf{r}'(1)=3\,\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+2\,\hat{\mathbf{k}}$

A normal vector for the surface at $\mathbf{r}(1)=( 1,1,1)$ is

${ \boldsymbol{\nabla} }(z^3+xyz)\big|_{(1,1,1)}=[yz\hat{\boldsymbol{\imath}}+xz\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}}]_{(1,1,1)} =\hat{\boldsymbol{\imath}}+\hat{\boldsymbol{\jmath}}+4\hat{\mathbf{k}}$

The angle $\theta$ between the curve and the normal vector to the surface is determined by

$\begin{aligned} \big|( 3,1,2)\big|\,\big|( 1,1,4)\big|\cos\theta =( 3,1,2) \cdot( 1,1,4) &\iff \sqrt{14}\sqrt{18}\cos\theta=12 \\ &\iff \sqrt{7\times 36}\cos\theta=12 \\ &\iff \cos\theta=\frac{2}{\sqrt{7}} \\ &\iff \theta=40.89^\circ \end{aligned}$

The angle between the curve and the surface is $90-40.89=49.11^\circ$ (to two decimal places).

한 번 더 — 비슷한 문제 풀기CLP-4 · 3.2절 · 16번

# 매개변수 곡면계산도전자동 분류

646.CLP-4 · 3.2절 · 18번대학 수준우리말 번역

방정식

$z=xy e^{-(x^2+y^2)/2}$

인 곡면에 접하는 수평인 평면을 모두 구하세요.

이 곡면 위에서 $z$ 의 최댓값과 최솟값은 각각 얼마입니까?

영어 원문 보기

Find all horizontal planes that are tangent to the surface with equation

$z=xy e^{-(x^2+y^2)/2}$

What are the largest and smallest values of $z$ on this surface?

정답·풀이 보기

$(x_0,y_0,z_0)$ 을 곡면 위의 임의의 점이라 하자. 점 $(x_0,y_0,z_0)$ 에서 곡면에 수직인 벡터는

$\begin{aligned} &{ \boldsymbol{\nabla} }\Big(xy e^{-(x^2+y^2)/2}-z\Big)\bigg|_{(x_0,y_0,z_0)} \\& =\left( y_0 e^{-(x_0^2+y_0^2)/2}-x_0^2y_0 e^{-(x_0^2+y_0^2)/2}, x_0 e^{-(x_0^2+y_0^2)/2}-x_0y_0^2 e^{-(x_0^2+y_0^2)/2},-1\right) \end{aligned}$

이다.

점 $(x_0,y_0,z_0)$ 에서 곡면의 접평면이 수평인 것은 이 벡터가 수직인 것과 필요충분조건이고, 이는 $x$ 성분과 $y$ 성분이 모두 0인 것과 필요충분조건이며, 이는 다시 다음과 필요충분조건이다.

$\begin{aligned} &y_0(1-x_0^2)=0\text{ 그리고 }x_0(1-y_0^2)=0\\ &\iff\big\{y_0=0\text{ 또는 }x_0=1\text{ 또는 }x_0=-1\big\} \text{ 그리고 }\big\{x_0=0\text{ 또는 }y_0=1\text{ 또는 }y_0=-1\big\}\\ &\iff (x_0,y_0)=(0,0)\text{ 또는 }(1,1)\text{ 또는 }(1,-1) \text{ 또는 }(-1,1)\text{ 또는 }(-1,-1) \end{aligned}$

이 점들에서 $z_0$ 의 값은 각각 $0$ , $e^{-1}$ , $-e^{-1}$ , $-e^{-1}$ , $e^{-1}$ 이다. 따라서 수평인 접평면은 $z=0$ , $z=e^{-1}$ , $z=-e^{-1}$ 이다. 곡면의 가장 높은 점과 가장 낮은 점에서 접평면은 수평이다. 따라서 $z$ 의 최댓값과 최솟값은 각각 $e^{-1}$ 과 $-e^{-1}$ 이다.

원본 교재 풀이(영어)

Let $(x_0,y_0,z_0)$ be any point on the surface. A vector normal to the surface at $(x_0,y_0,z_0)$ is

$\begin{aligned} &{ \boldsymbol{\nabla} }\Big(xy e^{-(x^2+y^2)/2}-z\Big)\bigg|_{(x_0,y_0,z_0)} \\& =\left( y_0 e^{-(x_0^2+y_0^2)/2}-x_0^2y_0 e^{-(x_0^2+y_0^2)/2}, x_0 e^{-(x_0^2+y_0^2)/2}-x_0y_0^2 e^{-(x_0^2+y_0^2)/2},-1\right) \end{aligned}$

The tangent plane to the surface at $(x_0,y_0,z_0)$ is horizontal if and only if this vector is vertical, which is the case if and only if its $x$ - and $y$ -components are zero, which in turn is the case if and only if

$\begin{aligned} &y_0(1-x_0^2)=0\text{ and }x_0(1-y_0^2)=0\\ &\iff\big\{y_0=0\text{ or }x_0=1\text{ or }x_0=-1\big\} \text{ and }\big\{x_0=0\text{ or }y_0=1\text{ or }y_0=-1\big\}\\ &\iff (x_0,y_0)=(0,0)\text{ or }(1,1)\text{ or }(1,-1) \text{ or }(-1,1)\text{ or }(-1,-1) \end{aligned}$

The values of $z_0$ at these points are $0$ , $e^{-1}$ , $-e^{-1}$ , $-e^{-1}$ and $e^{-1}$ , respectively. So the horizontal tangent planes are $z=0$ , $z=e^{-1}$ and $z=-e^{-1}$ . At the highest and lowest points of the surface, the tangent plane is horizontal. So the largest and smallest values of $z$ are $e^{-1}$ and $-e^{-1}$ , respectively.

한 번 더 — 비슷한 문제 풀기APEX · 14.5절 · 1번

# 매개변수 곡면계산도전자동 분류

647.CLP-4 · 3.3절 · 10번대학 수준우리말 번역

반구 $z=\sqrt{a^2-x^2-y^2}$ 중에서 원기둥 $\big(x-\frac{a}{2}\big)^2+y^2=\big(\frac{a}{2}\big)^2$ 의 내부에 놓인 부분의 겉넓이를 구하세요.

영어 원문 보기

Find the surface area of that part of the hemisphere $z=\sqrt{a^2-x^2-y^2}$ which lies within the cylinder $\big(x-\frac{a}{2}\big)^2+y^2=\big(\frac{a}{2}\big)^2$ .

답 쓰고 채점

정답·풀이 보기

반구의 일부분의 겉넓이를 구하려고 한다. 반구 위에서

$\begin{aligned} z=f(x,y)=\sqrt{a^2-x^2-y^2}\qquad f_x(x,y)=-\frac{x}{\sqrt{a^2-x^2-y^2}}\qquad f_y(x,y)=-\frac{y}{\sqrt{a^2-x^2-y^2}} \end{aligned}$

이므로

$\begin{aligned} \mathrm{d}S&=\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\,\mathrm{d}x\,\mathrm{d}y =\sqrt{1+\frac{x^2}{a^2-x^2-y^2}+\frac{y^2}{a^2-x^2-y^2}}\,\mathrm{d}x\,\mathrm{d}y \\ &=\sqrt{\frac{a^2}{a^2-x^2-y^2}}\,\mathrm{d}x\,\mathrm{d}y \end{aligned}$

이다. 극좌표로 나타내면 $\mathrm{d}S=\frac{a}{\sqrt{a^2-r^2}}\,r\,\mathrm{d}r\,\mathrm{d}\theta$ 이다. 원기둥 $x^2-ax+y^2=0$ 의 내부에 있는 반구 부분의 겉넓이를 구하려고 하는데, 이 원기둥의 방정식은 극좌표로 $r^2-ar\cos\theta=0$ 곧 $r=a\cos\theta$ 가 된다. 적분 영역의 위쪽 절반을 아래에 그렸다.

따라서

$\begin{aligned} {\rm Surface\ Area} &= 2\int_0^{\pi/2}\mathrm{d}\theta\int_0^{a\cos\theta}\mathrm{d}r\ r \frac{a}{\sqrt{a^2-r^2}} = 2a\int_0^{\pi/2}\mathrm{d}\theta\ \Big[-\sqrt{a^2-r^2}\,\Big]_0^{a\cos\theta} \\ &= 2a\int_0^{\pi/2}\mathrm{d}\theta\ \big[a-a\sin\theta\big] \\ &= 2a^2\Big[\theta+\cos\theta\Big]_0^{\pi/2} =a^2[\pi-2] \end{aligned}$

원본 교재 풀이(영어)

We are to find the surface area of part of a hemisphere. On the hemisphere

$\begin{aligned} z=f(x,y)=\sqrt{a^2-x^2-y^2}\qquad f_x(x,y)=-\frac{x}{\sqrt{a^2-x^2-y^2}}\qquad f_y(x,y)=-\frac{y}{\sqrt{a^2-x^2-y^2}} \end{aligned}$

so that

$\begin{aligned} \mathrm{d}S&=\sqrt{1+f_x(x,y)^2+f_y(x,y)^2}\,\mathrm{d}x\,\mathrm{d}y =\sqrt{1+\frac{x^2}{a^2-x^2-y^2}+\frac{y^2}{a^2-x^2-y^2}}\,\mathrm{d}x\,\mathrm{d}y \\ &=\sqrt{\frac{a^2}{a^2-x^2-y^2}}\,\mathrm{d}x\,\mathrm{d}y \end{aligned}$

In polar coordinates, this is $\mathrm{d}S=\frac{a}{\sqrt{a^2-r^2}}\,r\,\mathrm{d}r\,\mathrm{d}\theta$ . We are to find the surface area of the part of the hemisphere that is inside the cylinder, $x^2-ax+y^2=0$ , which in polar coordinates becomes $r^2-ar\cos\theta=0$ or $r=a\cos\theta$ . The top half of the domain of integration is sketched below.

So the

$\begin{aligned} {\rm Surface\ Area} &= 2\int_0^{\pi/2}\mathrm{d}\theta\int_0^{a\cos\theta}\mathrm{d}r\ r \frac{a}{\sqrt{a^2-r^2}} = 2a\int_0^{\pi/2}\mathrm{d}\theta\ \Big[-\sqrt{a^2-r^2}\,\Big]_0^{a\cos\theta} \\ &= 2a\int_0^{\pi/2}\mathrm{d}\theta\ \big[a-a\sin\theta\big] \\ &= 2a^2\Big[\theta+\cos\theta\Big]_0^{\pi/2} =a^2[\pi-2] \end{aligned}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.6절 · 3번

# 면적분과 플럭스활용보통자동 분류

648.CLP-4 · 3.3절 · 13번대학 수준우리말 번역

반지름이 $a$ 인 구 껍질이 원점에 중심을 두고 있다. 이 구에서 제1팔분공간에 놓인 부분의 무게중심(즉, 밀도가 상수일 때의 질량중심)을 구하세요.

영어 원문 보기

A spherical shell of radius $a$ is centred at the origin. Find the centroid (i.e. the centre of mass with constant density) of the part of the sphere that lies in the first octant.

답 쓰고 채점

정답·풀이 보기

대칭성에 의해 무게중심 $(\bar x,\bar y,\bar z)$ 는 $\bar x=\bar y=\bar z$ 를 만족한다. 구면좌표를 이용하여 구를 매개변수로 나타내면 다음과 같다.

$\mathbf{r}(\theta,\varphi) = a\sin\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +a\sin\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} +a\cos\varphi\,\hat{\mathbf{k}}$

그러면

$\begin{aligned} \frac{\partial \mathbf{r}}{\partial \theta} =-a\sin\varphi\sin\theta\,\hat{\boldsymbol{\imath}}+a\sin\varphi\cos\theta\,\hat{\boldsymbol{\jmath}}\qquad \frac{\partial \mathbf{r}}{\partial \varphi} =a\cos\varphi\cos\theta\,\hat{\boldsymbol{\imath}}+a\cos\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a\sin\varphi\,\hat{\mathbf{k}} \end{aligned}$

이므로

$\begin{aligned} \frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial \varphi} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ -a\sin\varphi\sin\theta & a\sin\varphi\cos\theta & 0 \\ a\cos\varphi\cos\theta & a\cos\varphi\sin\theta & -a\sin\varphi \end{matrix}\right] \\ &=-a^2\sin^2\varphi\cos\theta\,\hat{\boldsymbol{\imath}}-a^2\sin^2\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a^2\sin\varphi\cos\varphi\,\hat{\mathbf{k}}\\ \implies \mathrm{d}S&=\left|\frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial \varphi}\right| \,\mathrm{d}\theta\,\mathrm{d}\varphi =a^2\sin\varphi\,\mathrm{d}\theta\,\mathrm{d}\varphi \end{aligned}$

제1팔분공간에 있는 구면 부분의 겉넓이는 $\frac{1}{8}4\pi a^2=\frac{\pi a^2}{2}$ 이므로

$\begin{aligned} \bar x=\bar y=\bar z &=\frac{\iint _S z\,\mathrm{d}S}{\iint _S\,\mathrm{d}S} =\frac{1}{\pi a^2/2}\int_0^{\pi/2}\mathrm{d}\theta\int_0^{\pi/2}\mathrm{d}\varphi \ (a^2\sin\varphi)(\overbrace{a\cos\varphi}^{z}) \\ &=\frac{2a}{\pi}\frac{\pi}{2}\int_0^{\pi/2}\mathrm{d}\varphi \ \sin\varphi\cos\varphi =a\bigg[\frac{1}{2}\sin^2\varphi\bigg]_0^{\pi/2} =\frac{a}{2} \end{aligned}$

원본 교재 풀이(영어)

By symmetry, the centroid $(\bar x,\bar y,\bar z)$ obeys $\bar x=\bar y=\bar z$ . Parametrize the sphere using spherical coordinates.

$\mathbf{r}(\theta,\varphi) = a\sin\varphi\cos\theta\,\hat{\boldsymbol{\imath}} +a\sin\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} +a\cos\varphi\,\hat{\mathbf{k}}$

Then

$\begin{aligned} \frac{\partial \mathbf{r}}{\partial \theta} =-a\sin\varphi\sin\theta\,\hat{\boldsymbol{\imath}}+a\sin\varphi\cos\theta\,\hat{\boldsymbol{\jmath}}\qquad \frac{\partial \mathbf{r}}{\partial \varphi} =a\cos\varphi\cos\theta\,\hat{\boldsymbol{\imath}}+a\cos\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a\sin\varphi\,\hat{\mathbf{k}} \end{aligned}$

so that

$\begin{aligned} \frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial \varphi} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ -a\sin\varphi\sin\theta & a\sin\varphi\cos\theta & 0 \\ a\cos\varphi\cos\theta & a\cos\varphi\sin\theta & -a\sin\varphi \end{matrix}\right] \\ &=-a^2\sin^2\varphi\cos\theta\,\hat{\boldsymbol{\imath}}-a^2\sin^2\varphi\sin\theta\,\hat{\boldsymbol{\jmath}} -a^2\sin\varphi\cos\varphi\,\hat{\mathbf{k}}\\ \implies \mathrm{d}S&=\left|\frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial \varphi}\right| \,\mathrm{d}\theta\,\mathrm{d}\varphi =a^2\sin\varphi\,\mathrm{d}\theta\,\mathrm{d}\varphi \end{aligned}$

As the surface area of the part of the sphere in the first octant is $\frac{1}{8}4\pi a^2=\frac{\pi a^2}{2}$

$\begin{aligned} \bar x=\bar y=\bar z &=\frac{\iint _S z\,\mathrm{d}S}{\iint _S\,\mathrm{d}S} =\frac{1}{\pi a^2/2}\int_0^{\pi/2}\mathrm{d}\theta\int_0^{\pi/2}\mathrm{d}\varphi \ (a^2\sin\varphi)(\overbrace{a\cos\varphi}^{z}) \\ &=\frac{2a}{\pi}\frac{\pi}{2}\int_0^{\pi/2}\mathrm{d}\varphi \ \sin\varphi\cos\varphi =a\bigg[\frac{1}{2}\sin^2\varphi\bigg]_0^{\pi/2} =\frac{a}{2} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 3.3절 · 10번

# 면적분과 플럭스계산보통자동 분류

649.CLP-4 · 3.3절 · 14번대학 수준우리말 번역

원기둥 $x^2+y^2=2ay$ 에서 $z^2=x^2+y^2$ 의 바깥쪽에 놓인 부분의 넓이를 구하세요.

영어 원문 보기

Find the area of that part of the cylinder $x^2+y^2=2ay$ lying outside $z^2=x^2+y^2$ .

답 쓰고 채점

정답·풀이 보기

원기둥좌표계에서

$x=r\cos\theta\qquad y=r\sin\theta\qquad z=z$

이 좌표계에서 원기둥의 방정식 $x^2+y^2=2ay$ 는

$r^2=2ar\sin\theta\qquad\text{또는}\qquad r=2a\sin\theta$

가 된다. 즉 $r=f(\theta)$ 이고 $f(\theta) =2a\sin\theta$ 이다. 원기둥을 $\mathbf{r}(\theta,z) = x(\theta,z)\,\hat{\boldsymbol{\imath}} + y(\theta,z)\,\hat{\boldsymbol{\jmath}} + z(\theta,z)\,\hat{\mathbf{k}}$ 로 매개화하면 다음과 같다.

$\begin{aligned} x(\theta,z)&=f(\theta)\cos\theta=2a\sin\theta\cos\theta=a\sin2\theta\\ y(\theta,z)&=f(\theta)\sin\theta=2a\sin\theta\sin\theta=a(1-\cos2\theta)\\ z(\theta,z)&=z \end{aligned}$

이 매개화에서

$\begin{aligned} \frac{\partial \mathbf{r}}{\partial \theta}=2a\cos2\theta\,\hat{\boldsymbol{\imath}}+2a\sin2\theta\,\hat{\boldsymbol{\jmath}}\qquad \frac{\partial \mathbf{r}}{\partial z}=\hat{\mathbf{k}} &\implies \frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial z} =-2a\cos2\theta\,\hat{\boldsymbol{\jmath}}+2a\sin2\theta\,\hat{\boldsymbol{\imath}} \\ &\implies \mathrm{d}S=\left|\frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial z}\right|\, \mathrm{d}\theta\,\mathrm{d}z =2a\,\mathrm{d}\theta\,\mathrm{d}z \end{aligned}$

이제 적분 구간을 정해야 한다. 아래 왼쪽 그림은 원기둥을 위에서 본 모습이다.

그림에서 $0\le\theta\le\pi$ 임을 알 수 있다. 원뿔 $z^2=x^2+y^2=r^2$ (즉 $z=\pm r$ ) 과 원기둥 $r=2a\sin\theta$ 는 $z^2=r^2=4a^2\sin^2\theta$ 에서 만난다. 따라서 각각의 고정된 $\theta$ 에 대해 $z$ 는 $-2a\sin\theta$ 에서 $z=+2a\sin\theta$ 까지 변한다. (위 오른쪽 그림을 보라. $\theta$ 가 상수인 단면을 나타낸 것이다.) 마지막으로,

$\begin{aligned} {\rm Area} &= \int_{|z|\le 2a\sin\theta}2a\,\mathrm{d}\theta\,\mathrm{d}z =2a\int_0^\pi \!\!\mathrm{d}\theta\int_{-2a\sin\theta}^{2a\sin\theta} \!\!\!\mathrm{d}z =8a^2\int_0^\pi \!\!\mathrm{d}\theta\ \sin\theta =8a^2\Big[-\cos\theta\Big]_0^\pi \\ &=16a^2 \end{aligned}$

원본 교재 풀이(영어)

In cylindrical coordinates

$x=r\cos\theta\qquad y=r\sin\theta\qquad z=z$

In these coordinates the equation, $x^2+y^2=2ay$ , of the cylinder becomes

$r^2=2ar\sin\theta\qquad\text{or}\qquad r=2a\sin\theta$

That is, $r=f(\theta)$ with $f(\theta) =2a\sin\theta$ . Parametrize the cylinder by $\mathbf{r}(\theta,z) = x(\theta,z)\,\hat{\boldsymbol{\imath}} + y(\theta,z)\,\hat{\boldsymbol{\jmath}} + z(\theta,z)\,\hat{\mathbf{k}}$ with

$\begin{aligned} x(\theta,z)&=f(\theta)\cos\theta=2a\sin\theta\cos\theta=a\sin2\theta\\ y(\theta,z)&=f(\theta)\sin\theta=2a\sin\theta\sin\theta=a(1-\cos2\theta)\\ z(\theta,z)&=z \end{aligned}$

Under this parametrization,

$\begin{aligned} \frac{\partial \mathbf{r}}{\partial \theta}=2a\cos2\theta\,\hat{\boldsymbol{\imath}}+2a\sin2\theta\,\hat{\boldsymbol{\jmath}}\qquad \frac{\partial \mathbf{r}}{\partial z}=\hat{\mathbf{k}} &\implies \frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial z} =-2a\cos2\theta\,\hat{\boldsymbol{\jmath}}+2a\sin2\theta\,\hat{\boldsymbol{\imath}} \\ &\implies \mathrm{d}S=\left|\frac{\partial \mathbf{r}}{\partial \theta}\times\frac{\partial \mathbf{r}}{\partial z}\right|\, \mathrm{d}\theta\,\mathrm{d}z =2a\,\mathrm{d}\theta\,\mathrm{d}z \end{aligned}$

We still have to determine the limits of integration. The figure on the left below provides a top view of the cylinder.

From it we see that $0\le\theta\le\pi$ . The cone $z^2=x^2+y^2=r^2$ (i.e. $z=\pm r$ ) and the cylinder $r=2a\sin\theta$ intersect at $z^2=r^2=4a^2\sin^2\theta$ . So, for each fixed $\theta$ , $z$ runs from $-2a\sin\theta$ to $z=+2a\sin\theta$ . (See the figure on the right above. It shows a constant $\theta$ cross-section.) Finally,

$\begin{aligned} {\rm Area} &= \int_{|z|\le 2a\sin\theta}2a\,\mathrm{d}\theta\,\mathrm{d}z =2a\int_0^\pi \!\!\mathrm{d}\theta\int_{-2a\sin\theta}^{2a\sin\theta} \!\!\!\mathrm{d}z =8a^2\int_0^\pi \!\!\mathrm{d}\theta\ \sin\theta =8a^2\Big[-\cos\theta\Big]_0^\pi \\ &=16a^2 \end{aligned}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.6절 · 7번

# 면적분과 플럭스활용보통자동 분류

650.CLP-4 · 3.3절 · 21번대학 수준우리말 번역

(a) 꼭짓점이 점 $(0, 0, 3)$ 에 있고, 밑면이 $xy$ 평면 위의 원 $x^2 + y^2 = 4$ 인 원뿔의 곡면 $S$ 의 매개변수화를 구하세요. 원뿔의 옆면만 $S$ 에 속하고, 밑면은 속하지 않습니다. 매개변수의 정의역을 반드시 포함하세요. (b) (a)의 곡면 $S$ 의 질량중심의 $z$ 좌표를 구하세요.

영어 원문 보기

(a) Find a parametrization of the surface $S$ of the cone whose vertex is at the point $(0, 0, 3)$ , and whose base is the circle $x^2 + y^2 = 4$ in the $xy$ -plane. Only the cone surface belongs to $S$ , not the base. Be careful to include the domain for the parameters. (b) Find the $z$ -coordinate of the centre of mass of the surface $S$ from (a).

정답·풀이 보기

(a) $z$ 가 상수인 (수평) 단면은 각각 $z$ 축을 중심으로 하는 원이다. 반지름은 $z=0$ 일 때 $2$ 에서 $z=3$ 일 때 $0$ 까지 일차적으로 변한다. 따라서 높이 $z$ 에서의 반지름은 $\frac{2}{3}(3-z)$ 이고, 다음을 사용할 수 있다.

$\begin{aligned} \mathbf{r}(\theta,z) = \frac{2}{3}(3-z)\cos\theta\ \hat{\boldsymbol{\imath}} +\frac{2}{3}(3-z)\sin\theta\ \hat{\boldsymbol{\jmath}} +z\,\hat{\mathbf{k}}\qquad 0\le\theta<2\pi,\quad 0\le z\le 3 \end{aligned}$

이것이 매개변수화이다.

(b) 대칭성에 의해 질량중심은 $z$ 축 위에 있다. 어차피 $z$ 좌표만 구하면 된다. 질량중심의 $z$ 좌표는 원뿔 위에서 $z$ 의 가중평균이다. 밀도가 주어지지 않았으므로 상수라고 가정한다. 밀도를 $1$ 로 잡을 수 있으므로, 질량중심의 $z$ 좌표는 $\iint _S z\,\mathrm{d}S/ \iint _S \mathrm{d}S$ 이다.

$\begin{aligned} \tfrac{\partial\mathbf{r}}{\partial\theta} &=\big(-\tfrac{2}{3}(3-z)\sin\theta\,,\, \tfrac{2}{3}(3-z)\cos\theta\,,\, 0 \big) \\ \tfrac{\partial\mathbf{r}}{\partial z} &=\big(-\tfrac{2}{3}\cos\theta\,,\, -\tfrac{2}{3}\sin\theta\,,\, 1 \big) \\ \tfrac{\partial\mathbf{r}}{\partial\theta} \times \tfrac{\partial\mathbf{r}}{\partial z} &=\big(\tfrac{2}{3}(3-z)\cos\theta\,,\, \tfrac{2}{3}(3-z)\sin\theta\,,\, \tfrac{4}{9}(3-z)\big) \end{aligned}$

이므로, 이 매개변수화에 대한 겉넓이 요소는 다음과 같다.

$\begin{aligned} \mathrm{d}S &= \big|\tfrac{\partial\mathbf{r}}{\partial\theta}\times \tfrac{\partial\mathbf{r}}{\partial z} \big|\mathrm{d}\theta\mathrm{d}z =\tfrac{2}{3}(3-z)\big|(\cos\theta\,,\,\sin\theta\,,\,\tfrac{2}{3}\big)\big| \mathrm{d}\theta\mathrm{d}z \\ &=\tfrac{2\sqrt{13}}{9}(3-z) \mathrm{d}\theta\mathrm{d}z \end{aligned}$

따라서 원뿔의 겉넓이 $\iint _S \mathrm{d}S$ 는 다음과 같다.

$\begin{aligned} \int_0^3\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\ \tfrac{2\sqrt{13}}{9}(3-z) &=\tfrac{4\sqrt{13}}{9}\pi\int_0^3\mathrm{d}z\ (3-z) =-\tfrac{2\sqrt{13}}{9}\pi(3-z)^2\Big|_0^3 \\ &=2\sqrt{13}\pi \end{aligned}$

그리고 질량중심의 $z$ 좌표는 다음과 같다.

$\begin{aligned} \bar z &=\frac{1}{2\sqrt{13}\pi} \int_0^3\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\ \tfrac{2\sqrt{13}}{9}(3-z) z =\frac{2}{9}\int_0^3\mathrm{d}z\ (3z-z^2) =\frac{2}{9}{\Big[\frac{3z^2}{2}-\frac{z^3}{3}\Big]}_0^3 \\ &=\frac{2}{9}\frac{27}{6} =1 \end{aligned}$

이는 원뿔의 절반 높이보다 조금 낮은 값인데, 원뿔이 “아래쪽이 무겁기” 때문에 타당하다.

원본 교재 풀이(영어)

(a) Each (horizontal) constant $z$ cross-section is a circle centred on the $z$ -axis. The radius varies linearly from $2$ , when $z=0$ to $0$ , when $z=3$ . So the radius at height $z$ is $\frac{2}{3}(3-z)$ and we can use

$\begin{aligned} \mathbf{r}(\theta,z) = \frac{2}{3}(3-z)\cos\theta\ \hat{\boldsymbol{\imath}} +\frac{2}{3}(3-z)\sin\theta\ \hat{\boldsymbol{\jmath}} +z\,\hat{\mathbf{k}}\qquad 0\le\theta<2\pi,\quad 0\le z\le 3 \end{aligned}$

as the parametrization.

(b) By symmetry the centre of mass will lie on the $z$ -axis. We are only asked for the $z$ -coordinate anyway. The $z$ -coordinate of the centre of mass is the weighted average of $z$ over the cone. Since a density has not been specified, we assume that it is a constant. We may take the density to be $1$ , so the $z$ -coordinate of the centre of mass is $\iint _S z\,\mathrm{d}S/ \iint _S \mathrm{d}S$ .

Since

$\begin{aligned} \tfrac{\partial\mathbf{r}}{\partial\theta} &=\big(-\tfrac{2}{3}(3-z)\sin\theta\,,\, \tfrac{2}{3}(3-z)\cos\theta\,,\, 0 \big) \\ \tfrac{\partial\mathbf{r}}{\partial z} &=\big(-\tfrac{2}{3}\cos\theta\,,\, -\tfrac{2}{3}\sin\theta\,,\, 1 \big) \\ \tfrac{\partial\mathbf{r}}{\partial\theta} \times \tfrac{\partial\mathbf{r}}{\partial z} &=\big(\tfrac{2}{3}(3-z)\cos\theta\,,\, \tfrac{2}{3}(3-z)\sin\theta\,,\, \tfrac{4}{9}(3-z)\big) \end{aligned}$

the element of surface area for this parametrization is

$\begin{aligned} \mathrm{d}S &= \big|\tfrac{\partial\mathbf{r}}{\partial\theta}\times \tfrac{\partial\mathbf{r}}{\partial z} \big|\mathrm{d}\theta\mathrm{d}z =\tfrac{2}{3}(3-z)\big|(\cos\theta\,,\,\sin\theta\,,\,\tfrac{2}{3}\big)\big| \mathrm{d}\theta\mathrm{d}z \\ &=\tfrac{2\sqrt{13}}{9}(3-z) \mathrm{d}\theta\mathrm{d}z \end{aligned}$

So the surface area, $\iint _S \mathrm{d}S$ , of the cone is

$\begin{aligned} \int_0^3\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\ \tfrac{2\sqrt{13}}{9}(3-z) &=\tfrac{4\sqrt{13}}{9}\pi\int_0^3\mathrm{d}z\ (3-z) =-\tfrac{2\sqrt{13}}{9}\pi(3-z)^2\Big|_0^3 \\ &=2\sqrt{13}\pi \end{aligned}$

and the $z$ -coordinate of the centre of mass is

$\begin{aligned} \bar z &=\frac{1}{2\sqrt{13}\pi} \int_0^3\mathrm{d}z\int_0^{2\pi}\mathrm{d}\theta\ \tfrac{2\sqrt{13}}{9}(3-z) z =\frac{2}{9}\int_0^3\mathrm{d}z\ (3z-z^2) =\frac{2}{9}{\Big[\frac{3z^2}{2}-\frac{z^3}{3}\Big]}_0^3 \\ &=\frac{2}{9}\frac{27}{6} =1 \end{aligned}$

This is a little less than half way up the cone, which is reasonable since the cone is “bottom heavy”.

한 번 더 — 비슷한 문제 풀기APEX · 14.6절 · 4번

# 면적분과 플럭스계산보통자동 분류

651.CLP-4 · 3.3절 · 31번대학 수준우리말 번역

벡터장

$\mathbf{F} = \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}}{\sqrt{x^2+y^2+z^2}}$

의 순 바깥쪽 플럭스를, 원점을 중심으로 하는 반지름 $1$ 인 구와 반지름 $2$ 인 구 사이 영역의 경계면에 대해 구하세요.

영어 원문 보기

Compute the net outward flux of the vector field

$\mathbf{F} = \frac{\mathbf{r}}{|\mathbf{r}|} = \frac{x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}}{\sqrt{x^2+y^2+z^2}}$

across the boundary of the region between the spheres of radius $1$ and radius $2$ centred at the origin.

답 쓰고 채점

정답·풀이 보기

다음과 같이 놓는다.

$V=\big\{ \ (x,y,z)\ \big|\ 1\le x^2+y^2+z^2\le 4\ \big\}$

$V$ 의 경계는 두 부분으로 이루어진다. 하나는 원점을 중심으로 하는 반지름 $2$ 인 구 $S_2$ 로, (바깥쪽) 법선벡터는 $\hat{\mathbf{n}}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{\mathbf{r}}{2}$ 이다. 다른 하나는 원점을 중심으로 하는 반지름 $1$ 인 구 $S_1$ 로, (안쪽) 법선벡터는 $\hat{\mathbf{n}} = -\mathbf{r}$ 이다. 따라서

$\begin{aligned} \iint _{\partial V} \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\iint _{S_2} \frac{\mathbf{r}}{|\mathbf{r}|}\cdot\frac{\mathbf{r}}{2}\,\mathrm{d}S -\iint _{S_1} \frac{\mathbf{r}}{|\mathbf{r}|}\cdot\mathbf{r}\,\mathrm{d}S \\ &=\iint _{S_2} \mathrm{d}S -\iint _{S_1} \mathrm{d}S \\ &=4\pi(2)^2 -4\pi(1)^2 \\ &=12\pi \end{aligned}$

원본 교재 풀이(영어)

Write

$V=\big\{ \ (x,y,z)\ \big|\ 1\le x^2+y^2+z^2\le 4\ \big\}$

The boundary of $V$ consists of two parts — the sphere, $S_2$ , of radius $2$ , centred on the origin, with (outward) normal $\hat{\mathbf{n}}=\frac{\mathbf{r}}{|\mathbf{r}|}=\frac{\mathbf{r}}{2}$ , and the sphere $S_1$ of radius $1$ , centred on the origin, with (inward) normal $\hat{\mathbf{n}} = -\mathbf{r}$ , So,

$\begin{aligned} \iint _{\partial V} \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=\iint _{S_2} \frac{\mathbf{r}}{|\mathbf{r}|}\cdot\frac{\mathbf{r}}{2}\,\mathrm{d}S -\iint _{S_1} \frac{\mathbf{r}}{|\mathbf{r}|}\cdot\mathbf{r}\,\mathrm{d}S \\ &=\iint _{S_2} \mathrm{d}S -\iint _{S_1} \mathrm{d}S \\ &=4\pi(2)^2 -4\pi(1)^2 \\ &=12\pi \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 3.3절 · 21번

# 면적분과 플럭스계산보통자동 분류

652.CLP-4 · 3.3절 · 42번대학 수준우리말 번역

$\mathcal{S}$ 를 곡면 $(x+y+1)^2+z^2=4$ 중 제1팔분공간에 놓인 부분이라 하자. $\mathcal{S}$ 를 통과하는 $\mathbf{F}$ 의 아래쪽 방향 플럭스를 구하세요. 여기서 $\mathbf{F}=xy\,\hat{\boldsymbol{\imath}}+(z-xy)\,\hat{\boldsymbol{\jmath}}$

영어 원문 보기

Let $\mathcal{S}$ be the part of the surface $(x+y+1)^2+z^2=4$ which lies in the first octant. Find the flux of $\mathbf{F}$ downwards through $\mathcal{S}$ where $\mathbf{F}=xy\,\hat{\boldsymbol{\imath}}+(z-xy)\,\hat{\boldsymbol{\jmath}}$

답 쓰고 채점

정답·풀이 보기

먼저 다음을 관찰하자.

$(x+y+1)^2\ge 0$ 이므로, $(x+y+1)^2+z^2=4$ 위의 모든 점은 $|z|\le 2$ 이다. 그리고,
$|z_0|\le 2$ 일 때, 곡면 $(x+y+1)^2+z^2=4$ 는 수평면 $z=z_0$ 와 $(x+y+1)^2 = 4-z_0^2$ 에서 만난다. 즉 두 직선 $x+y=\pm\sqrt{4-z_0^2}-1$ , $z=z_0$ 에서 만난다.
직선 $x+y=\pm\sqrt{4-z_0^2}-1$ , $z=z_0$ 가 제1팔분공간과 만나는 것은 $z_0\ge0$ 이고 $\pm\sqrt{4-z_0^2}-1\ge 0$ 일 때, 그리고 그때에 한한다.
따라서 $x+y=-\sqrt{4-z_0^2}-1$ , $z=z_0$ 는 제1팔분공간과 절대 만나지 않고,
$x+y=\sqrt{4-z_0^2}-1$ , $z=z_0$ 가 제1팔분공간과 만나는 것은 $0\le z_0\le \sqrt{3}$ 일 때, 그리고 그때에 한한다.
$z_0=0$ 일 때, 직선 $x+y=\sqrt{4-z_0^2}-1$ , $z=z_0$ 는 $x+y=1$ , $z=0$ 이다.
$z_0=\sqrt{3}$ 일 때, 직선 $x+y=\sqrt{4-z_0^2}-1$ , $z=z_0$ 는 $x+y=0$ , $z=2$ 이다.
그러므로 $(x,y,z)$ 가 $\mathcal{S}$ 위를 움직일 때, $(x,y)$ 는 삼각형 $x\ge0$ , $y\ge 0$ , $x+y\le 1$ 위를 움직인다.

$G(x,y,z)=(x+y+1)^2+z^2$ 이라 하자. 그러면 $\hat{\mathbf{n}}\,\mathrm{d}S=\pm\frac{{ \boldsymbol{\nabla} } G}{{ \boldsymbol{\nabla} } G\cdot\hat{\mathbf{k}}}\ \mathrm{d}x\mathrm{d}y =\pm\frac{2(x+y+1)\hat{\boldsymbol{\imath}}+2(x+y+1)\hat{\boldsymbol{\jmath}}+2z\hat{\mathbf{k}}}{2z}\,\mathrm{d}x\mathrm{d}y$ 아래쪽 법선을 택하려면 음의 부호가 필요하므로

$\begin{aligned} \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=-\big[xy\,\hat{\boldsymbol{\imath}}+(z-xy)\,\hat{\boldsymbol{\jmath}}\big]\cdot \Big[\frac{(x+y+1)\hat{\boldsymbol{\imath}}+(x+y+1)\hat{\boldsymbol{\jmath}}+z\hat{\mathbf{k}}}{z}\Big]\,\mathrm{d}x\mathrm{d}y\\ &=-\frac{1}{z}\big[xy(x+y+1)+(z-xy)(x+y+1)\big]\,\mathrm{d}x\mathrm{d}y\\ &=-\frac{1}{z}\big[z(x+y+1)\big]\,\mathrm{d}x\mathrm{d}y\\ &=-(x+y+1)\,\mathrm{d}x\mathrm{d}y \end{aligned}$

적분 정의역은 $x\ge 0,\ y\ge 0,\ x+y\le 1$ 이므로

$\begin{aligned} \iint _\mathcal{S}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=-\int_0^1 \mathrm{d}x\int_0^{1-x} \mathrm{d}y \ (x+y+1) =-\int_0^1 \mathrm{d}x\ \Big[(1+x)(1-x)+\frac{1}{2}(1-x)^2\Big]\\ &=-\int_0^1 \mathrm{d}x\ \Big[\frac{3}{2}-x-\frac{1}{2} x^2\Big] =-\Big[\frac{3}{2}-\frac{1}{2}-\frac{1}{6}\Big] =-\frac{5}{6} \end{aligned}$

원본 교재 풀이(영어)

First observe that,

because $(x+y+1)^2\ge 0$ , all points on $(x+y+1)^2+z^2=4$ have $|z|\le 2$ and that,
for $|z_0|\le 2$ , the surface $(x+y+1)^2+z^2=4$ intersects the horizontal plane $z=z_0$ on $(x+y+1)^2 = 4-z_0^2$ , i.e. on the two lines $x+y=\pm\sqrt{4-z_0^2}-1$ , $z=z_0$ .
The line $x+y=\pm\sqrt{4-z_0^2}-1$ , $z=z_0$ intersects the first octant if and only if $z_0\ge0$ and $\pm\sqrt{4-z_0^2}-1\ge 0$ .
Thus $x+y=-\sqrt{4-z_0^2}-1$ , $z=z_0$ never intersects the first octant and
$x+y=\sqrt{4-z_0^2}-1$ , $z=z_0$ intersects the first octant if and only if $0\le z_0\le \sqrt{3}$ .
When $z_0=0$ , the line $x+y=\sqrt{4-z_0^2}-1$ , $z=z_0$ is $x+y=1$ , $z=0$ .
When $z_0=\sqrt{3}$ , the line $x+y=\sqrt{4-z_0^2}-1$ , $z=z_0$ is $x+y=0$ , $z=2$ .
So as $(x,y,z)$ runs over $\mathcal{S}$ , $(x,y)$ runs over the triangle $x\ge0$ , $y\ge 0$ , $x+y\le 1$ .

Let $G(x,y,z)=(x+y+1)^2+z^2$ . Then $\hat{\mathbf{n}}\,\mathrm{d}S=\pm\frac{{ \boldsymbol{\nabla} } G}{{ \boldsymbol{\nabla} } G\cdot\hat{\mathbf{k}}}\ \mathrm{d}x\mathrm{d}y =\pm\frac{2(x+y+1)\hat{\boldsymbol{\imath}}+2(x+y+1)\hat{\boldsymbol{\jmath}}+2z\hat{\mathbf{k}}}{2z}\,\mathrm{d}x\mathrm{d}y$ For the downward normal, we need the minus sign, so

$\begin{aligned} \mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=-\big[xy\,\hat{\boldsymbol{\imath}}+(z-xy)\,\hat{\boldsymbol{\jmath}}\big]\cdot \Big[\frac{(x+y+1)\hat{\boldsymbol{\imath}}+(x+y+1)\hat{\boldsymbol{\jmath}}+z\hat{\mathbf{k}}}{z}\Big]\,\mathrm{d}x\mathrm{d}y\\ &=-\frac{1}{z}\big[xy(x+y+1)+(z-xy)(x+y+1)\big]\,\mathrm{d}x\mathrm{d}y\\ &=-\frac{1}{z}\big[z(x+y+1)\big]\,\mathrm{d}x\mathrm{d}y\\ &=-(x+y+1)\,\mathrm{d}x\mathrm{d}y \end{aligned}$

The domain of integration is $x\ge 0,\ y\ge 0,\ x+y\le 1$ , so

$\begin{aligned} \iint _\mathcal{S}\mathbf{F}\cdot\hat{\mathbf{n}}\,\mathrm{d}S &=-\int_0^1 \mathrm{d}x\int_0^{1-x} \mathrm{d}y \ (x+y+1) =-\int_0^1 \mathrm{d}x\ \Big[(1+x)(1-x)+\frac{1}{2}(1-x)^2\Big]\\ &=-\int_0^1 \mathrm{d}x\ \Big[\frac{3}{2}-x-\frac{1}{2} x^2\Big] =-\Big[\frac{3}{2}-\frac{1}{2}-\frac{1}{6}\Big] =-\frac{5}{6} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 3.3절 · 31번

# 면적분과 플럭스계산도전자동 분류

653.CLP-4 · 4.1절 · 2번대학 수준우리말 번역

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

영어 원문 보기

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

답 쓰고 채점

정답·풀이 보기

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

$\begin{aligned} { \boldsymbol{\nabla} }\times \mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ 1 & 0 & y \end{matrix}\right] \\ &=\hat{\boldsymbol{\imath}} \end{aligned}$

이고, 이 벡터는 모든 $x$ , $y$ , $z$ 에 대해 $\mathbf{F}(x,y,z)$ 와의 내적이 $1$ 이므로 이 벡터장과 수직이 아니다.

원본 교재 풀이(영어)

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

$\begin{aligned} { \boldsymbol{\nabla} }\times \mathbf{F} &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\ 1 & 0 & y \end{matrix}\right] \\ &=\hat{\boldsymbol{\imath}} \end{aligned}$

has dot product $1$ with $\mathbf{F}(x,y,z)$ (for all $x$ , $y$ , $z$ ) and so is not perpendicular to it.

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 3번

# 발산과 회전계산기본자동 분류

654.CLP-4 · 4.1절 · 3번대학 수준우리말 번역

다음 벡터 항등식을 확인하세요.

(a) ${ \boldsymbol{\nabla} }\cdot(f\mathbf{F})=f{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f$ (b) ${ \boldsymbol{\nabla} }\cdot(\mathbf{F}\times\mathbf{G}) =\mathbf{G}\cdot({ \boldsymbol{\nabla} }\times\mathbf{F})- \mathbf{F}\cdot({ \boldsymbol{\nabla} }\times\mathbf{G})$ (c) ${ \boldsymbol{\nabla} }^2(fg)=f\,{ \boldsymbol{\nabla} }^2 g+2{ \boldsymbol{\nabla} } f\cdot{ \boldsymbol{\nabla} } g+g\,{ \boldsymbol{\nabla} }^2 f$

영어 원문 보기

Verify the vector identities

(a) ${ \boldsymbol{\nabla} }\cdot(f\mathbf{F})=f{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f$ (b) ${ \boldsymbol{\nabla} }\cdot(\mathbf{F}\times\mathbf{G}) =\mathbf{G}\cdot({ \boldsymbol{\nabla} }\times\mathbf{F})- \mathbf{F}\cdot({ \boldsymbol{\nabla} }\times\mathbf{G})$ (c) ${ \boldsymbol{\nabla} }^2(fg)=f\,{ \boldsymbol{\nabla} }^2 g+2{ \boldsymbol{\nabla} } f\cdot{ \boldsymbol{\nabla} } g+g\,{ \boldsymbol{\nabla} }^2 f$

정답·풀이 보기

(a) 곱의 미분법에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(f\mathbf{F}) &=\frac{\partial }{\partial x}(fF_1) +\frac{\partial }{\partial y}(fF_2) +\frac{\partial }{\partial z}(fF_3) \\ &=\phantom{+}f\frac{\partial F_1}{\partial x} +f\frac{\partial F_2}{\partial y} +f\frac{\partial F_3}{\partial z} \\ &\phantom{=}+F_1\frac{\partial f}{\partial x} +F_2\frac{\partial f}{\partial y} +F_3\frac{\partial f}{\partial z} \\ &=f\,{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f \end{aligned}$

(b) 다시 곱의 미분법에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(\mathbf{F}\times\mathbf{G}) &=\frac{\partial }{\partial x}(F_2G_3-F_3G_2) +\frac{\partial }{\partial y}(F_3G_1-F_1G_3) +\frac{\partial }{\partial z}(F_1G_2-F_2G_1) \\ &=\phantom{+}\frac{\partial F_2}{\partial x}G_3 -\frac{\partial F_3}{\partial x}G_2 +\frac{\partial F_3}{\partial y}G_1 - \frac{\partial F_1}{\partial y}G_3 +\frac{\partial F_1}{\partial z}G_2 - \frac{\partial F_2}{\partial z}G_1 \\ &\phantom{=}+F_2\frac{\partial G_3}{\partial x} -F_3\frac{\partial G_2}{\partial x} +F_3\frac{\partial G_1}{\partial y} -F_1\frac{\partial G_3}{\partial y} +F_1\frac{\partial G_2}{\partial z} - F_2\frac{\partial G_1}{\partial z} \\ &=\phantom{+} \left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)G_1 +\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)G_2 +\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)G_3 \\ &\phantom{=} -F_1\left(\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}\right) -F_2\left(\frac{\partial G_1}{\partial z}-\frac{\partial G_3}{\partial x}\right) -F_3\left(\frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y}\right) \\ &=\mathbf{G}\cdot({ \boldsymbol{\nabla} }\times\mathbf{F})- \mathbf{F}\cdot({ \boldsymbol{\nabla} }\times\mathbf{G}) \end{aligned}$

(c) ${ \boldsymbol{\nabla} }^2(fg) = { \boldsymbol{\nabla} }\cdot\big[{ \boldsymbol{\nabla} }(fg)\big]$ 임을 떠올리자. 먼저

$\begin{aligned} { \boldsymbol{\nabla} }(fg) &=\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x}(fg) +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y}(fg) +\hat{\mathbf{k}}\frac{\partial }{\partial z}(fg) \\ &=\phantom{+}\hat{\boldsymbol{\imath}} g\frac{\partial f}{\partial x} +\hat{\boldsymbol{\jmath}} g\frac{\partial f}{\partial y} +\hat{\mathbf{k}} g\frac{\partial f}{\partial z} \\ &\phantom{=}+\hat{\boldsymbol{\imath}} f\frac{\partial g}{\partial x} +\hat{\boldsymbol{\jmath}} f\frac{\partial g}{\partial y} +\hat{\mathbf{k}} f\frac{\partial g}{\partial z} \\ &=g{ \boldsymbol{\nabla} } f+f{ \boldsymbol{\nabla} } g \end{aligned}$

따라서 (a)를 두 번 이용하면

$\begin{aligned} { \boldsymbol{\nabla} }^2(fg) & = { \boldsymbol{\nabla} }\cdot\big(g{ \boldsymbol{\nabla} } f\big) +{ \boldsymbol{\nabla} }\cdot\big(f{ \boldsymbol{\nabla} } g\big) \\ &= g\big({ \boldsymbol{\nabla} }\cdot { \boldsymbol{\nabla} } f\big) + \big({ \boldsymbol{\nabla} } g\big)\cdot\big({ \boldsymbol{\nabla} } f\big) +f\big({ \boldsymbol{\nabla} }\cdot{ \boldsymbol{\nabla} } g\big) + \big({ \boldsymbol{\nabla} } f\big)\cdot\big({ \boldsymbol{\nabla} } g\big) \\ &=f\,{ \boldsymbol{\nabla} }^2 g+2{ \boldsymbol{\nabla} } f\cdot{ \boldsymbol{\nabla} } g+g\,{ \boldsymbol{\nabla} }^2 f \end{aligned}$

원본 교재 풀이(영어)

(a) By the product rule

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(f\mathbf{F}) &=\frac{\partial }{\partial x}(fF_1) +\frac{\partial }{\partial y}(fF_2) +\frac{\partial }{\partial z}(fF_3) \\ &=\phantom{+}f\frac{\partial F_1}{\partial x} +f\frac{\partial F_2}{\partial y} +f\frac{\partial F_3}{\partial z} \\ &\phantom{=}+F_1\frac{\partial f}{\partial x} +F_2\frac{\partial f}{\partial y} +F_3\frac{\partial f}{\partial z} \\ &=f\,{ \boldsymbol{\nabla} }\cdot\mathbf{F}+\mathbf{F}\cdot{ \boldsymbol{\nabla} } f \end{aligned}$

(b) Again by the product rule

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(\mathbf{F}\times\mathbf{G}) &=\frac{\partial }{\partial x}(F_2G_3-F_3G_2) +\frac{\partial }{\partial y}(F_3G_1-F_1G_3) +\frac{\partial }{\partial z}(F_1G_2-F_2G_1) \\ &=\phantom{+}\frac{\partial F_2}{\partial x}G_3 -\frac{\partial F_3}{\partial x}G_2 +\frac{\partial F_3}{\partial y}G_1 - \frac{\partial F_1}{\partial y}G_3 +\frac{\partial F_1}{\partial z}G_2 - \frac{\partial F_2}{\partial z}G_1 \\ &\phantom{=}+F_2\frac{\partial G_3}{\partial x} -F_3\frac{\partial G_2}{\partial x} +F_3\frac{\partial G_1}{\partial y} -F_1\frac{\partial G_3}{\partial y} +F_1\frac{\partial G_2}{\partial z} - F_2\frac{\partial G_1}{\partial z} \\ &=\phantom{+} \left(\frac{\partial F_3}{\partial y}-\frac{\partial F_2}{\partial z}\right)G_1 +\left(\frac{\partial F_1}{\partial z}-\frac{\partial F_3}{\partial x}\right)G_2 +\left(\frac{\partial F_2}{\partial x}-\frac{\partial F_1}{\partial y}\right)G_3 \\ &\phantom{=} -F_1\left(\frac{\partial G_3}{\partial y}-\frac{\partial G_2}{\partial z}\right) -F_2\left(\frac{\partial G_1}{\partial z}-\frac{\partial G_3}{\partial x}\right) -F_3\left(\frac{\partial G_2}{\partial x}-\frac{\partial G_1}{\partial y}\right) \\ &=\mathbf{G}\cdot({ \boldsymbol{\nabla} }\times\mathbf{F})- \mathbf{F}\cdot({ \boldsymbol{\nabla} }\times\mathbf{G}) \end{aligned}$

(c) Recall that ${ \boldsymbol{\nabla} }^2(fg) = { \boldsymbol{\nabla} }\cdot\big[{ \boldsymbol{\nabla} }(fg)\big]$ . First

$\begin{aligned} { \boldsymbol{\nabla} }(fg) &=\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x}(fg) +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y}(fg) +\hat{\mathbf{k}}\frac{\partial }{\partial z}(fg) \\ &=\phantom{+}\hat{\boldsymbol{\imath}} g\frac{\partial f}{\partial x} +\hat{\boldsymbol{\jmath}} g\frac{\partial f}{\partial y} +\hat{\mathbf{k}} g\frac{\partial f}{\partial z} \\ &\phantom{=}+\hat{\boldsymbol{\imath}} f\frac{\partial g}{\partial x} +\hat{\boldsymbol{\jmath}} f\frac{\partial g}{\partial y} +\hat{\mathbf{k}} f\frac{\partial g}{\partial z} \\ &=g{ \boldsymbol{\nabla} } f+f{ \boldsymbol{\nabla} } g \end{aligned}$

So by part (a), twice,

$\begin{aligned} { \boldsymbol{\nabla} }^2(fg) & = { \boldsymbol{\nabla} }\cdot\big(g{ \boldsymbol{\nabla} } f\big) +{ \boldsymbol{\nabla} }\cdot\big(f{ \boldsymbol{\nabla} } g\big) \\ &= g\big({ \boldsymbol{\nabla} }\cdot { \boldsymbol{\nabla} } f\big) + \big({ \boldsymbol{\nabla} } g\big)\cdot\big({ \boldsymbol{\nabla} } f\big) +f\big({ \boldsymbol{\nabla} }\cdot{ \boldsymbol{\nabla} } g\big) + \big({ \boldsymbol{\nabla} } f\big)\cdot\big({ \boldsymbol{\nabla} } g\big) \\ &=f\,{ \boldsymbol{\nabla} }^2 g+2{ \boldsymbol{\nabla} } f\cdot{ \boldsymbol{\nabla} } g+g\,{ \boldsymbol{\nabla} }^2 f \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 13번

# 발산과 회전계산기본자동 분류

655.CLP-4 · 4.1절 · 4번대학 수준우리말 번역

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

(a) $\mathbf{F}=x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}$ (b) $\mathbf{F}=xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}$ (c) $\mathbf{F}=\frac{x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}}$ (2차원에서의 극좌표 기저 벡터 $\hat{\bf r}$ ) (d) $\mathbf{F}=\frac{-y\hat{\boldsymbol{\imath}}+x\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}}$ (2차원에서의 극좌표 기저 벡터 $\hat{\boldsymbol{\theta}}$ )

영어 원문 보기

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

(a) $\mathbf{F}=x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}$ (b) $\mathbf{F}=xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}$ (c) $\mathbf{F}=\frac{x\hat{\boldsymbol{\imath}}+y\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}}$ (the polar basis vector $\hat{\bf r}$ in 2d) (d) $\mathbf{F}=\frac{-y\hat{\boldsymbol{\imath}}+x\hat{\boldsymbol{\jmath}}}{\sqrt{x^2+y^2}}$ (the polar basis vector $\hat{\boldsymbol{\theta}}$ in 2d)

정답·풀이 보기

(a) 정의에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}) &=\frac{\partial }{\partial x}\big(x\big) +\frac{\partial }{\partial y}\big(y\big) +\frac{\partial }{\partial z}\big(z\big) = 3 \\ { \boldsymbol{\nabla} }\times(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] x & y & z\end{matrix}\right] = \mathbf{0} \end{aligned}$

(b) 정의에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}) &=\frac{\partial }{\partial x}\big(xy^2\big) +\frac{\partial }{\partial y}\big(-yz^2\big) +\frac{\partial }{\partial z}\big(zx^2\big) =y^2-z^2+x^2 \\ { \boldsymbol{\nabla} }\times(xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] xy^2 & -yz^2 & zx^2\end{matrix}\right] = 2yz\,\hat{\boldsymbol{\imath}}-2xz\,\hat{\boldsymbol{\jmath}}-2xy\,\hat{\mathbf{k}} \end{aligned}$

(c) 정의에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\left(\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\frac{\partial }{\partial x}\left(\frac{x}{\sqrt{x^2+y^2}}\right) +\frac{\partial }{\partial y}\left(\frac{y}{\sqrt{x^2+y^2}}\right) \\ &= \frac{1}{\sqrt{x^2+y^2}} - \frac{x^2}{{[x^2+y^2]}^{3/2}} +\frac{1}{\sqrt{x^2+y^2}} - \frac{y^2}{{[x^2+y^2]}^{3/2}} \\ &=\frac{x^2+y^2\ -x^2\ +x^2+y^2\ -\ y^2}{{[x^2+y^2]}^{3/2}} \\ &=\frac{1}{\sqrt{x^2+y^2}} \\ { \boldsymbol{\nabla} }\times\left(\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} & 0\end{matrix}\right] \\[0.05in] &=\left(- \frac{xy}{{[x^2+y^2]}^{3/2}}+ \frac{xy}{{[x^2+y^2]}^{3/2}}\right)\hat{\mathbf{k}} = \mathbf{0} \end{aligned}$

(d) 정의에 의해

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\left(-\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\frac{\partial }{\partial x}\left(-\frac{y}{\sqrt{x^2+y^2}}\right) +\frac{\partial }{\partial y}\left(\frac{x}{\sqrt{x^2+y^2}}\right) \\ &= \frac{xy}{{[x^2+y^2]}^{3/2}} - \frac{xy}{{[x^2+y^2]}^{3/2}} =0 \\ { \boldsymbol{\nabla} }\times\left(-\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}} +\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] -\frac{y}{\sqrt{x^2+y^2}} & \frac{x}{\sqrt{x^2+y^2}} & 0\end{matrix}\right] \\[0.05in] &=\!\left(\!\frac{1}{\sqrt{x^2+y^2}} - \frac{x^2}{{[x^2+y^2]}^{3/2}} +\frac{1}{\sqrt{x^2+y^2}} - \frac{y^2}{{[x^2+y^2]}^{3/2}}\!\right)\!\hat{\mathbf{k}} \\ &= \frac{x^2+y^2\ -x^2\ +x^2+y^2\ -\ y^2}{{[x^2+y^2]}^{3/2}}\ \hat{\mathbf{k}} = \frac{\hat{\mathbf{k}}}{\sqrt{x^2+y^2}} \end{aligned}$

원본 교재 풀이(영어)

(a) By definition

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}) &=\frac{\partial }{\partial x}\big(x\big) +\frac{\partial }{\partial y}\big(y\big) +\frac{\partial }{\partial z}\big(z\big) = 3 \\ { \boldsymbol{\nabla} }\times(x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] x & y & z\end{matrix}\right] = \mathbf{0} \end{aligned}$

(b) By definition

$\begin{aligned} { \boldsymbol{\nabla} }\cdot(xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}) &=\frac{\partial }{\partial x}\big(xy^2\big) +\frac{\partial }{\partial y}\big(-yz^2\big) +\frac{\partial }{\partial z}\big(zx^2\big) =y^2-z^2+x^2 \\ { \boldsymbol{\nabla} }\times(xy^2\hat{\boldsymbol{\imath}}-yz^2\hat{\boldsymbol{\jmath}}+zx^2\hat{\mathbf{k}}) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] xy^2 & -yz^2 & zx^2\end{matrix}\right] = 2yz\,\hat{\boldsymbol{\imath}}-2xz\,\hat{\boldsymbol{\jmath}}-2xy\,\hat{\mathbf{k}} \end{aligned}$

(c) By definition

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\left(\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\frac{\partial }{\partial x}\left(\frac{x}{\sqrt{x^2+y^2}}\right) +\frac{\partial }{\partial y}\left(\frac{y}{\sqrt{x^2+y^2}}\right) \\ &= \frac{1}{\sqrt{x^2+y^2}} - \frac{x^2}{{[x^2+y^2]}^{3/2}} +\frac{1}{\sqrt{x^2+y^2}} - \frac{y^2}{{[x^2+y^2]}^{3/2}} \\ &=\frac{x^2+y^2\ -x^2\ +x^2+y^2\ -\ y^2}{{[x^2+y^2]}^{3/2}} \\ &=\frac{1}{\sqrt{x^2+y^2}} \\ { \boldsymbol{\nabla} }\times\left(\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] \frac{x}{\sqrt{x^2+y^2}} & \frac{y}{\sqrt{x^2+y^2}} & 0\end{matrix}\right] \\[0.05in] &=\left(- \frac{xy}{{[x^2+y^2]}^{3/2}}+ \frac{xy}{{[x^2+y^2]}^{3/2}}\right)\hat{\mathbf{k}} = \mathbf{0} \end{aligned}$

(d) By definition

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\left(-\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}}+\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\frac{\partial }{\partial x}\left(-\frac{y}{\sqrt{x^2+y^2}}\right) +\frac{\partial }{\partial y}\left(\frac{x}{\sqrt{x^2+y^2}}\right) \\ &= \frac{xy}{{[x^2+y^2]}^{3/2}} - \frac{xy}{{[x^2+y^2]}^{3/2}} =0 \\ { \boldsymbol{\nabla} }\times\left(-\frac{y}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\imath}} +\frac{x}{\sqrt{x^2+y^2}}\hat{\boldsymbol{\jmath}}\right) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\[0.05in] \frac{\partial }{\partial x} & \frac{\partial }{\partial y} & \frac{\partial }{\partial z} \\[0.05in] -\frac{y}{\sqrt{x^2+y^2}} & \frac{x}{\sqrt{x^2+y^2}} & 0\end{matrix}\right] \\[0.05in] &=\!\left(\!\frac{1}{\sqrt{x^2+y^2}} - \frac{x^2}{{[x^2+y^2]}^{3/2}} +\frac{1}{\sqrt{x^2+y^2}} - \frac{y^2}{{[x^2+y^2]}^{3/2}}\!\right)\!\hat{\mathbf{k}} \\ &= \frac{x^2+y^2\ -x^2\ +x^2+y^2\ -\ y^2}{{[x^2+y^2]}^{3/2}}\ \hat{\mathbf{k}} = \frac{\hat{\mathbf{k}}}{\sqrt{x^2+y^2}} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 5번

# 발산과 회전계산보통자동 분류

656.CLP-4 · 4.1절 · 5번대학 수준우리말 번역

(a) $\mathbf{r}=(x,y,z)$ , $r=|(x,y,z)|$ 일 때 ${ \boldsymbol{\nabla} }\cdot\big(\frac{\mathbf{r}}{r}\big)$ 를 계산하고 간단히 하세요. 답을 $r$ 에 관한 식으로 나타내세요. (b) ${ \boldsymbol{\nabla} }\times\big(yz\,\hat{\boldsymbol{\imath}} + 2xz\,\hat{\boldsymbol{\jmath}} + e^{xy}\,\hat{\mathbf{k}}\big)$ 를 계산하세요.

영어 원문 보기

(a) Compute and simplify ${ \boldsymbol{\nabla} }\cdot\big(\frac{\mathbf{r}}{r}\big)$ for $\mathbf{r}=(x,y,z)$ and $r=|(x,y,z)|$ . Express your answer in terms of $r$ . (b) Compute ${ \boldsymbol{\nabla} }\times\big(yz\,\hat{\boldsymbol{\imath}} + 2xz\,\hat{\boldsymbol{\jmath}} + e^{xy}\,\hat{\mathbf{k}}\big)$ .

정답·풀이 보기

(a) $\frac{\mathbf{r}}{r} =\frac{x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}}{[x^2+y^2+z^2]^{1/2}}$ 의 발산을 계산하면 된다.

$\begin{aligned} \frac{\partial }{\partial x}\frac{x}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{x(2x)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{y^2+z^2}{{[x^2+y^2+z^2]}^{3/2}} \\ \frac{\partial }{\partial y}\frac{y}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{y(2y)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{x^2+z^2}{{[x^2+y^2+z^2]}^{3/2}} \\ \frac{\partial }{\partial z}\frac{z}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{z(2z)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{x^2+y^2}{{[x^2+y^2+z^2]}^{3/2}} \end{aligned}$

이므로 구하는 발산은 다음과 같다.

$\begin{aligned} { \boldsymbol{\nabla} }\left(\frac{\mathbf{r}}{r}\right) &= \frac{2x^2+2y^2+2z^2}{{[x^2+y^2+z^2]}^{3/2}} =\frac{2r^2}{r^3} =\frac{2}{r} \end{aligned}$

(b)

$\begin{aligned} { \boldsymbol{\nabla} }\times\big(yz\,\hat{\boldsymbol{\imath}} + 2xz\,\hat{\boldsymbol{\jmath}} + e^{xy}\,\hat{\mathbf{k}}\big) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in] \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\[0.03in] yz&2xz&e^{xy}\end{matrix}\right] =\big(xe^{xy}-2x\big)\,\hat{\boldsymbol{\imath}}-\big(ye^{xy}-y\big)\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}} \end{aligned}$

원본 교재 풀이(영어)

(a) We are to compute the divergence of $\frac{\mathbf{r}}{r} =\frac{x\,\hat{\boldsymbol{\imath}}+y\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}}}{[x^2+y^2+z^2]^{1/2}}$ . Since

$\begin{aligned} \frac{\partial }{\partial x}\frac{x}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{x(2x)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{y^2+z^2}{{[x^2+y^2+z^2]}^{3/2}} \\ \frac{\partial }{\partial y}\frac{y}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{y(2y)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{x^2+z^2}{{[x^2+y^2+z^2]}^{3/2}} \\ \frac{\partial }{\partial z}\frac{z}{{[x^2+y^2+z^2]}^{1/2}} &=\frac{1}{{[x^2+y^2+z^2]}^{1/2}} -\frac{1}{2}\frac{z(2z)}{{[x^2+y^2+z^2]}^{3/2}} &\,=\,\frac{x^2+y^2}{{[x^2+y^2+z^2]}^{3/2}} \end{aligned}$

the specified divergence is

$\begin{aligned} { \boldsymbol{\nabla} }\left(\frac{\mathbf{r}}{r}\right) &= \frac{2x^2+2y^2+2z^2}{{[x^2+y^2+z^2]}^{3/2}} =\frac{2r^2}{r^3} =\frac{2}{r} \end{aligned}$

(b)

$\begin{aligned} { \boldsymbol{\nabla} }\times\big(yz\,\hat{\boldsymbol{\imath}} + 2xz\,\hat{\boldsymbol{\jmath}} + e^{xy}\,\hat{\mathbf{k}}\big) &=\det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in] \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\[0.03in] yz&2xz&e^{xy}\end{matrix}\right] =\big(xe^{xy}-2x\big)\,\hat{\boldsymbol{\imath}}-\big(ye^{xy}-y\big)\,\hat{\boldsymbol{\jmath}}+z\,\hat{\mathbf{k}} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 6번

# 발산과 회전계산보통자동 분류

657.CLP-4 · 4.1절 · 6번대학 수준우리말 번역

다음에서 기호 $\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}}$ , $r = |\mathbf{r}|$ 를 사용하고, $k$ 는 $k = 0, 1, -1, 2, -2, \dots$ 중 어떤 수이다.

(a) 다음을 만족하는 $k$ 의 값을 구하세요.

${ \boldsymbol{\nabla} } (r^k) = -3\frac{\mathbf{r}}{r^5}$ (b) 다음을 만족하는 $k$ 의 값을 구하세요.

${ \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) = 5r^2$ (c) 다음을 만족하는 $k$ 의 값을 구하세요.

${ \boldsymbol{\nabla} }^2 (r^k) = \frac{2}{r^4}$

영어 원문 보기

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

(a) Find the value $k$ for which

${ \boldsymbol{\nabla} } (r^k) = -3\frac{\mathbf{r}}{r^5}$ (b) Find the value $k$ for which

${ \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) = 5r^2$ (c) Find the value $k$ for which

${ \boldsymbol{\nabla} }^2 (r^k) = \frac{2}{r^4}$

답 쓰고 채점

(a)

(b)

(c)

정답·풀이 보기

(a) $r^k=\big(x^2+y^2+z^2\big)^{k/2}$ 이므로,

$\begin{aligned} \frac{\partial }{\partial x} r^k & = 2x\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\boldsymbol{\imath}})\ r^{k-2} \\ \frac{\partial }{\partial y} r^k & = 2y\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\boldsymbol{\jmath}})\ r^{k-2} \\ \frac{\partial }{\partial z} r^k & = 2z\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\mathbf{k}})\ r^{k-2} \end{aligned}$

구하는 값은 $k=-3$ 이다.

(b) (a)의 계산을 이용하면

$\begin{aligned} { \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) &= \frac{\partial }{\partial x} (x r^k) + \frac{\partial }{\partial y} (y r^k) + \frac{\partial }{\partial z} (z r^k) \\ &= 3r^k + x \frac{\partial }{\partial x} r^k + y \frac{\partial }{\partial y} r^k + z \frac{\partial }{\partial z} r^k \\ &= 3r^k + x \big(kx\ r^{k-2}\big) + y \big(ky\ r^{k-2}\big) + z \big(kz\ r^{k-2}\big) \\ &= \big(3+k\big)r^k \end{aligned}$

구하는 값은 $k=2$ 이다.

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

$\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 $k=-2$ 이다.

원본 교재 풀이(영어)

(a) Since $r^k=\big(x^2+y^2+z^2\big)^{k/2}$ ,

$\begin{aligned} \frac{\partial }{\partial x} r^k & = 2x\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\boldsymbol{\imath}})\ r^{k-2} \\ \frac{\partial }{\partial y} r^k & = 2y\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\boldsymbol{\jmath}})\ r^{k-2} \\ \frac{\partial }{\partial z} r^k & = 2z\ \frac{k}{2}\big(x^2+y^2+z^2\big)^{\frac{k}{2}-1} = k\ (\mathbf{r}\cdot\hat{\mathbf{k}})\ r^{k-2} \end{aligned}$

We want $k=-3$ .

(b) Using the computation in part (a)

$\begin{aligned} { \boldsymbol{\nabla} } \cdot (r^k\mathbf{r}) &= \frac{\partial }{\partial x} (x r^k) + \frac{\partial }{\partial y} (y r^k) + \frac{\partial }{\partial z} (z r^k) \\ &= 3r^k + x \frac{\partial }{\partial x} r^k + y \frac{\partial }{\partial y} r^k + z \frac{\partial }{\partial z} r^k \\ &= 3r^k + x \big(kx\ r^{k-2}\big) + y \big(ky\ r^{k-2}\big) + z \big(kz\ r^{k-2}\big) \\ &= \big(3+k\big)r^k \end{aligned}$

We want $k=2$ .

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

$\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 $k=-2$ .

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 7번

# 발산과 회전계산보통자동 분류

658.CLP-4 · 4.1절 · 7번대학 수준우리말 번역

$\mathbf{r}$ 를 벡터장 $\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}}$ 라 하고, $r$ 를 함수 $r = |\mathbf{r}|$ 라 하자. $\mathbf{a}$ 를 상수 벡터 $\mathbf{a} = a_1\,\hat{\boldsymbol{\imath}} + a_2\,\hat{\boldsymbol{\jmath}} + a_3\,\hat{\mathbf{k}}$ 라 하자. 다음 양을 계산하여 간단히 하세요. 답은 $\mathbf{a}$ , $\mathbf{r}$ , $r$ 로 나타내야 합니다. 답에 $x$ , $y$ , $z$ 가 들어 있으면 안 됩니다.

(a) ${ \boldsymbol{\nabla} }\cdot\mathbf{r}$ (b) ${ \boldsymbol{\nabla} }(r^2)$ (c) ${ \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a})$ (d) ${ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big)$

영어 원문 보기

Let $\mathbf{r}$ be the vector field $\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}}$ and let $r$ be the function $r = |\mathbf{r}|$ . Let $\mathbf{a}$ be the constant vector $\mathbf{a} = a_1\,\hat{\boldsymbol{\imath}} + a_2\,\hat{\boldsymbol{\jmath}} + a_3\,\hat{\mathbf{k}}$ . Compute and simplify the following quantities. Answers must be expressed in terms of $\mathbf{a}$ , $\mathbf{r}$ , and $r$ . There should be no $x$ 's, $y$ 's, or $z$ 's in your answers.

(a) ${ \boldsymbol{\nabla} }\cdot\mathbf{r}$ (b) ${ \boldsymbol{\nabla} }(r^2)$ (c) ${ \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a})$ (d) ${ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big)$

정답·풀이 보기

(a)

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

(b)

${ \boldsymbol{\nabla} }(r^2) =\left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big) =2x\,\hat{\boldsymbol{\imath}} +2y\,\hat{\boldsymbol{\jmath}} +2z\,\hat{\mathbf{k}} =2\mathbf{r}$

(c) 다음과 같으므로

$\mathbf{r}\times\mathbf{a} = \det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ x & y & z \\ a_1 & a_2 & a_3 \end{matrix}\right] =\hat{\boldsymbol{\imath}}\big(a_3y-a_2z\big) +\hat{\boldsymbol{\jmath}}\big(a_1z-a_3x\big) +\hat{\mathbf{k}}\big(a_2x-a_1y\big)$

다음을 얻는다.

$\begin{aligned} { \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a}) &= \det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in] \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\[0.03in] a_3y-a_2z& a_1z-a_3x & a_2x-a_1y\end{matrix}\right] =-2a_1\,\hat{\boldsymbol{\imath}} -2a_2\,\hat{\boldsymbol{\jmath}} -2a_3\,\hat{\mathbf{k}} \\ &=-2\mathbf{a} \end{aligned}$

(d) 다음과 같으므로

$\begin{aligned} { \boldsymbol{\nabla} }(r) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{1/2} \\ &=\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} +\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{1/2}} +\hat{\mathbf{k}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} \end{aligned}$

다음을 얻는다.

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big) &= \frac{\partial }{\partial x} \frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} +\frac{\partial }{\partial y} \frac{y}{\big(x^2+y^2+z^2\big)^{1/2}} +\frac{\partial }{\partial z} \frac{z}{\big(x^2+y^2+z^2\big)^{1/2}} \\ &= \frac{3}{\big(x^2+y^2+z^2\big)^{1/2}} -\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{3/2}} =\frac{2}{\big(x^2+y^2+z^2\big)^{1/2}} =\frac{2}{r} \end{aligned}$

원본 교재 풀이(영어)

(a)

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

(b)

${ \boldsymbol{\nabla} }(r^2) =\left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big) =2x\,\hat{\boldsymbol{\imath}} +2y\,\hat{\boldsymbol{\jmath}} +2z\,\hat{\mathbf{k}} =2\mathbf{r}$

(c) Since

$\mathbf{r}\times\mathbf{a} = \det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ x & y & z \\ a_1 & a_2 & a_3 \end{matrix}\right] =\hat{\boldsymbol{\imath}}\big(a_3y-a_2z\big) +\hat{\boldsymbol{\jmath}}\big(a_1z-a_3x\big) +\hat{\mathbf{k}}\big(a_2x-a_1y\big)$

we have

$\begin{aligned} { \boldsymbol{\nabla} }\times(\mathbf{r}\times\mathbf{a}) &= \det\left[\begin{matrix}\hat{\boldsymbol{\imath}}&\hat{\boldsymbol{\jmath}}&\hat{\mathbf{k}}\\[0.03in] \frac{\partial }{\partial x}& \frac{\partial }{\partial y}& \frac{\partial }{\partial z}\\[0.03in] a_3y-a_2z& a_1z-a_3x & a_2x-a_1y\end{matrix}\right] =-2a_1\,\hat{\boldsymbol{\imath}} -2a_2\,\hat{\boldsymbol{\jmath}} -2a_3\,\hat{\mathbf{k}} \\ &=-2\mathbf{a} \end{aligned}$

(d) Since

$\begin{aligned} { \boldsymbol{\nabla} }(r) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{1/2} \\ &=\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} +\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{1/2}} +\hat{\mathbf{k}}\frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} \end{aligned}$

we have

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r)\big) &= \frac{\partial }{\partial x} \frac{x}{\big(x^2+y^2+z^2\big)^{1/2}} +\frac{\partial }{\partial y} \frac{y}{\big(x^2+y^2+z^2\big)^{1/2}} +\frac{\partial }{\partial z} \frac{z}{\big(x^2+y^2+z^2\big)^{1/2}} \\ &= \frac{3}{\big(x^2+y^2+z^2\big)^{1/2}} -\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{3/2}} =\frac{2}{\big(x^2+y^2+z^2\big)^{1/2}} =\frac{2}{r} \end{aligned}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 6.5절 · 30번

# 발산과 회전계산보통자동 분류

659.CLP-4 · 4.1절 · 8번대학 수준우리말 번역

다음과 같이 두자.

$\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}},\qquad r = |\mathbf{r}|$

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

영어 원문 보기

Let

$\mathbf{r} = x\,\hat{\boldsymbol{\imath}} + y\,\hat{\boldsymbol{\jmath}} + z\,\hat{\mathbf{k}},\qquad r = |\mathbf{r}|$

(a) Compute $a$ where ${ \boldsymbol{\nabla} }\big(\frac{1}{r}\big) =- r^a\,\mathbf{r}$ . (b) Compute $a$ where ${ \boldsymbol{\nabla} }\cdot\big(r\,\mathbf{r}\big) = ar$ . (c) Compute $a$ where ${ \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^3)\big) = ar$ .

답 쓰고 채점

(a)

(b)

(c)

정답·풀이 보기

(a) 다음과 같으므로

$\begin{aligned} { \boldsymbol{\nabla} }\left(\frac{1}{r}\right) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{-1/2} \\ &=-\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{3/2}} -\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{3/2}} -\hat{\mathbf{k}}\frac{z}{\big(x^2+y^2+z^2\big)^{3/2}} \\ &=-\hat{\boldsymbol{\imath}}\frac{x}{r^3} -\hat{\boldsymbol{\jmath}}\frac{y}{r^3} -\hat{\mathbf{k}}\frac{x}{r^3} \end{aligned}$

$a=-3$ 이다.

(b) 다음과 같으므로

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big(r\,\mathbf{r}\big) &= \frac{\partial }{\partial x} \Big[\big(x^2+y^2+z^2\big)^{1/2}x\Big] +\frac{\partial }{\partial y}\Big[\big(x^2+y^2+z^2\big)^{1/2}y\Big] +\frac{\partial }{\partial z} \Big[\big(x^2+y^2+z^2\big)^{1/2}z\Big] \\ &= 3\big(x^2+y^2+z^2\big)^{1/2} +\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{1/2}} =4\big(x^2+y^2+z^2\big)^{1/2} =4r \end{aligned}$

$a=4$ 이다.

(c) 다음과 같으므로

$\begin{aligned} { \boldsymbol{\nabla} }(r^3) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{3/2} \\ &=\hat{\boldsymbol{\imath}}\,3x\big(x^2+y^2+z^2\big)^{1/2} +\hat{\boldsymbol{\jmath}}\,3y\big(x^2+y^2+z^2\big)^{1/2} +\hat{\mathbf{k}}\,3z\big(x^2+y^2+z^2\big)^{1/2} \\ &=3r\mathbf{r} \end{aligned}$

다음을 얻는다.

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^3)\big) &= { \boldsymbol{\nabla} }\cdot\big(3r\mathbf{r}\big) = 3\,{ \boldsymbol{\nabla} }\cdot\big(r\mathbf{r}\big) = 3\,\big(4r\big) \qquad\text{(b)에 의해} \\ &= 12 r \end{aligned}$

따라서 $a=12$ 이다.

원본 교재 풀이(영어)

(a) Since

$\begin{aligned} { \boldsymbol{\nabla} }\left(\frac{1}{r}\right) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{-1/2} \\ &=-\hat{\boldsymbol{\imath}}\frac{x}{\big(x^2+y^2+z^2\big)^{3/2}} -\hat{\boldsymbol{\jmath}}\frac{y}{\big(x^2+y^2+z^2\big)^{3/2}} -\hat{\mathbf{k}}\frac{z}{\big(x^2+y^2+z^2\big)^{3/2}} \\ &=-\hat{\boldsymbol{\imath}}\frac{x}{r^3} -\hat{\boldsymbol{\jmath}}\frac{y}{r^3} -\hat{\mathbf{k}}\frac{x}{r^3} \end{aligned}$

we have $a=-3$ .

(b) Since

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big(r\,\mathbf{r}\big) &= \frac{\partial }{\partial x} \Big[\big(x^2+y^2+z^2\big)^{1/2}x\Big] +\frac{\partial }{\partial y}\Big[\big(x^2+y^2+z^2\big)^{1/2}y\Big] +\frac{\partial }{\partial z} \Big[\big(x^2+y^2+z^2\big)^{1/2}z\Big] \\ &= 3\big(x^2+y^2+z^2\big)^{1/2} +\frac{1}{2}\frac{2x^2+2y^2+2z^2}{\big(x^2+y^2+z^2\big)^{1/2}} =4\big(x^2+y^2+z^2\big)^{1/2} =4r \end{aligned}$

we have $a=4$ .

(c) Since

$\begin{aligned} { \boldsymbol{\nabla} }(r^3) &= \left(\hat{\boldsymbol{\imath}}\frac{\partial }{\partial x} +\hat{\boldsymbol{\jmath}}\frac{\partial }{\partial y} +\hat{\mathbf{k}}\frac{\partial }{\partial z} \right) \big(x^2+y^2+z^2\big)^{3/2} \\ &=\hat{\boldsymbol{\imath}}\,3x\big(x^2+y^2+z^2\big)^{1/2} +\hat{\boldsymbol{\jmath}}\,3y\big(x^2+y^2+z^2\big)^{1/2} +\hat{\mathbf{k}}\,3z\big(x^2+y^2+z^2\big)^{1/2} \\ &=3r\mathbf{r} \end{aligned}$

we have

$\begin{aligned} { \boldsymbol{\nabla} }\cdot\big({ \boldsymbol{\nabla} }(r^3)\big) &= { \boldsymbol{\nabla} }\cdot\big(3r\mathbf{r}\big) = 3\,{ \boldsymbol{\nabla} }\cdot\big(r\mathbf{r}\big) = 3\,\big(4r\big) \qquad\text{by part (b)} \\ &= 12 r \end{aligned}$

so that $a=12$ .

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 9번

# 발산과 회전계산보통자동 분류

660.CLP-4 · 4.1절 · 9번대학 수준우리말 번역

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

(a) $\mathbf{F}=(1+yz)\hat{\boldsymbol{\imath}}+(2y+zx)\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}}$ (b) $\mathbf{G}= yz\hat{\boldsymbol{\imath}}+zx\hat{\boldsymbol{\jmath}}+xy\hat{\mathbf{k}}$

영어 원문 보기

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

(a) $\mathbf{F}=(1+yz)\hat{\boldsymbol{\imath}}+(2y+zx)\hat{\boldsymbol{\jmath}}+(3z^2+xy)\hat{\mathbf{k}}$ (b) $\mathbf{G}= yz\hat{\boldsymbol{\imath}}+zx\hat{\boldsymbol{\jmath}}+xy\hat{\mathbf{k}}$

정답·풀이 보기

(a) ${ \boldsymbol{\nabla} }\cdot\mathbf{F} =\frac{\partial }{\partial x}(1+yz)+\frac{\partial }{\partial y}(2y+zx)+\frac{\partial }{\partial z}(3z^2+xy)=2+6z\ne 0$ 이므로 $\mathbf{F}$ 는 판별 조건을 만족하지 못하여 벡터 퍼텐셜을 가질 수 없다.

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

$\begin{aligned} -\frac{\partial A_2}{\partial z}&= yz\quad & \iff&\quad& &A_2=-\frac{1}{2} yz^2+B_2(x,y)\\ \frac{\partial A_1}{\partial z}&=zx & \iff&\quad & &A_1=\phantom{-}\frac{1}{2} xz^2+B_1(x,y)\\ \frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y} &=xy & \iff& & & \frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy \end{aligned}$

$\frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy$ 의 해는 무수히 많다. 실제로 $B_2$ 는 완전히 임의로 정할 수 있다. $B_2=0$ 으로 택하면 $B_1=-\frac{1}{2} xy^2$ 이 조건을 만족한다. $B_1=0$ 으로 택하면 $B_2=\frac{1}{2} x^2y$ 가 조건을 만족한다. 따라서 두 가지 해는 $\mathbf{A}=\frac{1}{2}(z^2-y^2)x\hat{\boldsymbol{\imath}}-\frac{1}{2} yz^2\hat{\boldsymbol{\jmath}}$ 와 $\mathbf{A}=\frac{1}{2} xz^2\hat{\boldsymbol{\imath}}+\frac{1}{2}(x^2-z^2)y\hat{\boldsymbol{\jmath}}$ 이다.

원본 교재 풀이(영어)

(a) Since ${ \boldsymbol{\nabla} }\cdot\mathbf{F} =\frac{\partial }{\partial x}(1+yz)+\frac{\partial }{\partial y}(2y+zx)+\frac{\partial }{\partial z}(3z^2+xy)=2+6z\ne 0$ , $\mathbf{F}$ fails the screening test and cannot have a vector potential.

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

$\begin{aligned} -\frac{\partial A_2}{\partial z}&= yz\quad & \iff&\quad& &A_2=-\frac{1}{2} yz^2+B_2(x,y)\\ \frac{\partial A_1}{\partial z}&=zx & \iff&\quad & &A_1=\phantom{-}\frac{1}{2} xz^2+B_1(x,y)\\ \frac{\partial A_2}{\partial x}-\frac{\partial A_1}{\partial y} &=xy & \iff& & & \frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy \end{aligned}$

There are infinitely many solutions to $\frac{\partial B_2}{\partial x} -\frac{\partial B_1}{\partial y}=xy$ . In fact $B_2$ is completely arbitrary. If one chooses $B_2=0$ , then $B_1=-\frac{1}{2} xy^2$ does the job. If one chooses $B_1=0$ , then $B_2=\frac{1}{2} x^2y$ does the job. Thus two solutions are $\mathbf{A}=\frac{1}{2}(z^2-y^2)x\hat{\boldsymbol{\imath}}-\frac{1}{2} yz^2\hat{\boldsymbol{\jmath}}$ and $\mathbf{A}=\frac{1}{2} xz^2\hat{\boldsymbol{\imath}}+\frac{1}{2}(x^2-z^2)y\hat{\boldsymbol{\jmath}}$ .

한 번 더 — 비슷한 문제 풀기CLP-4 · 4.1절 · 2번

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