다변수 미분 전체 연습문제

621.CLP-3 · 2.4절 · 11번대학 수준우리말 번역

$F$ 를 $\mathbb{R}^2$ 위의 함수라 하자. $\mathbb{R}^2$ 의 점을 $(u, v)$ 로 나타내고, $F$ 의 대응하는 편도함수를 $F_u(u, v)$ , $F_v (u, v)$ , $F_{uu}(u, v)$ , $F_{uv}(u, v)$ 등으로 나타낸다. 이 편도함수들이 모두 연속이라고 가정한다.

$\begin{aligned} \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) \end{aligned}$

를 함수 $F$ 의 편도함수를 사용하여 나타내세요.

영어 원문 보기

Let $F$ be a function on $\mathbb{R}^2$ . Denote points in $\mathbb{R}^2$ by $(u, v)$ and the corresponding partial derivatives of $F$ by $F_u(u, v)$ , $F_v (u, v)$ , $F_{uu}(u, v)$ , $F_{uv}(u, v)$ , etc.. Assume those derivatives are all continuous. Express

$\begin{aligned} \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) \end{aligned}$

in terms of partial derivatives of the function $F$ .

정답·풀이 보기

$u(x,y) = x^2 - y^2$ , $v(x,y) = 2xy$ 라 하자. 그러면 $F(x^2 - y^2 , 2xy) =F\big(u(x,y),v(x,y)\big)$ 이다. 연쇄 법칙에 의해

$\begin{aligned} \frac{\partial }{\partial y}F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial y}F(u(x,y) , v(xy)) \\ &=F_u(u(x,y) , v(xy))\frac{\partial u}{\partial y}(x,y) + F_v(u(x,y) , v(xy))\frac{\partial v}{\partial y}(x,y) \\ &= F_u(x^2 - y^2 , 2xy)\ (-2y) + F_v(x^2 - y^2 , 2xy)\, (2x) \\ \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial x}\left\{-2y F_u(x^2 - y^2 , 2xy) + 2x F_v(x^2 - y^2 , 2xy) \right\} \\ &= -2y\frac{\partial }{\partial x}\left[ F_u(x^2 - y^2 , 2xy)\right] +2 F_v(x^2 - y^2 , 2xy) \\& + 2x \frac{\partial }{\partial x}\left[F_v(x^2 - y^2 , 2xy)\right] \\ &= -4xy\,F_{uu}(x^2 - y^2 , 2xy) -4y^2 F_{uv}(x^2 - y^2 , 2xy) +2 F_v(x^2 - y^2 , 2xy) \\& + 4x^2 F_{vu}(x^2 - y^2 , 2xy) +4xy\,F_{vv}(x^2 - y^2 , 2xy) \\ &= 2 \, F_v(x^2 - y^2 , 2xy) -4xy\, F_{uu}(x^2 - y^2 , 2xy) \\& +4(x^2-y^2)\, F_{uv}(x^2 - y^2 , 2xy) \\& +4xy\, F_{vv}(x^2 - y^2 , 2xy) \end{aligned}$

원본 교재 풀이(영어)

Let $u(x,y) = x^2 - y^2$ , and $v(x,y) = 2xy$ . Then $F(x^2 - y^2 , 2xy) =F\big(u(x,y),v(x,y)\big)$ . By the chain rule

$\begin{aligned} \frac{\partial }{\partial y}F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial y}F(u(x,y) , v(xy)) \\ &=F_u(u(x,y) , v(xy))\frac{\partial u}{\partial y}(x,y) + F_v(u(x,y) , v(xy))\frac{\partial v}{\partial y}(x,y) \\ &= F_u(x^2 - y^2 , 2xy)\ (-2y) + F_v(x^2 - y^2 , 2xy)\, (2x) \\ \frac{\partial^2}{\partial x\, \partial y} F(x^2 - y^2 , 2xy) &=\frac{\partial }{\partial x}\left\{-2y F_u(x^2 - y^2 , 2xy) + 2x F_v(x^2 - y^2 , 2xy) \right\} \\ &= -2y\frac{\partial }{\partial x}\left[ F_u(x^2 - y^2 , 2xy)\right] +2 F_v(x^2 - y^2 , 2xy) \\& + 2x \frac{\partial }{\partial x}\left[F_v(x^2 - y^2 , 2xy)\right] \\ &= -4xy\,F_{uu}(x^2 - y^2 , 2xy) -4y^2 F_{uv}(x^2 - y^2 , 2xy) +2 F_v(x^2 - y^2 , 2xy) \\& + 4x^2 F_{vu}(x^2 - y^2 , 2xy) +4xy\,F_{vv}(x^2 - y^2 , 2xy) \\ &= 2 \, F_v(x^2 - y^2 , 2xy) -4xy\, F_{uu}(x^2 - y^2 , 2xy) \\& +4(x^2-y^2)\, F_{uv}(x^2 - y^2 , 2xy) \\& +4xy\, F_{vv}(x^2 - y^2 , 2xy) \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 10번

# 다변수 연쇄법칙계산보통자동 분류

622.CLP-3 · 2.4절 · 12번대학 수준우리말 번역

$u(x,y)$ 가 다음과 같이 정의된다.

$u(x,y) = e^y\, F\big(xe^{-y^2}\big)$

여기서 $F(z)$ 는 임의의 함수이다.

(a) $F(z) = \ln(z)$ 일 때, $\frac{\partial u}{\partial x}$ 와 $\frac{\partial u}{\partial y}$ 를 구하세요. (b) 임의의 함수 $F(z)$ 에 대해 $u(x,y)$ 가 다음을 만족시킴을 보이세요.

$2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = u$

영어 원문 보기

$u(x,y)$ is defined as

$u(x,y) = e^y\, F\big(xe^{-y^2}\big)$

for an arbitrary function $F(z)$ .

(a) If $F(z) = \ln(z)$ , find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$ . (b) For an arbitrary $F(z)$ show that $u(x,y)$ satisfies

$2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} = u$

정답·풀이 보기

임의의 (미분가능한) 함수 $F$ 에 대해, 연쇄법칙과 곱의 미분법에 의해 다음이 성립한다.

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= \frac{\partial }{\partial x}\Big[e^y\,F\big(xe^{-y^2}\big)\Big] =e^y\,\frac{\partial }{\partial x}\Big[F\big(xe^{-y^2}\big)\Big] \\ &=e^y\,F'\big(xe^{-y^2}\big)\frac{\partial }{\partial x}\Big(xe^{-y^2}\Big) \\ &= e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} \\ \frac{\partial u}{\partial y}(x,y)&= \frac{\partial }{\partial y}\Big[e^y\,F\big(xe^{-y^2}\big)\Big]\\ &=e^y\,F\big(xe^{-y^2}\big) +e^y\,\frac{\partial }{\partial y}\Big[F\big(xe^{-y^2}\big)\Big] \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\,F'\big(xe^{-y^2}\big)\ \frac{\partial }{\partial y}\Big(xe^{-y^2}\Big) \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \end{aligned}$

(a) 특히 $F(z)=\ln(z)$ 일 때 $F'(z)=\frac{1}{z}$ 이므로

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= e^y\, \frac{1}{xe^{-y^2}}\ e^{-y^2} = \frac{e^y}{x}\\ \frac{\partial u}{\partial y}(x,y)&= e^y\, \ln\big(xe^{-y^2}\big) + e^y\, \frac{1}{xe^{-y^2}}\ (-2xy)e^{-y^2} = e^y\, \ln\big(xe^{-y^2}\big) -2y e^y \\ &= e^y\,\ln (x) -y^2\,e^y -2y e^y \end{aligned}$

(b) 일반적으로

$\begin{aligned} 2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} &=2xy\ e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} +e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \\ &=e^y\, F\big(xe^{-y^2}\big) \\ &=u \end{aligned}$

원본 교재 풀이(영어)

For any (differentiable) function $F$ , we have, by the chain and product rules,

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= \frac{\partial }{\partial x}\Big[e^y\,F\big(xe^{-y^2}\big)\Big] =e^y\,\frac{\partial }{\partial x}\Big[F\big(xe^{-y^2}\big)\Big] \\ &=e^y\,F'\big(xe^{-y^2}\big)\frac{\partial }{\partial x}\Big(xe^{-y^2}\Big) \\ &= e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} \\ \frac{\partial u}{\partial y}(x,y)&= \frac{\partial }{\partial y}\Big[e^y\,F\big(xe^{-y^2}\big)\Big]\\ &=e^y\,F\big(xe^{-y^2}\big) +e^y\,\frac{\partial }{\partial y}\Big[F\big(xe^{-y^2}\big)\Big] \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\,F'\big(xe^{-y^2}\big)\ \frac{\partial }{\partial y}\Big(xe^{-y^2}\Big) \\ &= e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \end{aligned}$

(a) In particular, when $F(z)=\ln(z)$ , $F'(z)=\frac{1}{z}$ and

$\begin{aligned} \frac{\partial u}{\partial x}(x,y)&= e^y\, \frac{1}{xe^{-y^2}}\ e^{-y^2} = \frac{e^y}{x}\\ \frac{\partial u}{\partial y}(x,y)&= e^y\, \ln\big(xe^{-y^2}\big) + e^y\, \frac{1}{xe^{-y^2}}\ (-2xy)e^{-y^2} = e^y\, \ln\big(xe^{-y^2}\big) -2y e^y \\ &= e^y\,\ln (x) -y^2\,e^y -2y e^y \end{aligned}$

(b) In general

$\begin{aligned} 2xy\frac{\partial u}{\partial x} + \frac{\partial u}{\partial y} &=2xy\ e^y\, F'\big(xe^{-y^2}\big)\ e^{-y^2} +e^y\, F\big(xe^{-y^2}\big) + e^y\, F'\big(xe^{-y^2}\big)\ (-2xy)e^{-y^2} \\ &=e^y\, F\big(xe^{-y^2}\big) \\ &=u \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 11번

# 다변수 연쇄법칙증명·논술보통자동 분류

623.CLP-3 · 2.4절 · 13번대학 수준우리말 번역

$f(x)$ 와 $g(x)$ 가 $x$ 에 대한 두 함수이고 $f''(7) = -2$ 이고 $g''(-4) = -1$ 을 만족시킨다고 하자. $z = h(s,t) = f(2s + 3t) + g(s - 6t)$ 가 $s$ 와 $t$ 에 대한 함수일 때, $s = 2$ 이고 $t = 1$ 일 때 $\frac{\partial^2 z}{\partial t^2}$ 의 값을 구하세요.

영어 원문 보기

Let $f(x)$ and $g(x)$ be two functions of $x$ satisfying $f''(7) = -2$ and $g''(-4) = -1$ . If $z = h(s,t) = f(2s + 3t) + g(s - 6t)$ is a function of $s$ and $t$ , find the value of $\frac{\partial^2 z}{\partial t^2}$ when $s = 2$ and $t = 1$ .

답 쓰고 채점

정답·풀이 보기

연쇄 법칙에 의해

$\begin{aligned} \frac{\partial h}{\partial t}(s,t) &=\frac{\partial }{\partial t}\big[f(2s + 3t)\big] + \frac{\partial }{\partial t}\big[g(s - 6t)\big] \\ &=f'(2s + 3t)\frac{\partial }{\partial t}(2s+3t) + g(s - 6t)\frac{\partial }{\partial t}(s-6t) \\ &= 3f'(2s + 3t) - 6g'(s - 6t) \\ \frac{\partial^2 h}{\partial t^2}(s,t) &= 3\frac{\partial }{\partial t}\big[f'(2s + 3t)\big] - 6\frac{\partial }{\partial t}\big[ g'(s - 6t) \big] \\ &= 3\,f''(2s + 3t)\frac{\partial }{\partial t}(2s+3t) - 6\,g''(s - 6t) \frac{\partial }{\partial t}(s-6t) \\ &= 9f''(2s+3t) + 36g''(s-6t) \end{aligned}$

특히

$\begin{aligned} \frac{\partial^2 h}{\partial t^2}(2,1) =9f''(7) +36g''(-4) =9(-2) +36(-1) =-54 \end{aligned}$

이다.

원본 교재 풀이(영어)

By the chain rule,

$\begin{aligned} \frac{\partial h}{\partial t}(s,t) &=\frac{\partial }{\partial t}\big[f(2s + 3t)\big] + \frac{\partial }{\partial t}\big[g(s - 6t)\big] \\ &=f'(2s + 3t)\frac{\partial }{\partial t}(2s+3t) + g(s - 6t)\frac{\partial }{\partial t}(s-6t) \\ &= 3f'(2s + 3t) - 6g'(s - 6t) \\ \frac{\partial^2 h}{\partial t^2}(s,t) &= 3\frac{\partial }{\partial t}\big[f'(2s + 3t)\big] - 6\frac{\partial }{\partial t}\big[ g'(s - 6t) \big] \\ &= 3\,f''(2s + 3t)\frac{\partial }{\partial t}(2s+3t) - 6\,g''(s - 6t) \frac{\partial }{\partial t}(s-6t) \\ &= 9f''(2s+3t) + 36g''(s-6t) \end{aligned}$

In particular

$\begin{aligned} \frac{\partial^2 h}{\partial t^2}(2,1) =9f''(7) +36g''(-4) =9(-2) +36(-1) =-54 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 12번

# 다변수 연쇄법칙계산보통자동 분류

624.CLP-3 · 2.4절 · 14번대학 수준우리말 번역

$w = f (xz, yz)$ 라고 하자. 여기서 $f$ 는 미분가능한 함수이다. 다음을 보여라.

$x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = z\frac{\partial w}{\partial z}$

영어 원문 보기

Suppose that $w = f (xz, yz)$ , where $f$ is a differentiable function. Show that

$x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} = z\frac{\partial w}{\partial z}$

정답·풀이 보기

먼저 $w(x,y,z)$ 의 1계 편도함수를 구한다. $u(x,y,z)=xz$ , $v(x,y,z)=yz$ 라 하면 $w(x,y,z) = f\big(u(x,y,z), v(x,y,z)\big)$ 이다. 연쇄 법칙에 의해

$\begin{aligned} \frac{\partial w}{\partial x}(x,y,z) &= \frac{\partial }{\partial x}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial x}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial x}(x,y,z) \\ &=z\frac{\partial f}{\partial u}(xz, yz) \\ \frac{\partial w}{\partial y}(x,y,z) &= \frac{\partial }{\partial y}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial y}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial y}(x,y,z) \\ &=z\frac{\partial f}{\partial v}(xz, yz) \\ \frac{\partial w}{\partial z}(x,y,z) &= \frac{\partial }{\partial z}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial z}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial z}(x,y,z) \\ &=x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz) \end{aligned}$

따라서

$\begin{aligned} x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} =xz\frac{\partial f}{\partial u}(xz, yz) + yz\frac{\partial f}{\partial v}(xz, yz) =z\left[x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz)\right] =z\frac{\partial w}{\partial z} \end{aligned}$

이므로 주어진 등식이 성립한다.

원본 교재 풀이(영어)

We'll first compute the first order partial derivatives of $w(x,y,z)$ . Write $u(x,y,z)=xz$ and $v(x,y,z)=yz$ so that $w(x,y,z) = f\big(u(x,y,z), v(x,y,z)\big)$ . By the chain rule,

$\begin{aligned} \frac{\partial w}{\partial x}(x,y,z) &= \frac{\partial }{\partial x}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial x}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial x}(x,y,z) \\ &=z\frac{\partial f}{\partial u}(xz, yz) \\ \frac{\partial w}{\partial y}(x,y,z) &= \frac{\partial }{\partial y}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial y}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial y}(x,y,z) \\ &=z\frac{\partial f}{\partial v}(xz, yz) \\ \frac{\partial w}{\partial z}(x,y,z) &= \frac{\partial }{\partial z}\big[f \big(u(x,y,z), v(x,y,z)\big)\big]\\ &=\frac{\partial f}{\partial u}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial u}{\partial z}(x,y,z) +\frac{\partial f}{\partial v}\big(u(x,y,z), v(x,y,z)\big)\frac{\partial v}{\partial z}(x,y,z) \\ &=x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz) \end{aligned}$

So

$\begin{aligned} x\frac{\partial w}{\partial x} + y\frac{\partial w}{\partial y} =xz\frac{\partial f}{\partial u}(xz, yz) + yz\frac{\partial f}{\partial v}(xz, yz) =z\left[x\frac{\partial f}{\partial u}(xz, yz) + y\frac{\partial f}{\partial v}(xz, yz)\right] =z\frac{\partial w}{\partial z} \end{aligned}$

as desired.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 22번

# 다변수 연쇄법칙증명·논술보통자동 분류

625.CLP-3 · 2.4절 · 15번대학 수준우리말 번역

$z = f (x, y)$ 가 연속인 이계 편도함수를 갖고, $x = r \cos t$ , $y = r \sin t$ 라 하자. 다음 편도함수를 $r$ , $t$ , 그리고 $f$ 의 편도함수로 나타내세요.

(a) $\frac{\partial z}{\partial t}$ (b) $\frac{\partial^2 z}{\partial t^2}$

영어 원문 보기

Suppose $z = f (x, y)$ has continuous second order partial derivatives, and $x = r \cos t$ , $y = r \sin t$ . Express the following partial derivatives in terms $r$ , $t$ , and partial derivatives of $f$ .

(a) $\frac{\partial z}{\partial t}$ (b) $\frac{\partial^2 z}{\partial t^2}$

정답·풀이 보기

정의에 의해 $z(r,t) = f(r\cos t\,,\,r\sin t)$ 이다.

(a) 연쇄 법칙에 의해

$\begin{aligned} \frac{\partial z}{\partial t}(r,t) &= \frac{\partial }{\partial t}\Big[f(r\cos t\,,\,r\sin t)\Big] \\ &= \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\frac{\partial }{\partial t}(r\cos t) +\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\frac{\partial }{\partial t}(r\sin t) \\ &= -r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \end{aligned}$

(b) 선형성, 곱의 미분법, 연쇄 법칙에 의해

$\begin{aligned} \frac{\partial^2 z}{\partial t^2}(r,t) &= -\frac{\partial }{\partial t}\left[r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] +\frac{\partial }{\partial t}\left[r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) -r\sin t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] \\ & -r\sin t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t \ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2}(r\cos t\,,\,r\sin t) -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial y\partial x}(r\cos t\,,\,r\sin t) \\ &\phantom{=}-r\sin t \ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \\ & -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y}(r\cos t\,,\,r\sin t) +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2}(r\cos t\,,\,r\sin t) \\ &= -r\cos t \ \frac{\partial f}{\partial x} -r\sin t \ \frac{\partial f}{\partial y} \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2} -2r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y} +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2} \end{aligned}$

여기서 $f$ 의 모든 편도함수는 $(r\cos t\,,\,r\sin t)$ 에서의 값이다.

원본 교재 풀이(영어)

By definition $z(r,t) = f(r\cos t\,,\,r\sin t)$ .

(a) By the chain rule

$\begin{aligned} \frac{\partial z}{\partial t}(r,t) &= \frac{\partial }{\partial t}\Big[f(r\cos t\,,\,r\sin t)\Big] \\ &= \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\frac{\partial }{\partial t}(r\cos t) +\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\frac{\partial }{\partial t}(r\sin t) \\ &= -r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \end{aligned}$

(b) By linearity, the product rule and the chain rule

$\begin{aligned} \frac{\partial^2 z}{\partial t^2}(r,t) &= -\frac{\partial }{\partial t}\left[r\sin t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] +\frac{\partial }{\partial t}\left[r\cos t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t\ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) -r\sin t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t)\right] \\ & -r\sin t\ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) +r\cos t\ \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t)\right] \\ &= -r\cos t \ \frac{\partial f}{\partial x}(r\cos t\,,\,r\sin t) \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2}(r\cos t\,,\,r\sin t) -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial y\partial x}(r\cos t\,,\,r\sin t) \\ &\phantom{=}-r\sin t \ \frac{\partial f}{\partial y}(r\cos t\,,\,r\sin t) \\ & -r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y}(r\cos t\,,\,r\sin t) +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2}(r\cos t\,,\,r\sin t) \\ &= -r\cos t \ \frac{\partial f}{\partial x} -r\sin t \ \frac{\partial f}{\partial y} \\ & +r^2\sin^2 t\ \frac{\partial^2 f}{\partial x^2} -2r^2\sin t\cos t\ \frac{\partial^2\ f}{\partial x\partial y} +r^2\cos^2 t\ \frac{\partial^2 f}{\partial y^2} \end{aligned}$

with all of the partial derivatives of $f$ evaluated at $(r\cos t\,,\,r\sin t)$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 25번

# 다변수 연쇄법칙활용보통자동 분류

626.CLP-3 · 2.4절 · 16번대학 수준우리말 번역

$z = f(x, y)$ 라 하자. 여기서 $f(x, y)$ 는 연속인 이계 편도함수를 가지며,

$f_x (2, 1) = 5, \qquad f_y(2, 1) =-2, \qquad f_{xx}(2, 1) = 2,\qquad f_{xy}(2, 1) = 1, \qquad f_{yy}(2, 1) = -4$

이다.

$x(t)=2t^2$ , $y(t)=t^3$ 이고 $t=1$ 일 때, $\frac{\mathrm{d^2}}{\mathrm{d}{t}^2} z\big(x(t),y(t)\big)$ 의 값을 구하세요.

영어 원문 보기

Let $z = f(x, y)$ , where $f(x, y)$ has continuous second-order partial derivatives, and

$f_x (2, 1) = 5, \qquad f_y(2, 1) =-2, \qquad f_{xx}(2, 1) = 2,\qquad f_{xy}(2, 1) = 1, \qquad f_{yy}(2, 1) = -4$

Find $\frac{\mathrm{d^2}}{\mathrm{d}{t}^2} z\big(x(t),y(t)\big)$ when $x(t)=2t^2$ , $y(t)=t^3$ and $t=1$ .

답 쓰고 채점

정답·풀이 보기

$x(t)=2t^2$ , $y(t)=t^3$ 에 대해 $w(t) = z\big(x(t),y(t)\big) = f\big(x(t),y(t)\big)$ 라 하자. $\frac{\mathrm{d^2}w}{\mathrm{d}{t}^2}(1)$ 을 계산하면 된다. 연쇄 법칙에 의해

$\begin{aligned} \frac{\mathrm{d}w}{\mathrm{d}t}(t)&= \frac{\mathrm{d}}{\mathrm{d}t} f\big(x(t),y(t)\big) \\ &= f_x(x(t)\,,\,y(t))\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) +f_y(x(t)\,,\,y(t))\,\frac{\mathrm{d}y}{\mathrm{d}t}(t) \\ &= 4t\,f_x(x(t)\,,\,y(t)) +3t^2\,f_y(x(t)\,,\,y(t)) \end{aligned}$

이다. 선형성, 곱의 미분법, 연쇄 법칙에 의해

$\begin{aligned} \frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) &= \frac{\mathrm{d}}{\mathrm{d}t}\left[4t\,f_x\big(x(t),y(t)\big) \right] +\frac{\mathrm{d}}{\mathrm{d}t}\left[3t^2\,f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x\big(x(t),y(t)\big) + 4t \frac{\mathrm{d}}{\mathrm{d}t}\left[f_x\big(x(t),y(t)\big) \right] \\& +6t\,f_y\big(x(t),y(t)\big) + 3t^2\frac{\mathrm{d}}{\mathrm{d}t}\left[f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x(2t^2\,,\,t^3) + 4t\Big[f_{xx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{xy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 3t^2\Big[f_{yx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{yy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &= 4\,f_x(2t^2\,,\,t^3) + 16t^2\,f_{xx}(2t^2\,,\,t^3) + 12t^3\,f_{xy}(2t^2\,,\,t^3) \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 12t^3\,f_{yx}(2t^2\,,\,t^3) + 9t^4\,f_{yy}(2t^2\,,\,t^3) \end{aligned}$

이다. 특히 $t=1$ 일 때, $f_{xy}(2, 1)=f_{yx}(2, 1)$ 이므로

$\begin{aligned} \left.\frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) \right|_{t=1} &= 4\,(5) + 16\,(2) + 12\,(1) \\ &\phantom{=} + 6\,(-2) + 12\,(1) + 9\,(-4) \\ &=28 \end{aligned}$

이다.

원본 교재 풀이(영어)

Write $w(t) = z\big(x(t),y(t)\big) = f\big(x(t),y(t)\big)$ with $x(t)=2t^2$ , $y(t)=t^3$ . We are to compute $\frac{\mathrm{d^2}w}{\mathrm{d}{t}^2}(1)$ . By the chain rule

$\begin{aligned} \frac{\mathrm{d}w}{\mathrm{d}t}(t)&= \frac{\mathrm{d}}{\mathrm{d}t} f\big(x(t),y(t)\big) \\ &= f_x(x(t)\,,\,y(t))\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) +f_y(x(t)\,,\,y(t))\,\frac{\mathrm{d}y}{\mathrm{d}t}(t) \\ &= 4t\,f_x(x(t)\,,\,y(t)) +3t^2\,f_y(x(t)\,,\,y(t)) \end{aligned}$

By linearity, the product rule, and the chain rule,

$\begin{aligned} \frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) &= \frac{\mathrm{d}}{\mathrm{d}t}\left[4t\,f_x\big(x(t),y(t)\big) \right] +\frac{\mathrm{d}}{\mathrm{d}t}\left[3t^2\,f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x\big(x(t),y(t)\big) + 4t \frac{\mathrm{d}}{\mathrm{d}t}\left[f_x\big(x(t),y(t)\big) \right] \\& +6t\,f_y\big(x(t),y(t)\big) + 3t^2\frac{\mathrm{d}}{\mathrm{d}t}\left[f_y\big(x(t),y(t)\big) \right] \\ &= 4\,f_x(2t^2\,,\,t^3) + 4t\Big[f_{xx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{xy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 3t^2\Big[f_{yx}\big(x(t),y(t)\big)\,\frac{\mathrm{d}x}{\mathrm{d}t}(t) + f_{yy}\big(x(t),y(t)\big)\,\frac{\mathrm{d}y}{\mathrm{d}t}(t)\Big] \\ &= 4\,f_x(2t^2\,,\,t^3) + 16t^2\,f_{xx}(2t^2\,,\,t^3) + 12t^3\,f_{xy}(2t^2\,,\,t^3) \\ &\phantom{=} + 6t\,f_y(2t^2\,,\,t^3) + 12t^3\,f_{yx}(2t^2\,,\,t^3) + 9t^4\,f_{yy}(2t^2\,,\,t^3) \end{aligned}$

In particular, when $t=1$ , and since $f_{xy}(2, 1)=f_{yx}(2, 1)$ ,

$\begin{aligned} \left.\frac{\mathrm{d^2}}{\mathrm{d}{t}^2} f\big(x(t),y(t)\big) \right|_{t=1} &= 4\,(5) + 16\,(2) + 12\,(1) \\ &\phantom{=} + 6\,(-2) + 12\,(1) + 9\,(-4) \\ &=28 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 24번

# 다변수 연쇄법칙활용보통자동 분류

627.CLP-3 · 2.4절 · 17번대학 수준우리말 번역

함수 $F(x,y,z)$ 가 등식 $\frac{\partial F}{\partial z} = \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2}$ 을 만족시키고, 혼합 편도함수 $\frac{\partial^2 F}{\partial x \partial y}$ 와 $\frac{\partial^2 F}{\partial y \partial x}$ 가 서로 같다고 하자. $A$ 를 어떤 상수라 하고 $G(\gamma, s, t) = F (\gamma + s, \gamma-s, At)$ 라 하자. $\frac{\partial G}{\partial t} = \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2}$ 이 되도록 하는 $A$ 의 값을 구하세요.

영어 원문 보기

Assume that the function $F(x,y,z)$ satisfies the equation $\frac{\partial F}{\partial z} = \frac{\partial^2 F}{\partial x^2} + \frac{\partial^2 F}{\partial y^2}$ and the mixed partial derivatives $\frac{\partial^2 F}{\partial x \partial y}$ and $\frac{\partial^2 F}{\partial y \partial x}$ are equal. Let $A$ be some constant and let $G(\gamma, s, t) = F (\gamma + s, \gamma-s, At)$ . Find the value of $A$ such that $\frac{\partial G}{\partial t} = \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2}$ .

답 쓰고 채점

정답·풀이 보기

연쇄 법칙에 의해

$\begin{aligned} \frac{\partial G}{\partial t}(\gamma ,s,t) &= \frac{\partial }{\partial t} \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial t}(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial t}(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial t}(At) \\ &= A\,\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At) \end{aligned}$

이고

$\begin{aligned} \frac{\partial G}{\partial \gamma }(\gamma ,s,t) &= \frac{\partial }{\partial \gamma } \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial \gamma }(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial \gamma }(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial \gamma }(At) \\ &= \frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \tag{E1} \end{aligned}$

이고

$\begin{aligned} \frac{\partial G}{\partial s}(\gamma ,s,t) &= \frac{\partial }{\partial s} \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial s}(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial s}(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial s}(At) \\ &= \frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) -\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \tag{E2} \end{aligned}$

이다.

다음 식의 우변에 있는 네 항에 연쇄 법칙을 적용하면 이계도함수를 구할 수 있다.

$\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &=\frac{\partial }{\partial \gamma }\Big[\frac{\partial G}{\partial \gamma }(\gamma , s, t)\Big] = \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] + \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] \\ \frac{\partial^2G}{\partial s^2}(\gamma ,s,t) &=\frac{\partial }{\partial s}\Big[\frac{\partial G}{\partial s}(\gamma , s, t)\Big] = \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] - \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] \end{aligned}$

다른 방법으로, (E1)과 (E2)에서 $F$ 를 $\frac{\partial F}{\partial x}$ 로 바꾸면 다음을 얻는다.

$\begin{aligned} \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) \Big] &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \end{aligned}$

(E1)과 (E2)에서 $F$ 를 $\frac{\partial F}{\partial y}$ 로 바꾸면 다음을 얻는다.

$\begin{aligned} \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s,At)\Big] &= \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s,At) \\ \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] &= \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}$

따라서

$\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}+ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}$

이고

$\begin{aligned} \frac{\partial^2G}{\partial s^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}-\Big[ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At)\Big] \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}$

이다.

그러므로 변수를 생략하고 나타내면,

$\begin{aligned} \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2} -\frac{\partial G}{\partial t} &=2\frac{\partial^2 F}{\partial x^2} + 2\frac{\partial^2 F}{\partial y^2} -A \frac{\partial F}{\partial z} =2\frac{\partial F}{\partial z}-A \frac{\partial F}{\partial z} =0 \end{aligned}$

이다. 이는 $A=2$ 이면 성립한다.

원본 교재 풀이(영어)

By the chain rule

$\begin{aligned} \frac{\partial G}{\partial t}(\gamma ,s,t) &= \frac{\partial }{\partial t} \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial t}(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial t}(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial t}(At) \\ &= A\,\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At) \end{aligned}$

and

$\begin{aligned} \frac{\partial G}{\partial \gamma }(\gamma ,s,t) &= \frac{\partial }{\partial \gamma } \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial \gamma }(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial \gamma }(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial \gamma }(At) \\ &= \frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \tag{E1} \end{aligned}$

and

$\begin{aligned} \frac{\partial G}{\partial s}(\gamma ,s,t) &= \frac{\partial }{\partial s} \Big[F (\gamma + s, \gamma-s, At)\Big] \\ &=\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial s}(\gamma + s) +\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At)\, \frac{\partial }{\partial s}(\gamma - s) \\& +\frac{\partial F}{\partial z}(\gamma + s, \gamma-s, At)\,\frac{\partial }{\partial s}(At) \\ &= \frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) -\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \tag{E2} \end{aligned}$

We can evaluate the second derivatives by applying the chain rule to the four terms on the right hand sides of

$\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &=\frac{\partial }{\partial \gamma }\Big[\frac{\partial G}{\partial \gamma }(\gamma , s, t)\Big] = \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] + \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] \\ \frac{\partial^2G}{\partial s^2}(\gamma ,s,t) &=\frac{\partial }{\partial s}\Big[\frac{\partial G}{\partial s}(\gamma , s, t)\Big] = \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] - \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] \end{aligned}$

Alternatively, we can observe that replacing $F$ by $\frac{\partial F}{\partial x}$ in (E1) and (E2) gives

$\begin{aligned} \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At)\Big] &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial x}(\gamma + s, \gamma-s, At) \Big] &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \end{aligned}$

replacing $F$ by $\frac{\partial F}{\partial y}$ in (E1) and (E2) gives

$\begin{aligned} \frac{\partial }{\partial \gamma }\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s,At)\Big] &= \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s,At) \\ \frac{\partial }{\partial s}\Big[\frac{\partial F}{\partial y}(\gamma + s, \gamma-s, At) \Big] &= \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}$

Consequently

$\begin{aligned} \frac{\partial^2G}{\partial\gamma ^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}+ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) +2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}$

and

$\begin{aligned} \frac{\partial^2G}{\partial s^2}(\gamma ,s,t) &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) \\ &\phantom{=}-\Big[ \frac{\partial^2 F}{\partial x\partial y}(\gamma + s, \gamma-s, At) -\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At)\Big] \\ &= \frac{\partial^2 F}{\partial x^2}(\gamma + s, \gamma-s, At) -2\frac{\partial^2 F}{\partial y\partial x}(\gamma + s, \gamma-s, At) +\frac{\partial^2 F}{\partial y^2}(\gamma + s, \gamma-s, At) \end{aligned}$

So, suppressing the arguments,

$\begin{aligned} \frac{\partial^2 G}{\partial \gamma^2} + \frac{\partial^2 G}{\partial s^2} -\frac{\partial G}{\partial t} &=2\frac{\partial^2 F}{\partial x^2} + 2\frac{\partial^2 F}{\partial y^2} -A \frac{\partial F}{\partial z} =2\frac{\partial F}{\partial z}-A \frac{\partial F}{\partial z} =0 \end{aligned}$

if $A=2$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 16번

# 다변수 연쇄법칙계산보통자동 분류

628.CLP-3 · 2.4절 · 18번대학 수준우리말 번역

$f(x)$ 를 미분가능한 함수라 하고, $f'(0) = 10$ 이 주어졌다고 하자. $g(s,t) = f (as - bt)$ 라 하자. 여기서 $a$ 와 $b$ 는 상수이다. 점 $(s,t) = (b,a)$ 에서 $\frac{\partial g}{\partial s}$ 의 값을 구하세요. 즉, $\frac{\partial g}{\partial s}\big|_{(b,a)}$ 를 구하세요.

영어 원문 보기

Let $f(x)$ be a differentiable function, and suppose it is given that $f'(0) = 10$ . Let $g(s,t) = f (as - bt)$ , where $a$ and $b$ are constants. Evaluate $\frac{\partial g}{\partial s}$ at the point $(s,t) = (b,a)$ , that is, find $\frac{\partial g}{\partial s}\big|_{(b,a)}$ .

답 쓰고 채점

정답·풀이 보기

연쇄 법칙에 의해

$\begin{aligned} \frac{\partial g}{\partial s}(s,t) = \frac{\partial }{\partial s}\big[f(as-bt)\big] =f'(as-bt)\frac{\partial }{\partial s}(as-bt) = af'(as-bt) \end{aligned}$

특히

$\begin{aligned} \frac{\partial g}{\partial s}(b,a) = af'(ab-ba) =af'(0) =10 a \end{aligned}$

원본 교재 풀이(영어)

By the chain rule

$\begin{aligned} \frac{\partial g}{\partial s}(s,t) = \frac{\partial }{\partial s}\big[f(as-bt)\big] =f'(as-bt)\frac{\partial }{\partial s}(as-bt) = af'(as-bt) \end{aligned}$

In particular

$\begin{aligned} \frac{\partial g}{\partial s}(b,a) = af'(ab-ba) =af'(0) =10 a \end{aligned}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 23번

# 다변수 연쇄법칙계산보통자동 분류

629.CLP-3 · 2.4절 · 19번대학 수준우리말 번역

$f(u,v)$ 를 두 변수의 미분가능한 함수라 하고, $z$ 를 $f(xz,yz) = 0$ 으로 음함수로 정의된 $x$ 와 $y$ 의 미분가능한 함수라 하자. 다음을 보이세요.

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z$

영어 원문 보기

Let $f(u,v)$ be a differentiable function of two variables, and let $z$ be a differentiable function of $x$ and $y$ defined implicitly by $f(xz,yz) = 0$ . Show that

$x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -z$

정답·풀이 보기

함수 $z(x,y)$ 는 모든 $x$ 와 $y$ 에 대해

$f\big(x\,z(x,y)\,,\, y\,z(x,y)\big) =0 \tag{$*$}$

을 만족한다. 연쇄 법칙에 의해

$\begin{aligned} &\frac{\partial }{\partial x}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y\,z_x(x,y) \\ &\frac{\partial }{\partial y}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] \end{aligned}$

이므로 $(*)$ 를 $x$ 에 대해, 그리고 $y$ 에 대해 미분하면

$\begin{aligned} f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y z_x(x,y) &=0 \\ f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] &=0 \end{aligned}$

이고, 인수를 생략하여 나타내면

$\begin{aligned} f_u\,\big[z+x\,z_x\big] + f_v\,y z_x &=0 \\ f_u\, x\,z_y + f_v\,\big[z+y\,z_y\big] &=0 \end{aligned}$

이다. 첫 번째 식을 $z_x$ 에 대해, 두 번째 식을 $z_y$ 에 대해 풀면

$\begin{aligned} z_x & = -\frac{z\,f_u}{x\,f_u+y\,f_v} \\ z_y & = -\frac{z\,f_v}{x\,f_u+y\,f_v} \end{aligned}$

이므로

$\begin{aligned} x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -\frac{xz\,f_u}{x\,f_u+y\,f_v} -\frac{yz\,f_v}{x\,f_u+y\,f_v} =-\frac{z\,(x\,f_u+y\,f_v)}{x\,f_u+y\,f_v} = -z \end{aligned}$

이다. 따라서 주어진 등식이 성립한다.

참고: 이는 물론 $x\,f_u+y\,f_v$ 가 0이 아니라는 가정 아래에서이다. 연쇄 법칙에 의해 이 가정은 $\frac{\partial }{\partial z}\big[f(xz,yz)\big]$ 가 0이 아니라는 가정과 같다. 이는 다시 $f(xz,yz)=0$ 을 $x$ 와 $y$ 의 함수로서 $z$ 에 대해 풀 수 있다는 진술과 거의 같지만, 완전히 같지는 않다.

원본 교재 풀이(영어)

We are told that the function $z(x,y)$ obeys

$f\big(x\,z(x,y)\,,\, y\,z(x,y)\big) =0 \tag{$*$}$

for all $x$ and $y$ . By the chain rule,

$\begin{aligned} &\frac{\partial }{\partial x}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial x}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y\,z_x(x,y) \\ &\frac{\partial }{\partial y}\big[f\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\big] \\[-0.05in] &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[x\,z(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\frac{\partial }{\partial y}\big[y\,z(x,y)\big] \\ &=f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] \end{aligned}$

so differentiating $(*)$ with respect to $x$ and with respect to $y$ gives

$\begin{aligned} f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,\big[z(x,y)+x\,z_x(x,y)\big] + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\,y z_x(x,y) &=0 \\ f_u\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, x\,z_y(x,y) + f_v\big(x\,z(x,y)\,,\, y\,z(x,y)\big)\, \big[z(x,y)+y\,z_y(x,y)\big] &=0 \end{aligned}$

or, leaving out the arguments,

$\begin{aligned} f_u\,\big[z+x\,z_x\big] + f_v\,y z_x &=0 \\ f_u\, x\,z_y + f_v\,\big[z+y\,z_y\big] &=0 \end{aligned}$

Solving the first equation for $z_x$ and the second for $z_y$ gives

$\begin{aligned} z_x & = -\frac{z\,f_u}{x\,f_u+y\,f_v} \\ z_y & = -\frac{z\,f_v}{x\,f_u+y\,f_v} \end{aligned}$

so that

$\begin{aligned} x\frac{\partial z}{\partial x}+y\frac{\partial z}{\partial y} = -\frac{xz\,f_u}{x\,f_u+y\,f_v} -\frac{yz\,f_v}{x\,f_u+y\,f_v} =-\frac{z\,(x\,f_u+y\,f_v)}{x\,f_u+y\,f_v} = -z \end{aligned}$

as desired.

Remark: This is of course under the assumption that $x\,f_u+y\,f_v$ is nonzero. That is equivalent, by the chain rule, to the assumption that $\frac{\partial }{\partial z}\big[f(xz,yz)\big]$ is non zero. That, in turn, is almost, but not quite, equivalent to the statement that $f(xz,yz)=0$ is can be solved for $z$ as a function of $x$ and $y$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 18번

# 다변수 연쇄법칙증명·논술보통자동 분류

630.CLP-3 · 2.4절 · 20번대학 수준우리말 번역

어떤 두 번 미분가능한 함수 $u = u(x, y)$ 에 대하여 $w(s,t) = u(2s + 3t, 3s - 2t)$ 라 하자.

(a) $w_{ss}$ 를 $u_{xx}$ , $u_{xy}$ , $u_{yy}$ 로 나타내세요 ( $u_{xy} = u_{yx}$ 라고 가정해도 됩니다). (b) $u_{xx} + u_{yy} = 0$ 이라 하자. 상수 $A$ 가 얼마일 때 $w_{ss} = Aw_{tt}$ 가 되는지 구하세요.

영어 원문 보기

Let $w(s,t) = u(2s + 3t, 3s - 2t)$ for some twice differentiable function $u = u(x, y)$ .

(a) Find $w_{ss}$ in terms of $u_{xx}$ , $u_{xy}$ , and $u_{yy}$ (you can assume that $u_{xy} = u_{yx}$ ). (b) Suppose $u_{xx} + u_{yy} = 0$ . For what constant $A$ will $w_{ss} = Aw_{tt}$ ?

정답·풀이 보기

(a) 연쇄 법칙에 의해

$\begin{aligned} w_s(s,t) &=\frac{\partial }{\partial s}\big[u(2s + 3t, 3s - 2t)\big] \\ &= u_x(2s + 3t, 3s - 2t)\frac{\partial }{\partial s}\big(2s + 3t\big) +u_y(2s + 3t, 3s - 2t)\frac{\partial }{\partial s}\big(3s - 2t\big) \\ &= 2\,u_x(2s + 3t, 3s - 2t) +3\,u_y(2s + 3t, 3s - 2t) \end{aligned}$

이고

$\begin{aligned} w_{ss}(s,t) &= 2\frac{\partial }{\partial s}\big[u_x(2s + 3t, 3s - 2t)\big] + 3\frac{\partial }{\partial s}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[4u_{xx}(2s + 3t, 3s - 2t) +6u_{xy}(2s + 3t, 3s - 2t) \big] \\& + \big[6u_{yx}(2s + 3t, 3s - 2t) +9u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=4\,u_{xx}(2s + 3t, 3s - 2t) +12\,u_{xy}(2s + 3t, 3s - 2t) +9\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}$

이다.

(b) 다시 연쇄 법칙에 의해

$\begin{aligned} w_t(s,t) &=\frac{\partial }{\partial t}\big[u(2s + 3t, 3s - 2t)\big] \\ &= u_x(2s + 3t, 3s - 2t)\frac{\partial }{\partial t}\big(2s + 3t\big) +u_y(2s + 3t, 3s - 2t)\frac{\partial }{\partial t}\big(3s - 2t\big) \\ &= 3\,u_x(2s + 3t, 3s - 2t) -2\, u_y(2s + 3t, 3s - 2t) \end{aligned}$

이고

$\begin{aligned} w_{tt}(s,t) &= 3\frac{\partial }{\partial t}\big[u_x(2s + 3t, 3s - 2t)\big] - 2\frac{\partial }{\partial t}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[9u_{xx}(2s + 3t, 3s - 2t) -6u_{xy}(2s + 3t, 3s - 2t) \big] \\& - \big[6u_{yx}(2s + 3t, 3s - 2t) -4u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=9\,u_{xx}(2s + 3t, 3s - 2t) -12\,u_{xy}(2s + 3t, 3s - 2t) +4\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}$

이다.

따라서 임의의 상수 $A$ 에 대해

$\begin{aligned} w_{ss} - Aw_{tt} &= (4-9A) u_{xx} +(12+12A) u_{xy} + (9-4A) u_{yy} \end{aligned}$

이다.

$u_{xx} + u_{yy}=0$ 이 주어졌으므로, $A=-1$ 이면 이 값은 원하는 대로 0이 된다. (이때 $(4-9A)=(9-4A)=13$ 이다.)

원본 교재 풀이(영어)

(a) By the chain rule

$\begin{aligned} w_s(s,t) &=\frac{\partial }{\partial s}\big[u(2s + 3t, 3s - 2t)\big] \\ &= u_x(2s + 3t, 3s - 2t)\frac{\partial }{\partial s}\big(2s + 3t\big) +u_y(2s + 3t, 3s - 2t)\frac{\partial }{\partial s}\big(3s - 2t\big) \\ &= 2\,u_x(2s + 3t, 3s - 2t) +3\,u_y(2s + 3t, 3s - 2t) \end{aligned}$

and

$\begin{aligned} w_{ss}(s,t) &= 2\frac{\partial }{\partial s}\big[u_x(2s + 3t, 3s - 2t)\big] + 3\frac{\partial }{\partial s}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[4u_{xx}(2s + 3t, 3s - 2t) +6u_{xy}(2s + 3t, 3s - 2t) \big] \\& + \big[6u_{yx}(2s + 3t, 3s - 2t) +9u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=4\,u_{xx}(2s + 3t, 3s - 2t) +12\,u_{xy}(2s + 3t, 3s - 2t) +9\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}$

(b) Again by the chain rule

$\begin{aligned} w_t(s,t) &=\frac{\partial }{\partial t}\big[u(2s + 3t, 3s - 2t)\big] \\ &= u_x(2s + 3t, 3s - 2t)\frac{\partial }{\partial t}\big(2s + 3t\big) +u_y(2s + 3t, 3s - 2t)\frac{\partial }{\partial t}\big(3s - 2t\big) \\ &= 3\,u_x(2s + 3t, 3s - 2t) -2\, u_y(2s + 3t, 3s - 2t) \end{aligned}$

and

$\begin{aligned} w_{tt}(s,t) &= 3\frac{\partial }{\partial t}\big[u_x(2s + 3t, 3s - 2t)\big] - 2\frac{\partial }{\partial t}\big[u_y(2s + 3t, 3s - 2t)\big] \\ &= \big[9u_{xx}(2s + 3t, 3s - 2t) -6u_{xy}(2s + 3t, 3s - 2t) \big] \\& - \big[6u_{yx}(2s + 3t, 3s - 2t) -4u_{yy}(2s + 3t, 3s - 2t) \big] \\ &=9\,u_{xx}(2s + 3t, 3s - 2t) -12\,u_{xy}(2s + 3t, 3s - 2t) +4\,u_{yy}(2s + 3t, 3s - 2t) \end{aligned}$

Consquently, for any constant $A$ ,

$\begin{aligned} w_{ss} - Aw_{tt} &= (4-9A) u_{xx} +(12+12A) u_{xy} + (9-4A) u_{yy} \end{aligned}$

Given that $u_{xx} + u_{yy}=0$ , this will be zero, as desired, if $A=-1$ . (Then $(4-9A)=(9-4A)=13$ .)

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 15번

# 다변수 연쇄법칙계산보통자동 분류

631.CLP-3 · 2.4절 · 22번대학 수준우리말 번역

정의에 따르면, 미분가능한 함수 $f(x,y)$ 의 점 $\big( x_0\,,\,y_0\big)$ 에서의 기울기 벡터는 다음과 같다.

${ \boldsymbol{\nabla} } f(x_0,y_0) =\left< \frac{\partial f}{\partial x}\big( x_0\,,\,y_0\big)\,,\, \frac{\partial f}{\partial y}\big( x_0\,,\,y_0\big)\right>$

다음을 알고 있다고 하자.

${ \boldsymbol{\nabla} } f(3,6)=\left< 7,8\right>$

또한

${ \boldsymbol{\nabla} } g(1,2)=\left< -1,4\right>,$

이고

${ \boldsymbol{\nabla} } h(1,2)=\left< -5,10\right>.$

이라고 하자.

$g(1,2)=3$ , $h(1,2)=6$ 이고 $z(s,t)=f\big(g(s,t),h(s,t)\big)$ 일 때, 다음을 구하세요.

${ \boldsymbol{\nabla} } z(1,2)$

영어 원문 보기

By definition, the gradient of the differentiable function $f(x,y)$ at the point $\big( x_0\,,\,y_0\big)$ is

${ \boldsymbol{\nabla} } f(x_0,y_0) =\left< \frac{\partial f}{\partial x}\big( x_0\,,\,y_0\big)\,,\, \frac{\partial f}{\partial y}\big( x_0\,,\,y_0\big)\right>$

Suppose that we know

${ \boldsymbol{\nabla} } f(3,6)=\left< 7,8\right>$

Suppose also that

${ \boldsymbol{\nabla} } g(1,2)=\left< -1,4\right>,$

and

${ \boldsymbol{\nabla} } h(1,2)=\left< -5,10\right>.$

Assuming $g(1,2)=3$ , $h(1,2)=6$ , and $z(s,t)=f\big(g(s,t),h(s,t)\big)$ , find

${ \boldsymbol{\nabla} } z(1,2)$

정답·풀이 보기

연쇄 법칙에 의해

$\begin{aligned} \frac{\partial z}{\partial s}(s,t) &=\frac{\partial }{\partial s}f\big(g(s,t),h(s,t)\big) &&=\frac{\partial f}{\partial x}\big(g(s,t),h(s,t)\big) \frac{\partial g}{\partial s}(s,t) +\frac{\partial f}{\partial y}\big(g(s,t),h(s,t)\big) \frac{\partial h}{\partial s}(s,t) \\ \frac{\partial z}{\partial t}(s,t) &=\frac{\partial }{\partial t}f\big(g(s,t),h(s,t)\big) &&=\frac{\partial f}{\partial x}\big(g(s,t),h(s,t)\big) \frac{\partial g}{\partial t}(s,t) +\frac{\partial f}{\partial y}\big(g(s,t),h(s,t)\big) \frac{\partial h}{\partial t}(s,t) \end{aligned}$

특히

$\begin{aligned} \frac{\partial z}{\partial s}(1,2) &=\frac{\partial f}{\partial x}\big(g(1,2),h(1,2)\big) \frac{\partial g}{\partial s}(1,2) +\frac{\partial f}{\partial y}\big(g(1,2),h(1,2)\big) \frac{\partial h}{\partial s}(1,2) \\ &=\frac{\partial f}{\partial x}(3,6) \frac{\partial g}{\partial s}(1,2) +\frac{\partial f}{\partial y}(3,6)) \frac{\partial h}{\partial s}(1,2) \\ &=7\times(-1)+8\times(-5) =-47\\ \frac{\partial z}{\partial t}(1,2) &=\frac{\partial f}{\partial x}\big(g(1,2),h(1,2)\big) \frac{\partial g}{\partial t}(1,2) +\frac{\partial f}{\partial y}\big(g(1,2),h(1,2)\big) \frac{\partial h}{\partial t}(1,2) \\ &=7\times4+8\times 10 =108 \end{aligned}$

따라서 ${ \boldsymbol{\nabla} } z(1,2)=\left< -47,108\right>$ 이다.

원본 교재 풀이(영어)

By the chain rule

$\begin{aligned} \frac{\partial z}{\partial s}(s,t) &=\frac{\partial }{\partial s}f\big(g(s,t),h(s,t)\big) &&=\frac{\partial f}{\partial x}\big(g(s,t),h(s,t)\big) \frac{\partial g}{\partial s}(s,t) +\frac{\partial f}{\partial y}\big(g(s,t),h(s,t)\big) \frac{\partial h}{\partial s}(s,t) \\ \frac{\partial z}{\partial t}(s,t) &=\frac{\partial }{\partial t}f\big(g(s,t),h(s,t)\big) &&=\frac{\partial f}{\partial x}\big(g(s,t),h(s,t)\big) \frac{\partial g}{\partial t}(s,t) +\frac{\partial f}{\partial y}\big(g(s,t),h(s,t)\big) \frac{\partial h}{\partial t}(s,t) \end{aligned}$

In particular

$\begin{aligned} \frac{\partial z}{\partial s}(1,2) &=\frac{\partial f}{\partial x}\big(g(1,2),h(1,2)\big) \frac{\partial g}{\partial s}(1,2) +\frac{\partial f}{\partial y}\big(g(1,2),h(1,2)\big) \frac{\partial h}{\partial s}(1,2) \\ &=\frac{\partial f}{\partial x}(3,6) \frac{\partial g}{\partial s}(1,2) +\frac{\partial f}{\partial y}(3,6)) \frac{\partial h}{\partial s}(1,2) \\ &=7\times(-1)+8\times(-5) =-47\\ \frac{\partial z}{\partial t}(1,2) &=\frac{\partial f}{\partial x}\big(g(1,2),h(1,2)\big) \frac{\partial g}{\partial t}(1,2) +\frac{\partial f}{\partial y}\big(g(1,2),h(1,2)\big) \frac{\partial h}{\partial t}(1,2) \\ &=7\times4+8\times 10 =108 \end{aligned}$

Hence ${ \boldsymbol{\nabla} } z(1,2)=\left< -47,108\right>$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 23번

# 다변수 연쇄법칙계산보통자동 분류

632.CLP-3 · 2.4절 · 23번대학 수준우리말 번역

(a) $f$ 를 실수 전체에서 정의된 임의의 미분가능한 함수라 하자. 평면 전체에서 다음과 같이 정의된 함수 $w$ 가

$w(x,y)=e^{-y}f(x-y)$

편미분방정식

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}=0$

을 만족시킴을 보이세요. (b) 방정식 $x=u^3-3uv^2$ , $y=3u^2v-v^3$ , $z=u^2-v^2$ 은 $z$ 를 $x$ 와 $y$ 의 함수로 정의한다. 점 $(x,y)=(2,11)$ 에 대응하는 점 $(u,v)=(2,1)$ 에서 $\frac{\partial z}{\partial x}$ 의 값을 구하세요.

영어 원문 보기

(a) Let $f$ be an arbitrary differentiable function defined on the entire real line. Show that the function $w$ defined on the entire plane as

$w(x,y)=e^{-y}f(x-y)$

satisfies the partial differential equation:

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y}=0$ (b) The equations $x=u^3-3uv^2$ , $y=3u^2v-v^3$ and $z=u^2-v^2$ define $z$ as a function of $x$ and $y$ . Determine $\frac{\partial z}{\partial x}$ at the point $(u,v)=(2,1)$ which corresponds to the point $(x,y)=(2,11)$ .

정답·풀이 보기

(a) 곱의 미분법과 연쇄 법칙에 의해

$\begin{aligned} w_x(x,y) &= \frac{\partial }{\partial x}\big[e^{-y}f(x-y)\big] = e^{-y} \frac{\partial }{\partial x}\big[f(x-y)\big] = e^{-y} f'(x-y)\frac{\partial }{\partial x}(x-y) \\ &= e^{-y}f'(x-y) \\% w_y(x,y) &= \frac{\partial }{\partial y}\big[e^{-y}f(x-y)\big] = -e^{-y}f(x-y) + e^{-y} \frac{\partial }{\partial y}\big[f(x-y)\big] \\ &= -e^{-y}f(x-y) + e^{-y} f'(x-y)\frac{\partial }{\partial y}(x-y) \\ &= -e^{-y}f(x-y)-e^{-y}f'(x-y) \end{aligned}$

이므로

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y} =e^{-y}f(x-y)+e^{-y}f'(x-y)-e^{-y}f(x-y)-e^{-y}f'(x-y) =0$

이다. 따라서 성립한다.

(b) $x=u^3-3uv^2$ , $y=3u^2v-v^3$ 을 $x$ , $y$ 가 주어진 매개변수이고 $u$ , $v$ 가 미지수인 두 방정식으로 생각하자. 문제는 적어도 $(u,v)=(2,1)$ , $(x,y)=(2,11)$ 근처에서 이 두 방정식을 $u$ , $v$ 에 대해 $x$ , $y$ 로 풀 수 있음을 암묵적으로 알려 준다. 즉, 함수 $u(x,y)$ 와 $v(x,y)$ 가 다음 식으로 정해진다는 것을 알려 준다.

$x=u(x,y)^3-3u(x,y)\,v(x,y)^2\qquad y=3u(x,y)^2v(x,y)-v(x,y)^3$

등식 $x=u(x,y)^3-3u(x,y)v(x,y)^2$ 의 양변에 $\frac{\partial }{\partial x}$ 를 취하면

$\begin{aligned} 1&=3\,u(x,y)^2\,\frac{\partial u}{\partial x}(x,y) -3\frac{\partial u}{\partial x}(x,y)\,v(x,y)^2 -6\,u(x,y)\,v(x,y)\,\frac{\partial v}{\partial x}(x,y) \\[0.05in] \end{aligned}$

이다. 다음으로 $y=3u(x,y)^2v(x,y)-v(x,y)^3$ 의 양변에 $\frac{\partial }{\partial x}$ 를 취하면

$\begin{aligned} 0&=6\,u(x,y)\,\frac{\partial u}{\partial x}(x,y)\,v(x,y) +3\,u(x,y)^2\,\frac{\partial v}{\partial x}(x,y) -3\,v(x,y)^2\,\frac{\partial v}{\partial x}(x,y) \end{aligned}$

이다. $x=2$ , $y=11$ , $u=2$ , $v=1$ 을 대입하면

$\begin{aligned} 1&=12\frac{\partial u}{\partial x}(2,11) -3\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11) &&=9\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11)\\[0.05in] 0&=12\frac{\partial u}{\partial x}(2,11) +12\frac{\partial v}{\partial x}(2,11) -3\frac{\partial v}{\partial x}(2,11) &&=12\frac{\partial u}{\partial x}(2,11) +9\frac{\partial v}{\partial x}(2,11) \end{aligned}$

이다. 두 번째 식에서 $\frac{\partial v}{\partial x}(2,11) =-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)$ 이다. 이를 첫 번째 식에 대입하면

$1=9\frac{\partial u}{\partial x}(2,11) -12\left[-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)\right] = 25\frac{\partial u}{\partial x}(2,11)$

이므로 $\frac{\partial u}{\partial x}(2,11)=\frac{1}{25}$ 이고 $\frac{\partial v}{\partial x}(2,11)=-\frac{4}{75}$ 이다. 또한 문제에서 $z(x,y)=u(x,y)^2-v(x,y)^2$ 임을 알 수 있다. 따라서

$\begin{aligned} &\frac{\partial z}{\partial x}(x,y) =2u(x,y)\frac{\partial u}{\partial x}(x,y) -2v(x,y)\frac{\partial v}{\partial x}(x,y)\\ \implies & \frac{\partial z}{\partial x}(2,11) =4\frac{\partial u}{\partial x}(2,11) -2\frac{\partial v}{\partial x}(2,11) =4\frac{1}{25}+2\frac{4}{75}=\frac{20}{75} =\frac{4}{15} \end{aligned}$

이다.

원본 교재 풀이(영어)

(a) By the product and chain rules

$\begin{aligned} w_x(x,y) &= \frac{\partial }{\partial x}\big[e^{-y}f(x-y)\big] = e^{-y} \frac{\partial }{\partial x}\big[f(x-y)\big] = e^{-y} f'(x-y)\frac{\partial }{\partial x}(x-y) \\ &= e^{-y}f'(x-y) \\% w_y(x,y) &= \frac{\partial }{\partial y}\big[e^{-y}f(x-y)\big] = -e^{-y}f(x-y) + e^{-y} \frac{\partial }{\partial y}\big[f(x-y)\big] \\ &= -e^{-y}f(x-y) + e^{-y} f'(x-y)\frac{\partial }{\partial y}(x-y) \\ &= -e^{-y}f(x-y)-e^{-y}f'(x-y) \end{aligned}$

Hence

$w+\frac{\partial w}{\partial x}+\frac{\partial w}{\partial y} =e^{-y}f(x-y)+e^{-y}f'(x-y)-e^{-y}f(x-y)-e^{-y}f'(x-y) =0$

as desired.

(b) Think of $x=u^3-3uv^2$ , $y=3u^2v-v^3$ as two equations in the two unknowns $u$ , $v$ with $x$ , $y$ just being given parameters. The question implicitly tells us that those two equations can be solved for $u$ , $v$ in terms of $x$ , $y$ , at least near $(u,v)=(2,1)$ , $(x,y)=(2,11)$ . That is, the question implicitly tells us that the functions $u(x,y)$ and $v(x,y)$ are determined by

$x=u(x,y)^3-3u(x,y)\,v(x,y)^2\qquad y=3u(x,y)^2v(x,y)-v(x,y)^3$

Applying $\frac{\partial }{\partial x}$ to both sides of the equation $x=u(x,y)^3-3u(x,y)v(x,y)^2$ gives

$\begin{aligned} 1&=3\,u(x,y)^2\,\frac{\partial u}{\partial x}(x,y) -3\frac{\partial u}{\partial x}(x,y)\,v(x,y)^2 -6\,u(x,y)\,v(x,y)\,\frac{\partial v}{\partial x}(x,y) \\[0.05in] \end{aligned}$

Then applying $\frac{\partial }{\partial x}$ to both sides of $y=3u(x,y)^2v(x,y)-v(x,y)^3$ gives

$\begin{aligned} 0&=6\,u(x,y)\,\frac{\partial u}{\partial x}(x,y)\,v(x,y) +3\,u(x,y)^2\,\frac{\partial v}{\partial x}(x,y) -3\,v(x,y)^2\,\frac{\partial v}{\partial x}(x,y) \end{aligned}$

Substituting in $x=2$ , $y=11$ , $u=2$ , $v=1$ gives

$\begin{aligned} 1&=12\frac{\partial u}{\partial x}(2,11) -3\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11) &&=9\frac{\partial u}{\partial x}(2,11) -12\frac{\partial v}{\partial x}(2,11)\\[0.05in] 0&=12\frac{\partial u}{\partial x}(2,11) +12\frac{\partial v}{\partial x}(2,11) -3\frac{\partial v}{\partial x}(2,11) &&=12\frac{\partial u}{\partial x}(2,11) +9\frac{\partial v}{\partial x}(2,11) \end{aligned}$

From the second equation $\frac{\partial v}{\partial x}(2,11) =-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)$ . Substituting into the first equation gives

$1=9\frac{\partial u}{\partial x}(2,11) -12\left[-\frac{4}{3}\frac{\partial u}{\partial x}(2,11)\right] = 25\frac{\partial u}{\partial x}(2,11)$

so that $\frac{\partial u}{\partial x}(2,11)=\frac{1}{25}$ and $\frac{\partial v}{\partial x}(2,11)=-\frac{4}{75}$ . The question also tells us that $z(x,y)=u(x,y)^2-v(x,y)^2$ . Hence

$\begin{aligned} &\frac{\partial z}{\partial x}(x,y) =2u(x,y)\frac{\partial u}{\partial x}(x,y) -2v(x,y)\frac{\partial v}{\partial x}(x,y)\\ \implies & \frac{\partial z}{\partial x}(2,11) =4\frac{\partial u}{\partial x}(2,11) -2\frac{\partial v}{\partial x}(2,11) =4\frac{1}{25}+2\frac{4}{75}=\frac{20}{75} =\frac{4}{15} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 20번

# 다변수 연쇄법칙증명·논술보통자동 분류

633.CLP-3 · 2.4절 · 24번대학 수준우리말 번역

방정식

$\begin{aligned} x^2-y\cos(uv)&=v\\ x^2+y^2-\sin(uv)&=\frac{4}{\pi}u \end{aligned}$

은 $(u,v)=\big(\frac{\pi}{2},0\big)$ 일 때의 점 $(x,y)=(1,1)$ 근처에서 $x$ 와 $y$ 를 $u$ 와 $v$ 의 함수로 암시적으로 정의한다(즉, $x=x(u,v)$ , $y=y(u,v)$ ).

(a) $(u,v)=\big(\frac{\pi}{2},0\big)$ 에서

$\frac{\partial x}{\partial u}\text{ 와 } \frac{\partial y}{\partial u}$

의 값을 구하세요. (b) $z=x^4+y^4$ 일 때, 점 $(u,v)=\big(\frac{\pi}{2},0\big)$ 에서 $\frac{\partial z}{\partial u}$ 의 값을 구하세요.

영어 원문 보기

The equations

$\begin{aligned} x^2-y\cos(uv)&=v\\ x^2+y^2-\sin(uv)&=\frac{4}{\pi}u \end{aligned}$

define $x$ and $y$ implicitly as functions of $u$ and $v$ (i.e. $x=x(u,v)$ , and $y=y(u,v)$ ) near the point $(x,y)=(1,1)$ at which $(u,v)=\big(\frac{\pi}{2},0\big)$ .

(a) Find

$\frac{\partial x}{\partial u}\text{ and } \frac{\partial y}{\partial u}$

at $(u,v)=\big(\frac{\pi}{2},0\big)$ . (b) If $z=x^4+y^4$ , determine $\frac{\partial z}{\partial u}$ at the point $(u,v)=\big(\frac{\pi}{2},0\big)$ .

정답·풀이 보기

(a) 주어진 식은

$x(u,v)^2-y(u,v)\cos(uv)=v\qquad x(u,v)^2+y(u,v)^2-\sin(uv)=\frac{4}{\pi}u$

이다. 두 방정식에 $\frac{\partial }{\partial u}$ 를 적용하면

$\begin{aligned} 2x(u,v)\frac{\partial x}{\partial u}(u,v) -\frac{\partial y}{\partial u}(u,v)\cos(uv)+v\,y(u,v)\sin(uv)&=0 \\ 2x(u,v)\frac{\partial x}{\partial u}(u,v) +2y(u,v)\frac{\partial y}{\partial u}(u,v) -v\cos(uv)&=\frac{4}{\pi} \end{aligned}$

이다. $u=\frac{\pi}{2}$ , $v=0$ , $x\big(\frac{\pi}{2},0\big)=1$ , $y\big(\frac{\pi}{2},0\big)=1$ 을 대입하면

$\begin{aligned} 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) -\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right)&=0 \\[0.1in] 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) +2\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right) &=\frac{4}{\pi} \end{aligned}$

이다. 첫째 방정식에서 얻은 $\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=2\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)$ 를 둘째 방정식에 대입하면 $6\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{\pi}$ 이므로 $\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{2}{3\pi}$ 이고 $\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{3\pi}$ 이다.

(b) $z(u,v)=x(u,v)^4+y(u,v)^4$ 으로 주어져 있다. 따라서

$\begin{aligned} \frac{\partial z}{\partial u}(u,v) &=4x(u,v)^3\ \frac{\partial x}{\partial u}(u,v) +4y(u,v)^3\ \frac{\partial y}{\partial u}(u,v) \end{aligned}$

이다. $u=\frac{\pi}{2}$ , $v=0$ , $x\big(\frac{\pi}{2},0\big)=1$ , $y\big(\frac{\pi}{2},0\big)=1$ 을 대입하고 (a)의 결과를 이용하면

$\begin{aligned} \frac{\partial z}{\partial u}\left(\frac{\pi}{2},0\right) &=4\,x\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial x}{\partial u}\!\left(\frac{\pi}{2},0\right) +4\,y\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial y}{\partial u}\!\left(\frac{\pi}{2},0\right) \\ &=4\left(\frac{2}{3\pi}\right) +4\left(\frac{4}{3\pi}\right) \\ &=\frac{8}{\pi} \end{aligned}$

이다.

원본 교재 풀이(영어)

(a) We are told that

$x(u,v)^2-y(u,v)\cos(uv)=v\qquad x(u,v)^2+y(u,v)^2-\sin(uv)=\frac{4}{\pi}u$

Applying $\frac{\partial }{\partial u}$ to both equations gives

$\begin{aligned} 2x(u,v)\frac{\partial x}{\partial u}(u,v) -\frac{\partial y}{\partial u}(u,v)\cos(uv)+v\,y(u,v)\sin(uv)&=0 \\ 2x(u,v)\frac{\partial x}{\partial u}(u,v) +2y(u,v)\frac{\partial y}{\partial u}(u,v) -v\cos(uv)&=\frac{4}{\pi} \end{aligned}$

Setting $u=\frac{\pi}{2}$ , $v=0$ , $x\big(\frac{\pi}{2},0\big)=1$ , $y\big(\frac{\pi}{2},0\big)=1$ gives

$\begin{aligned} 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) -\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right)&=0 \\[0.1in] 2\frac{\partial x}{\partial u}\left(\frac{\pi}{2},0\right) +2\frac{\partial y}{\partial u}\left(\frac{\pi}{2},0\right) &=\frac{4}{\pi} \end{aligned}$

Substituting $\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=2\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)$ , from the first equation, into the second equation gives $6\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{\pi}$ so that $\frac{\partial x}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{2}{3\pi}$ and $\frac{\partial y}{\partial u}\big(\frac{\pi}{2},0\big)=\frac{4}{3\pi}$ .

(b) We are told that $z(u,v)=x(u,v)^4+y(u,v)^4$ . So

$\begin{aligned} \frac{\partial z}{\partial u}(u,v) &=4x(u,v)^3\ \frac{\partial x}{\partial u}(u,v) +4y(u,v)^3\ \frac{\partial y}{\partial u}(u,v) \end{aligned}$

Substituting in $u=\frac{\pi}{2}$ , $v=0$ , $x\big(\frac{\pi}{2},0\big)=1$ , $y\big(\frac{\pi}{2},0\big)=1$ and using the results of part (a),

$\begin{aligned} \frac{\partial z}{\partial u}\left(\frac{\pi}{2},0\right) &=4\,x\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial x}{\partial u}\!\left(\frac{\pi}{2},0\right) +4\,y\!\left(\frac{\pi}{2},0\right)^3\ \frac{\partial y}{\partial u}\!\left(\frac{\pi}{2},0\right) \\ &=4\left(\frac{2}{3\pi}\right) +4\left(\frac{4}{3\pi}\right) \\ &=\frac{8}{\pi} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 15번

# 다변수 연쇄법칙계산보통자동 분류

634.CLP-3 · 2.4절 · 25번대학 수준우리말 번역

$f(u,v)$ 를 미분가능한 함수라 하고, $u=x+y$ , $v=x-y$ 라 하자. 다음을 만족하는 상수 $\alpha$ 를 구하세요.

$\begin{aligned} (f_x)^2+(f_y)^2=\alpha \big((f_u)^2+(f_v)^2\big) \end{aligned}$

영어 원문 보기

Let $f(u,v)$ be a differentiable function, and let $u=x+y$ and $v=x-y$ . Find a constant, $\alpha$ , such that

$\begin{aligned} (f_x)^2+(f_y)^2=\alpha \big((f_u)^2+(f_v)^2\big) \end{aligned}$

정답·풀이 보기

이 문제는 나쁜(그러나 흔히 쓰는) 표기를 사용하고 있다. 하나의 기호 $f$ 가 서로 다른 두 함수, 즉 $f(u,v)$ 와 $f(x,y)=f(u,v)\big|_{u=x+y,v=x-y}$ 에 쓰였기 때문이다. 더 나은 표현은 다음과 같다.

$f(u,v)$ 와 $F(x,y)$ 를 $F(x,y)=f(x+y,x-y)$ 를 만족하는 미분가능한 함수라 하자. 다음을 만족하는 상수 $\alpha$ 를 구하세요.

$F_x(x,y)^2+F_y(x,y)^2=\alpha \big\{f_u(x+y,x-y)^2+f_v(x+y,x-y)^2\big\}$

연쇄 법칙에 의해

$\begin{aligned} \frac{\partial F}{\partial x}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial x}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial x}(x-y) \\ &=f_u(x+y,x-y)+f_v(x+y,x-y)\\ \frac{\partial F}{\partial y}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial y}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial y}(x-y) \\ &=f_u(x+y,x-y)-f_v(x+y,x-y) \end{aligned}$

따라서

$\begin{aligned} F_x(x,y)^2+F_y(x,y)^2 &=\Big[f_u(x+y,x-y)+f_v(x+y,x-y)\Big]^2 \\ & +\Big[f_u(x+y,x-y)-f_v(x+y,x-y)\Big]^2\\ &=2f_u(x+y,x-y)^2+2f_v(x+y,x-y)^2 \end{aligned}$

그러므로 $\alpha =2$ 이면 된다.

원본 교재 풀이(영어)

This question uses bad (but standard) notation, in that the one symbol $f$ is used for two different functions, namely $f(u,v)$ and $f(x,y)=f(u,v)\big|_{u=x+y,v=x-y}$ . A better wording is

Let $f(u,v)$ and $F(x,y)$ be differentiable functions such that $F(x,y)=f(x+y,x-y)$ . Find a constant, $\alpha$ , such that

$F_x(x,y)^2+F_y(x,y)^2=\alpha \big\{f_u(x+y,x-y)^2+f_v(x+y,x-y)^2\big\}$

By the chain rule

$\begin{aligned} \frac{\partial F}{\partial x}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial x}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial x}(x-y) \\ &=f_u(x+y,x-y)+f_v(x+y,x-y)\\ \frac{\partial F}{\partial y}(x,y) &=f_u(x+y,x-y)\frac{\partial }{\partial y}(x+y) +f_v(x+y,x-y)\frac{\partial }{\partial y}(x-y) \\ &=f_u(x+y,x-y)-f_v(x+y,x-y) \end{aligned}$

Hence

$\begin{aligned} F_x(x,y)^2+F_y(x,y)^2 &=\Big[f_u(x+y,x-y)+f_v(x+y,x-y)\Big]^2 \\ & +\Big[f_u(x+y,x-y)-f_v(x+y,x-y)\Big]^2\\ &=2f_u(x+y,x-y)^2+2f_v(x+y,x-y)^2 \end{aligned}$

So $\alpha =2$ does the job.

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 14번

# 다변수 연쇄법칙계산보통자동 분류

635.CLP-3 · 2.4절 · 26번대학 수준우리말 번역

파동 방정식

$\frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\,\frac{\partial^2u}{\partial t^2} =0$

은 파동과 같은 현상을 다루는 많은 모형에서 나타난다. $u(x,t)$ 와 $v(\xi,\eta)$ 가 다음의 변수 변환으로 서로 연결되어 있다고 하자.

$\begin{aligned} u(x,t)&=v\big(\xi(x,t),\eta(x,t)\big)\\ \xi(x,t)&=x-ct\\ \eta(x,t)&=x+ct \end{aligned}$

(a) $\ \frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\frac{\partial^2u}{\partial t^2} =0\$ 일 필요충분조건이 $\ \frac{\partial^2 v}{\partial\xi\partial\eta}=0$ 임을 보이세요. (b) $\ \frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\frac{\partial^2u}{\partial t^2} =0\$ 일 필요충분조건이 어떤 함수 $F$ 와 $G$ 에 대해 $\ u(x,t)=F(x-ct)+G(x+ct)\$ 임을 보이세요. (c) $\ F(x-ct)+G(x+ct)\$ 를 진행파의 관점에서 해석하세요. $u(x,t)$ 를 $x$ 축을 따라 진행하는 파동의, 위치 $x$ 와 시각 $t$ 에서의 높이라고 생각하세요.

참고: 낯선 기호 $\xi$ 와 $\eta$ 에 당황하지 마세요. 이들은 그리스 문자 중 해롭지 않은 두 글자일 뿐이며, 각각 “크사이”와 “에타”라고 읽습니다.

영어 원문 보기

The wave equation

$\frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\,\frac{\partial^2u}{\partial t^2} =0$

arises in many models involving wave-like phenomena. Let $u(x,t)$ and $v(\xi,\eta)$ be related by the change of variables

$\begin{aligned} u(x,t)&=v\big(\xi(x,t),\eta(x,t)\big)\\ \xi(x,t)&=x-ct\\ \eta(x,t)&=x+ct \end{aligned}$

(a) Show that $\ \frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\frac{\partial^2u}{\partial t^2} =0\$ if and only if $\ \frac{\partial^2 v}{\partial\xi\partial\eta}=0$ . (b) Show that $\ \frac{\partial^2u}{\partial x^2} -\frac{1}{c^2}\frac{\partial^2u}{\partial t^2} =0\$ if and only if $\ u(x,t)=F(x-ct)+G(x+ct)\$ for some functions $F$ and $G$ . (c) Interpret $\ F(x-ct)+G(x+ct)\$ in terms of travelling waves. Think of $u(x,t)$ as the height, at position $x$ and time $t$ , of a wave that is travelling along the $x$ -axis.

Remark: Don't be thrown by the strange symbols $\xi$ and $\eta$ . They are just two harmless letters from the Greek alphabet, called “xi” and “eta” respectively.

정답·풀이 보기

$u(x,t)=v\big(\xi(x,t),\eta(x,t)\big)$ 임을 기억하자. 연쇄 법칙에 의해

$\begin{aligned} \frac{\partial u}{\partial x}(x,t) &= \frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial x} +\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial x} \\ &= \frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) +\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \\ \frac{\partial u}{\partial t}(x,t) &= \frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial t} +\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial t} \\ &= -c\frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) +c\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \end{aligned}$

연쇄 법칙을 다시 적용하면

$\begin{aligned} \frac{\partial^2 u}{\partial x^2}(x,t) &= \frac{\partial }{\partial x} \Big[\frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big)\Big] + \frac{\partial }{\partial x} \Big[\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big)\Big]\\ &= \frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial x} +\frac{\partial^2 v\,}{\partial\eta\partial \xi} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial x}\\& + \frac{\partial^2 v\,}{\partial\xi\partial \eta} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial x} +\frac{\partial^2 v}{\partial \eta^2}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial x}\\ &=\frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) +2\frac{\partial^2 v\,}{\partial\xi\partial \eta} \big(\xi(x,t),\eta(x,t)\big) +\frac{\partial^2 v}{\partial \eta^2}\big(\xi(x,t),\eta(x,t)\big) \end{aligned}$

이고

$\begin{aligned} \frac{\partial^2 u}{\partial t^2}(x,t) &=-c \frac{\partial }{\partial t} \Big[\frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big)\Big] +c \frac{\partial }{\partial t} \Big[ \frac{\partial v}{\partial\eta} \big(\xi(x,t),\eta(x,t)\big)\Big] \\ &=-c \Big[\frac{\partial^2 v}{\partial \xi^2} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial t} +\frac{\partial^2 v\,}{\partial\eta\partial \xi} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial t}\Big] \\& +c \Big[\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial t} +\frac{\partial^2 v}{\partial\eta^2}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial t}\Big]\\ &=c^2\frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) -2c^2\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big) +c^2\frac{\partial^2 v}{\partial\eta^2}\big(\xi(x,t),\eta(x,t)\big) \end{aligned}$

이므로

$\frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t) =4\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big)$

이다. 따라서

$\begin{aligned} \frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t)=0 \text{ 모든 }(x,t) &\iff 4\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big)=0\text{ 모든 }(x,t)\\ &\iff \frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi,\eta\big)=0\text{ 모든 }(\xi,\eta) \end{aligned}$

(b) 이제 $\frac{\partial^2 v\,}{\partial\xi\partial\eta}\big(\xi,\eta\big) =\frac{\partial}{\partial\xi} \Big[\frac{\partial v}{\partial\eta}\Big]=0$ 이다. 잠시 $\frac{\partial v}{\partial \eta}=w$ 라고 놓자. 등식 $\frac{\partial w}{\partial \xi}(\xi,\eta)=0$ 은, 각각의 고정된 $\eta$ 에 대해 $w(\xi,\eta)$ 가 상수임을 뜻한다. 이 상수의 값은 $\eta$ 에 따라 달라질 수 있다. 즉 어떤 함수 $H$ 에 대해 $\frac{\partial v}{\partial \eta}(\xi,\eta)=w(\xi,\eta) =H(\eta)$ 이다. (확인해 보면 $\frac{\partial }{\partial \xi}H(\eta)=0$ 이다.) 따라서 $v$ 의 $\eta$ 에 대한 도함수는 ( $\xi$ 를 상수로 보면) $H(\eta)$ 이다.

$G(\eta)$ 를 도함수가 $H(\eta)$ 인 임의의 함수 (즉 $H(\eta)$ 의 부정적분)라고 하자. 그러면 $\frac{\partial }{\partial \eta}\big[v(\xi,\eta)-G(\eta)]=H(\eta)-H(\eta)=0$ 이다. 이것은 각각의 고정된 $\xi$ 에 대해 $v(\xi,\eta)-G(\xi,\eta)$ 가 $\eta$ 와 무관한 상수일 때, 그리고 그때에 한해 성립한다. 즉 다음과 필요충분조건이다.

$v(\xi,\eta)-G(\eta)=F(\xi)$

(어떤 함수 $F$ 에 대해). 따라서

$\begin{aligned} \frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t)=0 &\iff \frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi,\eta\big)=0\text{ 모든 }(\xi,\eta)\\ &\iff v(\xi,\eta)=F(\xi)+G(\eta)\text{ 어떤 함수 $F$ 와 $G$ 에 대해}\\ &\iff u(x,t)=v\big(\xi(x,t),\eta(x,t)\big) =F\big(\xi(x,t)\big)+G\big(\eta(x,t)\big)\\ &\phantom{\iff u(x,t)\,}=F(x-ct)+G(x+ct) \end{aligned}$

(c) $F(x-ct)$ 의 해석을 제시하겠다. $G(x+ct)$ 의 경우도 비슷하다. $u(x,t)=F(x-ct)$ 라고 하자. $u(x,t)$ 를 위치 $x$ , 시각 $t$ 에서의 물의 높이라고 생각하자. 임의의 수 $z$ 를 잡자. 시공간에서 $x-ct=z$ 를 만족하는 모든 점 $(x,t)$ 는 $u$ 의 값이 $F(z)$ 로 모두 같다. 그러므로 위치가 $x=z+ct$ 가 되도록 움직이면 (즉 속력 $c$ 로 오른쪽으로 움직이면) 항상 같은 파동의 높이를 보게 된다. 따라서 $F(x-ct)$ 는 속력 $c$ 로 오른쪽으로 진행하는 파동을 나타낸다.

마찬가지로 $G(x+ct)$ 는 속력 $c$ 로 왼쪽으로 진행하는 파동을 나타낸다.

원본 교재 풀이(영어)

Recall that $u(x,t)=v\big(\xi(x,t),\eta(x,t)\big)$ . By the chain rule

$\begin{aligned} \frac{\partial u}{\partial x}(x,t) &= \frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial x} +\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial x} \\ &= \frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) +\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \\ \frac{\partial u}{\partial t}(x,t) &= \frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial t} +\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial t} \\ &= -c\frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big) +c\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big) \end{aligned}$

Again by the chain rule

$\begin{aligned} \frac{\partial^2 u}{\partial x^2}(x,t) &= \frac{\partial }{\partial x} \Big[\frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big)\Big] + \frac{\partial }{\partial x} \Big[\frac{\partial v}{\partial \eta}\big(\xi(x,t),\eta(x,t)\big)\Big]\\ &= \frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial x} +\frac{\partial^2 v\,}{\partial\eta\partial \xi} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial x}\\& + \frac{\partial^2 v\,}{\partial\xi\partial \eta} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial x} +\frac{\partial^2 v}{\partial \eta^2}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial x}\\ &=\frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) +2\frac{\partial^2 v\,}{\partial\xi\partial \eta} \big(\xi(x,t),\eta(x,t)\big) +\frac{\partial^2 v}{\partial \eta^2}\big(\xi(x,t),\eta(x,t)\big) \end{aligned}$

and

$\begin{aligned} \frac{\partial^2 u}{\partial t^2}(x,t) &=-c \frac{\partial }{\partial t} \Big[\frac{\partial v}{\partial \xi}\big(\xi(x,t),\eta(x,t)\big)\Big] +c \frac{\partial }{\partial t} \Big[ \frac{\partial v}{\partial\eta} \big(\xi(x,t),\eta(x,t)\big)\Big] \\ &=-c \Big[\frac{\partial^2 v}{\partial \xi^2} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial t} +\frac{\partial^2 v\,}{\partial\eta\partial \xi} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial t}\Big] \\& +c \Big[\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big) \frac{\partial \xi}{\partial t} +\frac{\partial^2 v}{\partial\eta^2}\big(\xi(x,t),\eta(x,t)\big) \frac{\partial \eta}{\partial t}\Big]\\ &=c^2\frac{\partial^2 v}{\partial \xi^2}\big(\xi(x,t),\eta(x,t)\big) -2c^2\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big) +c^2\frac{\partial^2 v}{\partial\eta^2}\big(\xi(x,t),\eta(x,t)\big) \end{aligned}$

so that

$\frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t) =4\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big)$

Hence

$\begin{aligned} \frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t)=0 \text{ for all }(x,t) &\iff 4\frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi(x,t),\eta(x,t)\big)=0\text{ for all }(x,t)\\ &\iff \frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi,\eta\big)=0\text{ for all }(\xi,\eta) \end{aligned}$

(b) Now $\frac{\partial^2 v\,}{\partial\xi\partial\eta}\big(\xi,\eta\big) =\frac{\partial}{\partial\xi} \Big[\frac{\partial v}{\partial\eta}\Big]=0$ . Temporarily rename $\frac{\partial v}{\partial \eta}=w$ . The equation $\frac{\partial w}{\partial \xi}(\xi,\eta)=0$ says that, for each fixed $\eta$ , $w(\xi,\eta)$ is a constant. The value of the constant may depend on $\eta$ . That is, $\frac{\partial v}{\partial \eta}(\xi,\eta)=w(\xi,\eta) =H(\eta)$ , for some function $H$ . (As a check, observe that $\frac{\partial }{\partial \xi}H(\eta)=0$ .) So the derivative of $v$ with respect to $\eta$ , (viewing $\xi$ as a constant) is $H(\eta)$ .

Let $G(\eta)$ be any function whose derivative is $H(\eta)$ (i.e. an indefinite integral of $H(\eta)$ ). Then $\frac{\partial }{\partial \eta}\big[v(\xi,\eta)-G(\eta)]=H(\eta)-H(\eta)=0$ . This is the case if and only if, for each fixed $\xi$ , $v(\xi,\eta)-G(\xi,\eta)$ is a constant, independent of $\eta$ . That is, if and only if

$v(\xi,\eta)-G(\eta)=F(\xi)$

for some function $F$ . Hence

$\begin{aligned} \frac{\partial^2 u}{\partial x^2}(x,t) -\frac{1}{c^2}\frac{\partial^2 u}{\partial t^2}(x,t)=0 &\iff \frac{\partial^2 v\,}{\partial\xi\partial\eta} \big(\xi,\eta\big)=0\text{ for all }(\xi,\eta)\\ &\iff v(\xi,\eta)=F(\xi)+G(\eta)\text{ for some functions $F$ and $G$}\\ &\iff u(x,t)=v\big(\xi(x,t),\eta(x,t)\big) =F\big(\xi(x,t)\big)+G\big(\eta(x,t)\big)\\ &\phantom{\iff u(x,t)\,}=F(x-ct)+G(x+ct) \end{aligned}$

(c) We'll give the interpretation of $F(x-ct)$ . The case $G(x+ct)$ is similar. Suppose that $u(x,t)=F(x-ct)$ . Think of $u(x,t)$ as the height of water at position $x$ and time $t$ . Pick any number $z$ . All points $(x,t)$ in space time for which $x-ct=z$ have the same value of $u$ , namely $F(z)$ . So if you move so that your position is $x=z+ct$ (i.e. you move the right with speed $c$ ) you always see the same wave height. Thus $F(x-ct)$ represents a wave moving to the right with speed $c$ .

Similarly, $G(x+ct)$ represents a wave moving to the left with speed $c$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 17번

# 다변수 연쇄법칙증명·논술도전자동 분류

636.CLP-3 · 2.5절 · 1번대학 수준우리말 번역

곡면 $x^2+y^2+(z-1)^2=1$ 과 $x^2+y^2+(z+1)^2=1$ 이 $(0,0,0)$ 에서 서로 접한다고 말하는 것이 타당한가요?

영어 원문 보기

Is it reasonable to say that the surfaces $x^2+y^2+(z-1)^2=1$ and $x^2+y^2+(z+1)^2=1$ are tangent to each other at $(0,0,0)$ ?

정답·풀이 보기

$F(x,y,z) = x^2+y^2+(z-1)^2-1$ , $G(x,y,z) = x^2+y^2+(z+1)^2-1$ 이라 하자. 곡면 $F(x,y,z)=0$ 을 $S_1$ , 곡면 $G(x,y,z)=0$ 을 $S_2$ 라 하자. 먼저 $F(0,0,0)=G(0,0,0)=0$ 이므로 점 $(0,0,0)$ 은 $S_1$ 과 $S_2$ 위에 모두 있다. $F$ 와 $G$ 의 그래디언트는 다음과 같다.

$\begin{aligned} { \boldsymbol{\nabla} } F(x,y,z) &=\left<\frac{\partial F}{\partial x}(x,y,z)\,,\, \frac{\partial F}{\partial y}(x,y,z)\,,\, \frac{\partial F}{\partial z}(x,y,z)\right> =\left< 2x\,,\,2y\,,\,2(z-1)\right> \\ { \boldsymbol{\nabla} } G(x,y,z) &=\left<\frac{\partial G}{\partial x}(x,y,z)\,,\, \frac{\partial G}{\partial y}(x,y,z)\,,\, \frac{\partial G}{\partial z}(x,y,z)\right> =\left< 2x\,,\,2y\,,\,2(z+1)\right> \end{aligned}$

특히,

${ \boldsymbol{\nabla} } F(0,0,0)=\left< 0,0,-2\right>\qquad { \boldsymbol{\nabla} } G(0,0,0)=\left< 0,0,2\right>$

이므로 벡터 $\hat{\mathbf{k}}=-\frac{1}{2}{ \boldsymbol{\nabla} } F(0,0,0) =\frac{1}{2}{ \boldsymbol{\nabla} } G(0,0,0)$ 은 $(0,0,0)$ 에서 두 곡면에 모두 수직이다. 따라서 $(0,0,0)$ 에서 $S_1$ 과 $S_2$ 의 접평면은 모두

$\hat{\mathbf{k}}\cdot\left< x-0,y-0,z-0\right>=0\qquad\text{또는}\qquad z=0$

이다.

평면 $z=0$ 을 $P$ 라 하자. 그러면 $S_1$ 은 $(0,0,0)$ 에서 $P$ 에 접하고, $P$ 는 $(0,0,0)$ 에서 $S_2$ 에 접한다. 따라서 $S_1$ 과 $S_2$ 가 $(0,0,0)$ 에서 접한다고 말하는 것은 타당하다.

원본 교재 풀이(영어)

Write $F(x,y,z) = x^2+y^2+(z-1)^2-1$ and $G(x,y,z) = x^2+y^2+(z+1)^2-1$ . Let $S_1$ denote the surface $F(x,y,z)=0$ and $S_2$ denote the surface $G(x,y,z)=0$ . First note that $F(0,0,0)=G(0,0,0)=0$ so that the point $(0,0,0)$ lies on both $S_1$ and $S_2$ . The gradients of $F$ and $G$ are

$\begin{aligned} { \boldsymbol{\nabla} } F(x,y,z) &=\left<\frac{\partial F}{\partial x}(x,y,z)\,,\, \frac{\partial F}{\partial y}(x,y,z)\,,\, \frac{\partial F}{\partial z}(x,y,z)\right> =\left< 2x\,,\,2y\,,\,2(z-1)\right> \\ { \boldsymbol{\nabla} } G(x,y,z) &=\left<\frac{\partial G}{\partial x}(x,y,z)\,,\, \frac{\partial G}{\partial y}(x,y,z)\,,\, \frac{\partial G}{\partial z}(x,y,z)\right> =\left< 2x\,,\,2y\,,\,2(z+1)\right> \end{aligned}$

In particular,

${ \boldsymbol{\nabla} } F(0,0,0)=\left< 0,0,-2\right>\qquad { \boldsymbol{\nabla} } G(0,0,0)=\left< 0,0,2\right>$

so that the vector $\hat{\mathbf{k}}=-\frac{1}{2}{ \boldsymbol{\nabla} } F(0,0,0) =\frac{1}{2}{ \boldsymbol{\nabla} } G(0,0,0)$ is normal to both surfaces at $(0,0,0)$ . So the tangent plane to both $S_1$ and $S_2$ at $(0,0,0)$ is

$\hat{\mathbf{k}}\cdot\left< x-0,y-0,z-0\right>=0\qquad\text{or}\qquad z=0$

Denote by $P$ the plane $z=0$ . Thus $S_1$ is tangent to $P$ at $(0,0,0)$ and $P$ is tangent to $S_2$ at $(0,0,0)$ . So it is reasonable to say that $S_1$ and $S_2$ are tangent at $(0,0,0)$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.5절 · 2번

# 접평면과 선형근사계산기본자동 분류

637.CLP-3 · 2.5절 · 2번대학 수준우리말 번역

점 $\mathbf{r}_0= (x_0,y_0,z_0)$ 이 곡면 $G(x,y,z)=0$ 위에 있다고 하자. ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)\ne\mathbf{0}$ 이라고 가정하자. 매개변수로 나타낸 곡선 $\mathbf{r}(t)=\big(x(t),y(t),z(t)\big)$ 가 이 곡면에 포함되어 있고 $\mathbf{r}(t_0)=\mathbf{r}_0$ 이라고 하자. 이때 $\mathbf{r}_0$ 에서 곡선의 접선이 $\mathbf{r}_0$ 에서 $G=0$ 의 접평면 위에 놓임을 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 $S$ 라 하고, 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 매개변수 방정식은

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

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

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

곧

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

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

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 어떤 $t$ 가 존재하여

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

이다. 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

$x-x_0 = t x'(t_0)\qquad y-y_0 = t y'(t_0)\qquad z-z_0 = t z'(t_0) \tag{$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-3 · 2.6절 · 10번

# 접평면과 선형근사증명·논술기본자동 분류

638.CLP-3 · 2.5절 · 7번대학 수준우리말 번역

지정된 점에서 지정된 함수의 그래프에 대한 접평면의 방정식과 법선의 방정식을 구하세요.

(a) $f(x,y)=x^2-y^2$ , 점 $(-2,1)$ (b) $f(x,y)=e^{xy}$ , 점 $(2,0)$

영어 원문 보기

Find the equations of the tangent plane and the normal line to the graph of the specified function at the specified point.

(a) $f(x,y)=x^2-y^2$ at $(-2,1)$ (b) $f(x,y)=e^{xy}$ at $(2,0)$

정답·풀이 보기

(a) 주어진 그래프는 $z=f(x,y)=x^2-y^2$ 또는 $F(x,y,z)=x^2-y^2-z=0$ 이다. $f(-2,1)=3$ 임을 알 수 있다. 벡터

$\begin{aligned} { \boldsymbol{\nabla} } F(-2,1,3) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< 2x,-2y,-1\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< -4,-2,-1\right> \end{aligned}$

는 $(-2,1,3)$ 에서 그래프의 법선 벡터이다. 따라서 접평면은

$-4(x+2)-2(y-1)-(z-3)=0\text{ 또는 } 4x+2y+z=-3$

이고, 법선은

$\left< x,y,z\right>=\left< -2,1,3\right>+t\left< 4,2,1\right>$

이다.

(b) 주어진 그래프는 $z=f(x,y)=e^{xy}$ 또는 $F(x,y,z)=e^{xy}-z=0$ 이다. $f(2,0)=1$ 임을 알 수 있다. 벡터

$\begin{aligned} { \boldsymbol{\nabla} } F(2,0,1) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< ye^{xy},xe^{xy},-1\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< 0,2,-1\right> \end{aligned}$

는 $(2,0,1)$ 에서 그래프의 법선 벡터이다. 따라서 접평면은

$0(x-2)+2(y-0)-(z-1)=0\text{ 또는 } 2y-z=-1$

이고, 법선은

$\left< x,y,z\right>=\left< 2,0,1\right>+t\left< 0,2,-1\right>$

이다.

원본 교재 풀이(영어)

(a) The specified graph is $z=f(x,y)=x^2-y^2$ or $F(x,y,z)=x^2-y^2-z=0$ . Observe that $f(-2,1)=3$ . The vector

$\begin{aligned} { \boldsymbol{\nabla} } F(-2,1,3) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< 2x,-2y,-1\right>\Big|_{(x,y,z)=(-2,1,3)} \\ &= \left< -4,-2,-1\right> \end{aligned}$

is a normal vector to the graph at $(-2,1,3)$ . So the tangent plane is

$-4(x+2)-2(y-1)-(z-3)=0\text{ or } 4x+2y+z=-3$

and the normal line is

$\left< x,y,z\right>=\left< -2,1,3\right>+t\left< 4,2,1\right>$

(b) The specified graph is $z=f(x,y)=e^{xy}$ or $F(x,y,z)=e^{xy}-z=0$ . Observe that $f(2,0)=1$ . The vector

$\begin{aligned} { \boldsymbol{\nabla} } F(2,0,1) &= \left< F_x(x,y,z),F_y(x,y,z),F_z(x,y,z)\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< ye^{xy},xe^{xy},-1\right>\Big|_{(x,y,z)=(2,0,1)} \\ &= \left< 0,2,-1\right> \end{aligned}$

is a normal vector to the graph at $(2,0,1)$ . So the tangent plane is

$0(x-2)+2(y-0)-(z-1)=0\text{ or } 2y-z=-1$

and the normal line is

$\left< x,y,z\right>=\left< 2,0,1\right>+t\left< 0,2,-1\right>$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.5절 · 1번

# 접평면과 선형근사그래프·표 해석보통자동 분류

639.CLP-3 · 2.5절 · 11번대학 수준우리말 번역

곡면 $x^2 + 9y^2 + 4z^2 = 17$ 위의 점 중에서 접평면이 평면 $x - 8z = 0$ 과 평행한 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 $\left< 2x\,,\, 18y\,,\,8z\right>$ 이다. 평면 $x - 8z = 0$ 의 법선벡터는 $\left< 1\,,\,0\,,\,-8\right>$ 이다. 따라서 $\left< 2x\,,\, 18y\,,\,8z\right>$ 가 $\left< 1\,,\,0\,,\,-8\right>$ 과 평행해야 하며, 즉 $\left< 1\,,\,0\,,\,-8\right>$ 의 0이 아닌 상수배여야 한다. 이는 $x\ne 0$ 이고 $y=0$ , $z=-2x$ 일 때마다 성립한다. 또한 $(x,y,z)$ 가 곡면 $x^2 + 9y^2 + 4z^2 = 17$ 위에 있어야 한다. 따라서 $y=0$ , $z=-2x$ 이고

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

이어야 한다.

따라서 조건을 만족하는 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 $\left< 2x\,,\, 18y\,,\,8z\right>$ . A normal vector to the plane $x - 8z = 0$ is $\left< 1\,,\,0\,,\,-8\right>$ . So we want $\left< 2x\,,\, 18y\,,\,8z\right>$ to be parallel to $\left< 1\,,\,0\,,\,-8\right>$ , i.e. to be a nonzero constant times $\left< 1\,,\,0\,,\,-8\right>$ . 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 풀기APEX · 12.7절 · 2번

# 접평면과 선형근사계산보통자동 분류

640.CLP-3 · 2.5절 · 13번대학 수준우리말 번역

쌍곡면 $z^2=4x^2+y^2-1$ 위의 점 중에서 접평면이 평면 $2x-y+z=0$ 과 평행한 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 점으로서 접평면이 평면 $2x-y+z=0$ 과 평행한 점이라 하자. 평면 $2x-y+z=0$ 의 법선벡터는 $\left< 2,-1,1\right>$ 이다. 쌍곡면은 $G(x,y,z)=4x^2+y^2-z^2-1$ 이고 ${ \boldsymbol{\nabla} } G(x,y,z) = \left< 8x,2y,-2z\right>$ 이므로, $(x_0,y_0,z_0)$ 에서 쌍곡면의 법선벡터는 ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)=\left< 8x_0,2y_0,-2z_0\right>$ 이다. 따라서 $(x_0,y_0,z_0)$ 이 주어진 조건을 만족할 필요충분조건은 다음을 만족하는 0이 아닌 $t$ 가 존재하는 것이다.

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

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 $\left< 2,-1,1\right>$ . Because the hyperboloid is $G(x,y,z)=4x^2+y^2-z^2-1$ and ${ \boldsymbol{\nabla} } G(x,y,z) = \left< 8x,2y,-2z\right>$ , a normal vector to the hyperboloid at $(x_0,y_0,z_0)$ is ${ \boldsymbol{\nabla} } G(x_0,y_0,z_0)=\left< 8x_0,2y_0,-2z_0\right>$ . So $(x_0,y_0,z_0)$ satisfies the required conditions if and only if there is a nonzero $t$ obeying

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

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.5절 · 11번

# 접평면과 선형근사계산보통자동 분류

다변수 미분 전체 문제