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문제 은행 · 미적분 · 대학 미적분 · 다변수 미분

다변수 연쇄법칙

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

CLP-3 · 2.4절 · 9번대학 수준우리말 번역

f(x,y)f(x,y) 가 라플라스 방정식 ∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0 을 만족시킨다고 가정하세요. 합성함수 g(s,t)=f(s−t,s+t)g(s,t) = f (s - t, s + t) 에 대해서도 같은 식이 성립함을 보이세요. 즉, ∂2g∂s2+∂2g∂t2=0\frac{\partial^2 g}{\partial s^2}+\frac{\partial^2 g}{\partial t^2}=0 임을 보이세요. f(x,y)f(x,y) 는 매끄러운 함수여서 연쇄 법칙과 혼합 편도함수의 같음에 관한 클레로의 정리를 적용할 수 있다고 가정해도 됩니다.

영어 원문 보기

Assume that f(x,y)f(x,y) satisfies Laplace's equation ∂2f∂x2+∂2f∂y2=0\frac{\partial^2 f}{\partial x^2}+\frac{\partial^2 f}{\partial y^2}=0. Show that this is also the case for the composite function g(s,t)=f(s−t,s+t)g(s,t) = f (s - t, s + t). That is, show that ∂2g∂s2+∂2g∂t2=0\frac{\partial^2 g}{\partial s^2}+\frac{\partial^2 g}{\partial t^2}=0. You may assume that f(x,y)f(x,y) is a smooth function so that the Chain Rule and Clairaut's Theorem on the equality of the mixed partial derivatives apply.

정답·풀이 보기

연쇄 법칙에 의해,

∂g∂s(s,t)=∂∂sf(s−t,s+t)=∂f∂x(s−t , s+t)∂∂s(s−t)+∂f∂y(s−t , s+t)∂∂s(s+t)=∂f∂x(s−t , s+t)+∂f∂y(s−t , s+t)∂2g∂s2(s,t)=∂∂s[∂f∂x(s−t , s+t)]+∂∂s[∂f∂y(s−t , s+t)]=∂2f∂x2(s−t , s+t)+∂2f∂y∂x(s−t , s+t)+∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)={∂2f∂x2(s−t , s+t)+2∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)}\begin{aligned} \frac{\partial g}{\partial s}(s,t) &=\frac{\partial }{\partial s} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s+t\big) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial s^2}(s,t) &= \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big) \\ & + \frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) + 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}

이고

∂g∂t(s,t)=∂∂tf(s−t,s+t)=∂f∂x(s−t , s+t)∂∂t(s−t)+∂f∂y(s−t , s+t)∂∂t(s+t)=−∂f∂x(s−t , s+t)+∂f∂y(s−t , s+t)∂2g∂t2(s,t)=−∂∂t[∂f∂x(s−t , s+t)]+∂∂t[∂f∂y(s−t , s+t)]=−[−∂2f∂x2(s−t , s+t)+∂2f∂y∂x(s−t , s+t)]+[−∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)]={∂2f∂x2(s−t , s+t)−2∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)}\begin{aligned} \frac{\partial g}{\partial t}(s,t)&=\frac{\partial }{\partial t} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s+t\big) \\ &= -\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial t^2}(s,t) &=- \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= - \Big[-\frac{\partial^2 f}{\partial x^2} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big)\Big] \\ & + \Big[-\frac{\partial^2 f}{\partial x\partial y} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \Big] \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) - 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}

이다. 인수를 생략하면

∂2g∂s2+∂2g∂t2={∂2f∂x2+2∂2f∂x∂y+∂2f∂y2}+{∂2f∂x2−2∂2f∂x∂y+∂2f∂y2}=2[∂2f∂x2+∂2f∂y2]=0\begin{aligned} \frac{\partial^2 g}{\partial s^2} + \frac{\partial^2 g}{\partial t^2} &=\left\{\frac{\partial^2 f}{\partial x^2} + 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} +\left\{\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} \\ &=2\left[\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}\right] \\ &=0 \end{aligned}

이므로 원하는 결과가 얻어진다.

원본 교재 풀이(영어)

By the chain rule,

∂g∂s(s,t)=∂∂sf(s−t,s+t)=∂f∂x(s−t , s+t)∂∂s(s−t)+∂f∂y(s−t , s+t)∂∂s(s+t)=∂f∂x(s−t , s+t)+∂f∂y(s−t , s+t)∂2g∂s2(s,t)=∂∂s[∂f∂x(s−t , s+t)]+∂∂s[∂f∂y(s−t , s+t)]=∂2f∂x2(s−t , s+t)+∂2f∂y∂x(s−t , s+t)+∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)={∂2f∂x2(s−t , s+t)+2∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)}\begin{aligned} \frac{\partial g}{\partial s}(s,t) &=\frac{\partial }{\partial s} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial s}\big(s+t\big) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial s^2}(s,t) &= \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big) \\ & + \frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) + 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}

and

∂g∂t(s,t)=∂∂tf(s−t,s+t)=∂f∂x(s−t , s+t)∂∂t(s−t)+∂f∂y(s−t , s+t)∂∂t(s+t)=−∂f∂x(s−t , s+t)+∂f∂y(s−t , s+t)∂2g∂t2(s,t)=−∂∂t[∂f∂x(s−t , s+t)]+∂∂t[∂f∂y(s−t , s+t)]=−[−∂2f∂x2(s−t , s+t)+∂2f∂y∂x(s−t , s+t)]+[−∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)]={∂2f∂x2(s−t , s+t)−2∂2f∂x∂y(s−t , s+t)+∂2f∂y2(s−t , s+t)}\begin{aligned} \frac{\partial g}{\partial t}(s,t)&=\frac{\partial }{\partial t} f (s - t, s + t) \\ &= \frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s-t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\frac{\partial }{\partial t}\big(s+t\big) \\ &= -\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big) +\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big) \\ \frac{\partial^2 g}{\partial t^2}(s,t) &=- \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(s-t\,,\,s+t\big)\right] + \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(s-t\,,\,s+t\big)\right] \\ &= - \Big[-\frac{\partial^2 f}{\partial x^2} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y\partial x}\big(s-t\,,\,s+t\big)\Big] \\ & + \Big[-\frac{\partial^2 f}{\partial x\partial y} \big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big) \Big] \\ &=\left\{\frac{\partial^2 f}{\partial x^2}\big(s-t\,,\,s+t\big) - 2\frac{\partial^2 f}{\partial x\partial y}\big(s-t\,,\,s+t\big) +\frac{\partial^2 f}{\partial y^2}\big(s-t\,,\,s+t\big)\right\} \end{aligned}

Suppressing the arguments

∂2g∂s2+∂2g∂t2={∂2f∂x2+2∂2f∂x∂y+∂2f∂y2}+{∂2f∂x2−2∂2f∂x∂y+∂2f∂y2}=2[∂2f∂x2+∂2f∂y2]=0\begin{aligned} \frac{\partial^2 g}{\partial s^2} + \frac{\partial^2 g}{\partial t^2} &=\left\{\frac{\partial^2 f}{\partial x^2} + 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} +\left\{\frac{\partial^2 f}{\partial x^2} - 2\frac{\partial^2 f}{\partial x\partial y} +\frac{\partial^2 f}{\partial y^2}\right\} \\ &=2\left[\frac{\partial^2 f}{\partial x^2} +\frac{\partial^2 f}{\partial y^2}\right] \\ &=0 \end{aligned}

as desired.

한 번 더 — 비슷한 문제OpenStax 3권 · 4.5절 · 37번우리말 번역

원기둥의 부피는 V(x,y)=πx2yV(x,y)=\pi {x}^{2}y 로 주어집니다. 여기서 xx 는 원기둥의 반지름이고 y 는 원기둥의 높이입니다. xx 와 yy 가 tt 의 함수로서 x=12tx=\frac{1}{2}t, y=13ty=\frac{1}{3}t 이고, 따라서 x 와 yx\,\text{와}\,y 가 모두 시간에 따라 증가한다고 합니다. x=2x=2 이고 y=43y=\frac{4}{3} 일 때, 부피는 얼마나 빠르게 증가하는지 구하세요.

다 풀었으면 정답 확인

V′=4πV'=4\pi

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