다변수 연쇄법칙 연습문제 — CLP-3 · 2.4절 · 24번

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

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

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),

한 번 더 — 비슷한 문제CLP-3 · 2.4절 · 15번우리말 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) $\frac{\partial^2 z}{\partial t^2}$

다 풀었으면 정답 확인

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.

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