다변수 연쇄법칙 연습문제

1.CLP-3 · 2.4절 · 8번대학 수준우리말 번역

$g(s,t)=f(2s+3t,3s-2t)$ 의 모든 이계도함수를 구하세요. $f(x,y)$ 는 매끄러운 함수이므로 연쇄 법칙과 혼합 편도함수의 같음에 관한 클레로의 정리가 적용된다고 가정해도 됩니다.

영어 원문 보기

Find all second order derivatives of $g(s,t)=f(2s+3t,3s-2t)$ . You may assume that $f(x,y)$ 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

주어진 함수는

$g(s,t)=f(2s+3t,3s-2t)$

이다.

일계도함수는

$\begin{aligned} g_s(s,t)&=2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\\ g_t(s,t)&=3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t) \end{aligned}$

이다.

이계도함수는

$\begin{aligned} g_{ss}(s,t) &=\frac{\partial }{\partial s}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(2f_{11}+3f_{12}\Big) +3\Big(2f_{21}+3f_{22}\Big)\\ &=4f_{11}+6f_{12}+6f_{21}+9f_{22}\\ &=4f_{11}(2s+3t,3s-2t)+12f_{12}(2s+3t,3s-2t)+9f_{22}(2s+3t,3s-2t)\\ g_{st}(s,t)&=\frac{\partial }{\partial t}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(3f_{11}-2f_{12}\Big) +3\Big(3f_{21}-2f_{22}\Big)\\ &=6f_{11}(2s+3t,3s-2t)+5f_{12}(2s+3t,3s-2t)-6f_{22}(2s+3t,3s-2t)\\ g_{tt}(s,t)&=\frac{\partial }{\partial t}\Big(3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t)\Big)\\ &=3\Big(3f_{11}-2f_{12}\Big) -2\Big(3f_{21}-2f_{22}\Big)\\ &=9f_{11}(2s+3t,3s-2t)-12f_{12}(2s+3t,3s-2t)+4f_{22}(2s+3t,3s-2t) \end{aligned}$

이다.

여기서 $f_1$ 은 $f$ 를 첫 번째 변수에 대해 편미분한 편도함수를 뜻하고, $f_{12}$ 는 $f$ 를 먼저 첫 번째 변수에 대해 편미분한 다음 그 결과를 두 번째 변수에 대해 편미분한 것이며, 나머지도 같은 방식이다.

원본 교재 풀이(영어)

The given function is

$g(s,t)=f(2s+3t,3s-2t)$

The first order derivatives are

$\begin{aligned} g_s(s,t)&=2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\\ g_t(s,t)&=3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t) \end{aligned}$

The second order derivatives are

$\begin{aligned} g_{ss}(s,t) &=\frac{\partial }{\partial s}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(2f_{11}+3f_{12}\Big) +3\Big(2f_{21}+3f_{22}\Big)\\ &=4f_{11}+6f_{12}+6f_{21}+9f_{22}\\ &=4f_{11}(2s+3t,3s-2t)+12f_{12}(2s+3t,3s-2t)+9f_{22}(2s+3t,3s-2t)\\ g_{st}(s,t)&=\frac{\partial }{\partial t}\Big(2f_1(2s+3t,3s-2t)+3f_2(2s+3t,3s-2t)\Big)\\ &=2\Big(3f_{11}-2f_{12}\Big) +3\Big(3f_{21}-2f_{22}\Big)\\ &=6f_{11}(2s+3t,3s-2t)+5f_{12}(2s+3t,3s-2t)-6f_{22}(2s+3t,3s-2t)\\ g_{tt}(s,t)&=\frac{\partial }{\partial t}\Big(3f_1(2s+3t,3s-2t)-2f_2(2s+3t,3s-2t)\Big)\\ &=3\Big(3f_{11}-2f_{12}\Big) -2\Big(3f_{21}-2f_{22}\Big)\\ &=9f_{11}(2s+3t,3s-2t)-12f_{12}(2s+3t,3s-2t)+4f_{22}(2s+3t,3s-2t) \end{aligned}$

Here $f_1$ denotes the partial derivative of $f$ with respect to its first argument, $f_{12}$ is the result of first taking one partial derivative of $f$ with respect to its first argument and then taking a partial derivative with respect to its second argument, and so on.

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.5절 · 39번

# 다변수 연쇄법칙활용보통자동 분류

2.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

3.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