다변수 미분 전체 연습문제

21.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

22.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

23.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

24.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

25.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

26.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

27.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

28.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

29.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

30.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

31.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

32.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

33.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

34.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

35.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

36.CLP-3 · 2.5절 · 14번대학 수준우리말 번역

두 곡면 $\ z^2=4x^2+9y^2\$ 과 $\ 6x+3y+2z=5\$ 의 교선에 접하고 길이가 $\sqrt{3}$ 인 벡터를 점 $\ (2,1,-5)$ 에서 구하세요.

영어 원문 보기

Find a vector of length $\sqrt{3}$ which is tangent to the curve of intersection of the surfaces $\ z^2=4x^2+9y^2\$ and $\ 6x+3y+2z=5\$ at $\ (2,1,-5)$ .

정답·풀이 보기

점 $(2,1,-5)$ 에서 곡면 $F(x,y,z)=4x^2+9y^2-z^2=0$ 에 수직인 벡터 하나는 다음과 같다.

$\begin{aligned} { \boldsymbol{\nabla} } F(2,1,-5) = \left< 8x,18y,-2z\right>\Big|_{(2,1,-5)}=\left< 16,18,10\right> \end{aligned}$

점 $(2,1,-5)$ 에서 곡면 $G(x,y,z)=6x+3y+2z=5$ 에 수직인 벡터 하나는 다음과 같다.

$\begin{aligned} { \boldsymbol{\nabla} } G(2,1,-5) = \left< 6,3,2\right> \end{aligned}$

이제

곡선은 곡면 $z^2=4x^2+9y^2$ 위에 있다. 따라서 점 $(2,1,-5)$ 에서 곡선의 접벡터는 법선벡터 $\frac{1}{2}\left< 16,18,10\right>=\left< 8,9,5\right>$ 에 수직이다.
곡선은 곡면 $6x+3y+2z=5$ 위에도 있다. 따라서 점 $(2,1,-5)$ 에서 곡선의 접벡터는 법선벡터 $\left< 6,3,2\right>$ 에도 수직이다.
따라서 점 $(2,1,-5)$ 에서 곡선의 접벡터는 다음 벡터와 평행하다.

$\begin{aligned} \left< 8,9,5\right>\times \left< 6,3,2\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 8 & 9 & 5 \\ 6 & 3 & 2 \end{matrix}\right] =\left< 3,14,-30\right> \end{aligned}$

구하는 벡터는 다음과 같다.

$\pm\sqrt{3}\frac{\left< 3,14,-30\right>}{|\left< 3,14,-30\right>|} =\pm\sqrt{\frac{3}{1105}}\left< 3,14,-30\right>$

원본 교재 풀이(영어)

One vector normal to the surface $F(x,y,z)=4x^2+9y^2-z^2=0$ at $(2,1,-5)$ is

$\begin{aligned} { \boldsymbol{\nabla} } F(2,1,-5) = \left< 8x,18y,-2z\right>\Big|_{(2,1,-5)}=\left< 16,18,10\right> \end{aligned}$

One vector normal to the surface $G(x,y,z)=6x+3y+2z=5$ at $(2,1,-5)$ is

$\begin{aligned} { \boldsymbol{\nabla} } G(2,1,-5) = \left< 6,3,2\right> \end{aligned}$

Now

The curve lies in the surface $z^2=4x^2+9y^2$ . So the tangent vector to the curve at $(2,1,-5)$ is perpendicular to the normal vector $\frac{1}{2}\left< 16,18,10\right>=\left< 8,9,5\right>$ .
The curve also lies in the surface $6x+3y+2z=5$ . So the tangent vector to the curve at $(2,1,-5)$ is also perpendicular to the normal vector $\left< 6,3,2\right>$ .
So the tangent vector to the curve at $(2,1,-5)$ is parallel to

$\begin{aligned} \left< 8,9,5\right>\times \left< 6,3,2\right> =\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ 8 & 9 & 5 \\ 6 & 3 & 2 \end{matrix}\right] =\left< 3,14,-30\right> \end{aligned}$

The desired vectors are

$\pm\sqrt{3}\frac{\left< 3,14,-30\right>}{|\left< 3,14,-30\right>|} =\pm\sqrt{\frac{3}{1105}}\left< 3,14,-30\right>$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.5절 · 15번

# 접평면과 선형근사계산보통자동 분류

37.CLP-3 · 2.6절 · 3번대학 수준우리말 번역

계산기나 컴퓨터를 사용하지 않고 $f(x,y)=\sin(\pi xy+\ln y)$ 의 $(0.01,1.05)$ 에서의 근삿값을 구하세요.

영어 원문 보기

Find an approximate value for $f(x,y)=\sin(\pi xy+\ln y)$ at $(0.01,1.05)$ without using a calculator or computer.

답 쓰고 채점

정답·풀이 보기

선형 근사 $f(0.01,1.05)\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05)$ 를 적용한다. 이때

$\begin{aligned} f(x,y)&=\sin(\pi xy+\ln y) & f(0,1)&=\sin 0=0\\ f_x(x,y)&=\pi y\cos(\pi xy+\ln y) & f_x(0,1)&=\pi\cos 0=\pi\\ f_y(x,y)&=\left(\pi x+\frac{1}{y}\right)\cos(\pi xy+\ln y) & f_y(0,1)&=\cos 0=1 \end{aligned}$

이다. 따라서

$\begin{aligned} f(0.01,1.05)&\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05) =0+\pi(0.01)+1(0.05) \\ &=0.01\,\pi + 0.05 \approx 0.0814 \end{aligned}$

이다.

원본 교재 풀이(영어)

Apply the linear approximation $f(0.01,1.05)\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05)$ , with

$\begin{aligned} f(x,y)&=\sin(\pi xy+\ln y) & f(0,1)&=\sin 0=0\\ f_x(x,y)&=\pi y\cos(\pi xy+\ln y) & f_x(0,1)&=\pi\cos 0=\pi\\ f_y(x,y)&=\left(\pi x+\frac{1}{y}\right)\cos(\pi xy+\ln y) & f_y(0,1)&=\cos 0=1 \end{aligned}$

This gives

$\begin{aligned} f(0.01,1.05)&\approx f(0,1)+f_x(0,1)(0.01)+f_y(0,1)(0.05) =0+\pi(0.01)+1(0.05) \\ &=0.01\,\pi + 0.05 \approx 0.0814 \end{aligned}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.4절 · 39번

# 접평면과 선형근사계산보통자동 분류

38.CLP-3 · 2.6절 · 4번대학 수준우리말 번역

$\displaystyle f(x,y)=\frac{x^2y}{x^4+2y^2}$ 라 하자. 계산기나 컴퓨터를 사용하지 않고 $f (-0.9\,,\, 1.1)$ 의 근삿값을 구하세요.

영어 원문 보기

Let $\displaystyle f(x,y)=\frac{x^2y}{x^4+2y^2}$ . Find an approximate value for $f (-0.9\,,\, 1.1)$ without using a calculator or computer.

답 쓰고 채점

정답·풀이 보기

$(x,y)=(-1,1)$ 에서 $f(x,y)$ 의 1계 미분계수가 필요하다. 그래서 먼저 이를 구한다.

$\begin{aligned} f_x(x,y)&=\frac{2xy}{x^4+2y^2}-\frac{x^2y(4x^3)}{{(x^4+2y^2)}^2}\qquad & f_x(-1,1)&=-\frac{2}{3} +\frac{4}{3^2}=-\frac{2}{9} \\ f_y(x,y)&=\frac{x^2}{x^4+2y^2}-\frac{x^2y(4y)}{{(x^4+2y^2)}^2}\qquad & f_y(-1,1)&=\frac{1}{3} -\frac{4}{3^2}=-\frac{1}{9} \end{aligned}$

$(-1,1)$ 근처에서 $f(x,y)$ 의 선형 근사는 다음과 같다.

$\begin{aligned} f(x,y)\approx=f(-1,1) + f_x(-1,1)\,(x+1) + f_y(-1,1)\,(y-1) =\frac{1}{3} -\frac{2}{9}\,(x+1) -\frac{1}{9}\,(y-1) \end{aligned}$

특히

$\begin{aligned} f(-0.9,1.1)\approx \frac{1}{3} -\frac{2}{9}\,(0.1) -\frac{1}{9}\,(0.1) =\frac{27}{90}=0.3 \end{aligned}$

원본 교재 풀이(영어)

We are going to need the first order derivatives of $f(x,y)$ at $(x,y)=(-1,1)$ . So we find them first.

$\begin{aligned} f_x(x,y)&=\frac{2xy}{x^4+2y^2}-\frac{x^2y(4x^3)}{{(x^4+2y^2)}^2}\qquad & f_x(-1,1)&=-\frac{2}{3} +\frac{4}{3^2}=-\frac{2}{9} \\ f_y(x,y)&=\frac{x^2}{x^4+2y^2}-\frac{x^2y(4y)}{{(x^4+2y^2)}^2}\qquad & f_y(-1,1)&=\frac{1}{3} -\frac{4}{3^2}=-\frac{1}{9} \end{aligned}$

The linear approximation to $f(x,y)$ about $(-1,1)$ is

$\begin{aligned} f(x,y)\approx=f(-1,1) + f_x(-1,1)\,(x+1) + f_y(-1,1)\,(y-1) =\frac{1}{3} -\frac{2}{9}\,(x+1) -\frac{1}{9}\,(y-1) \end{aligned}$

In particular

$\begin{aligned} f(-0.9,1.1)\approx \frac{1}{3} -\frac{2}{9}\,(0.1) -\frac{1}{9}\,(0.1) =\frac{27}{90}=0.3 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.6절 · 3번

# 접평면과 선형근사계산보통자동 분류

39.CLP-3 · 2.6절 · 6번대학 수준우리말 번역

직각삼각형의 한 변의 길이를 측정했더니 $3$ 이고 최대 오차는 $\pm 0.1$ 이며, 다른 한 변의 길이는 $4$ 이고 최대 오차는 $\pm 0.2$ 이다. 선형 근사를 이용하여 이 직각삼각형의 빗변의 길이를 계산할 때 생길 수 있는 최대 오차를 어림하세요.

영어 원문 보기

One side of a right triangle is measured to be $3$ with a maximum possible error of $\pm 0.1$ , and the other side is measured to be $4$ with a maximum possible error of $\pm 0.2$ . Use the linear approximation to estimate the maximum possible error in calculating the length of the hypotenuse of the right triangle.

답 쓰고 채점

정답·풀이 보기

두 변의 길이를 $x$ , $y$ 라 하면 $x=3\pm 0.1$ , $y=4\pm 0.2$ 이다. 그러면 빗변의 길이는 $f(x,y)=\sqrt{x^2+y^2}$ 이다. 다음과 같다.

$\begin{aligned} f(x,y)&=\sqrt{x^2+y^2} & f(3,4)&=5 \\ f_x(x,y)&=\frac{x}{\sqrt{x^2+y^2}}\qquad & f_x(3,4)&=\frac{3}{5} \\ f_y(x,y)&=\frac{y}{\sqrt{x^2+y^2}} & f_y(3,4)&=\frac{4}{5} \end{aligned}$

선형 근사에 의해

$\begin{aligned} f(x,y)\approx f(3,4) + f_x(3,4)\,(x-3) + f_y(3,4)\,(y-4) =5 +\frac{3}{5}\,(x-3) +\frac{4}{5}\,(y-4) \end{aligned}$

따라서 빗변의 길이를 계산할 때 생기는 최대 오차의 근삿값은

$\begin{aligned} \frac{3}{5}\,(0.1) +\frac{4}{5}\,(0.2) =\frac{1.1}{5} = 0.22 \end{aligned}$

이다.

원본 교재 풀이(영어)

Denote by $x$ and $y$ the lengths of sides with $x=3\pm 0.1$ and $y=4\pm 0.2$ . Then the length of the hypotenuse is $f(x,y)=\sqrt{x^2+y^2}$ . Note that

$\begin{aligned} f(x,y)&=\sqrt{x^2+y^2} & f(3,4)&=5 \\ f_x(x,y)&=\frac{x}{\sqrt{x^2+y^2}}\qquad & f_x(3,4)&=\frac{3}{5} \\ f_y(x,y)&=\frac{y}{\sqrt{x^2+y^2}} & f_y(3,4)&=\frac{4}{5} \end{aligned}$

By the linear approximation

$\begin{aligned} f(x,y)\approx f(3,4) + f_x(3,4)\,(x-3) + f_y(3,4)\,(y-4) =5 +\frac{3}{5}\,(x-3) +\frac{4}{5}\,(y-4) \end{aligned}$

So the approximate maximum error in calculating the length of the hypotenuse is

$\begin{aligned} \frac{3}{5}\,(0.1) +\frac{4}{5}\,(0.2) =\frac{1.1}{5} = 0.22 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.6절 · 7번

# 접평면과 선형근사계산보통자동 분류

40.CLP-3 · 2.6절 · 7번대학 수준우리말 번역

저항이 $R_1$ , $R_2$ 인 두 저항을 병렬로 연결하면, 합성 저항 R 은 방정식 $\frac{1}{R} =\frac{1}{R_1}+\frac{1}{R_2}$ 을 만족시킨다. 선형 근사를 이용하여 $R_1$ 이 $2$ 옴에서 $1.9$ 옴으로 감소하고 $R_2$ 가 8 옴에서 8.1 옴으로 증가할 때 $R$ 의 변화량을 어림하세요.

영어 원문 보기

If two resistors of resistance $R_1$ and $R_2$ are wired in parallel, then the resulting resistance R satisfies the equation $\frac{1}{R} =\frac{1}{R_1}+\frac{1}{R_2}$ . Use the linear approximation to estimate the change in $R$ if $R_1$ decreases from $2$ to $1.9$ ohms and $R_2$ increases from 8 to 8.1 ohms.

답 쓰고 채점

정답·풀이 보기

함수 $R(R_1,R_2)$ 는 다음 식으로 음함수로 정의된다.

$\frac{1}{R(R_1,R_2)} =\frac{1}{R_1}+\frac{1}{R_2} \tag{$*$}$

특히

$\frac{1}{R(2,8)} = \frac{1}{2}+\frac{1}{8} = \frac{5}{8} \implies R(2,8) = \frac{8}{5}$

다음 선형 근사를 이용하고자 한다.

$R(R_1,R_2) \approx R(2,8) + \frac{\partial R}{\partial R_1}(2,8)\,(R_1-2) + \frac{\partial R}{\partial R_2}(2,8)\,(R_2-8)$

그러기 위해서는 편도함수 $\frac{\partial R}{\partial R_1}(2,8)$ 와 $\frac{\partial R}{\partial R_2}(2,8)$ 가 필요하다. 이를 구하기 위해 $(*)$ 를 $R_1$ 과 $R_2$ 에 대해 각각 미분한다.

$\begin{aligned} -\frac{1}{R(R_1,R_2)^2}\frac{\partial R}{\partial R_1}(R_1,R_2) &= -\frac{1}{R_1^2} \\ -\frac{1}{R(R_1,R_2)^2}\frac{\partial R}{\partial R_2}(R_1,R_2) &= -\frac{1}{R_2^2} \\ \end{aligned}$

$R_1=2$ , $R_2=8$ 을 대입하면

$\begin{aligned} -\frac{1}{(8/5)^2}\frac{\partial R}{\partial R_1}(2,8) &= -\frac{1}{4} &\quad\implies\quad & &\frac{\partial R}{\partial R_1}(2,8) &=\frac{16}{25} \\ -\frac{1}{(8/5)^2}\frac{\partial R}{\partial R_2}(2,8) &= -\frac{1}{64} &\quad\implies\quad & &\frac{\partial R}{\partial R_2}(2,8) &=\frac{1}{25} \end{aligned}$

따라서 $R$ 의 변화량은 다음과 같다.

$\begin{aligned} R(1.9,8.1)-R(2,8) \approx \frac{16}{25} (-0.1) + \frac{1}{25}(0.1) =-\frac{15}{250} = -0.06 \end{aligned}$

원본 교재 풀이(영어)

The function $R(R_1,R_2)$ is defined implictly by

$\frac{1}{R(R_1,R_2)} =\frac{1}{R_1}+\frac{1}{R_2} \tag{$*$}$

In particular

$\frac{1}{R(2,8)} = \frac{1}{2}+\frac{1}{8} = \frac{5}{8} \implies R(2,8) = \frac{8}{5}$

We wish to use the linear approximation

$R(R_1,R_2) \approx R(2,8) + \frac{\partial R}{\partial R_1}(2,8)\,(R_1-2) + \frac{\partial R}{\partial R_2}(2,8)\,(R_2-8)$

To do so, we need the partial derivatives $\frac{\partial R}{\partial R_1}(2,8)$ and $\frac{\partial R}{\partial R_2}(2,8)$ . To find them, we differentiate $(*)$ with respect to $R_1$ and $R_2$ :

$\begin{aligned} -\frac{1}{R(R_1,R_2)^2}\frac{\partial R}{\partial R_1}(R_1,R_2) &= -\frac{1}{R_1^2} \\ -\frac{1}{R(R_1,R_2)^2}\frac{\partial R}{\partial R_2}(R_1,R_2) &= -\frac{1}{R_2^2} \\ \end{aligned}$

Setting $R_1=2$ and $R_2=8$ gives

$\begin{aligned} -\frac{1}{(8/5)^2}\frac{\partial R}{\partial R_1}(2,8) &= -\frac{1}{4} &\quad\implies\quad & &\frac{\partial R}{\partial R_1}(2,8) &=\frac{16}{25} \\ -\frac{1}{(8/5)^2}\frac{\partial R}{\partial R_2}(2,8) &= -\frac{1}{64} &\quad\implies\quad & &\frac{\partial R}{\partial R_2}(2,8) &=\frac{1}{25} \end{aligned}$

So the specified change in $R$ is

$\begin{aligned} R(1.9,8.1)-R(2,8) \approx \frac{16}{25} (-0.1) + \frac{1}{25}(0.1) =-\frac{15}{250} = -0.06 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.6절 · 4번

# 접평면과 선형근사계산보통자동 분류

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