다변수 미분 전체 연습문제

1.CLP-3 · 2.1절 · 6번대학 수준우리말 번역

가능하다면 다음 극한값을 구하세요.

(a) $\displaystyle \lim_{(x,y)\rightarrow(2,-1)}\ \big(xy+x^2\big)$ (b) $\displaystyle \lim_{(x,y)\rightarrow(0,0)}\ \frac{x}{x^2+y^2}$ (c) $\displaystyle \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2}{x^2+y^2}$ (d) $\displaystyle \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^3}{x^2+y^2}$ (e) $\displaystyle \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2y^2}{x^2+y^4}$ (f) $\displaystyle \lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy}$

영어 원문 보기

Evaluate, if possible,

(a) $\displaystyle \lim_{(x,y)\rightarrow(2,-1)}\ \big(xy+x^2\big)$ (b) $\displaystyle \lim_{(x,y)\rightarrow(0,0)}\ \frac{x}{x^2+y^2}$ (c) $\displaystyle \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2}{x^2+y^2}$ (d) $\displaystyle \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^3}{x^2+y^2}$ (e) $\displaystyle \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2y^2}{x^2+y^4}$ (f) $\displaystyle \lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy}$

정답·풀이 보기

(a) $\displaystyle \lim_{(x,y)\rightarrow(2,-1)}\ \big(xy+x^2\big)=2(-1)+2^2=2$

(b) 극좌표로 바꾸면,

$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x}{x^2+y^2} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r\cos\theta}{r^2} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{\cos\theta}{r} \end{aligned}$

이 극한은 존재하지 않는다. 예를 들어,

$\theta=0$ 이면,

$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=0}}\ \frac{\cos\theta}{r} =\lim_{r\rightarrow0^+}\ \frac{1}{r} =+\infty$

반면 $\theta=\pi$ 이면,

$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=\pi}}\ \frac{\cos\theta}{r} =\lim_{r\rightarrow0^+}\ \frac{-1}{r} =-\infty$

이기 때문이다.

(c) 극좌표로 바꾸면,

$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2}{x^2+y^2} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^2\cos^2\theta}{r^2} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \cos^2\theta \end{aligned}$

이 극한은 존재하지 않는다. 예를 들어,

$\theta=0$ 이면,

$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=0}}\ \cos^2\theta =\lim_{r\rightarrow0^+}\ 1 = 1$

반면 $\theta=\frac{\pi}{2}$ 이면,

$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=\pi/2}}\ \cos^2\theta =\lim_{r\rightarrow0^+}\ 0 =0$

이기 때문이다.

(d) 극좌표로 바꾸면,

$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^3}{x^2+y^2} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^3\cos^3\theta}{r^2} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ r\cos^3\theta =0 \end{aligned}$

모든 $\theta$ 에 대해 $|\cos\theta|\le 1$ 이기 때문이다.

(e) 극좌표로 바꾸면,

$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2y^2}{x^2+y^4} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^2\cos^2\theta\ r^2\sin^2\theta} {r^2\cos^2\theta+r^4\sin^4\theta} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ r^2\sin^2\theta \frac{\cos^2\theta}{\cos^2\theta+r^2\sin^4\theta} \\ &= 0 \end{aligned}$

여기서 모든 $r>0$ 에 대해 다음이 성립함을 이용하였다.

$\left|\sin^2\theta\frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta}\right| \le \frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta} \le \left.\begin{cases} \frac{\cos^2\theta} {\cos^2\theta}&\text{(}\cos\theta\ne 0\text{ 일 때)}\\ 0 &\text{(}\cos\theta =0\text{ 일 때)}\end{cases} \right\} \le 1$

(f) 먼저 다음을 관찰하자.

$\begin{aligned} \lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy} =\left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right] \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right] \end{aligned}$

$\displaystyle \left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right]$ 은 로피탈 법칙 또는 미분계수의 정의를 이용하여 구할 수 있으며, 다음과 같다.

$\lim_{x\rightarrow 0}\ \frac{\sin x}{x} =\lim_{x\rightarrow 0}\ \frac{\sin x-\sin 0}{x-0} =\frac{\mathrm{d}}{\mathrm{d}x}\sin x\bigg|_{x=0} =\cos x\Big|_{x=0}=1$

마찬가지로 $\displaystyle \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right]$ 도 로피탈 법칙 또는 미분계수의 정의를 이용하여 구할 수 있으며, 다음과 같다.

$\lim_{y\rightarrow 0}\ \frac{e^y-1}{y} =\lim_{y\rightarrow 0}\ \frac{e^y-e^0}{y-0} =\frac{\mathrm{d}}{\mathrm{d}y}e^y\bigg|_{y=0} =e^y\Big|_{y=0}=1$

따라서 모두 합하면

$\begin{aligned} \lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy} =\left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right] \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right] =[1]\ [1]=1 \end{aligned}$

이다.

원본 교재 풀이(영어)

(a) $\displaystyle \lim_{(x,y)\rightarrow(2,-1)}\ \big(xy+x^2\big)=2(-1)+2^2=2$

(b) Switching to polar coordinates,

$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x}{x^2+y^2} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r\cos\theta}{r^2} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{\cos\theta}{r} \end{aligned}$

which does not exist, since, for example,

if $\theta=0$ , then

$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=0}}\ \frac{\cos\theta}{r} =\lim_{r\rightarrow0^+}\ \frac{1}{r} =+\infty$

while if $\theta=\pi$ , then

$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=\pi}}\ \frac{\cos\theta}{r} =\lim_{r\rightarrow0^+}\ \frac{-1}{r} =-\infty$

(c) Switching to polar coordinates,

$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2}{x^2+y^2} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^2\cos^2\theta}{r^2} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \cos^2\theta \end{aligned}$

which does not exist, since, for example,

if $\theta=0$ , then

$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=0}}\ \cos^2\theta =\lim_{r\rightarrow0^+}\ 1 = 1$

while if $\theta=\frac{\pi}{2}$ , then

$\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{\theta=\pi/2}}\ \cos^2\theta =\lim_{r\rightarrow0^+}\ 0 =0$

(d) Switching to polar coordinates,

$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^3}{x^2+y^2} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^3\cos^3\theta}{r^2} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ r\cos^3\theta =0 \end{aligned}$

since $|\cos\theta|\le 1$ for all $\theta$ .

(e) Switching to polar coordinates,

$\begin{aligned} \lim_{(x,y)\rightarrow(0,0)}\ \frac{x^2y^2}{x^2+y^4} &=\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ \frac{r^2\cos^2\theta\ r^2\sin^2\theta} {r^2\cos^2\theta+r^4\sin^4\theta} =\lim_{\genfrac{}{}{0pt}{}{r\rightarrow0^+}{0\le\theta<2\pi}}\ r^2\sin^2\theta \frac{\cos^2\theta}{\cos^2\theta+r^2\sin^4\theta} \\ &= 0 \end{aligned}$

Here, we used that

$\left|\sin^2\theta\frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta}\right| \le \frac{\cos^2\theta} {\cos^2\theta+r^2\sin^4\theta} \le \left.\begin{cases} \frac{\cos^2\theta} {\cos^2\theta}&\text{ if }\cos\theta\ne 0 \\ 0 &\text { if }\cos\theta =0 \end{cases} \right\} \le 1$

for all $r>0$ .

(f) To start, observe that

$\begin{aligned} \lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy} =\left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right] \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right] \end{aligned}$

We may evaluate $\displaystyle \left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right]$ by l'Hopital's rule or by using the definition of the derivative to give

$\lim_{x\rightarrow 0}\ \frac{\sin x}{x} =\lim_{x\rightarrow 0}\ \frac{\sin x-\sin 0}{x-0} =\frac{\mathrm{d}}{\mathrm{d}x}\sin x\bigg|_{x=0} =\cos x\Big|_{x=0}=1$

Similarly, we may evaluate $\displaystyle \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right]$ by l'Hopital's rule or by using the definition of the derivative to give

$\lim_{y\rightarrow 0}\ \frac{e^y-1}{y} =\lim_{y\rightarrow 0}\ \frac{e^y-e^0}{y-0} =\frac{\mathrm{d}}{\mathrm{d}y}e^y\bigg|_{y=0} =e^y\Big|_{y=0}=1$

So all together

$\begin{aligned} \lim_{(x,y)\rightarrow (0,0)}\ \frac{(\sin x)\left(e^y-1\right)}{xy} =\left[\lim_{x\rightarrow 0}\ \frac{\sin x}{x}\right] \left[\lim_{y\rightarrow 0}\ \frac{e^y-1}{y}\right] =[1]\ [1]=1 \end{aligned}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.2절 · 10번

# 다변수함수의 극한과 연속계산보통자동 분류

2.CLP-3 · 2.2절 · 4번대학 수준우리말 번역

다음 함수의 모든 1계 편도함수를 구하고, 주어진 점에서의 값을 구하세요.

(a) $f(x,y,z)=x^3y^4z^5\qquad (0,-1,-1)$ (b) $w(x,y,z)=\ln\left(1+e^{xyz}\right)\qquad (2,0,-1)$ (c) $f(x,y)=\frac{1}{\sqrt{x^2+y^2}}\qquad (-3,4)$

영어 원문 보기

Find all first partial derivatives of the following functions and evaluate them at the given point.

(a) $f(x,y,z)=x^3y^4z^5\qquad (0,-1,-1)$ (b) $w(x,y,z)=\ln\left(1+e^{xyz}\right)\qquad (2,0,-1)$ (c) $f(x,y)=\frac{1}{\sqrt{x^2+y^2}}\qquad (-3,4)$

정답·풀이 보기

(a)

$\begin{aligned} f_x(x,y,z)&=3x^2y^4z^5 & f_x(0,-1,-1)&=0\\ f_y(x,y,z)&=4x^3y^3z^5 & f_y(0,-1,-1)&=0\\ f_z(x,y,z)&=5x^3y^4z^4 & f_z(0,-1,-1)&=0 \end{aligned}$

(b)

$\begin{aligned} w_x(x,y,z)&=\frac{yz e^{xyz}}{1+e^{xyz}} & w_x(2,0,-1)&=0\\ w_y(x,y,z)&=\frac{xz e^{xyz}}{1+e^{xyz}} & w_y(2,0,-1)&=-1\\ w_z(x,y,z)&=\frac{xy e^{xyz}}{1+e^{xyz}} & w_z(2,0,-1)&=0 \end{aligned}$

(c)

$\begin{aligned} f_x(x,y)&=-\frac{x}{(x^2+y^2)^{3/2}} & f_x(-3,4)&=\frac{3}{125}\\ f_y(x,y)&=-\frac{y}{(x^2+y^2)^{3/2}} & f_y(-3,4)&=-\frac{4}{125} \end{aligned}$

원본 교재 풀이(영어)

(a)

$\begin{aligned} f_x(x,y,z)&=3x^2y^4z^5 & f_x(0,-1,-1)&=0\\ f_y(x,y,z)&=4x^3y^3z^5 & f_y(0,-1,-1)&=0\\ f_z(x,y,z)&=5x^3y^4z^4 & f_z(0,-1,-1)&=0 \end{aligned}$

(b)

$\begin{aligned} w_x(x,y,z)&=\frac{yz e^{xyz}}{1+e^{xyz}} & w_x(2,0,-1)&=0\\ w_y(x,y,z)&=\frac{xz e^{xyz}}{1+e^{xyz}} & w_y(2,0,-1)&=-1\\ w_z(x,y,z)&=\frac{xy e^{xyz}}{1+e^{xyz}} & w_z(2,0,-1)&=0 \end{aligned}$

(c)

$\begin{aligned} f_x(x,y)&=-\frac{x}{(x^2+y^2)^{3/2}} & f_x(-3,4)&=\frac{3}{125}\\ f_y(x,y)&=-\frac{y}{(x^2+y^2)^{3/2}} & f_y(-3,4)&=-\frac{4}{125} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.2절 · 5번

# 편도함수계산보통자동 분류

3.CLP-3 · 2.2절 · 6번대학 수준우리말 번역

곡면 $z(x, y)$ 가 $zy - y + x = \ln(xyz)$ 로 정의된다.

(a) $\frac{\partial z}{\partial x}$ , $\frac{\partial z}{\partial y}$ 를 $x$ , $y$ , $z$ 로 나타내어 구하세요. (b) $(x, y, z) = (-1, -2, 1/2)$ 에서 $\frac{\partial z}{\partial x}$ 와 $\frac{\partial z}{\partial y}$ 의 값을 구하세요.

영어 원문 보기

A surface $z(x, y)$ is defined by $zy - y + x = \ln(xyz)$ .

(a) Compute $\frac{\partial z}{\partial x}$ , $\frac{\partial z}{\partial y}$ in terms of $x$ , $y$ , $z$ . (b) Evaluate $\frac{\partial z}{\partial x}$ and $\frac{\partial z}{\partial y}$ at $(x, y, z) = (-1, -2, 1/2)$ .

정답·풀이 보기

(a) $z(x,y)$ 는 다음을 만족시킨다고 주어졌다.

$\begin{aligned} z(x,y)\, y - y + x = \ln\big(xy\,z(x,y)\big) \tag{$*$}\end{aligned}$

이는 모든 $(x,y)$ ( $(-1,-2)$ 근방)에 대해 성립한다. $(*)$ 를 $x$ 에 대해 미분하면

$\begin{aligned} y\,\frac{\partial z}{\partial x}(x,y) + 1 = \frac{1}{x} + \frac{\frac{\partial z}{\partial x}(x,y)}{z(x,y)} \implies \frac{\partial z}{\partial x}(x,y) = \frac{\frac{1}{x}-1}{y-\frac{1}{z(x,y)}} \end{aligned}$

이고, 인수 $(x,y)$ 를 생략한 뒤 분자와 분모에 모두 $xz$ 를 곱하면

$\begin{aligned} \frac{\partial z}{\partial x} = \frac{z-xz}{xyz-x} = \frac{z(1-x)}{x(yz-1)} \end{aligned}$

$(*)$ 를 $y$ 에 대해 미분하면

$\begin{aligned} z(x,y)+y\,\frac{\partial z}{\partial y}(x,y) - 1 = \frac{1}{y} + \frac{\frac{\partial z}{\partial y}(x,y)}{z(x,y)} \implies \frac{\partial z}{\partial y}(x,y) = \frac{\frac{1}{y}+1-z(x,y)}{y-\frac{1}{z(x,y)}} \end{aligned}$

이고, 인수 $(x,y)$ 를 생략한 뒤 분자와 분모에 모두 $yz$ 를 곱하면

$\begin{aligned} \frac{\partial z}{\partial y} = \frac{z+yz-yz^2}{y^2z-y} = \frac{z(1+y-yz)}{y(yz-1)} \end{aligned}$

(b) $(x,y,z) = (-1, -2, 1/2)$ 일 때,

$\begin{aligned} \frac{\partial z}{\partial x}(-1,-2) &= \left.\frac{\frac{1}{x}-1}{y-\frac{1}{z}} \right|_{(x,y,z) = (-1, -2, 1/2)} =\frac{\frac{1}{-1}-1}{-2-2} =\frac{1}{2} \\ \frac{\partial z}{\partial y}(-1,-2) &= \left.\frac{\frac{1}{y}+1-z}{y-\frac{1}{z}} \right|_{(x,y,z) = (-1, -2, 1/2)} =\frac{\frac{1}{-2}+1-\frac{1}{2}}{-2-2} =0 \end{aligned}$

원본 교재 풀이(영어)

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

$\begin{aligned} z(x,y)\, y - y + x = \ln\big(xy\,z(x,y)\big) \tag{$*$}\end{aligned}$

for all $(x,y)$ (near $(-1,-2)$ ). Differentiating $(*)$ with respect to $x$ gives

$\begin{aligned} y\,\frac{\partial z}{\partial x}(x,y) + 1 = \frac{1}{x} + \frac{\frac{\partial z}{\partial x}(x,y)}{z(x,y)} \implies \frac{\partial z}{\partial x}(x,y) = \frac{\frac{1}{x}-1}{y-\frac{1}{z(x,y)}} \end{aligned}$

or, dropping the arguments $(x,y)$ and multiplying both the numerator and denominator by $xz$ ,

$\begin{aligned} \frac{\partial z}{\partial x} = \frac{z-xz}{xyz-x} = \frac{z(1-x)}{x(yz-1)} \end{aligned}$

Differentiating $(*)$ with respect to $y$ gives

$\begin{aligned} z(x,y)+y\,\frac{\partial z}{\partial y}(x,y) - 1 = \frac{1}{y} + \frac{\frac{\partial z}{\partial y}(x,y)}{z(x,y)} \implies \frac{\partial z}{\partial y}(x,y) = \frac{\frac{1}{y}+1-z(x,y)}{y-\frac{1}{z(x,y)}} \end{aligned}$

or, dropping the arguments $(x,y)$ and multiplying both the numerator and denominator by $yz$ ,

$\begin{aligned} \frac{\partial z}{\partial y} = \frac{z+yz-yz^2}{y^2z-y} = \frac{z(1+y-yz)}{y(yz-1)} \end{aligned}$

(b) When $(x,y,z) = (-1, -2, 1/2)$ ,

$\begin{aligned} \frac{\partial z}{\partial x}(-1,-2) &= \left.\frac{\frac{1}{x}-1}{y-\frac{1}{z}} \right|_{(x,y,z) = (-1, -2, 1/2)} =\frac{\frac{1}{-1}-1}{-2-2} =\frac{1}{2} \\ \frac{\partial z}{\partial y}(-1,-2) &= \left.\frac{\frac{1}{y}+1-z}{y-\frac{1}{z}} \right|_{(x,y,z) = (-1, -2, 1/2)} =\frac{\frac{1}{-2}+1-\frac{1}{2}}{-2-2} =0 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.2절 · 7번

# 편도함수계산보통자동 분류

4.CLP-3 · 2.2절 · 7번대학 수준우리말 번역

$(T, U, V, W)$ 가 다음 식으로 서로 관계되어 있을 때, $(1, 1, 2, 4)$ 에서 $\frac{\partial U}{\partial T}$ 와 $\frac{\partial T}{\partial V}$ 를 구하세요.

$(TU-V)^2 \ln(W-UV) = \ln 2$

영어 원문 보기

Find $\frac{\partial U}{\partial T}$ and $\frac{\partial T}{\partial V}$ at $(1, 1, 2, 4)$ if $(T, U, V, W)$ are related by

$(TU-V)^2 \ln(W-UV) = \ln 2$

정답·풀이 보기

네 변수 $T$ , $U$ , $V$ , $W$ 는 하나의 방정식 $(TU-V)^2 \ln(W-UV) = \ln 2$ 를 만족한다고 하였다. 따라서 이 변수들이 모두 독립인 것은 아니다. 대략 말하면, 이 중 세 변수를 독립변수로 보고, 주어진 방정식을 나머지 한 변수에 대해 선택한 세 독립변수의 함수로 풀 수 있다.

먼저 $\frac{\partial U}{\partial T}$ 를 구하라고 하였다. 이는 $T$ , $V$ , $W$ 를 독립변수로 보고 $U$ 를 다음을 만족하는 함수 $U(T,V,W)$ 로 보라는 뜻이다.

$\big(T\, U(T,V,W)-V\big)^2 \ln\big(W-U(T,V,W)\,V\big) = \ln 2 \tag{E1}$

이는 $(1, 1, 2, 4)$ 에 충분히 가까운 모든 $(T, U, V, W)$ 에 대해 성립한다. (E1)의 양변을 $T$ 에 대해 미분하면 다음과 같다.

$\begin{aligned} &2\big(T\, U(T,V,W)-V\big) \left[ U(T,V,W) +T\ \frac{\partial U}{\partial T}(T,V,W)\right] \ln\big(W-U(T,V,W)\,V\big) \\ & -\big(T\, U(T,V,W)-V\big)^2 \frac{1}{W-U(T,V,W)\,V}\frac{\partial U}{\partial T}(T,V,W)\,V = 0 \end{aligned}$

특히 $(T, U, V, W)=(1, 1, 2, 4)$ 에서는 다음과 같다.

$\begin{aligned} &2\big((1)(1)-2\big) \left[ 1 +(1)\frac{\partial U}{\partial T}(1,2,4)\right] \ln\big(4-(1)(2)\big) \\ & -\big((1)(1)-2\big)^2 \frac{1}{4-(1)(2)}\frac{\partial U}{\partial T}(1,2,4)\,(2) = 0 \end{aligned}$

이를 간단히 하면 다음과 같다.

$\begin{aligned} -2\left[ 1 +\frac{\partial U}{\partial T}(1,2,4)\right] \ln(2) -\frac{\partial U}{\partial T}(1,2,4)=0 \implies \frac{\partial U}{\partial T}(1,2,4) = -\frac{2\ln(2)}{1+2\ln(2)} \end{aligned}$

다음으로 $\frac{\partial T}{\partial V}$ 를 구하라고 하였다. 이는 $U$ , $V$ , $W$ 를 독립변수로 보고 $T$ 를 다음을 만족하는 함수 $T(U,V,W)$ 로 보라는 뜻이다.

$\big(T(U,V,W)\, U-V\big)^2 \ln\big(W-U\,V\big) = \ln 2 \tag{E2}$

이는 $(1, 1, 2, 4)$ 에 충분히 가까운 모든 $(T, U, V, W)$ 에 대해 성립한다. (E2)의 양변을 $V$ 에 대해 미분하면 다음과 같다.

$\begin{aligned} &2\big(T(U,V,W)\, U-V\big)\ \left[\frac{\partial T}{\partial V}(U,V,W)\ U-1\right] \ln\big(W-U\,V\big) \\ & -\big(T(U,V,W)\, U-V\big)^2 \frac{U}{W-U\,V} = 0 \end{aligned}$

특히 $(T, U, V, W)=(1, 1, 2, 4)$ 에서는 다음과 같다.

$\begin{aligned} &2\big((1)(1)-2\big) \left[ (1)\frac{\partial T}{\partial V}(1,2,4)-1\right] \ln\big(4-(1)(2)\big) \\ & -\big((1)(1)-2\big)^2 \frac{1}{4-(1)(2)} = 0 \end{aligned}$

이를 간단히 하면 다음과 같다.

$\begin{aligned} -2\left[\frac{\partial T}{\partial V}(1,2,4)-1\right] \ln(2) -\frac{1}{2}=0 \implies \frac{\partial T}{\partial V}(1,2,4) = 1 -\frac{1}{4\ln(2)} \end{aligned}$

원본 교재 풀이(영어)

We are told that the four variables $T$ , $U$ , $V$ , $W$ obey the single equation $(TU-V)^2 \ln(W-UV) = \ln 2$ . So they are not all independent variables. Roughly speaking, we can treat any three of them as independent variables and solve the given equation for the fourth as a function of the three chosen independent variables.

We are first asked to find $\frac{\partial U}{\partial T}$ . This implicitly tells to treat $T$ , $V$ and $W$ as independent variables and to view $U$ as a function $U(T,V,W)$ that obeys

$\big(T\, U(T,V,W)-V\big)^2 \ln\big(W-U(T,V,W)\,V\big) = \ln 2 \tag{E1}$

for all $(T, U, V, W)$ sufficiently near $(1, 1, 2, 4)$ . Differentiating (E1) with respect to $T$ gives

$\begin{aligned} &2\big(T\, U(T,V,W)-V\big) \left[ U(T,V,W) +T\ \frac{\partial U}{\partial T}(T,V,W)\right] \ln\big(W-U(T,V,W)\,V\big) \\ & -\big(T\, U(T,V,W)-V\big)^2 \frac{1}{W-U(T,V,W)\,V}\frac{\partial U}{\partial T}(T,V,W)\,V = 0 \end{aligned}$

In particular, for $(T, U, V, W)=(1, 1, 2, 4)$ ,

$\begin{aligned} &2\big((1)(1)-2\big) \left[ 1 +(1)\frac{\partial U}{\partial T}(1,2,4)\right] \ln\big(4-(1)(2)\big) \\ & -\big((1)(1)-2\big)^2 \frac{1}{4-(1)(2)}\frac{\partial U}{\partial T}(1,2,4)\,(2) = 0 \end{aligned}$

This simplifies to

$\begin{aligned} -2\left[ 1 +\frac{\partial U}{\partial T}(1,2,4)\right] \ln(2) -\frac{\partial U}{\partial T}(1,2,4)=0 \implies \frac{\partial U}{\partial T}(1,2,4) = -\frac{2\ln(2)}{1+2\ln(2)} \end{aligned}$

We are then asked to find $\frac{\partial T}{\partial V}$ . This implicitly tells to treat $U$ , $V$ and $W$ as independent variables and to view $T$ as a function $T(U,V,W)$ that obeys

$\big(T(U,V,W)\, U-V\big)^2 \ln\big(W-U\,V\big) = \ln 2 \tag{E2}$

for all $(T, U, V, W)$ sufficiently near $(1, 1, 2, 4)$ . Differentiating (E2) with respect to $V$ gives

$\begin{aligned} &2\big(T(U,V,W)\, U-V\big)\ \left[\frac{\partial T}{\partial V}(U,V,W)\ U-1\right] \ln\big(W-U\,V\big) \\ & -\big(T(U,V,W)\, U-V\big)^2 \frac{U}{W-U\,V} = 0 \end{aligned}$

In particular, for $(T, U, V, W)=(1, 1, 2, 4)$ ,

$\begin{aligned} &2\big((1)(1)-2\big) \left[ (1)\frac{\partial T}{\partial V}(1,2,4)-1\right] \ln\big(4-(1)(2)\big) \\ & -\big((1)(1)-2\big)^2 \frac{1}{4-(1)(2)} = 0 \end{aligned}$

This simplifies to

$\begin{aligned} -2\left[\frac{\partial T}{\partial V}(1,2,4)-1\right] \ln(2) -\frac{1}{2}=0 \implies \frac{\partial T}{\partial V}(1,2,4) = 1 -\frac{1}{4\ln(2)} \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.2절 · 4번

# 편도함수계산보통자동 분류

5.CLP-3 · 2.2절 · 8번대학 수준우리말 번역

$u = x^2 + yz$ , $x = \rho r \cos(\theta)$ , $y = \rho r \sin(\theta)$ , $z = \rho r$ 라고 하자. 점 $(\rho_0 , r_0 , \theta_0) = (2, 3, \pi/2)$ 에서 $\frac{\partial u}{\partial r}$ 를 구하세요.

영어 원문 보기

Suppose that $u = x^2 + yz$ , $x = \rho r \cos(\theta)$ , $y = \rho r \sin(\theta)$ and $z = \rho r$ . Find $\frac{\partial u}{\partial r}$ at the point $(\rho_0 , r_0 , \theta_0) = (2, 3, \pi/2)$ .

답 쓰고 채점

정답·풀이 보기

함수

$\begin{aligned} u(\rho, r,\theta) &= \big[\rho r\cos\theta\big]^2 +\big[\rho r\sin\theta\big] \rho r \\ &=\rho^2 r^2\cos^2\theta +\rho^2 r^2\sin\theta \end{aligned}$

이므로

$\begin{aligned} \frac{\partial u}{\partial r}(\rho, r,\theta) &=2 \rho^2 r\cos^2\theta +2 \rho^2 r\sin\theta \end{aligned}$

이고

$\begin{aligned} \frac{\partial u}{\partial r}(2, 3,\pi/2) &=2 (2^2) (3) (0)^2 +2 (2^2) (3) (1) =24 \end{aligned}$

이다.

원본 교재 풀이(영어)

The function

$\begin{aligned} u(\rho, r,\theta) &= \big[\rho r\cos\theta\big]^2 +\big[\rho r\sin\theta\big] \rho r \\ &=\rho^2 r^2\cos^2\theta +\rho^2 r^2\sin\theta \end{aligned}$

So

$\begin{aligned} \frac{\partial u}{\partial r}(\rho, r,\theta) &=2 \rho^2 r\cos^2\theta +2 \rho^2 r\sin\theta \end{aligned}$

and

$\begin{aligned} \frac{\partial u}{\partial r}(2, 3,\pi/2) &=2 (2^2) (3) (0)^2 +2 (2^2) (3) (1) =24 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.2절 · 9번

# 편도함수계산보통자동 분류

6.CLP-3 · 2.2절 · 9번대학 수준우리말 번역

미분계수의 정의를 이용하여 다음 함수에 대해 $f_x(0,0)$ 과 $f_y(0,0)$ 의 값을 구하세요.

$f(x,y)=\begin{cases} \frac{x^2-2y^2}{x-y}&\text{$x\ne y$ 이면}\\ 0&\text{$x=y$ 이면} \end{cases}$

영어 원문 보기

Use the definition of the derivative to evaluate $f_x(0,0)$ and $f_y(0,0)$ for

$f(x,y)=\begin{cases} \frac{x^2-2y^2}{x-y}&\text{if $x\ne y$}\\ 0&\text{if $x=y$} \end{cases}$

정답·풀이 보기

정의에 의해

$f_x(x_0,y_0)=\lim_{\Delta x\rightarrow 0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x} \qquad f_y(x_0,y_0)=\lim_{\Delta y\rightarrow 0}\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}$

이다. $x_0=y_0=0$ 을 대입하면

$\begin{aligned} f_x(0,0)&=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)-f(0,0)}{\Delta x} &=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)}{\Delta x} &=\lim_{\Delta x\rightarrow 0}\frac{((\Delta x)^2-2\times0^2)/(\Delta x-0)}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}1 =1\\ f_y(0,0)&=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)-f(0,0)}{\Delta y} &=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)}{\Delta y} &=\lim_{\Delta y\rightarrow 0}\frac{(0^2-2(\Delta y)^2)/(0-\Delta y)}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}2 =2 \end{aligned}$

이다.

원본 교재 풀이(영어)

By definition

$f_x(x_0,y_0)=\lim_{\Delta x\rightarrow 0}\frac{f(x_0+\Delta x,y_0)-f(x_0,y_0)}{\Delta x} \qquad f_y(x_0,y_0)=\lim_{\Delta y\rightarrow 0}\frac{f(x_0,y_0+\Delta y)-f(x_0,y_0)}{\Delta y}$

Setting $x_0=y_0=0$ ,

$\begin{aligned} f_x(0,0)&=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)-f(0,0)}{\Delta x} &=\lim_{\Delta x\rightarrow 0}\frac{f(\Delta x,0)}{\Delta x} &=\lim_{\Delta x\rightarrow 0}\frac{((\Delta x)^2-2\times0^2)/(\Delta x-0)}{\Delta x} \\ &=\lim_{\Delta x\rightarrow 0}1 =1\\ f_y(0,0)&=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)-f(0,0)}{\Delta y} &=\lim_{\Delta y\rightarrow 0}\frac{f(0,\Delta y)}{\Delta y} &=\lim_{\Delta y\rightarrow 0}\frac{(0^2-2(\Delta y)^2)/(0-\Delta y)}{\Delta y} \\ &=\lim_{\Delta y\rightarrow 0}2 =2 \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.2절 · 2번

# 편도함수계산보통자동 분류

7.CLP-3 · 2.3절 · 3번대학 수준우리말 번역

주어진 편도함수를 구하세요.

(a) $f(x,y) = x^2y^3$ ; $f_{xx}(x,y)$ , $f_{xyy}(x,y)$ , $f_{yxy}(x,y)$ (b) $f(x,y) = e^{xy^2}$ ; $f_{xx}(x,y)$ , $f_{xy}(x,y)$ , $f_{xxy}(x,y)$ , $f_{xyy}(x,y)$ (c) $\displaystyle f(u,v,w) = \frac{1}{u+2v+3w}\$ , $\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(u,v,w)\$ , $\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(3,2,1)$

영어 원문 보기

Find the specified partial derivatives.

(a) $f(x,y) = x^2y^3$ ; $f_{xx}(x,y)$ , $f_{xyy}(x,y)$ , $f_{yxy}(x,y)$ (b) $f(x,y) = e^{xy^2}$ ; $f_{xx}(x,y)$ , $f_{xy}(x,y)$ , $f_{xxy}(x,y)$ , $f_{xyy}(x,y)$ (c) $\displaystyle f(u,v,w) = \frac{1}{u+2v+3w}\$ , $\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(u,v,w)\$ , $\displaystyle \frac{\partial^3 f}{\partial u\partial v\partial w}(3,2,1)$

정답·풀이 보기

(a) 다음과 같다.

$\begin{aligned} f_x(x,y) &= 2xy^3 & f_{xx}(x,y) &= 2y^3 \\ & & f_{xy}(x,y) &= 6xy^2 & f_{yxy}(x,y) = f_{xyy}(x,y) &= 12xy \end{aligned}$

(b) 다음과 같다.

$\begin{aligned} f_x(x,y) &= y^2e^{xy^2} & f_{xx}(x,y) &= y^4e^{xy^2} & f_{xxy}(x,y) &= 4y^3e^{xy^2} + 2xy^5e^{xy^2} \\ & & f_{xy}(x,y) &= 2ye^{xy^2}+2xy^3e^{xy^2} & f_{xyy}(x,y) &= \big(2+4xy^2+6xy^2+4x^2y^4\big)e^{xy^2}\\ & & & & &= \big(2+10xy^2+4x^2y^4\big)e^{xy^2} \end{aligned}$

(c) 다음과 같다.

$\begin{aligned} \frac{\partial f}{\partial u}(u,v,w) &= -\frac{1}{(u+2v+3w)^2} \\ \frac{\partial^2 f}{\partial u\,\partial v}(u,v,w) &= \frac{4}{(u+2v+3w)^3} \\ \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(u,v,w) &= -\frac{36}{(u+2v+3w)^4} \end{aligned}$

특히

$\begin{aligned} \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(3,2,1) &= -\frac{36}{(3+2\times 2+3\times 1)^4} = -\frac{36}{10^4} = -\frac{9}{2500} \end{aligned}$

원본 교재 풀이(영어)

(a) We have

$\begin{aligned} f_x(x,y) &= 2xy^3 & f_{xx}(x,y) &= 2y^3 \\ & & f_{xy}(x,y) &= 6xy^2 & f_{yxy}(x,y) = f_{xyy}(x,y) &= 12xy \end{aligned}$

(b) We have

$\begin{aligned} f_x(x,y) &= y^2e^{xy^2} & f_{xx}(x,y) &= y^4e^{xy^2} & f_{xxy}(x,y) &= 4y^3e^{xy^2} + 2xy^5e^{xy^2} \\ & & f_{xy}(x,y) &= 2ye^{xy^2}+2xy^3e^{xy^2} & f_{xyy}(x,y) &= \big(2+4xy^2+6xy^2+4x^2y^4\big)e^{xy^2}\\ & & & & &= \big(2+10xy^2+4x^2y^4\big)e^{xy^2} \end{aligned}$

(c) We have

$\begin{aligned} \frac{\partial f}{\partial u}(u,v,w) &= -\frac{1}{(u+2v+3w)^2} \\ \frac{\partial^2 f}{\partial u\,\partial v}(u,v,w) &= \frac{4}{(u+2v+3w)^3} \\ \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(u,v,w) &= -\frac{36}{(u+2v+3w)^4} \end{aligned}$

In particular

$\begin{aligned} \frac{\partial^3 f}{\partial u\,\partial v\,\partial w}(3,2,1) &= -\frac{36}{(3+2\times 2+3\times 1)^4} = -\frac{36}{10^4} = -\frac{9}{2500} \end{aligned}$

한 번 더 — 비슷한 문제 풀기OpenStax 3권 · 4.3절 · 18번

# 편도함수계산보통자동 분류

8.CLP-3 · 2.3절 · 5번대학 수준우리말 번역

주어진 편도함수를 구하세요.

(a) $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ ; $f_{xyz}(x,y,z)$ (b) $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big) +\arctan\big(e^{\sqrt{xz}}\big) +\arctan\big(e^{\sqrt{yz}}\big)$ ; $f_{xyz}(x,y,z)$ (c) $f(x,y,z) = \arctan\big(e^{\sqrt{xyz}}\big)$ ; $f_{xx}(1,0,0)$

영어 원문 보기

Find the specified partial derivatives.

(a) $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ ; $f_{xyz}(x,y,z)$ (b) $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big) +\arctan\big(e^{\sqrt{xz}}\big) +\arctan\big(e^{\sqrt{yz}}\big)$ ; $f_{xyz}(x,y,z)$ (c) $f(x,y,z) = \arctan\big(e^{\sqrt{xyz}}\big)$ ; $f_{xx}(1,0,0)$

정답·풀이 보기

(a) $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ 는 $z$ 에 무관하므로 $f_z(x,y,z) = 0$ 이고, 따라서

$f_{xyz}(x,y,z) =f_{zxy}(x,y,z) =0$

이다.

(b) $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ , $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ , $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$ 라 하자. 그러면

$u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ 는 $z$ 에 무관하므로 $u_z(x,y,z) = 0$ 이고, 따라서 $u_{xyz}(x,y,z) =u_{zxy}(x,y,z) =0$
$v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ 는 $y$ 에 무관하므로 $v_y(x,y,z) = 0$ 이고, 따라서 $v_{xyz}(x,y,z) =v_{yxz}(x,y,z) =0$
$w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$ 는 $x$ 에 무관하므로 $w_x(x,y,z) = 0$ 이고, 따라서 $w_{xyz}(x,y,z) =0$

$f(x,y,z)=u(x,y,z)+v(x,y,z)+w(x,y,z)$ 이므로

$f_{xyz}(x,y,z)=u_{xyz}(x,y,z)+v_{xyz}(x,y,z)+w_{xyz}(x,y,z)=0$

이다.

(c) $f_{xx}(x,0,0)$ 을 구하는 과정에서 $y$ 와 $z$ 는 모두 $0$ 으로 고정된다. 따라서 $g(x) = f(x,0,0)$ 이라 하면 $f_{xx}(x,0,0)=g''(x)$ 이다. 이제

$g(x) = f(x,0,0) = \arctan\big(e^{\sqrt{xyz}}\big)\Big|_{y=z=0} =\arctan(1) =\frac{\pi}{4}$

이고 이는 모든 $x$ 에 대해 성립한다. 그러므로 모든 $x$ 에 대해 $g'(x)=0$ 이고 $g''(x)=0$ 이다. 특히

$f_{xx}(1,0,0) = g''(1) = 0$

이다.

원본 교재 풀이(영어)

(a) As $f(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ is independent of $z$ , we have $f_z(x,y,z) = 0$ and hence

$f_{xyz}(x,y,z) =f_{zxy}(x,y,z) =0$

(b) Write $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ , $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ and $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$ . Then

As $u(x,y,z) = \arctan\big(e^{\sqrt{xy}}\big)$ is independent of $z$ , we have $u_z(x,y,z) = 0$ and hence $u_{xyz}(x,y,z) =u_{zxy}(x,y,z) =0$
As $v(x,y,z) = \arctan\big(e^{\sqrt{xz}}\big)$ is independent of $y$ , we have $v_y(x,y,z) = 0$ and hence $v_{xyz}(x,y,z) =v_{yxz}(x,y,z) =0$
As $w(x,y,z) = \arctan\big(e^{\sqrt{yz}}\big)$ is independent of $x$ , we have $w_x(x,y,z) = 0$ and hence $w_{xyz}(x,y,z) =0$

As $f(x,y,z)=u(x,y,z)+v(x,y,z)+w(x,y,z)$ , we have

$f_{xyz}(x,y,z)=u_{xyz}(x,y,z)+v_{xyz}(x,y,z)+w_{xyz}(x,y,z)=0$

(c) In the course of evaluating $f_{xx}(x,0,0)$ , both $y$ and $z$ are held fixed at $0$ . Thus, if we set $g(x) = f(x,0,0)$ , then $f_{xx}(x,0,0)=g''(x)$ . Now

$g(x) = f(x,0,0) = \arctan\big(e^{\sqrt{xyz}}\big)\Big|_{y=z=0} =\arctan(1) =\frac{\pi}{4}$

for all $x$ . So $g'(x)=0$ and $g''(x)=0$ for all $x$ . In particular,

$f_{xx}(1,0,0) = g''(1) = 0$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.3절 · 4번

# 편도함수계산보통자동 분류

9.CLP-3 · 2.4절 · 6번대학 수준우리말 번역

연쇄 법칙을 이용하는 방법을 포함하여 두 가지 방법으로 $\frac{\partial w}{\partial s}$ 와 $\frac{\partial w}{\partial t}$ 의 값을 구하세요. 단, 함수 $w=x^2+y^2+z^2$ 이고 $x=st,\ y=s\cos t$ , $z=s\sin t$ 입니다.

영어 원문 보기

Use two methods (one using the chain rule) to evaluate $\frac{\partial w}{\partial s}$ and $\frac{\partial w}{\partial t}$ given that the function $w=x^2+y^2+z^2$ , with $x=st,\ y=s\cos t$ and $z=s\sin t$ .

정답·풀이 보기

방법 1: $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ 이고 $x(s,t)=st$ , $y(s,t)=s\cos t$ , $z(s,t)=s\sin t$ 이므로 $w(s,t)$ 를 명시적으로 나타낼 수 있다:

$\begin{aligned} w(s,t)&=(st)^2+(s\cos t)^2+(s\sin t)^2 =s^2(t^2+1)\\ \implies w_s(s,t)&=2s(t^2+1)\qquad\text{그리고}\qquad w_t(s,t)=s^2(2t) \end{aligned}$

방법 2:* 문제에서 $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ 이고 $x(s,t)=st$ , $y(s,t)=s\cos t$ , $z(s,t)=s\sin t$ 로 주어졌다. 즉, $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ 이며 $W(x,y,z)= x^2+y^2+z^2$ 이다. $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ 에 연쇄 법칙을 적용하고 $\frac{\partial W}{\partial x}=2x$ , $\frac{\partial W}{\partial y}=2y$ , $\frac{\partial W}{\partial z}=2z$ 임을 이용하면 다음을 얻는다.

$\begin{aligned} \frac{\partial w}{\partial s}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial s}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial s}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial s}(s,t) \\ &=2 x(s,t)\ x_s(s,t)+2 y(s,t)\ y_s(s,t)+2 z(s,t)\ z_s(s,t) \\ &=2 (st)\ t+2 (s\cos t)\ \cos t+2 (s\sin t)\ \sin t\\ &=2 st^2+2 s\\ \frac{\partial w}{\partial t}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial t}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial t}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial t}(s,t) \\ &=2 x(s,t)\ x_t(s,t)+2 y(s,t)\ y_t(s,t)+2 z(s,t)\ z_t(s,t)\\ &=2 (st)\ s+2 (s\cos t)\ (-s\sin t)+2 (s\sin t)\ (s\cos t)\\ &=2 s^2t \end{aligned}$

원본 교재 풀이(영어)

Method 1: Since $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ with $x(s,t)=st$ , $y(s,t)=s\cos t$ and $z(s,t)=s\sin t$ we can write out $w(s,t)$ explicitly:

$\begin{aligned} w(s,t)&=(st)^2+(s\cos t)^2+(s\sin t)^2 =s^2(t^2+1)\\ \implies w_s(s,t)&=2s(t^2+1)\qquad\text{and}\qquad w_t(s,t)=s^2(2t) \end{aligned}$

Method 2:* The question specifies that $w(s,t)=x(s,t)^2+y(s,t)^2+z(s,t)^2$ with $x(s,t)=st$ , $y(s,t)=s\cos t$ and $z(s,t)=s\sin t$ . That is, $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ with $W(x,y,z)= x^2+y^2+z^2$ . Applying the chain rule to $w(s,t) = W\big(x(s,t), y(s,t), z(s,t)\big)$ and noting that $\frac{\partial W}{\partial x}=2x$ , $\frac{\partial W}{\partial y}=2y$ , $\frac{\partial W}{\partial z}=2z$ , gives

$\begin{aligned} \frac{\partial w}{\partial s}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial s}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial s}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial s}(s,t) \\ &=2 x(s,t)\ x_s(s,t)+2 y(s,t)\ y_s(s,t)+2 z(s,t)\ z_s(s,t) \\ &=2 (st)\ t+2 (s\cos t)\ \cos t+2 (s\sin t)\ \sin t\\ &=2 st^2+2 s\\ \frac{\partial w}{\partial t}(s,t) &= \frac{\partial W}{\partial x}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial x}{\partial t}(s,t) + \frac{\partial W}{\partial y}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial y}{\partial t}(s,t)\\& + \frac{\partial W}{\partial z}\big(x(s,t), y(s,t), z(s,t)\big)\frac{\partial z}{\partial t}(s,t) \\ &=2 x(s,t)\ x_t(s,t)+2 y(s,t)\ y_t(s,t)+2 z(s,t)\ z_t(s,t)\\ &=2 (st)\ s+2 (s\cos t)\ (-s\sin t)+2 (s\sin t)\ (s\cos t)\\ &=2 s^2t \end{aligned}$

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 7번

# 다변수 연쇄법칙계산보통자동 분류

10.CLP-3 · 2.4절 · 7번대학 수준우리말 번역

$\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy)$ 의 값을 $f$ 의 편도함수로 나타내어 구하세요. 연쇄 법칙과, 혼합 편도함수가 서로 같음을 나타내는 클레로의 정리를 적용할 수 있도록 $f$ 는 매끄러운 함수라고 가정해도 됩니다.

영어 원문 보기

Evaluate $\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy)$ in terms of partial derivatives of $f$ . You may assume that $f$ 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

정의에 따라

$\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) =\frac{\partial }{\partial x}\left[\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right) \right]$

이다.

안쪽부터 차례로 편도함수를 계산한다. $F(x,y)=f(2x+3y,xy)$ 라 하면 가장 안쪽의 편도함수는 $G(x,y)=\frac{\partial }{\partial y}f(2x+3y,xy)=\frac{\partial }{\partial y}F(x,y)$ 이다. 연쇄 법칙에 의해

$\begin{aligned} G(x,y)=\frac{\partial }{\partial y}F(x,y) &=f_1(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_2(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_1(2x+3y,xy)+xf_2(2x+3y,xy) \end{aligned}$

이다.

여기서 아래첨자 $1$ 은 두 번째 변수를 고정한 채 $f$ 를 첫 번째 변수에 대해 편미분한다는 뜻이고, 아래첨자 $2$ 는 첫 번째 변수를 고정한 채 $f$ 를 두 번째 변수에 대해 편미분한다는 뜻이다. 다음으로 가운데 편도함수를 $H(x,y)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right)$ 라 하면

$\begin{aligned} H(x,y) &= \frac{\partial }{\partial y} G(x,y) \\ &=\frac{\partial }{\partial y}\Big(3f_1(2x+3y,xy)+xf_2(2x+3y,xy)\Big) \\ &=3\frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) +x\frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) \end{aligned}$

이다.

연쇄 법칙을 (두 번) 쓰면

$\begin{aligned} \frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) &=f_{11}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{12}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \\ \frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) &=f_{21}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{22}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy) \end{aligned}$

이므로

$\begin{aligned} H(x,y)&=3\Big(3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \Big) \\[-0.05in] & +x\Big(3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy)\Big) \\ &=9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy) \end{aligned}$

이다.

마지막 등호에서는 $f_{21}(2x+3y,xy)=f_{12}(2x+3y,xy)$ 임을 이용했다. 기호 $f_{21}$ 은 먼저 두 번째 변수에 대해 미분한 다음 첫 번째 변수에 대해 미분한다는 뜻이다. 예를 들어 $f(x,y)=e^{2y}\sin x$ 이면

$\begin{aligned} f_{21}(x,y) =\frac{\partial }{\partial x} \Big[\frac{\partial }{\partial y}\big(e^{2y}\sin x\big)\Big] =\frac{\partial }{\partial x} \Big[2e^{2y}\sin x\Big] =2e^y\cos x \end{aligned}$

이다.

마지막으로

$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=\frac{\partial }{\partial x} H(x,y) \\ &=\frac{\partial }{\partial x}\Big(9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy)\Big)\\ &=9\frac{\partial }{\partial x}\Big(f_{11}(2x+3y,xy)\Big)\\& +6f_{12}(2x+3y,xy) +6x\frac{\partial }{\partial x}\Big(f_{12}(2x+3y,xy)\Big) \\& +2xf_{22}(2x+3y,xy) +x^2\frac{\partial }{\partial x}\Big(f_{22}(2x+3y,xy)\Big) \end{aligned}$

이다.

연쇄 법칙을 세 번 적용하면

$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=9\Big(2f_{111}+yf_{112}\Big)\\& +6f_{12}+6x\Big(2f_{121}+yf_{122}\Big)\\& +2xf_{22}+x^2\Big(2f_{221}+yf_{222}\Big)\\ &=6\,f_{12}+2x\,f_{22}+18\,f_{111}+(9y+12x)\,f_{112} +(6xy+2x^2)\,f_{122}+x^2y\,f_{222} \end{aligned}$

이다.

우변의 모든 함수의 인수는 $(2x+3y,xy)$ 이다.

원본 교재 풀이(영어)

By definition,

$\frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) =\frac{\partial }{\partial x}\left[\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right) \right]$

We'll compute the derivatives from the inside out. Let's call $F(x,y)=f(2x+3y,xy)$ so that the innermost derivative is $G(x,y)=\frac{\partial }{\partial y}f(2x+3y,xy)=\frac{\partial }{\partial y}F(x,y)$ . By the chain rule

$\begin{aligned} G(x,y)=\frac{\partial }{\partial y}F(x,y) &=f_1(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_2(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_1(2x+3y,xy)+xf_2(2x+3y,xy) \end{aligned}$

Here the subscript $1$ means take the partial derivative of $f$ with respect to the first argument while holding the second argument fixed, and the subscript $2$ means take the partial derivative of $f$ with respect to the second argument while holding the first argument fixed. Next call the middle derivative $H(x,y)=\frac{\partial }{\partial y}\left(\frac{\partial }{\partial y}f(2x+3y,xy)\right)$ so that

$\begin{aligned} H(x,y) &= \frac{\partial }{\partial y} G(x,y) \\ &=\frac{\partial }{\partial y}\Big(3f_1(2x+3y,xy)+xf_2(2x+3y,xy)\Big) \\ &=3\frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) +x\frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) \end{aligned}$

By the chain rule (twice),

$\begin{aligned} \frac{\partial }{\partial y}\Big(f_1(2x+3y,xy)\Big) &=f_{11}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{12}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \\ \frac{\partial }{\partial y}\Big(f_2(2x+3y,xy)\Big) &=f_{21}(2x+3y,xy)\frac{\partial }{\partial y}(2x+3y)+f_{22}(2x+3y,xy)\frac{\partial }{\partial y}(xy) \\ &= 3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy) \end{aligned}$

so that

$\begin{aligned} H(x,y)&=3\Big(3f_{11}(2x+3y,xy)+xf_{12}(2x+3y,xy) \Big) \\[-0.05in] & +x\Big(3f_{21}(2x+3y,xy)+xf_{22}(2x+3y,xy)\Big) \\ &=9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy) \end{aligned}$

In the last equality we used that $f_{21}(2x+3y,xy)=f_{12}(2x+3y,xy)$ . The notation $f_{21}$ means first differentiate with respect to the second argument and then differentiate with respect to the first argument. For example, if $f(x,y)=e^{2y}\sin x$ , then

$\begin{aligned} f_{21}(x,y) =\frac{\partial }{\partial x} \Big[\frac{\partial }{\partial y}\big(e^{2y}\sin x\big)\Big] =\frac{\partial }{\partial x} \Big[2e^{2y}\sin x\Big] =2e^y\cos x \end{aligned}$

Finally, we get to

$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=\frac{\partial }{\partial x} H(x,y) \\ &=\frac{\partial }{\partial x}\Big(9f_{11}(2x+3y,xy)+6xf_{12}(2x+3y,xy)+x^2f_{22}(2x+3y,xy)\Big)\\ &=9\frac{\partial }{\partial x}\Big(f_{11}(2x+3y,xy)\Big)\\& +6f_{12}(2x+3y,xy) +6x\frac{\partial }{\partial x}\Big(f_{12}(2x+3y,xy)\Big) \\& +2xf_{22}(2x+3y,xy) +x^2\frac{\partial }{\partial x}\Big(f_{22}(2x+3y,xy)\Big) \end{aligned}$

By three applications of the chain rule

$\begin{aligned} \frac{\partial^3}{\partial x\partial y^2}f(2x+3y,xy) &=9\Big(2f_{111}+yf_{112}\Big)\\& +6f_{12}+6x\Big(2f_{121}+yf_{122}\Big)\\& +2xf_{22}+x^2\Big(2f_{221}+yf_{222}\Big)\\ &=6\,f_{12}+2x\,f_{22}+18\,f_{111}+(9y+12x)\,f_{112} +(6xy+2x^2)\,f_{122}+x^2y\,f_{222} \end{aligned}$

All functions on the right hand side have arguments $(2x+3y,xy)$ .

한 번 더 — 비슷한 문제 풀기APEX · 12.5절 · 2번

# 다변수 연쇄법칙계산보통자동 분류

11.CLP-3 · 2.4절 · 10번대학 수준우리말 번역

$z = f(x,y)$ 이고 $x = 2s + t$ , $y = s - t$ 라 하자. 다음 등식이 성립하도록 하는 상수 $a$ , $b$ , $c$ 의 값을 구하세요.

$a\frac{\partial^2 z}{\partial x^2} +b\frac{\partial^2 z}{\partial x\,\partial y} +c\frac{\partial^2 z}{\partial y^2} =\frac{\partial^2 z}{\partial s^2} +\frac{\partial^2 z}{\partial t^2}$

$z = f(x,y)$ 는 매끄러운 함수이므로 연쇄 법칙과 혼합 편도함수의 같음에 대한 클레로의 정리를 적용할 수 있다고 가정해도 됩니다.

영어 원문 보기

Let $z = f(x,y)$ where $x = 2s + t$ and $y = s - t$ . Find the values of the constants $a$ , $b$ and $c$ such that

$a\frac{\partial^2 z}{\partial x^2} +b\frac{\partial^2 z}{\partial x\,\partial y} +c\frac{\partial^2 z}{\partial y^2} =\frac{\partial^2 z}{\partial s^2} +\frac{\partial^2 z}{\partial t^2}$

You may assume that $z = f(x,y)$ 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

이 문제의 표기는 매우 좋지 않다. 기호 $z$ 가 한 등식 안에서 두 가지 다른 뜻으로 쓰였기 때문이다. 좌변에서는 $x$ 와 $y$ 의 함수이고, 우변에서는 $s$ 와 $t$ 의 함수이다. 안타깝게도 이러한 표기의 남용은 매우 흔하다. $s$ 와 $t$ 의 함수를 $F(s,t)$ 로 바꾸어 표기의 충돌을 없애자. 즉,

$F(s,t) = f\big(2s+t\,,\,s-t\big)$

이 새 표기에서는 다음이 성립하도록 $a$ , $b$ , $c$ 를 구하면 된다.

$\begin{aligned} a\frac{\partial^2 f}{\partial x^2} +b\frac{\partial^2 f}{\partial x\partial y} +c\frac{\partial^2 f}{\partial y^2} &=\frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} \end{aligned}$

여기서 우변의 인수는 $(s,t)$ 이고 좌변의 인수는 $\big(2s+t\,,\,s-t\big)$ 이다.

연쇄 법칙에 의해

$\begin{aligned} \frac{\partial F}{\partial s}(s,t)&= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial s}(2s+t) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial s}(s-t) \\ &= 2\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big) \\ \frac{\partial^2 F}{\partial s^2}(s,t) &= 2\frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\right]\\ &= 4\frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) + 2\frac{\partial^2 f}{\partial y\partial x} \big(2s+t\,,\,s-t\big) \\ & + 2\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big) \end{aligned}$

이고

$\begin{aligned} \frac{\partial F}{\partial t}(s,t)&= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial t}(2s+t) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial t}(s-t)\\ &= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big) -\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big) \\ \frac{\partial^2 F}{\partial t^2}(s,t) &= \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\right] -\frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) -\frac{\partial^2 f}{\partial y\partial x}\big(2s+t\,,\,s-t\big) \\ & -\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big) \end{aligned}$

이다. 인수를 생략하면

$\begin{aligned} \frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} &=5\frac{\partial^2 f}{\partial x^2} +2\frac{\partial^2 f}{\partial x\partial y} +2\frac{\partial^2 f}{\partial y^2} \end{aligned}$

이다. 마지막으로 문제의 (좋지 않은) 표기로 되돌리면

$\begin{aligned} \frac{\partial^2 z}{\partial s^2} + \frac{\partial^2 z}{\partial t^2} &=5\frac{\partial^2 z}{\partial x^2} +2\frac{\partial^2 z}{\partial x\partial y} +2\frac{\partial^2 z}{\partial y^2} \end{aligned}$

이므로 $a=5$ 이고 $b=c=2$ 이다.

원본 교재 풀이(영어)

The notation in the statement of this question is horrendous — the symbol $z$ is used with two different meanings in one equation. On the left hand side, it is a function of $x$ and $y$ , and on the right hand side, it is a function of $s$ and $t$ . Unfortunately that abuse of notation is also very common. Let us undo the notation conflict by renaming the function of $s$ and $t$ to $F(s,t)$ . That is,

$F(s,t) = f\big(2s+t\,,\,s-t\big)$

In this new notation, we are being asked to find $a$ , $b$ and $c$ so that

$\begin{aligned} a\frac{\partial^2 f}{\partial x^2} +b\frac{\partial^2 f}{\partial x\partial y} +c\frac{\partial^2 f}{\partial y^2} &=\frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} \end{aligned}$

with the arguments on the right hand side being $(s,t)$ and the arguments on the left hand side being $\big(2s+t\,,\,s-t\big)$ .

By the chain rule,

$\begin{aligned} \frac{\partial F}{\partial s}(s,t)&= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial s}(2s+t) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial s}(s-t) \\ &= 2\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big) \\ \frac{\partial^2 F}{\partial s^2}(s,t) &= 2\frac{\partial }{\partial s}\left[\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\right] + \frac{\partial }{\partial s}\left[\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\right]\\ &= 4\frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) + 2\frac{\partial^2 f}{\partial y\partial x} \big(2s+t\,,\,s-t\big) \\ & + 2\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big) \end{aligned}$

and

$\begin{aligned} \frac{\partial F}{\partial t}(s,t)&= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial t}(2s+t) +\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\frac{\partial }{\partial t}(s-t)\\ &= \frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big) -\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big) \\ \frac{\partial^2 F}{\partial t^2}(s,t) &= \frac{\partial }{\partial t}\left[\frac{\partial f}{\partial x}\big(2s+t\,,\,s-t\big)\right] -\frac{\partial }{\partial t}\left[\frac{\partial f}{\partial y}\big(2s+t\,,\,s-t\big)\right] \\ &= \frac{\partial^2 f}{\partial x^2}\big(2s+t\,,\,s-t\big) -\frac{\partial^2 f}{\partial y\partial x}\big(2s+t\,,\,s-t\big) \\ & -\frac{\partial^2 f}{\partial x\partial y} \big(2s+t\,,\,s-t\big) +\frac{\partial^2 f}{\partial y^2}\big(2s+t\,,\,s-t\big) \end{aligned}$

Suppressing the arguments

$\begin{aligned} \frac{\partial^2 F}{\partial s^2} + \frac{\partial^2 F}{\partial t^2} &=5\frac{\partial^2 f}{\partial x^2} +2\frac{\partial^2 f}{\partial x\partial y} +2\frac{\partial^2 f}{\partial y^2} \end{aligned}$

Finally, translating back into the (horrendous) notation of the question

$\begin{aligned} \frac{\partial^2 z}{\partial s^2} + \frac{\partial^2 z}{\partial t^2} &=5\frac{\partial^2 z}{\partial x^2} +2\frac{\partial^2 z}{\partial x\partial y} +2\frac{\partial^2 z}{\partial y^2} \end{aligned}$

so that $a=5$ and $b=c=2$ .

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 26번

# 다변수 연쇄법칙계산보통자동 분류

12.CLP-3 · 2.4절 · 11번대학 수준우리말 번역

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

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

를 함수 $F$ 의 편도함수를 사용하여 나타내세요.

영어 원문 보기

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

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

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

정답·풀이 보기

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

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

원본 교재 풀이(영어)

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

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

한 번 더 — 비슷한 문제 풀기CLP-3 · 2.4절 · 10번

# 다변수 연쇄법칙계산보통자동 분류

13.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

14.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

15.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

16.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

17.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

18.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

19.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

20.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

다변수 미분 전체 문제