벡터함수의 미분과 적분 연습문제 — CLP-3 · 1.6절 · 23번

CLP-3 · 1.6절 · 23번대학 수준우리말 번역

입자가 곡면 $z^2=12y$ 와 $18x=yz$ 의 교선 $\mathcal{C}$ 를 따라 위쪽 방향으로 움직인다. 입자가 $(1,3,6)$ 에 있을 때 속도 $\mathbf{v}$ 와 가속도 $\mathbf{a}$ 는 다음과 같다. $\mathbf{v} =6\,\hat{\boldsymbol{\imath}}+12\,\hat{\boldsymbol{\jmath}}+12\,\hat{\mathbf{k}}\qquad \mathbf{a} = 27\,\hat{\boldsymbol{\imath}}+30\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}$

(a) $u=\frac{z}{6}$ 를 매개변수로 사용하여 $\mathcal{C}$ 의 벡터 매개변수 방정식을 쓰세요. (b) $(0,0,0)$ 에서 $(1,3,6)$ 까지 $\mathcal{C}$ 의 길이를 구하세요. (c) $u=u(t)$ 가 시각 $t$ 에서 입자의 위치에 해당하는 매개변수 값일 때, 입자가 $(1,3,6)$ 에 있을 때의 $\frac{\mathrm{d}u}{\mathrm{d}t}$ 를 구하세요. (d) 입자가 $(1,3,6)$ 에 있을 때의 $\frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}$ 를 구하세요.

영어 원문 보기

A particle moves along the curve $\mathcal{C}$ of intersection of the surfaces $z^2=12y$ and $18x=yz$ in the upward direction. When the particle is at $(1,3,6)$ its velocity $\mathbf{v}$ and acceleration $\mathbf{a}$ are given by $\mathbf{v} =6\,\hat{\boldsymbol{\imath}}+12\,\hat{\boldsymbol{\jmath}}+12\,\hat{\mathbf{k}}\qquad \mathbf{a} = 27\,\hat{\boldsymbol{\imath}}+30\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}$

(a) Write a vector parametric equation for $\mathcal{C}$ using $u=\frac{z}{6}$ as a parameter. (b) Find the length of $\mathcal{C}$ from $(0,0,0)$ to $(1,3,6)$ . (c) If $u=u(t)$ is the parameter value for the particle's position at time $t$ , find $\frac{\mathrm{d}u}{\mathrm{d}t}$ when the particle is at $(1,3,6)$ . (d) Find $\frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}$ when the particle is at $(1,3,6)$ .

정답·풀이 보기

(a) $z=6u$ , $y=\frac{z^2}{12}=3u^2$ , $x=\frac{yz}{18}=u^3$ 이므로

$\begin{aligned} \mathbf{r}(u)&=u^3\,\hat{\boldsymbol{\imath}}+3u^2\,\hat{\boldsymbol{\jmath}}+6u\,\hat{\mathbf{k}}\\ \end{aligned}$

(b)

$\begin{aligned} \mathbf{r}'(u)&=3u^2\,\hat{\boldsymbol{\imath}}+6u\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}\\ \mathbf{r}''(u)&=6u\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}\\ \frac{\mathrm{d}s}{\mathrm{d}u}(u)=|\mathbf{r}'(u)| &=\sqrt{9u^4+36u^2+36}=3\big(u^2+2\big)\\ \end{aligned}$

$\int_\mathcal{C} \ ds =\int_0^1 \frac{\mathrm{d}s}{\mathrm{d}u}\ \mathrm{d}u =\int_0^1 3\big(u^2+2\big)\ \mathrm{d}u =\big[u^3+6u\big]_0^1 =7$

(c) 시각 $t$ 에서 입자의 위치를 $\mathbf{R}(t)$ 라 하자. 그러면 $\mathbf{R}(t)=\mathbf{r}\big(u(t)\big)\implies \mathbf{R}'(t)=\mathbf{r}'\big(u(t)\big)\frac{\mathrm{d}u}{\mathrm{d}t}$ 특히 입자가 시각 $t_1$ 에 $(1,3,6)$ 에 있으면 $u(t_1)=1$ 이고 $6\,\hat{\boldsymbol{\imath}}+12\,\hat{\boldsymbol{\jmath}}+12\,\hat{\mathbf{k}}=\mathbf{R}'(t_1)=\mathbf{r}'(1)\frac{\mathrm{d}u}{\mathrm{d}t}(t_1) =\big(3\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}\big)\frac{\mathrm{d}u}{\mathrm{d}t}(t_1)$ 이므로 $\frac{\mathrm{d}u}{\mathrm{d}t}(t_1)=2$ 이다.

(d) 곱의 미분법과 합성함수의 미분법에 의해 $\mathbf{R}'(t)=\mathbf{r}'\big(u(t)\big)\frac{\mathrm{d}u}{\mathrm{d}t}\implies \mathbf{R}''(t)=\mathbf{r}''\big(u(t)\big)\Big(\frac{\mathrm{d}u}{\mathrm{d}t}\Big)^2 +\mathbf{r}'\big(u(t)\big)\frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}$ 특히

$\begin{aligned} 27\,\hat{\boldsymbol{\imath}}+30\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}} &=\mathbf{R}''(t_1)=\mathbf{r}''(1)\Big(\frac{\mathrm{d}u}{\mathrm{d}t}(t_1)\Big)^2 +\mathbf{r}'\big(1\big)\frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}(t_1) \\ &=\big(6\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}\big)2^2+\big(3\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}\big)\frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}(t_1) \end{aligned}$

간단히 하면 $3\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}} =\big(3\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}\big)\frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}(t_1) \implies \frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}(t_1)=1$

원본 교재 풀이(영어)

(a) Since $z=6u$ , $y=\frac{z^2}{12}=3u^2$ and $x=\frac{yz}{18}=u^3$ ,

$\begin{aligned} \mathbf{r}(u)&=u^3\,\hat{\boldsymbol{\imath}}+3u^2\,\hat{\boldsymbol{\jmath}}+6u\,\hat{\mathbf{k}}\\ \end{aligned}$

(b)

$\int_\mathcal{C} \ ds =\int_0^1 \frac{\mathrm{d}s}{\mathrm{d}u}\ \mathrm{d}u =\int_0^1 3\big(u^2+2\big)\ \mathrm{d}u =\big[u^3+6u\big]_0^1 =7$

(c) Denote by $\mathbf{R}(t)$ the position of the particle at time $t$ . Then $\mathbf{R}(t)=\mathbf{r}\big(u(t)\big)\implies \mathbf{R}'(t)=\mathbf{r}'\big(u(t)\big)\frac{\mathrm{d}u}{\mathrm{d}t}$ In particular, if the particle is at $(1,3,6)$ at time $t_1$ , then $u(t_1)=1$ and $6\,\hat{\boldsymbol{\imath}}+12\,\hat{\boldsymbol{\jmath}}+12\,\hat{\mathbf{k}}=\mathbf{R}'(t_1)=\mathbf{r}'(1)\frac{\mathrm{d}u}{\mathrm{d}t}(t_1) =\big(3\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}\big)\frac{\mathrm{d}u}{\mathrm{d}t}(t_1)$ which implies that $\frac{\mathrm{d}u}{\mathrm{d}t}(t_1)=2$ .

(d) By the product and chain rules, $\mathbf{R}'(t)=\mathbf{r}'\big(u(t)\big)\frac{\mathrm{d}u}{\mathrm{d}t}\implies \mathbf{R}''(t)=\mathbf{r}''\big(u(t)\big)\Big(\frac{\mathrm{d}u}{\mathrm{d}t}\Big)^2 +\mathbf{r}'\big(u(t)\big)\frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}$ In particular,

Simplifying $3\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}} =\big(3\,\hat{\boldsymbol{\imath}}+6\,\hat{\boldsymbol{\jmath}}+6\,\hat{\mathbf{k}}\big)\frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}(t_1) \implies \frac{\mathrm{d^2}u}{\mathrm{d}{t}^2}(t_1)=1$

한 번 더 — 비슷한 문제CLP-3 · 1.6절 · 22번우리말 번역

3차원 공간에서 한 입자가 위치 벡터 $\mathbf{r}(t)$ 로 움직이고, $\mathbf{r}''(t)=-\mathbf{r}(t)$ 를 만족한다고 하자. "에너지" $|\mathbf{r}(t)|^2+|\mathbf{r}'(t)|^2$ 이 상수임(즉, $t$ 에 무관함)을 보이세요.

다 풀었으면 정답 확인

$\begin{aligned} \frac{\mathrm{d}}{\mathrm{d}t}\big[|\mathbf{r}(t)|^2+|\mathbf{r}'(t)|^2\big] &=\frac{\mathrm{d}}{\mathrm{d}t}\big[\mathbf{r}(t)\cdot\mathbf{r}(t)+\mathbf{r}'(t)\cdot\mathbf{r}'(t)\big]\\ &=2\mathbf{r}(t)\cdot\mathbf{r}'(t)+2\mathbf{r}'(t)\cdot\mathbf{r}''(t)\\ &=2\mathbf{r}'(t)\cdot\big[\mathbf{r}(t)+\mathbf{r}''(t)\big]\\ &=0\qquad \text{ since $\mathbf{r}''(t)=-\mathbf{r}(t)$ } \end{aligned}$

모든 $t$ 에 대해 $\frac{\mathrm{d}}{\mathrm{d}t}\big[|\mathbf{r}(t)|^2+|\mathbf{r}'(t)|^2\big]=0$ 이므로, $|\mathbf{r}(t)|^2+|\mathbf{r}'(t)|^2$ 은 $t$ 에 무관하다.

이 문제로 이어서 풀기

# 벡터함수의 미분과 적분활용도전자동 분류