단위접선·법선벡터 연습문제

1.CLP-4 · 1.4절 · 31번대학 수준우리말 번역

두 언덕 사이의 경주 트랙이 매개변수 곡선

$\mathbf{r}(\theta) = \Big(4 \cos\theta\,,\, 2\sin\theta\,,\, \tfrac{1}{4}\cos(2\theta)\Big),\qquad 0 \le \theta \le 2\pi$

로 나타내어진다.

(a) 점 $\big(-4, 0, \frac{1}{4}\big)$ 에서 트랙의 곡률을 구하세요. (b) 점 $\big(-4, 0, \frac{1}{4}\big)$ 에서 굽은 부분을 가장 잘 근사하는 원의 반지름(즉, 그 점에서의 접촉원의 반지름)을 구하세요. (c) 자동차가 트랙을 따라 달리며, 시각 $t$ 에서의 위치가 $\mathbf{r}(t^2)$ 으로 주어진다. ( $t$ 와 $\theta$ 의 관계는 $\theta = t^2$ 임에 유의하세요.) 다음 양을 구하세요.

(i) 점 $\big(-4, 0, \frac{1}{4}\big)$ 에서의 속력 (ii) 점 $\big(-4, 0, \frac{1}{4}\big)$ 에서의 가속도 (iii) 점 $\big(-4, 0, \frac{1}{4}\big)$ 에서 가속도의 법선 성분의 크기

영어 원문 보기

A race track between two hills is described by the parametric curve

$\mathbf{r}(\theta) = \Big(4 \cos\theta\,,\, 2\sin\theta\,,\, \tfrac{1}{4}\cos(2\theta)\Big),\qquad 0 \le \theta \le 2\pi$

(a) Compute the curvature of the track at the point $\big(-4, 0, \frac{1}{4}\big)$ . (b) Compute the radius of the circle that best approximates the bend at the point $\big(-4, 0, \frac{1}{4}\big)$ (that is, the radius of the osculating circle at that point). (c) A car drives down the track so that its position at time $t$ is given by $\mathbf{r}(t^2)$ . (Note the relationship between $t$ and $\theta$ is $\theta = t^2$ ). Compute the following quantities.

(i) The speed at the point $\big(-4, 0, \frac{1}{4}\big)$ . (ii) The acceleration at the point $\big(-4, 0, \frac{1}{4}\big)$ . (iii) The magnitude of the normal component of the acceleration at the point

$\big(-4, 0, \frac{1}{4}\big)$ .

정답·풀이 보기

(a) 주어진 곡선에서 $\mathbf{r}(\pi) = \big(-4,0,\frac{1}{4}\big)$ 이고

$\begin{aligned} \mathbf{r}(\theta) &= \Big(4 \cos\theta\,,\, 2\sin\theta\,,\, \frac{1}{4}\cos(2\theta)\Big) \\ \mathbf{v}(\theta)=\mathbf{r}'(\theta) &= \Big(-4 \sin\theta\,,\, 2\cos\theta\,,\, -\frac{1}{2}\sin(2\theta)\Big) \\ \mathbf{a}(\theta) = \mathbf{r}''(\theta) &= \Big(-4 \cos\theta\,,\, -2\sin\theta\,,\, -\cos(2\theta)\Big) \\ \mathbf{v}(\pi) &= \big(0\,,\, -2\,,\, 0\big) \\ \mathbf{a}(\pi) &= \big(4\,,\, 0\,,\, -1\big) \\ \mathbf{v}(\pi) \times \mathbf{a}(\pi) &= \big(2\,,\, 0\,,\, 8\big) \end{aligned}$

따라서 $\theta=\pi$ 에서의 곡률은

$\begin{aligned} \kappa (\pi)&=\frac{|\mathbf{v}(\pi) \times \mathbf{a}(\pi)|}{|\mathbf{v}(\pi)|^3} =\frac{|(2,0,8)|}{|(0,-2,0)|^3} =\frac{\sqrt{17}}{4} \end{aligned}$

이다.

(b) 반지름은

$\frac{1}{\kappa (\pi)} = \frac{4}{\sqrt{17}}$

이다.

$\begin{aligned} \mathbf{R}'(t)&=2t\,\mathbf{r}'(t^2) \\ \mathbf{R}''(t) &= 2\,\mathbf{r}'(t^2) +4t^2\,\mathbf{r}''(t^2) \end{aligned}$

이다. 특히,

$\begin{aligned} \mathbf{R}(\sqrt{\pi}) &= \Big(-4\,,\, 0\,,\, \frac{1}{4}\Big) \\ \mathbf{R}'(\sqrt{\pi}) &= 2\sqrt{\pi}\,\mathbf{v}(\pi) =\big(0\,,\, -4\sqrt{\pi}\,,\, 0\big) \\ \text{속력} = \big|\mathbf{R}'(\sqrt{\pi})\big| &=4\sqrt{\pi} \\ \text{가속도}=\mathbf{R}''(\sqrt{\pi}) &= 2\,\mathbf{v}(\pi)+4\pi\,\mathbf{a}(\pi) = \big( 16\pi\,,\, -4\,,\, -4\pi\big) \end{aligned}$

이다.

가속도의 법선 성분의 크기는

$\begin{aligned} \kappa \Big(\frac{\mathrm{d}s}{\mathrm{d}t}\Big)^2 = \frac{\sqrt{17}}{4}\big(4\sqrt{\pi}\big)^2 =4\sqrt{17}\,\pi \end{aligned}$

이다.

원본 교재 풀이(영어)

(a) For the specified curve $\mathbf{r}(\pi) = \big(-4,0,\frac{1}{4}\big)$ and

So the curvature at $\theta=\pi$ is

$\begin{aligned} \kappa (\pi)&=\frac{|\mathbf{v}(\pi) \times \mathbf{a}(\pi)|}{|\mathbf{v}(\pi)|^3} =\frac{|(2,0,8)|}{|(0,-2,0)|^3} =\frac{\sqrt{17}}{4} \end{aligned}$

(b) The radius is

$\frac{1}{\kappa (\pi)} = \frac{4}{\sqrt{17}}$

$\begin{aligned} \mathbf{R}'(t)&=2t\,\mathbf{r}'(t^2) \\ \mathbf{R}''(t) &= 2\,\mathbf{r}'(t^2) +4t^2\,\mathbf{r}''(t^2) \end{aligned}$

In particular,

$\begin{aligned} \mathbf{R}(\sqrt{\pi}) &= \Big(-4\,,\, 0\,,\, \frac{1}{4}\Big) \\ \mathbf{R}'(\sqrt{\pi}) &= 2\sqrt{\pi}\,\mathbf{v}(\pi) =\big(0\,,\, -4\sqrt{\pi}\,,\, 0\big) \\ \text{speed} = \big|\mathbf{R}'(\sqrt{\pi})\big| &=4\sqrt{\pi} \\ \text{acceleration}=\mathbf{R}''(\sqrt{\pi}) &= 2\,\mathbf{v}(\pi)+4\pi\,\mathbf{a}(\pi) = \big( 16\pi\,,\, -4\,,\, -4\pi\big) \end{aligned}$

The normal component of the acceleration has magnitude

$\begin{aligned} \kappa \Big(\frac{\mathrm{d}s}{\mathrm{d}t}\Big)^2 = \frac{\sqrt{17}}{4}\big(4\sqrt{\pi}\big)^2 =4\sqrt{17}\,\pi \end{aligned}$

한 번 더 — 비슷한 문제 풀기APEX · 11.4절 · 4번

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