벡터와 공간 전체 문제

(a) Parametrize $C$ by $x$. Since $y=x^2$ and $z=\frac{2}{3}x^3$,

$$\begin{aligned} \mathbf{r}(x)&=x\,\hat{\boldsymbol{\imath}}+x^2\,\hat{\boldsymbol{\jmath}}+\frac{2}{3}x^3\,\hat{\mathbf{k}}\\ \mathbf{r}'(x)&=\hat{\boldsymbol{\imath}}+2x\,\hat{\boldsymbol{\jmath}}+2x^2\,\hat{\mathbf{k}}\\ \mathbf{r}''(x)&=2\,\hat{\boldsymbol{\jmath}}+4x\,\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}x} &=|\mathbf{r}'(x)| =\sqrt{1+4x^2+4x^4}=1+2x^2 \end{aligned}$$

and $$\int_C \ \mathrm{d}s =\int_0^3 \frac{\mathrm{d}s}{\mathrm{d}x}\ \mathrm{d}x =\int_0^3 \big(1+2x^2\big)\ \mathrm{d}x ={\Big[x+\frac{2}{3}x^3\Big]}_0^3 =21$$

(b) The particle travelled a distance of 21 units in $\frac{7}{2}$ time units. This corresponds to a speed of $\frac{21}{7/2}=6$.

(c) Denote by $\mathbf{R}(t)$ the position of the particle at time $t$. Then $$\mathbf{R}(t)=\mathbf{r}\big(x(t)\big)\implies \mathbf{R}'(t)=\mathbf{r}'\big(x(t)\big)\frac{\mathrm{d}x}{\mathrm{d}t}$$ By parts (a) and (b) and the chain rule $$6=\frac{\mathrm{d}s}{\mathrm{d}t}=\frac{\mathrm{d}s}{\mathrm{d}x}\frac{\mathrm{d}x}{\mathrm{d}t}=(1+2x^2)\frac{\mathrm{d}x}{\mathrm{d}t} \implies \frac{\mathrm{d}x}{\mathrm{d}t}=\frac{6}{1+2x^2}$$ In particular, the particle is at $\big(1,1,\frac{2}{3}\big)$ at $x=1$. At this time $\frac{\mathrm{d}x}{\mathrm{d}t}=\frac{6}{1+2\times 1}=2$ and $$\mathbf{R}'=\mathbf{r}'\big(1\big)\frac{\mathrm{d}x}{\mathrm{d}t} =\big(\hat{\boldsymbol{\imath}}+2\,\hat{\boldsymbol{\jmath}}+2\,\hat{\mathbf{k}}\big)2 =2\hat{\boldsymbol{\imath}}+4\,\hat{\boldsymbol{\jmath}}+4\,\hat{\mathbf{k}}$$

(d) By the product and chain rules, $$\mathbf{R}'(t)=\mathbf{r}'\big(x(t)\big)\frac{\mathrm{d}x}{\mathrm{d}t}\implies \mathbf{R}''(t)=\mathbf{r}''\big(x(t)\big){\Big(\frac{\mathrm{d}x}{\mathrm{d}t}\Big)}^2 +\mathbf{r}'\big(x(t)\big)\frac{\mathrm{d^2}x}{\mathrm{d}{t}^2}$$ Applying $\frac{\mathrm{d}}{\mathrm{d}t}$ to $6=\big(1+2x(t)^2\big)\frac{\mathrm{d}x}{\mathrm{d}t}(t)$ gives $$0=4x{\Big(\frac{\mathrm{d}x}{\mathrm{d}t}\Big)}^2+(1+2x^2)\frac{\mathrm{d^2}x}{\mathrm{d}{t}^2}$$ In particular, when $x=1$ and $\frac{\mathrm{d}x}{\mathrm{d}t}=2$, $0=4\times 1\big(2\big)^2+(3)\frac{\mathrm{d^2}x}{\mathrm{d}{t}^2}$ gives $\frac{\mathrm{d^2}x}{\mathrm{d}{t}^2}=-\frac{16}{3}$ and $$\mathbf{R}''=\big(2\,\hat{\boldsymbol{\jmath}}+4\,\hat{\mathbf{k}}\big)\big(2\big)^2 -\big(\hat{\boldsymbol{\imath}}+2\,\hat{\boldsymbol{\jmath}}+2\,\hat{\mathbf{k}}\big)\frac{16}{3} = -\frac{8}{3}\big(2\hat{\boldsymbol{\imath}}+\,\hat{\boldsymbol{\jmath}}-2\,\hat{\mathbf{k}}\big)$$

The question is already set up as an $xy$-plane, with the camera at the origin, so the vector in the direction the camera is pointing is $(x(t),y(t))$. Let $\theta$ be the angle the camera makes with the positive $x$-axis (due east). The camera, the object, and the due-east direction (positive $x$-axis) make a right triangle.

$$\begin{aligned} \tan\theta&=\frac{y}{x}\\ \text{Differentiating implicitly with respect to $t$:} \\ \sec^2\theta\,\frac{\mathrm{d}\theta}{\mathrm{d}t}&=\frac{xy'-yx'}{x^2}\\ \frac{\mathrm{d}\theta}{\mathrm{d}t}&=\cos^2\theta\left(\frac{xy'-yx'}{x^2}\right)=\left(\frac{x}{\sqrt{x^2+y^2}}\right)^2\left(\frac{xy'-yx'}{x^2}\right)=\frac{xy'-yx'}{{x^2+y^2}} \end{aligned}$$

Using the Theorem of Pappus, we can calculate the surface area and volume of a pipe with the same length and radius as this pipe. So, we need to find the length of the pipe, $L$.

$$\begin{aligned} \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}&=(\sqrt{2t},t,1)\\ \left| \frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}\right|&=\sqrt{2t+t^2+1}=|t+1|\\ L&=\int_0^{10}(t+1)\mathrm{d}t=60 \end{aligned}$$

A pipe with radius 3 and length 60 has surface area $60(2\pi\cdot 3)=360\pi$ and volume $60(\pi \cdot3^2)=540\pi$.

Define $\mathbf{u}(t)=e^{\alpha t}\frac{d\mathbf{r}}{dt}(t)$. Then

$$\begin{aligned} \frac{\mathrm{d}\mathbf{u}}{\mathrm{d}t}(t)&=\alpha e^{\alpha t}\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t) +e^{\alpha t}\frac{d^2\mathbf{r}}{dt^2}(t)\\ &=\alpha e^{\alpha t}\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t) -ge^{\alpha t}\hat{\mathbf{k}} -\alpha e^{\alpha t}\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(t)\\ &=-ge^{\alpha t}\hat{\mathbf{k}} \end{aligned}$$

Integrating both sides of this equation from $t=0$ to $t=T$ gives

$$\mathbf{u}(T)-\mathbf{u}(0)=-g\frac{e^{\alpha T}-1}{\alpha}\hat{\mathbf{k}}$$

so that

$$\mathbf{u}(T)=\mathbf{u}(0)-g\frac{e^{\alpha T}-1}{\alpha}\hat{\mathbf{k}} =\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(0)-g\frac{e^{\alpha T}-1}{\alpha}\hat{\mathbf{k}} =\mathbf{v}_0-g\frac{e^{\alpha T}-1}{\alpha}\hat{\mathbf{k}}$$

Substituting in $\mathbf{u}(T)=e^{\alpha t}\frac{d\mathbf{r}}{dt}(T)$ and multiplying through by $e^{-\alpha T}$ gives

$$\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}(T) =e^{-\alpha T}\mathbf{v}_0-g\frac{1-e^{-\alpha T}}{\alpha}\hat{\mathbf{k}}$$

Integrating both sides of this equation from $T=0$ to $T=t$ gives

$$\begin{aligned} \mathbf{r}(t)-\mathbf{r}(0) =\frac{e^{-\alpha t}-1}{-\alpha}\mathbf{v}_0-g\frac{ t}{\alpha}\hat{\mathbf{k}} +g\frac{e^{-\alpha t}-1}{-\alpha^2}\hat{\mathbf{k}} \end{aligned}$$

so that

$$\begin{aligned} \mathbf{r}(t)=\mathbf{r}_0-\frac{e^{-\alpha t}-1}{\alpha}\mathbf{v}_0 +g\frac{1-\alpha t-e^{-\alpha t}}{\alpha^2}\hat{\mathbf{k}} \end{aligned}$$

In this equation, $s$ stands for arclength.

When we take a very small interval from $t$ to $t+h$, the change in arclength $s(t+h)-s(t)$ is approximately $|\mathbf{r}(t+h)-\mathbf{r}(t)|$, because our curve is approximated by a straight line. So, $\frac{s(t+h)-s(t)}{h} \approx \frac{|\mathbf{r}(t+h)-\mathbf{r}(t)|}{h}$, leading to $\frac{\mathrm{d}s}{\mathrm{d}t}=|\frac{\mathrm{d}\mathbf{r}}{\mathrm{d}t}|=|\mathbf{v}(t)|$.

The magnitude of velocity is speed; in this text we generally call this $v$. That is, $v=|\mathbf{v}(t)|$. This leads to the potentially confusing (but standard) convention that $s$ stands for arclength, while $v$ stands for speed.

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

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

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

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

(c) Set $\mathbf{R}(t) = \mathbf{r}(t^2)$. Then

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

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

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

From the text, the skateboarder will make it all the way around when $\frac{5}{2}(5) \le\frac{E}{mg}$. Energy $E$ is given by $E=\frac12m|\mathbf{v}|^2+mgy$, the sum of the kinetic and potential energy of the system. At $y=0$, all the energy is kinetic, so $E=\frac12m|\mathbf{v}|^2$, where $|\mathbf{v}|$ is the skater's velocity at the bottom of the culvert.

So, we solve:

$$\begin{aligned} \frac{25}{2}&\le\frac{E}{mg}=\frac{\frac12m|\mathbf{v}|^2}{m\cdot9.8}\\ |\mathbf{v}|&\ge5\sqrt{9.8} \end{aligned}$$

So, a speed of $5\sqrt{9.8}$ m/s or higher is needed. (That's about 56 kph.)