호의 길이와 곡률

• Solution 1: We consider the situation geometrically. If we plot $\mathbf{R}$ in space (of the relevant dimension), regardless of its parametrization, the derivative at a point will give a vector tangent to $\mathbf{R}$, in the direction the curve moves when the parameter is increasing. Since $\mathbf{a}(t_0)$ and $\mathbf{b}(s_0)$ describe the same spot on the curve, $\mathbf{a}'(t_0)$ and $\mathbf{b}'(s_0)$ will be parallel (Since we specified the derivatives are nonzero, there's no messiness about vectors being parallel to a zero vector.) — they're both tangent to the same piece of curve. Furthermore, as $t$ increases, so does $s$, so the direction of increasing $t$ is the same as the direction of increasing $s$. Therefore, A. holds.

Now we consider the magnitudes of the vectors, to rule out E. Recall $|\mathbf{a}'(t)|$ is the speed at which the curve changes relative to $t$; this could be any (nonnegative) number. By the same token, $|\mathbf{b}'(s)|=1$. So, $\mathbf{b}'(s_0)$ is a unit vector, while $\mathbf{a}'(t_0)$ may or may not be. Then the two vectors are not necessarily equal (although they could be).

So, the best answer is A.

• Solution 2: The chain rule gives us a relationship between $\mathbf{b}'(s)$ and $\mathbf{a}'(t)$.

$$\begin{aligned} \frac{\mathrm{d}\mathbf{b}}{\mathrm{d}s}&=\frac{\mathrm{d}}{\mathrm{d}s}[\mathbf{a}(t(s))]=\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}\,\frac{\mathrm{d}t}{\mathrm{d}s} \end{aligned}$$

So, the vectors $\frac{\mathrm{d}\mathbf{b}}{\mathrm{d}s}$ and $\frac{\mathrm{d}\mathbf{a}}{\mathrm{d}t}$ differ only by the scalar function $\frac{\mathrm{d}t}{\mathrm{d}s}$. So, at any point along the curve, these vectors are parallel.

Furthermore, we know that $t$ and $s$ are positively correlated: as $t$ increases, so does $s$, because we're covering more arclength. So, $\frac{\mathrm{d}t}{\mathrm{d}s}$ is nonnegative. Furthermore, since the derivatives are nonzero, $\frac{\mathrm{d}t}{\mathrm{d}s}$ is nonzero. So, $\mathbf{b}'(s_0)$ and $\mathbf{a}'(t_0)$ are positive scalar multiples of each other. That is, they are parallel, and pointing in the same direction. However, unless $\frac{\mathrm{d}t}{\mathrm{d}s}=1$ (that is, $t(s)=s+C$ for some constant $C$), the vectors do not have the same magnitude, and hence are not equal.

So, A is the best solution.

Using $\arctan t = z$, and so $t=\tan z$:

$$\begin{aligned} \mathbf{r}(t)&=\left(\frac{1}{\sqrt{1+t^2}},~ \frac{\arctan t}{\sqrt{1+t^{-2}}},~\arctan t\right)\\ &=\left(\frac{1}{\sqrt{1+\tan^2z}},~ \frac{z}{\sqrt{1+\cot^{2}z}},~z\right)\\ &=\left(\frac{1}{|\sec z|},~ \frac{z}{|\csc z|},~z\right)\\ &=\left(|\cos z|,~ z|\sin z|,~z\right) \\ \text{Since $0 \le t $, and $\arctan t < \pi/2$ we have $0 \le z < \pi/2$, so $\cos z$ and $\sin z$ are both nonnegative.} \\ &=\left(\cos z,~ z\sin z,~z\right) \end{aligned}$$

If we didn't have the restricted domain, this would make a spiral going up: $z$ is both the height of the spiral and a radian measure. The $\hat{\boldsymbol{\imath}}$-component of the spiral stays between $-1$ and $1$, while the $\hat{\boldsymbol{\jmath}}$-component increases. So, our spiral gets increasingly “wide," while staying the same “thickness."

Due to the restricted domain, our actual curve is only one-quarter of a “turn" of this spiral, indicated in red above.

The parameter $z$ is a measure of height, and it is also a radian measure as the spiral turns.

$$\begin{aligned} \mathbf{r}(t)&=(\tfrac12 t^2 , \tfrac13 t^3)\\ \mathbf{r}'(t)&=(t,t^2)\\ |\mathbf{r}'(t)|&=\sqrt{t^2+t^4}=|t|\sqrt{1+t^2}\\ s(t)&=\int_{-1}^t|x|\sqrt{1+x^2}\mathrm{d}x=\begin{cases} \int_{-1}^t -x\sqrt{1+x^2}\mathrm{d}x & \text{when } t \le 0\\ \int_{-1}^0 -x\sqrt{1+x^2}\mathrm{d}x +\int_{0}^t x\sqrt{1+x^2}\mathrm{d}x & \text{when } t > 0\\ \end{cases} \\ \text{Let $u=1+x^2$, $\frac12\mathrm{d}u=x\mathrm{d}x$} \\ &=\begin{cases} -\int_{2}^{1+t^2} \frac{1}{2}\sqrt{u}\mathrm{d}u & \text{when } t \le 0\\ -\int_{2}^1 \frac12\sqrt{u}\mathrm{d}u+ \int_{1}^{1+t^2} \frac12\sqrt{u}\mathrm{d}u & \text{when } t > 0\\ \end{cases} \\&=\begin{cases} -\frac{1}{3}u^{3/2}|_{2}^{1+t^2} & \text{when } t \le 0\\ -\frac{1}{3}u^{3/2}|_{2}^{1}+ \frac13u^{3/2}|_{1}^{1+t^2} & \text{when } t > 0\\ \end{cases} \\&=\begin{cases} \frac{2^{3/2}}{3}-\frac{1}{3}(1+t^2)^{3/2} & \text{when } t \le 0\\ -\frac23+\frac{2^{3/2}}{3}+\frac13(1+t^2)^{3/2} & \text{when } t > 0\\ \end{cases} \\ \text{Solving for $t$ in terms of $s$:} \\ 1+t^2&=\begin{cases}(2\sqrt2-3s)^{2/3} &\text{when } t \le 0\\ (3s+2-2\sqrt{2})^{2/3} &\text{when }t>0\end{cases}\\ t^2&=\begin{cases}(2\sqrt2-3s)^{2/3}-1 &\text{when } t \le 0\\ (3s+2-2\sqrt{2})^{2/3}-1 &\text{when }t>0 \end{cases} \\ \text{Remembering that $\sqrt{t^2}=|t|$:} \\ t&=\begin{cases}-\sqrt{(2\sqrt2-3s)^{2/3}-1} &\text{when } t \le 0\\ \sqrt{(3s+2-2\sqrt{2})^{2/3}-1} &\text{when }t>0 \end{cases} \\ \text{Noting that $t=0$ when $s=\tfrac{1}{3}(2\sqrt2-1)$, we find our reparametrization of $(\tfrac12t^2,\tfrac13t^3)$.} \\ \mathbf{R}(s)&=\begin{cases} \Big(\frac12\left[ (2\sqrt2-3s)^{2/3}-1\right],-\frac13\left[(2\sqrt2-3s)^{2/3}-1\right]^{3/2}\Big)&\text{ when }s\le \frac13(2\sqrt2-1)\\ \Big(\frac12\left[(3s+2-2\sqrt2)^{2/3}-1 \right],\frac13\left[(3s+2-2\sqrt2)^{2/3}-1\right]^{3/2}\Big)&\text{ when }s>\frac13(2\sqrt2-1) \end{cases} \end{aligned}$$

Remark: after a computation with this much detail, it's nice to find a few points to check, to verify that our answer is reasonable. For instance, when $s=0$, $t$ should be $-1$, and vice-versa. Also, we found that $t=0$ corresponds to $s=\frac13(2\sqrt2-1)$. So, we should be able to verify that $\mathbf{r}(0)=\mathbf{R}\left(\frac13(2\sqrt2-1)\right)$ and $\mathbf{r}(-1)=\mathbf{R}(0)$.

• Solution 1: Note that $(\cos t+\sin t )^2+(\sin t - \cos t)^2=2$ for all $t$. So, the points $(x,y)$ of our curve lie on $x^2+y^2=2$, which is a circle of radius $\sqrt{2}$. Indeed

$$\begin{aligned} x(t)&= \cos t+\sin t =\sqrt{2}\big[\cos t \cos\tfrac{\pi}{4} +\sin t \sin\tfrac{\pi}{4}\big] \\ &=\sqrt{2}\cos\big(t-\tfrac{\pi}{4}\big) \\ y(t)&= \sin t-\cos t =\sqrt{2}\big[\sin t \cos\tfrac{\pi}{4}-\cos t\sin \tfrac{\pi}{4}\big] \\ &=\sqrt{2}\sin\big(t-\tfrac{\pi}{4}\big) \end{aligned}$$

So, $\mathbf{r}(t)$ circumnavigates a circle of radius $\sqrt{2}$ and consequently has curvature $\kappa =\frac{1}{\sqrt{2}}$.

• Solution 2: We use the formula $\kappa = \dfrac{|\mathbf{v}(t) \times \mathbf{a}(t)|}{\left| \left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^3\right|}$, remembering that $\mathbf{v}(t)=\mathbf{r}'(t)$, $\mathbf{a}(t)=\mathbf{r}''(t)$, and $\frac{\mathrm{d}s}{\mathrm{d}t}=\left| \mathbf{r}'(t)\right|$.

$$\begin{aligned} \mathbf{v}(t)&=\mathbf{r}'(t)=(-\sin t +\cos t, \cos t + \sin t)\\ \mathbf{a}(t)&=\mathbf{r}''(t)=(-\cos t -\sin t, -\sin t + \cos t)\\ \mathbf{v}(t) \times \mathbf{a}(t)&=\big[(-\sin t + \cos t)^2+(\cos t + \sin t)^2\big]\hat{\mathbf{k}}=2\hat{\mathbf{k}}\\ \frac{\mathrm{d}s}{\mathrm{d}t}&=\left|\frac{\mathrm{d}\mathbf{v}}{\mathrm{d}t} \right|=\sqrt{(-\sin t + \cos t)^2+(\cos t + \sin t )^2}=\sqrt2\\ \kappa &=\left|\frac{\mathbf{v}(t) \times \mathbf{a}(t)}{\left(\frac{\mathrm{d}s}{\mathrm{d}t}\right)^3}\right|=\left|\frac{2\hat{\mathbf{k}}}{\sqrt{2}^3}\right|=\frac{1}{\sqrt 2} \end{aligned}$$

For the given ellipse

$$\begin{aligned} \mathbf{r}(t)&= a\cos t\ \hat{\boldsymbol{\imath}} +b\sin t\ \hat{\boldsymbol{\jmath}} \\ \mathbf{v}(t)&= -a\sin t\ \hat{\boldsymbol{\imath}} +b\cos t\ \hat{\boldsymbol{\jmath}} \\ |\mathbf{v}(t)|& = \sqrt{a^2\sin^2t+b^2\cos^2 t}\\ \mathbf{a}(t)&= -a\cos t\ \hat{\boldsymbol{\imath}} -b\sin t\ \hat{\boldsymbol{\jmath}} \\ \mathbf{v}(t)\times\mathbf{a}(t) &=\det\left[\begin{matrix} \hat{\boldsymbol{\imath}} & \hat{\boldsymbol{\jmath}} & \hat{\mathbf{k}} \\ -a\sin t & b\cos t & 0 \\ -a\cos t & -b\sin t & 0 \end{matrix}\right] = ab\,\hat{\mathbf{k}}\\ \kappa (t)&=\frac{|\mathbf{v}(t)\times\mathbf{a}(t)|}{|\mathbf{v}(t)|^3} =\frac{ab}{{[a^2\sin^2t+b^2\cos^2 t]}^{3/2}} \end{aligned}$$

Hence the maximum (minimum) curvature is achieved when the denominator is a minimum (maximum) which is the case when $\sin t =0$ ($\cos t=0$). So $\kappa _{\rm max}=\frac{a}{b^2}$ and $\kappa _{\rm min}=\frac{b}{a^2}$.

The velocity vector is

$$\mathbf{v}(\theta)=x'(\theta)\,\hat{\boldsymbol{\imath}}+y'(\theta)\,\hat{\boldsymbol{\jmath}} = \cos\big(\tfrac{1}{2}\pi\theta^2\big)\,\hat{\boldsymbol{\imath}} + \sin\big(\tfrac{1}{2}\pi\theta^2\big)\,\hat{\boldsymbol{\jmath}}$$

Consequently the speed

$$\frac{\mathrm{d}s}{\mathrm{d}\theta}(\theta)= |\mathbf{v}(\theta)|=1 \implies s(\theta)=\theta+s(0)$$

Since $s(\theta)$ is zero when $\theta=0$, we have $s(\theta)=\theta$ and hence

$$\hat{\mathbf{T}}(s)=\mathbf{v}(s) = \cos\big(\tfrac{1}{2}\pi s^2\big)\,\hat{\boldsymbol{\imath}} + \sin\big(\tfrac{1}{2}\pi s^2\big)\,\hat{\boldsymbol{\jmath}}$$

so that

$$\kappa (s)=\left|\frac{\mathrm{d}\hat{\mathbf{T}}}{\mathrm{d}s}(s)\right| =\big|-\pi s\sin\big(\tfrac{1}{2}\pi s^2\big)\,\hat{\boldsymbol{\imath}} +\pi s \cos\big(\tfrac{1}{2}\pi s^2\big)\,\hat{\boldsymbol{\jmath}}\big| =\pi s$$