호의 길이와 곡률

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 $\mathbf{r}(-1)=\mathbf{R}(0)$.

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

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

The curve is $y=y(x)={x^3}/{3}$. Since $y'(x) = x^2$ and $y''(x) = 2x$, the curvature is

$$\begin{aligned} \kappa (x) &=\frac{\big|\frac{\mathrm{d^2}y}{\mathrm{d}{x}^2}(x)\big|} {\Big[1+\big(\frac{\mathrm{d}y}{\mathrm{d}x}(x)\big)^2\Big]^{3/2}} =\frac{\big|2x\big|} {\big[1+x^4\big]^{3/2}} \end{aligned}$$

We'd like to find the critical points of $\kappa (x)$, but differentiating it looks messy. Since $\kappa (x)$ has only nonnegative values, its maxima correspond the maxima of the function $\kappa ^2(x)$. So, we find the critical points of $\kappa ^2(x)$ instead, to save ourselves some computational toil.

$$\begin{aligned} 0&=\frac{\mathrm{d}}{\mathrm{d}x}\kappa (x)^2 =\frac{\mathrm{d}}{\mathrm{d}x}\frac{4x^2}{{(1+x^4)}^3} =\frac{8x}{{(1+x^4)}^3} - 3\frac{16x^5}{{(1+x^4)}^4} =\frac{8x(1+x^4)-3\times 16x^5}{{(1+x^4)}^4} \\ &=\frac{8x(1-5x^4)}{{(1+x^4)}^4} \end{aligned}$$

Note that $\kappa (0)=0$ and $\kappa (x)\rightarrow 0$ as $x\rightarrow\pm \infty$. So the maximum occurs when $x=\pm 1/\sqrt[4]{5}$.