호의 길이와 곡률

• 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 $\kappa _{\rm min}=\frac{b}{a^2}$.