5.2 解答

1. 曲線 $\vec{r}(t)$ において，点 $\vec{r}(a)$ を通る接線の方程式を考える．接線上に任意の点 $\vec{r}(t)$ を取り，既知の点 $\vec{r}(a)$ と結ぶと， $\vec{t}(t) - \vec{r}(a)$ は接線ベクトル $\vec{r}'(a)$ のスカラー倍で表わせる．したがって，接線の方程式は

$\displaystyle \vec{r}(t) = \vec{r}(a) + t\vec{r}'(a)$

$\displaystyle \vec{r}'(1) = (e^{t},-e^{-t},-\frac{1}{t})\mid_{t=1} = (e,-e^{-1},-1)$

$\displaystyle \vec{r}(t) = \vec{r}(1) + t\vec{r}'(1) = (e,e^{-1},0) + t(e,-e^{-1},-1)$

(b) $\vec{r}(t) = (\cos{\pi t}, \sin{\pi t}, t)$ より $t=2$

での接線ベクトルは

$\displaystyle \vec{r}'(2) = (-\pi \sin{\pi t}, \pi \cos{\pi t}, 1)\mid_{t=2} = (0,\pi,1)$

$\displaystyle \vec{r}(t) = \vec{r}(2) + t\vec{r}'(2) = (1,0,2) + t(0,\pi,1)$

(a) $\vec{r}(t) = (at, bt^2)$ より $x = at, y = bt^2$

．これより

を消去する． $t = \frac{x}{a}$ より $y = bt^2 = b(\frac{x^2}{a^2})$ ．

(b) $\vec{r}(t) = (t^3, t^2)$ より $x = t^3, y = t^2$

．これより

を消去する． $t = x^{1/3}$ より $y = t^2 = x^{2/3}$ ．

を消去する．まず， $x^2 + y^2 = \cos^{2}{2t} + \sin^{2}{2t} = 1$ ．次に， $x = \cos{2z}, y = \sin{2z}$ となるので， $\frac{y}{x} = \tan{2z}$ ．よって， $z = \frac{1}{2}\tan^{-1}{(\frac{y}{x})}$ ．

3. 微小時間 $\Delta t$ に対する弧長 $\Delta s$ は速さ $\Vert\frac{d\vec{r}(t)}{dt}\Vert$ に微小時間 $\Delta t$ をかけたものである．したがって， $\vec{r}(a)$ から $\vec{r}(t)$ までの弧長 $s$

は

$\displaystyle s(t) = \int_{a}^{t} \Vert\frac{d\vec{r}(t)}{dt}\Vert dt$

$\displaystyle \Vert\frac{d\vec{r}(t)}{dt}\Vert = \Vert(1,\frac{\sec{t}\tan{t}}{\sec{t}},0)\Vert = \sqrt{1 + \tan^{2}{t}} = \sec{t}$

$\displaystyle s$	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{4}}\Vert\frac{d\vec{r}(t)}{dt}\Vert dt$
	$\displaystyle =$	$\displaystyle \int_{0}^{\frac{\pi}{4}}\sec{t}dt$
	$\displaystyle =$	$\displaystyle \left[\log\vert\sec{t} + \tan{t}\vert\right]_{0}^{\frac{\pi}{4}}$
	$\displaystyle =$	$\displaystyle \log\vert sec{\frac{\pi}{4}} + \tan{\frac{\pi}{4}}\vert - \log\vert\sec{0} + \tan{0}\vert$
	$\displaystyle =$	$\displaystyle \log\vert\sqrt{2} + 1\vert - \log{1} = \log(1 + \sqrt{2})$

$\displaystyle \Vert\frac{d\vec{r}(t)}{dt}\Vert$	$\displaystyle =$	$\displaystyle \Vert e^{t}(\cos{t}-\sin{t},\sin{t}+\cos{t})\Vert$
	$\displaystyle =$	$\displaystyle e^{t}\sqrt{(\cos{t}-\sin{t})^2 + (\sin{t}+\cos{t})^2} = e^{t}\sqrt{2}$

$\displaystyle s$	$\displaystyle =$	$\displaystyle \int_{0}^{\pi}\Vert\frac{d\vec{r}(t)}{dt}\Vert dt$
	$\displaystyle =$	$\displaystyle \int_{0}^{\pi}\sqrt{2}e^{t}dt$
	$\displaystyle =$	$\displaystyle \left[\sqrt{2}e^{t}\vert\right]_{0}^{\pi}$
	$\displaystyle =$	$\displaystyle \sqrt{2}(e^{\pi} - 1)$

$\displaystyle x = \cos^{3}{t}, y = \sin^{3}{t}, 0 \leq t \leq \pi$

$\displaystyle \Vert\frac{d\vec{r}(t)}{dt}\Vert$	$\displaystyle =$	$\displaystyle \Vert(-3\cos^{2}{t}\sin{t}, 3\sin^{2}{t}\cos{t})\Vert$
	$\displaystyle =$	$\displaystyle 3\sqrt{\cos^{4}(t)\sin^{2}{t} + \sin^{4}{t}\cos^{2}{t}} = 3\cos{t}\sin{t}$

$\displaystyle s$	$\displaystyle =$	$\displaystyle 2\int_{0}^{\frac{\pi}{2}}\Vert\frac{d\vec{r}(t)}{dt}\Vert dt$
	$\displaystyle =$	$\displaystyle 2\int_{0}^{\frac{\pi}{2}}3\cos{t}\sin{t}dt$
	$\displaystyle =$	$\displaystyle 6\left[\frac{\sin^{2}{t}}{2}\vert\right]_{0}^{\frac{\pi}{2}} = 3$

4. 曲線が

で与えられるとき，曲率 $\kappa$ は $\kappa = \vert\frac{d\phi}{ds}\vert$ で表わされる．ここで， $\phi$ は曲線と $x$

軸のなす角で，

は曲線の長さを表わす．つまり，曲線の長さが $\Delta s$ だけ変化したとき，角 $\phi$ はどれだけ変化するかを測っている．

$\tan{\phi}$ は曲線の接線の傾きなので， $y' = \tan{\phi}$ ．つまり， $\phi = \tan^{-1}(y')$ ．また，

$\displaystyle \Delta s = \sqrt{(\Delta x)^2 + (\Delta y)^2} = \sqrt{1 + (y')^2}\Delta x$

$\displaystyle \frac{d\phi}{ds} = \frac{d\phi}{dx}\frac{dx}{ds} = \frac{y''}{1 + (y')^2}\frac{1}{ds/dx} = \frac{y''}{(1 + (y')^2)^{3/2}}$

$\displaystyle \kappa = \vert\frac{d\phi}{ds} = \frac{\vert y''\vert}{(1 + (y')^2)^{3/2}}\vert$

$\displaystyle y' = -e^{-x}, y'' = e^{-x}$

$\displaystyle \kappa = \frac{e^{-x}}{(1 + (e^{-x})^{2})^{3/2}} = \frac{e^{-x}}{(1 + e^{-2x})^{3/2}}$

$\displaystyle \kappa = \vert\frac{d\phi}{ds} = \frac{\vert y''\vert}{(1 + (y')^2)^{3/2}}\vert$

$\displaystyle y' = \frac{\sec{x}\tan{x}}{\sec{x}} = \tan{x}, y'' = \sec^{2}{x}$

$\displaystyle \kappa = \frac{\sec^{2}{x}}{(1 + (\tan{x})^{2})^{3/2}} = \frac{\sec^{2}{x}}{\sec^{3}{x}} = \frac{1}{\sec{x}} = \cos{x}$

$\displaystyle \kappa = \Vert\frac{d\hat{t}}{ds}\Vert = \Vert\frac{d\hat{t}/dt}{ds/dt}\Vert$

$\displaystyle \hat{t}$	$\displaystyle =$	$\displaystyle \frac{\vec{r}'(t)}{\Vert\vec{r}'(t)\Vert}$
	$\displaystyle =$	$\displaystyle \frac{(-3\sin{t},3\cos{t})}{\Vert(-3\sin{t},3\cos{t})\Vert} = (-\sin{t},\cos{t})$

$\displaystyle \frac{ds}{dt} = \Vert\frac{d\vec{r}}{dt}\Vert = \Vert\sqrt{(-3\sin{t},3\cos{t})}\Vert = 3$

$\displaystyle \kappa = \Vert\frac{d\hat{t}}{ds}\Vert = \Vert\frac{(-\cos{t},-\sin{t})}{3}\Vert = \frac{1}{3}$