\(L\) を正定数とする. 座標平面の \(x\) 軸上の正の部分にある点 P \(( t , 0 )\) に対し, 原点 O を中心とし点 P を通る円周上を, P から出発して反時計回りに道のり \(L\) だけ進んだ点を Q \(\left( u(t) , v(t) \right)\) と表す.

1. (1)　\(u(t)\) , \(v(t)\) を求めよ.

2. (2)　\(0 \lt a \lt 1\) の範囲の実数 \(a\) に対し, 積分 \[ f(a) = \displaystyle\int _ a^1 \sqrt{\{ u'(t) \}^2 +\{ v'(t) \}^2} \, dt \] を求めよ.

3. (3)　極限 \(\displaystyle\lim _ {a \rightarrow +0} \dfrac{f(a)}{\log a}\) を求めよ.

【 解 答 】

(1)

\(\angle\text{POQ} = \dfrac{L}{t}\) なので \[ u(t) = \underline{t \cos \dfrac{L}{t}} , \quad v(t) = \underline{t \sin \dfrac{L}{t}} \]

(2)

\[\begin{align} u'(t) & = \cos \dfrac{L}{t} +t \left( -\dfrac{L}{t^2} \right) \left( -\sin \dfrac{L}{t} \right) \\ & = \cos \dfrac{L}{t} +\dfrac{L}{t} \sin \dfrac{L}{t} \\ v'(t) & = \sin \dfrac{L}{t} +t \left( -\dfrac{L}{t^2} \right) \cos \dfrac{L}{t} \\ & = \sin \dfrac{L}{t} -\dfrac{L}{t} \cos \dfrac{L}{t} \end{align}\] なので \[\begin{align} \{ u'(t) \}^2 & = \cos^2 \dfrac{L}{t} +\dfrac{2L}{t} \cos \dfrac{L}{t} \sin \dfrac{L}{t} +\dfrac{L^2}{t^2} \sin^2 \dfrac{L}{t} \\ \{ v'(t) \}^2 & = \sin^2 \dfrac{L}{t} -\dfrac{2L}{t} \cos \dfrac{L}{t}\sin \dfrac{L}{t} +\dfrac{L^2}{t^2} \cos^2 \dfrac{L}{t} \end{align}\] \(\sin^2 \dfrac{L}{t} +\cos^2 \dfrac{L}{t} = 1\) に着目すれば \[\begin{align} \sqrt{\{ u'(t) \}^2 +\{ v'(t) \}^2} & = \sqrt{1 +\dfrac{L^2}{t^2}} \\ & = \dfrac{\sqrt{L^2 +t^2}}{t} \quad ( \ \text{∵} \ t \gt 0 \end{align}\] したがって \[ f(a) = \displaystyle\int _ a^1 ! \dfrac{\sqrt{L^2 +t^2}}{t} \, dt \] ここで \(t = L \tan \theta\) と置換すると \[ dt = \dfrac{L}{\cos^2 \theta} d \theta \\ t \ : \ a \rightarrow 1 \text{のとき} \quad \theta \ : \ \theta _ 1 \rightarrow \theta _ 2 \\ \left( \ \text{ただし, } \tan \theta _ 1 = \dfrac{a}{L} , \ \tan \theta _ 2 = \dfrac{1}{L} \quad ...[1] \ \right) \] これを用いて \[\begin{align} f(a) & = \displaystyle\int _ {\theta _ 1}^{\theta _ 2} \dfrac{L \sqrt{1 +\tan^2 \theta}}{L \tan \theta} \cdot \dfrac{L}{\cos^2 \theta} \, d\theta \\ & = L \displaystyle\int _ {\theta _ 1}^{\theta _ 2} \dfrac{1}{\sin \theta \cos^2 \theta} \, d\theta \\ & = L \displaystyle\int _ {\theta _ 1}^{\theta _ 2} \dfrac{\sin \theta}{\cos^2 \theta \left( 1 -\cos^2 \theta \right)} \, d\theta \\ & = -L \displaystyle\int _ {\theta _ 1}^{\theta _ 2} \left( \dfrac{1}{\cos^2 \theta} +\dfrac{1}{2( 1 -\cos \theta )} +\dfrac{1}{2( 1 +\cos \theta )} \right) \, ( \cos \theta)' d\theta \\ & = -L \left[ -\dfrac{1}{\cos \theta} -\dfrac{1}{2} \log( 1 -\cos \theta ) +\dfrac{1}{2} \log( 1 +\cos \theta ) \right] _ {\theta _ 1}^{\theta _ 2} \\ & = L \left( \dfrac{1}{\cos \theta _ 2} -\dfrac{1}{\cos \theta _ 1} +\dfrac{1}{2} \log \dfrac{1 -\cos \theta _ 2}{1 +\cos \theta _ 2} -\dfrac{1}{2} \log \dfrac{1 -\cos \theta _ 1}{1 +\cos \theta _ 1} \right) \end{align}\] ここで [1] より \[\begin{align} \dfrac{1}{\cos \theta _ 1} & = \sqrt{1 +\tan^2 \theta _ 1} = \dfrac{\sqrt{L^2 +a^2}}{L} , \\ \dfrac{1}{\cos \theta _ 2} & = \sqrt{1 +\tan^2 \theta _ 2} = \dfrac{\sqrt{L^2 +1}}{L} \end{align}\] これを用いると \[\begin{align} f(a) & = L \left( \dfrac{\sqrt{L^2 +1}}{L} -\dfrac{\sqrt{L^2 +a^2}}{L} \right. \\ & \qquad \left. +\dfrac{1}{2} \log \dfrac{\sqrt{L^2 +1} -L}{\sqrt{L^2 +1} +L} -\dfrac{1}{2} \log \dfrac{\sqrt{L^2 +a^2} -L}{\sqrt{L^2 +a^2} +L} \right) \\ & = \underline{\sqrt{L^2 +1} -\sqrt{L^2 +a^2}} \\ & \qquad \underline{+L \log \left( \sqrt{L^2 +1} -L \right) -L \log \dfrac{\sqrt{L^2 +a^2} -L}{a}} \end{align}\]

(3)

\(A = \sqrt{L^2 +1} -\sqrt{L^2 +a^2} +L \log \left( \sqrt{L^2 +1} -L \right)\) , \(B = L \log \dfrac{\sqrt{L^2 +a^2} -L}{a}\) とおく.
\(\displaystyle\lim _ {a \rightarrow +0} \log a = -\infty\) であることから, \(a \rightarrow +0\) のとき \[\begin{align} \dfrac{A}{\log a} & \rightarrow 0 \\ \dfrac{B}{\log a} & = L \cdot \dfrac{\log \frac{a^2}{a \left( \sqrt{L^2 +a^2} +L \right)}}{\log a} \\ & = L \left\{ 1 -\dfrac{\log \left( \sqrt{L^2 +a^2} +L \right)}{\log a} \right\} \\ & \rightarrow L \end{align}\] ゆえに \[ \displaystyle\lim _ {a \rightarrow +0} \dfrac{f(a)}{\log a} = \displaystyle\lim _ {a \rightarrow +0} \dfrac{A-B}{\log a} = \underline{-L} \]