東北大学(理系) 2026年　問題5


$座標平面上の曲線Cは、媒介変数 \ t\ を用いて、x=e^t\cos t,\ \ y=e^t\sin t \ \ \big(0 \leqq t \leqq \dfrac{\pi}{2}\big)\ \ と表されている。$
$曲線 \ C\ の \ y\ 軸に平行な接線を \ \ell\ とし、曲線 \ C,直線 \ \ell \ および \ x\ 軸で囲まれる図形を \ D\ とする。以下の$
$問いに答えよ。$
$(1)\ \ 直線 \ \ell\ の方程式を求めよ。$
$(2)\ \ 正の実数 \ \alpha,\ \beta \ に対し、次の \ 2\ つの不定積分を求めよ。$
\[\qquad I=\int e^{\alpha t}\cos \beta tdt,\quad J=\int e^{\alpha t}\sin \beta tdt\] $(3)\ \ D\ を \ x\ 軸の周りに \ 1\ 回転させてできる立体の体積 \ V\ の値を求めよ。$


(1)


$曲線 \ C\ の \ y\ 軸に平行な接線は \ \ \dfrac{dx}{dy}=0 \ \ だから$

$\dfrac{dx}{dt}=e^t \cos t -e^t \sin t=0$

$\cos t-\sin t=0$

$\sqrt{2}\cos \big(t+\dfrac{\pi}{4}\big)=0$

$0 \leqq t \leqq \dfrac{\pi}{2}\ \ より \quad \dfrac{\pi}{4} \leqq t+\dfrac{\pi}{4} \leqq \dfrac{3}{4}\pi \ \ だから$

$t+\dfrac{\pi}{4}=\dfrac{\pi}{2} \qquad \therefore \ \ t=\dfrac{\pi}{4}$

$よって \ 直線 \ \ell\ は \quad x=e^{\scriptsize{\dfrac{\pi}{4}}}\cos \dfrac{\pi}{4}=\dfrac{1}{\sqrt{2}}e^{\scriptsize{\dfrac{\pi}{4}}}$


(2)


$解 \ 1\ \ (部分積分法を \ 2\ 回行う方法)$

\begin{eqnarray*} I &=&\int e^{\alpha t}\cos \beta tdt\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\cos \beta t +\dfrac{\beta}{\alpha}\int e^{\alpha t}\sin \beta tdt\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\cos \beta t +\dfrac{\beta}{\alpha}\big(\dfrac{1}{\alpha} e^{\alpha t}\sin \beta t- \dfrac{\beta}{\alpha} \int e^{\alpha t}\cos \beta tdt\big)\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\cos \beta t +\dfrac{\beta}{\alpha ^2} e^{\alpha t}\sin \beta t- \dfrac{\beta^2}{\alpha^2} I\\ \end{eqnarray*} $\big(1+\dfrac{\beta^2}{\alpha^2}\big)I= e^{\alpha t}\big(\dfrac{1}{\alpha}\cos \beta t + \dfrac{\beta}{\alpha ^2} \sin \beta t\big)$

$\dfrac{\alpha^2+\beta^2}{\alpha^2}I= e^{\alpha t}\big(\dfrac{1}{\alpha}\cos \beta t + \dfrac{\beta}{\alpha ^2} \sin \beta t\big)$

$\therefore \ \ I=\dfrac{e^{\alpha t}}{\alpha^2+\beta^2}(\alpha\cos \beta t + \beta \sin \beta t)$


\begin{eqnarray*} J &=&\int e^{\alpha t}\sin \beta tdt\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\sin \beta t -\dfrac{\beta}{\alpha}\int e^{\alpha t}\cos \beta tdt\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\sin \beta t -\dfrac{\beta}{\alpha}\big(\dfrac{1}{\alpha} e^{\alpha t}\cos \beta t+ \dfrac{\beta}{\alpha} \int e^{\alpha t}\sin \beta tdt\big)\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\sin \beta t -\dfrac{\beta}{\alpha ^2} e^{\alpha t}\cos \beta t- \dfrac{\beta^2}{\alpha^2} J\\ \end{eqnarray*} $\big(1+\dfrac{\beta^2}{\alpha^2}\big)J= e^{\alpha t}\big(\dfrac{1}{\alpha}\sin \beta t - \dfrac{\beta}{\alpha ^2} \cos \beta t\big)$

$\dfrac{\alpha^2+\beta^2}{\alpha^2}J= e^{\alpha t}\big(\dfrac{1}{\alpha}\sin \beta t - \dfrac{\beta}{\alpha ^2} \cos \beta t\big)$

$\therefore \ \ J=\dfrac{e^{\alpha t}}{\alpha^2+\beta^2}(\alpha\sin \beta t - \beta \cos \beta t)$


$解 \ 2\ \ (方程式を解く方法)$

\begin{eqnarray*} I &=&\int e^{\alpha t}\cos \beta tdt\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\cos \beta t +\dfrac{\beta}{\alpha}\int e^{\alpha t}\sin \beta tdt\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\cos \beta t +\dfrac{\beta}{\alpha}J \hspace{5em}① \end{eqnarray*}
\begin{eqnarray*} J &=&\int e^{\alpha t}\sin \beta tdt\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\sin \beta t -\dfrac{\beta}{\alpha}\int e^{\alpha t}\cos \beta tdt\\ \\ &=&\dfrac{1}{\alpha} e^{\alpha t}\sin \beta t -\dfrac{\beta}{\alpha}I \hspace{5em}②\\ \end{eqnarray*}
$②を①に代入して$

$I=\dfrac{1}{\alpha} e^{\alpha t}\cos \beta t +\dfrac{\beta}{\alpha}\big(\dfrac{1}{\alpha} e^{\alpha t}\sin \beta t -\dfrac{\beta}{\alpha}I \big)$

$I=\dfrac{1}{\alpha} e^{\alpha t}\cos \beta t +\dfrac{\beta}{\alpha^2}e^{\alpha t}\sin \beta t -\dfrac{\beta^2}{\alpha^2}I$

$\alpha^2 I=\alpha e^{\alpha t}\cos \beta t +\beta e^{\alpha t}\sin \beta t -\beta^2 I$

$(\alpha^2 +\beta^2)I= e^{\alpha t}(\alpha \cos \beta t +\beta \sin \beta t )$

$\therefore \ \ I=\dfrac{e^{\alpha t}}{\alpha^2+\beta^2}(\alpha \cos \beta t +\beta \sin \beta t )$

$これを②に代入して \ J\ を求める。$


(3)


$\dfrac{dx}{dt}=e^t \cos t -e^t \sin t=e^t(\cos t-\sin t)=\sqrt{2}e^t\cos \big(t+\dfrac{\pi}{4}\big)$

$\dfrac{dy}{dt}=e^t \sin t +e^t \cos t=e^t(\cos t+\sin t)=\sqrt{2}e^t\cos \big(t-\dfrac{\pi}{4}\big)$

$(1)より \ \ \dfrac{dx}{dt}=0 \ の解は \ \ t=\dfrac{\pi}{4}$

$0 \leqq t \leqq \dfrac{\pi}{2} \ \ り \quad -\dfrac{\pi}{4} \leqq t -\dfrac{\pi}{4} \leqq \dfrac{\pi}{4} \ \ だから \ \ \dfrac{dy}{dt} > 0$

　

$x,\ y\ の増減表は$
\[ \begin{array}{c||c|c|c|c|c} t & 0 & \cdots & \dfrac{\pi}{4} & \cdots & \cfrac{\pi}{2}\\ \hline \cfrac{dx}{dt} & & + & 0 & - & \\ \hline \cfrac{dy}{dt} & & + & + & + & \\ \hline x & 1 & \nearrow & 極大 & \searrow & 0 \\ \hline y & 0 & \nearrow & \nearrow & \nearrow & \\ \end{array} \]
$x\ は \ t=\dfrac{\pi}{4}\ で極大、極大値は \quad x=\dfrac{1}{\sqrt{2}}e^{\scriptsize{\dfrac{\pi}{4}}}$

$x,\ y\ のグラフは右のとおり$
\[ x=e^t \cos t,\ \ y=e^t \sin t , \quad dx=(e^t \cos t-e^t \sin t)dt \qquad \begin{array}{c|c} x & 1\ \rightarrow \scriptsize{\dfrac{1}{\sqrt{2}}}e^{\scriptsize{\dfrac{\pi}{4}}}\\ \hline t & \ 0 \rightarrow \dfrac{\pi}{4}\\ \end{array} \] \begin{eqnarray*} V &=&\pi \int_1^{\scriptsize{\dfrac{1}{\sqrt{2}}}}e^{\scriptsize{\dfrac{\pi}{4}}} \ y^2dx\\ \\ &=&\pi \int_0^{\scriptsize{\dfrac{\pi}{4}}}(e^t \sin t)^2(e^t\cos t-e^t\sin t)dt\\ \\ &=&\pi \int_0^{\scriptsize{\dfrac{\pi}{4}}}(e^{3t} \sin ^2t\cos t-e^{3t}\sin ^3t)dt\\ \\ &=&\pi \int_0^{\scriptsize{\dfrac{\pi}{4}}}(e^{3t} (1-\cos^2t)\cos t-e^{3t}\sin ^3t)dt\\ \\ &=&\pi \int_0^{\scriptsize{\dfrac{\pi}{4}}}(e^{3t}\cos t - e^{3t}\cos^3t-e^{3t}\sin ^3t)dt\\ \\ &=&\pi \int_0^{\scriptsize{\dfrac{\pi}{4}}}\big\{e^{3t}\cos t - \dfrac{e^{3t}}{4}(\cos 3t+3\cos t)-\dfrac{e^{3t}}{4}(3\sin t- \sin 3t)\big\}dt\\ \\ &=&\dfrac{\pi}{4} \int_0^{\scriptsize{\dfrac{\pi}{4}}}\big(e^{3t}\cos t -3e^{3t}\sin t -e^{3t}\cos 3t +e^{3t}\sin 3t\big)dt\\ \\ &=&\dfrac{\pi}{4} \big[\dfrac{e^{3t}}{10}(3\cos t +\sin t)- \dfrac{3e^{3t}}{10}(3\sin t - \cos t)-\dfrac{e^{3t}}{18}(3\cos 3t +3\sin 3t)+\dfrac{e^{3t}}{18}(3\sin 3t -3\cos 3t)\big]_0^{\scriptsize{\dfrac{\pi}{4}}}\\ \\ &=&\dfrac{\pi}{4} \big\{\dfrac{e^{\scriptsize{\dfrac{3\pi}{4}}}}{10}\big(3\cos \dfrac{\pi}{4}+\sin \dfrac{\pi}{4}\big)- \dfrac{3e^{\scriptsize{\dfrac{\pi}{4}}}}{10}\big(3\sin \dfrac{\pi}{4}-\cos \dfrac{\pi}{4}\big) - \dfrac{e^{\scriptsize{\dfrac{3\pi}{4}}}}{18}\big(3\cos \dfrac{3\pi}{4}+3\sin \dfrac{3\pi}{4}\big)+ \dfrac{e^{\scriptsize{\dfrac{\pi}{4}}}}{18}\big(3\sin \dfrac{3\pi}{4}-3\cos \dfrac{3\pi}{4}\big)\big\}\\ \\ & & - \big(\dfrac{1}{10} \times 3 +\dfrac{3}{10}-\dfrac{1}{18} \times 3 +\dfrac{1}{18} \times (-3)\big)\big\}\\ \\ &=&\dfrac{\pi}{4} \big\{\dfrac{e^{\scriptsize{\dfrac{3\pi}{4}}}}{10} \times 2\sqrt{2} - \dfrac{3e^{\scriptsize{\dfrac{3\pi}{4}}}}{10} \times \sqrt{2} + \dfrac{e^{\scriptsize{\dfrac{\pi}{4}}}}{18} \times 3\sqrt{2} - \big(\dfrac{3}{10} +\dfrac{3}{10}- \dfrac{1}{3}\big)\big\}\\ \\ &=&\dfrac{\pi}{4} \big\{e^{\scriptsize{\dfrac{3\pi}{4}}} \big(\dfrac{\sqrt{2}}{5} - \dfrac{3\sqrt{2}}{10} + \dfrac{\sqrt{2}}{6}\big) - \dfrac{4}{15}\big\}\\ \\ &=&\dfrac{\pi}{60}\big(\sqrt{2} e^{\scriptsize{\dfrac{3\pi}{4}}} - 4\big) \end{eqnarray*}

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