単位円の半分に収まる(?)正方形の面積は有理数(一辺の長さは無理数(sin(arctan(2))=2/√5))
点\(C\)の位置
$$\begin{eqnarray} &&C(\overline{OF},\overline{CF}) \\&=&C(\cos{(\theta)},\sin{(\theta)}) \end{eqnarray}$$角\(\theta\)との関係
$$\begin{eqnarray} \frac{\overline{CF}}{\overline{OF}}&=&2\;\cdots\;正方形となるため \\&=&\frac{\sin{(\theta)}}{\cos{(\theta)}} \\&=&\tan{(\theta)} \\\tan^{-1}{(2)}&=&\theta \end{eqnarray}$$線分\(OF\)の長さ
$$\begin{eqnarray} \overline{OF}&=&\cos{\left(\tan^{-1}{\left(2\right)}\right)} \\&=&\frac{1}{\sqrt{\tan(\tan^{-1}{(2)})^2+1}} \\&&\;\cdots\;\cos{(\theta)}=\frac{1}{\sqrt{\tan^2{\left(\theta\right)}+1}} \\&=&\frac{1}{\sqrt{2^2+1}} \\&=&\frac{1}{\sqrt{5}} \end{eqnarray}$$線分\(CF\)の長さ
$$\begin{eqnarray} \overline{CF} \\&=&\sin{(\theta)} \\&=&2\cos{(\theta)} \\&=&2\frac{1}{\sqrt{5}} \\&=&\frac{2}{\sqrt{5}} \end{eqnarray}$$面積\(A_1\)
$$\begin{eqnarray} A_1&=&\overline{CF}^2 \\&=&\left(\frac{2}{\sqrt{5}} \right)^2 \\&=&\frac{2^2}{\left(\sqrt{5}\right)^2} \\&=&\frac{4}{5}\;(=0.8) \end{eqnarray}$$1/(1+x^2)の[0, t]での定積分がarctan(t)となる話
問
original: https://www.youtube.com/watch?v=PvL2RDp1H4Y
中心を点\(P\)に持つ円と直線\(g\)との交点\(Q\)を求める．
\(x\)座標を求める． $$\begin{eqnarray} x^2+\left(y+1\right)^2&=&1 \\x^2+\left(\color{red}{\frac{x}{t}-1}\color{black}{}+1\right)^2&=&1\;\cdots\;g:y=\frac{x}{t}-1を代入 \\x^2+\left(\frac{x}{t}\right)^2&=&1 \\x^2\left(1+\frac{1}{t^2}\right)&=&1 \\x^2\left(\frac{t^2+1}{t^2}\right)&=&1 \\x^2&=&\frac{t^2}{t^2+1} \\x&=&\sqrt{\frac{t^2}{t^2+1}}\;\cdots\;x\ge0 \\&=&\frac{t}{\sqrt{t^2+1}} \end{eqnarray}$$ \(y\)座標は\(g\)を用いて求める． $$\begin{eqnarray} y&=&\frac{x}{t}-1 \\&=&\frac{\frac{t}{\sqrt{t^2+1}}}{t}-1 \\&=&\frac{1}{\sqrt{t^2+1}}-1 \end{eqnarray}$$面積\(A_1\)(=四角形\(ABCO\))を求める．
$$\begin{eqnarray} A_1&=&\overline{OA}\cdot\overline{AB}\;\cdots\;長方形の面積 \\&=&t\cdot\frac{1}{2}\frac{1}{1+t^2} \\&=&\frac{t}{2\left(1+t^2\right)} \end{eqnarray}$$面積\(A_2\)(=凾数\(f\)と線分\(\overline{CB}\)及び\(y\)軸で囲まれる領域(図の\(x\lt0\)の領域の着色は間違い))を求める．
$$\begin{eqnarray} 2y&=&\frac{1}{1+x^2} \\1+x^2&=&\frac{1}{2y} \\x^2&=&\frac{1}{2y}-1 \\x^2&=&\frac{1}{2y}-1 \\&=&\frac{1-2y}{2y} \\x&=&\sqrt{\frac{1-2y}{2y}} \end{eqnarray}$$ $$\begin{eqnarray} A_2&=&\int_{\frac{1}{2\left(1+t^2\right)}}^{\frac{1}{2}} x\mathrm{d}y \\&=&\int_{\frac{1}{2\left(1+t^2\right)}}^{\frac{1}{2}} \sqrt{\frac{1-2y}{2y}} \mathrm{d}y \\&=&\int_{\sqrt{\frac{1}{\left(1+t^2\right)}}-1}^{0} \sqrt{\frac{1-\left(u+1\right)^2}{\cancel{2y}}}\cancel{\sqrt{2y}}\mathrm{d}u \\&&\;\cdots\;u=\sqrt{2y}-1,\;\frac{\mathrm{d}u}{\mathrm{d}y} =\frac{1}{2}\frac{1}{\sqrt{2y}}\cdot2=\frac{1}{\sqrt{2y}},\;\mathrm{d}y=\sqrt{2y}\mathrm{d}u \\&&\;\cdots\;\frac{1}{2}\rightarrow\sqrt{2\cdot \frac{1}{2}}-1=0 \\&&\;\cdots\;\frac{1}{2\left(1+t^2\right)}\rightarrow\sqrt{2\cdot \frac{1}{2\left(1+t^2\right)}}-1 =\frac{1}{\sqrt{1+t^2}}-1 \\&=&\int_{\frac{1}{\sqrt{1+t^2}}-1}^{0} \sqrt{1-\left(u+1\right)^2}\mathrm{d}u \end{eqnarray}$$面積\(B_1\)(=三角形\(PQR\))を求める．
$$\begin{eqnarray} B_1&=&\frac{1}{2}\overline{QR}\cdot\overline{PR}\;\cdots\;三角形の面積 \\&=&\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+1}}\cdot\left\{1+\left(\frac{1}{\sqrt{t^2+1}}-1\right)\right\} \\&=&\frac{1}{2}\cdot\frac{t}{\sqrt{t^2+1}}\cdot\frac{1}{\sqrt{t^2+1}} \\&=&\frac{t}{2\left(1+t^2\right)} \end{eqnarray}$$面積\(B_2\)(=円\(x^2+(y-1)^2=1\)と線分\(\overline{QR}\)及び\(y\)軸で囲まれる領域)を求める．
$$\begin{eqnarray} x^2+\left(y+1\right)^2&=&1 \\x^2&=&1-\left(y+1\right)^2 \\x&=&\sqrt{1-\left(y+1\right)^2} \end{eqnarray}$$ $$\begin{eqnarray} B_2&=&\int_{\frac{1}{\sqrt{t^2+1}}-1}^{0} x \mathrm{d}y \\&=&\int_{\frac{1}{\sqrt{t^2+1}}-1}^{0} \sqrt{1-\left(y+1\right)^2} \mathrm{d}y \end{eqnarray}$$扇PQRの面積を求める．
$$\begin{eqnarray} 扇PQR&=&半径1の面積\cdot比率=\left(1\cdot1\cdot\pi\right)\cdot\frac{\theta}{2\pi} &=&\frac{\theta}{2} \end{eqnarray}$$\(\theta\)と\(t\)の関係式を求める．
$$\begin{eqnarray} \tan{\left(\theta\right)}&=&\frac{t-0}{0-(-1)}=\frac{t}{1}=t \\\tan^{-1}{\left(t\right)}&=&\theta \end{eqnarray}$$扇PQRの面積を介して．
$$\begin{eqnarray} 扇PQR&=&\frac{\theta}{2} \\&=&\frac{\tan^{-1}\left(t\right)}{2} \\&=&B_1+B_2 \\&=&A_1+A_2\;\cdots\;B_1=A_1,\;B_2=A_2 \\&=&\int_0^t \frac{1}{2}\frac{1}{1+x^2}\mathrm{d}x =凾数fの区間[0,t]における定積分 \end{eqnarray}$$ $$\begin{eqnarray} \int_0^t \frac{1}{1+x^2}\mathrm{d}x&=&\tan^{-1}\left(t\right) \end{eqnarray}$$
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