高専の数学解説

第9章 9.32

(1)

接線の方程式は y−y0=y′(x−x0)y - y_0 = y'(x - x_0)

点Qのとき −y0=y′(x−x0)-y_0 = y'(x - x_0)

点PはQRQRの中点であるので

y0=y2y_0 = \frac{y}{2} x0=x2x_0 = \frac{x}{2} として代入すると

−y2=y′(x−x2)-\frac{y}{2} = y'(x - \frac{x}{2}) xy′+y=0xy' + y = 0

(2)

xy′+y=0xy' + y = 0 y′+1xy=0y' + \frac{1}{x}y = 0

P=1xP = \frac{1}{x}、Q=0Q = 0 とおくと ∫1xdx=log⁡x\int\frac{1}{x}dx = \log x これから

y=1x{∫0dx+C}=1xCy = \frac{1}{x}\{\int 0dx + C\} = \frac{1}{x}C

(3,7)(\sqrt{3}, \sqrt{7}) のとき C=21C = \sqrt{21} これから

y=1x21y = \frac{1}{x}\sqrt{21} xy=21xy = \sqrt{21}

(3)

y′+1xy=0y' + \frac{1}{x}y = 0 xy=21xy = \sqrt{21} から y′=−1x221y' = -\frac{1}{x^2}\sqrt{21}

S=2π∫2101x211+21x4dx=2π21∫2101xx4+21x4dx=2π21∫210x4+21x3dxS = 2\pi\int_{\sqrt{2}}^{\sqrt{10}} \frac{1}{x}\sqrt{21}\sqrt{1 + \frac{21}{x^4}}dx = 2\pi\sqrt{21}\int_{\sqrt{2}}^{\sqrt{10}} \frac{1}{x}\sqrt{\frac{x^4 + 21}{x^4}}dx = 2\pi\sqrt{21}\int_{\sqrt{2}}^{\sqrt{10}} \frac{\sqrt{x^4 + 21}}{x^3}dx

x2=tx^2 = t とおくと 2xdx=dt2xdx = dt dx=12tdtdx = \frac{1}{2\sqrt{t}}dt x:2→10x: \sqrt{2} \to \sqrt{10} t:2→10t: 2 \to 10 これから

=2π21∫210t2+21t3⋅12tdt=π21∫210t2+21t4dt=π21∫210t2+21t2dt= 2\pi\sqrt{21}\int_2^{10} \sqrt{\frac{t^2 + 21}{t^3}} \cdot \frac{1}{2\sqrt{t}}dt = \pi\sqrt{21}\int_2^{10} \sqrt{\frac{t^2 + 21}{t^4}}dt = \pi\sqrt{21}\int_2^{10} \frac{\sqrt{t^2 + 21}}{t^2}dt

t=21tan⁡θt = \sqrt{21}\tan\theta とおくと dt=211cos⁡2θdθdt = \sqrt{21}\frac{1}{\cos^2\theta}d\theta t:2→10t: 2 \to 10 θ:tan⁡−1221→tan⁡−11021\theta: \tan^{-1}\frac{2}{\sqrt{21}} \to \tan^{-1}\frac{10}{\sqrt{21}} これから

=π21∫tan⁡−110212121sec⁡θ⋅212l⋅tan⁡2θ⋅cos⁡2θdθ=π21∫tan⁡−11021211cos⁡θ⋅sin⁡2θdθ= \pi\sqrt{21}\int_{\tan^{-1}\frac{10}{\sqrt{21}}}^{\sqrt{21}} \frac{\sqrt{21}\sec\theta \cdot \sqrt{21}}{2l \cdot \tan^2 \theta \cdot \cos^2 \theta} d\theta = \pi\sqrt{21}\int_{\tan^{-1}\frac{10}{\sqrt{21}}}^{\sqrt{21}} \frac{1}{\cos\theta \cdot \sin^2 \theta} d\theta

=π21∫tan⁡−1102121(cos⁡θsin⁡2θ+1cos⁡θ)dθ= \pi\sqrt{21}\int_{\tan^{-1}\frac{10}{\sqrt{21}}}^{\sqrt{21}} \left(\frac{\cos\theta}{\sin^2 \theta} + \frac{1}{\cos \theta}\right)d\theta

第１項の式を sin⁡θ=k\sin\theta = k、第２項の式を tan⁡θ2=u\tan\frac{\theta}{2} = uとおくと

=π21(∫1k2dk+∫21−u2du)=π21[[−1k]+[log⁡1+u1−u]]= \pi\sqrt{21}\left(\int \frac{1}{k^2} dk + \int \frac{2}{1-u^2} du\right) = \pi\sqrt{21}\left[\left[-\frac{1}{k}\right] + \left[\log\frac{1+u}{1-u}\right]\right] k,uk,uを元に戻して

=π21[−1sin⁡θ+log⁡∣1+tan⁡θ21−tan⁡θ2∣]tan⁡−11021π2−tan⁡−1221=π21[−1sin⁡θ+log⁡1+sin⁡θcos⁡θ]tan⁡−11021tan⁡−1221= \pi\sqrt{21}\left[-\frac{1}{\sin \theta} + \log\left|\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}}\right|\right]_{\tan^{-1}\frac{10}{\sqrt{21}}}^{\frac{\pi}{2}-\tan^{-1}\frac{2}{\sqrt{21}}} = \pi\sqrt{21}\left[-\frac{1}{\sin \theta} + \log\frac{1+\sin \theta}{\cos \theta}\right]_{\tan^{-1}\frac{10}{\sqrt{21}}}^{\tan^{-1}\frac{2}{\sqrt{21}}}

=π21{(−1110+log⁡1+10112111)−(−52+log⁡1+25215)}=21π(75+log⁡3)= \pi\sqrt{21}\left\{\left(-\frac{11}{10} + \log\frac{1+\frac{10}{11}}{\frac{\sqrt{21}}{11}}\right) - \left(-\frac{5}{2} + \log\frac{1+\frac{2}{5}}{\frac{\sqrt{21}}{5}}\right)\right\} = \sqrt{21}\pi\left(\frac{7}{5} + \log 3\right)

下から2行目の log⁡1+tan⁡θ21−tan⁡θ2\log\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}} の変換

1+tan⁡θ21−tan⁡θ2=1+sin⁡θ2cos⁡θ21−sin⁡θ2cos⁡θ2=cos⁡θ2+sin⁡θ2cos⁡θ2⋅cos⁡θ2cos⁡θ2−sin⁡θ2\frac{1+\tan\frac{\theta}{2}}{1-\tan\frac{\theta}{2}} = \frac{1+\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}}{1-\frac{\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}}} = \frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}} \cdot \frac{\cos\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}}

=cos⁡θ2+sin⁡θ2cos⁡θ2−sin⁡θ2=(cos⁡θ2+sin⁡θ2)(cos⁡θ2+sin⁡θ2)(cos⁡θ2−sin⁡θ2)(cos⁡θ2+sin⁡θ2)= \frac{\cos\frac{\theta}{2}+\sin\frac{\theta}{2}}{\cos\frac{\theta}{2}-\sin\frac{\theta}{2}} = \frac{(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})}{(\cos\frac{\theta}{2}-\sin\frac{\theta}{2})(\cos\frac{\theta}{2}+\sin\frac{\theta}{2})}

=1+2cos⁡θ2sin⁡θ2cos⁡2θ2−sin⁡2θ2= \frac{1+2\cos\frac{\theta}{2}\sin\frac{\theta}{2}}{\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}}

cos⁡(θ2+θ2)=cos⁡θ2cos⁡θ2−sin⁡θ2sin⁡θ2=cos⁡2θ2−sin⁡2θ2\cos(\frac{\theta}{2}+\frac{\theta}{2}) = \cos\frac{\theta}{2}\cos\frac{\theta}{2}-\sin\frac{\theta}{2}\sin\frac{\theta}{2} = \cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}

sin⁡(θ2+θ2)=sin⁡θ2cos⁡θ2+cos⁡θ2sin⁡θ2=2cos⁡θ2sin⁡θ2\sin(\frac{\theta}{2}+\frac{\theta}{2}) = \sin\frac{\theta}{2}\cos\frac{\theta}{2}+\cos\frac{\theta}{2}\sin\frac{\theta}{2} = 2\cos\frac{\theta}{2}\sin\frac{\theta}{2}

これから

1+2cos⁡θ2sin⁡θ2cos⁡2θ2−sin⁡2θ2=1+sin⁡θcos⁡θとなる\frac{1+2\cos\frac{\theta}{2}\sin\frac{\theta}{2}}{\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}} = \frac{1+\sin\theta}{\cos\theta} \text{となる}

解説: r31bn1z