PDF p11印刷問6のr<a、同一接線、円下半分の領域を確認。表紙が参照する裏表紙選択表は収録末尾p19,p20白紙で未収録。医学部対応は固定書誌による留保を保持。
(1) 曲線 D D D A A A 1 / a 1/a 1/ a K = ( u , v ) K=(u,v) K = ( u , v ) u = ± r u=\pm r u = ± r K A KA K A v − log a = − a ( u − a ) = a ( a − u ) , r 2 = ( a − u ) 2 + ( v − log a ) 2 = ( a − u ) 2 ( 1 + a 2 ) . v-\log a=-a(u-a)=a(a-u),\qquad r^2=(a-u)^2+(v-\log a)^2=(a-u)^2(1+a^2). v − log a = − a ( u − a ) = a ( a − u ) , r 2 = ( a − u ) 2 + ( v − log a ) 2 = ( a − u ) 2 ( 1 + a 2 ) . S = 1 + a 2 > 1 S=\sqrt{1+a^2}>1 S = 1 + a 2 > 1 r = ∣ a − u ∣ S r=|a-u|S r = ∣ a − u ∣ S u = − r u=-r u = − r r = ( a + r ) S > r r=(a+r)S>r r = ( a + r ) S > r u = r u=r u = r r < a r<a r < a r = ( a − r ) S r=(a-r)S r = ( a − r ) S r = a S 1 + S = a 1 + a 2 1 + 1 + a 2 . \displaystyle
r=\frac{aS}{1+S}=\frac{a\sqrt{1+a^2}}{1+\sqrt{1+a^2}}. r = 1 + S a S = 1 + 1 + a 2 a 1 + a 2 .
(2) S = 1 + a 2 > 1 S=\sqrt{1+a^2}>1 S = 1 + a 2 > 1 r a = S 1 + S > 1 2 , \displaystyle
\frac{r}{a}=\frac{S}{1+S}>\frac12, a r = 1 + S S > 2 1 , r > a / 2 r>a/2 r > a /2
(3) a = 3 a=\sqrt3 a = 3 S = 2 S=2 S = 2 r = 2 3 / 3 r=2\sqrt3/3 r = 2 3 /3 y y y k k k k = log a + a ( a − r ) = 1 + log 3 k=\log a+a(a-r)=1+\log\sqrt3 k = log a + a ( a − r ) = 1 + log 3 g ( x ) = k − r 2 − ( x − r ) 2 ( 0 ≦ x ≦ 2 r ) g(x)=k-\sqrt{r^2-(x-r)^2}\qquad(0\le x\le 2r) g ( x ) = k − r 2 − ( x − r ) 2 ( 0 ≦ x ≦ 2 r ) log 3 = ∫ 1 3 d t t > 1 2 + 1 3 = 5 6 \displaystyle \log3=\int_1^3\frac{dt}{t}>\frac12+\frac13=\frac56 log 3 = ∫ 1 3 t d t > 2 1 + 3 1 = 6 5 k − r > 17 12 − 2 3 3 > 0 \displaystyle k-r>\frac{17}{12}-\frac{2\sqrt3}{3}>0 k − r > 12 17 − 3 2 3 > 0 17 2 > 192 17^2>192 1 7 2 > 192 x x x g ( a ) = log a g(a)=\log a g ( a ) = log a g ′ ( a ) = 1 / a g'(a)=1/a g ′ ( a ) = 1/ a h ( x ) = g ( x ) − log x h(x)=g(x)-\log x h ( x ) = g ( x ) − log x h ′ ′ ( x ) = r 2 ( r 2 − ( x − r ) 2 ) 3 / 2 + 1 x 2 > 0. \displaystyle
h''(x)=\frac{r^2}{\bigl(r^2-(x-r)^2\bigr)^{3/2}}+\frac1{x^2}>0. h ′′ ( x ) = ( r 2 − ( x − r ) 2 ) 3/2 r 2 + x 2 1 > 0. h ′ ( a ) = h ( a ) = 0 h'(a)=h(a)=0 h ′ ( a ) = h ( a ) = 0 h ( x ) > 0 h(x)>0 h ( x ) > 0 1 ≦ x < a 1\le x<a 1 ≦ x < a 0 < x ≦ 1 0<x\le1 0 < x ≦ 1 log x ≦ 0 < g ( x ) \log x\le0<g(x) log x ≦ 0 < g ( x ) A = ∫ 0 a g ( x ) d x − ∫ 1 a log x d x = a k − J − ( a log a − a + 1 ) , \displaystyle
A=\int_0^a g(x)\,dx-\int_1^a \log x\,dx
=ak-J-(a\log a-a+1), A = ∫ 0 a g ( x ) d x − ∫ 1 a log x d x = ak − J − ( a log a − a + 1 ) , J = ∫ 0 a r 2 − ( x − r ) 2 d x \displaystyle J=\int_0^a\sqrt{r^2-(x-r)^2}\,dx J = ∫ 0 a r 2 − ( x − r ) 2 d x a − r = r / 2 a-r=r/2 a − r = r /2 J J J 2 π / 3 2\pi/3 2 π /3 r / 2 r/2 r /2 3 r / 2 \sqrt3r/2 3 r /2 J = 1 2 r 2 ⋅ 2 π 3 + 1 2 ⋅ r 2 ⋅ 3 r 2 = π r 2 3 + 3 r 2 8 = 4 π 9 + 3 6 . \displaystyle
J=\frac12r^2\cdot\frac{2\pi}{3}+\frac12\cdot\frac r2\cdot\frac{\sqrt3r}{2}
=\frac{\pi r^2}{3}+\frac{\sqrt3r^2}{8}
=\frac{4\pi}{9}+\frac{\sqrt3}{6}. J = 2 1 r 2 ⋅ 3 2 π + 2 1 ⋅ 2 r ⋅ 2 3 r = 3 π r 2 + 8 3 r 2 = 9 4 π + 6 3 . k = 1 + log a k=1+\log a k = 1 + log a A = 2 a − 1 − J = 11 3 6 − 1 − 4 π 9 . \displaystyle
A=2a-1-J=\boxed{\frac{11\sqrt3}{6}-1-\frac{4\pi}{9}}. A = 2 a − 1 − J = 6 11 3 − 1 − 9 4 π .
