(1)
x = e t − e − t 2 \displaystyle x=\dfrac{e^{t}-e^{-t}}{2} x = 2 e t − e − t y = e t + e − t 2 \displaystyle y=\dfrac{e^{t}+e^{-t}}{2} y = 2 e t + e − t y 2 − x 2 = ( e t + e − t ) 2 − ( e t − e − t ) 2 4 = 4 e t e − t 4 = 1 \displaystyle
y^{2}-x^{2}=\frac{(e^{t}+e^{-t})^{2}-(e^{t}-e^{-t})^{2}}{4}=\frac{4e^{t}e^{-t}}{4}=1 y 2 − x 2 = 4 ( e t + e − t ) 2 − ( e t − e − t ) 2 = 4 4 e t e − t = 1 t t t y = e t + e − t 2 = ( e t / 2 − e − t / 2 ) 2 + 2 2 ≧ 1 > 0 \displaystyle
y=\frac{e^{t}+e^{-t}}{2}=\frac{(e^{t/2}-e^{-t/2})^{2}+2}{2}\geqq1>0 y = 2 e t + e − t = 2 ( e t /2 − e − t /2 ) 2 + 2 ≧ 1 > 0 y 2 = x 2 + 1 y^{2}=x^{2}+1 y 2 = x 2 + 1 y = x 2 + 1 . y=\sqrt{x^{2}+1}. y = x 2 + 1 . x ( t ) x(t) x ( t ) d x d t = e t + e − t 2 = y ( t ) > 0 \displaystyle \dfrac{dx}{dt}=\dfrac{e^{t}+e^{-t}}{2}=y(t)>0 d t d x = 2 e t + e − t = y ( t ) > 0 x ( t ) x(t) x ( t ) t t t x ( 0 ) = 0 x(0)=0 x ( 0 ) = 0 t → ∞ t\to\infty t → ∞ x → ∞ x\to\infty x → ∞ t ≧ 0 t\geqq0 t ≧ 0 x x x 0 0 0 C C C y = x 2 + 1 ( x ≧ 0 ) y=\sqrt{x^{2}+1}\quad(x\geqq0) y = x 2 + 1 ( x ≧ 0 )
(2)
t = 0 t=0 t = 0 P = M P=M P = M A ( 0 ) = S ( 0 ) = 0 A(0)=S(0)=0 A ( 0 ) = S ( 0 ) = 0 t > 0 t>0 t > 0 X = x ( t ) = e t − e − t 2 ( > 0 ) \displaystyle X=x(t)=\dfrac{e^{t}-e^{-t}}{2}\,(>0) X = x ( t ) = 2 e t − e − t ( > 0 ) Y = y ( t ) = e t + e − t 2 ( > 1 ) \displaystyle Y=y(t)=\dfrac{e^{t}+e^{-t}}{2}\,(>1) Y = y ( t ) = 2 e t + e − t ( > 1 ) C C C 0 ≦ x ≦ X 0\leqq x\leqq X 0 ≦ x ≦ X y = f ( x ) : = x 2 + 1 y=f(x):=\sqrt{x^{2}+1} y = f ( x ) := x 2 + 1
【直線 O P OP O P O P OP O P y = Y X x \displaystyle y=\dfrac{Y}{X}x y = X Y x f ( x ) = Y X x \displaystyle f(x)=\dfrac{Y}{X}x f ( x ) = X Y x 2 2 2 Y 2 − X 2 = 1 Y^{2}-X^{2}=1 Y 2 − X 2 = 1 1 + x 2 = Y 2 X 2 x 2 であるための必要十分条件は x 2 ⋅ Y 2 − X 2 X 2 = 1 、すなわち x 2 = X 2 . \displaystyle
1+x^{2}=\frac{Y^{2}}{X^{2}}x^{2}\text{であるための必要十分条件は} x^{2}\cdot\frac{Y^{2}-X^{2}}{X^{2}}=1\text{、すなわち} x^{2}=X^{2}. 1 + x 2 = X 2 Y 2 x 2 であるための必要十分条件は x 2 ⋅ X 2 Y 2 − X 2 = 1 、すなわち x 2 = X 2 . 0 ≦ x ≦ X 0\leqq x\leqq X 0 ≦ x ≦ X f ( x ) = Y X x \displaystyle f(x)=\dfrac{Y}{X}x f ( x ) = X Y x x = 0 , X x=0,X x = 0 , X x = 0 x=0 x = 0 f ( 0 ) = 1 > 0 f(0)=1>0 f ( 0 ) = 1 > 0 f ( x ) > Y X x ( 0 < x < X ) \displaystyle
f(x)>\frac{Y}{X}x\qquad(0<x<X) f ( x ) > X Y x ( 0 < x < X ) O M OM O M C C C M M M P P P P O PO P O 0 ≦ x ≦ X 0\leqq x\leqq X 0 ≦ x ≦ X O P OP O P A ( t ) = ∫ 0 X { f ( x ) − Y X x } d x = ∫ 0 X f ( x ) d x − X Y 2 . \displaystyle
A(t)=\int_{0}^{X}\Bigl\{f(x)-\frac{Y}{X}x\Bigr\}dx=\int_{0}^{X}f(x)\,dx-\frac{XY}{2}. A ( t ) = ∫ 0 X { f ( x ) − X Y x } d x = ∫ 0 X f ( x ) d x − 2 X Y . x = x ( t ′ ) = e t ′ − e − t ′ 2 ( 0 ≦ t ′ ≦ t ) \displaystyle x=x(t')=\dfrac{e^{t'}-e^{-t'}}{2}\,(0\leqq t'\leqq t) x = x ( t ′ ) = 2 e t ′ − e − t ′ ( 0 ≦ t ′ ≦ t ) d x = y ( t ′ ) d t ′ dx=y(t')\,dt' d x = y ( t ′ ) d t ′ f ( x ) = y ( t ′ ) f(x)=y(t') f ( x ) = y ( t ′ ) ∫ 0 X f ( x ) d x = ∫ 0 t y ( t ′ ) 2 d t ′ = ∫ 0 t e 2 t ′ + 2 + e − 2 t ′ 4 d t ′ = [ e 2 t ′ − e − 2 t ′ 8 + t ′ 2 ] 0 t = e 2 t − e − 2 t 8 + t 2 . \displaystyle
\int_{0}^{X}f(x)\,dx=\int_{0}^{t}y(t')^{2}\,dt'=\int_{0}^{t}\frac{e^{2t'}+2+e^{-2t'}}{4}\,dt'=\left[\frac{e^{2t'}-e^{-2t'}}{8}+\frac{t'}{2}\right]_{0}^{t}=\frac{e^{2t}-e^{-2t}}{8}+\frac{t}{2}. ∫ 0 X f ( x ) d x = ∫ 0 t y ( t ′ ) 2 d t ′ = ∫ 0 t 4 e 2 t ′ + 2 + e − 2 t ′ d t ′ = [ 8 e 2 t ′ − e − 2 t ′ + 2 t ′ ] 0 t = 8 e 2 t − e − 2 t + 2 t . e 2 t − e − 2 t 8 = ( e t − e − t ) ( e t + e − t ) 8 = X Y 2 \displaystyle \dfrac{e^{2t}-e^{-2t}}{8}=\dfrac{(e^{t}-e^{-t})(e^{t}+e^{-t})}{8}=\dfrac{XY}{2} 8 e 2 t − e − 2 t = 8 ( e t − e − t ) ( e t + e − t ) = 2 X Y ∫ 0 X f ( x ) d x = X Y 2 + t 2 , ゆえに A ( t ) = ( X Y 2 + t 2 ) − X Y 2 = t 2 . \displaystyle
\int_{0}^{X}f(x)\,dx=\frac{XY}{2}+\frac{t}{2},\qquad\text{ゆえに}\qquad A(t)=\left(\frac{XY}{2}+\frac{t}{2}\right)-\frac{XY}{2}=\frac{t}{2}. ∫ 0 X f ( x ) d x = 2 X Y + 2 t , ゆえに A ( t ) = ( 2 X Y + 2 t ) − 2 X Y = 2 t .
【弦 M P MP M P M P MP M P M ( 0 , 1 ) M(0,1) M ( 0 , 1 ) P ( X , Y ) P(X,Y) P ( X , Y ) y = 1 + Y − 1 X x = : g ( x ) \displaystyle y=1+\dfrac{Y-1}{X}x=:g(x) y = 1 + X Y − 1 x =: g ( x ) f ( x ) = g ( x ) f(x)=g(x) f ( x ) = g ( x ) 2 ⋅ Y − 1 X x + ( Y − 1 X ) 2 x 2 − x 2 = 0 であるための必要十分条件は x { 2 ⋅ Y − 1 X + x ⋅ ( Y − 1 ) 2 − X 2 X 2 } = 0. \displaystyle
2\cdot\frac{Y-1}{X}x+\left(\frac{Y-1}{X}\right)^{2}x^{2}-x^{2}=0\text{であるための必要十分条件は} x\left\{2\cdot\frac{Y-1}{X}+x\cdot\frac{(Y-1)^{2}-X^{2}}{X^{2}}\right\}=0. 2 ⋅ X Y − 1 x + ( X Y − 1 ) 2 x 2 − x 2 = 0 であるための必要十分条件は x { 2 ⋅ X Y − 1 + x ⋅ X 2 ( Y − 1 ) 2 − X 2 } = 0. ( Y − 1 ) 2 − X 2 = ( Y 2 − X 2 ) − 2 Y + 1 = 2 − 2 Y = − 2 ( Y − 1 ) (Y-1)^{2}-X^{2}=(Y^{2}-X^{2})-2Y+1=2-2Y=-2(Y-1) ( Y − 1 ) 2 − X 2 = ( Y 2 − X 2 ) − 2 Y + 1 = 2 − 2 Y = − 2 ( Y − 1 ) t > 0 t>0 t > 0 Y > 1 Y>1 Y > 1 x ≠ 0 x\neq0 x = 0 2 ⋅ Y − 1 X − 2 ⋅ Y − 1 X 2 x = 0 であるための必要十分条件は x = X . \displaystyle
2\cdot\frac{Y-1}{X}-2\cdot\frac{Y-1}{X^{2}}x=0\text{であるための必要十分条件は} x=X. 2 ⋅ X Y − 1 − 2 ⋅ X 2 Y − 1 x = 0 であるための必要十分条件は x = X . f ( x ) = g ( x ) f(x)=g(x) f ( x ) = g ( x ) x = 0 , X x=0,X x = 0 , X x = 0 x=0 x = 0 g g g Y − 1 X > 0 \displaystyle \dfrac{Y-1}{X}>0 X Y − 1 > 0 f ′ ( 0 ) = 0 f'(0)=0 f ′ ( 0 ) = 0 x x x g ( x ) > f ( x ) g(x)>f(x) g ( x ) > f ( x ) g ( x ) > f ( x ) ( 0 < x < X ) g(x)>f(x)\qquad(0<x<X) g ( x ) > f ( x ) ( 0 < x < X ) C C C M M M P P P M P MP M P S ( t ) = ∫ 0 X { g ( x ) − f ( x ) } d x = ∫ 0 X g ( x ) d x − ∫ 0 X f ( x ) d x . \displaystyle
S(t)=\int_{0}^{X}\{g(x)-f(x)\}\,dx=\int_{0}^{X}g(x)\,dx-\int_{0}^{X}f(x)\,dx. S ( t ) = ∫ 0 X { g ( x ) − f ( x )} d x = ∫ 0 X g ( x ) d x − ∫ 0 X f ( x ) d x . ∫ 0 X g ( x ) d x = X + Y − 1 X ⋅ X 2 2 = X 2 + X Y 2 . \displaystyle
\int_{0}^{X}g(x)\,dx=X+\frac{Y-1}{X}\cdot\frac{X^{2}}{2}=\frac{X}{2}+\frac{XY}{2}. ∫ 0 X g ( x ) d x = X + X Y − 1 ⋅ 2 X 2 = 2 X + 2 X Y . ∫ 0 X f ( x ) d x = X Y 2 + t 2 \displaystyle\int_{0}^{X}f(x)\,dx=\frac{XY}{2}+\frac{t}{2} ∫ 0 X f ( x ) d x = 2 X Y + 2 t S ( t ) = ( X 2 + X Y 2 ) − ( X Y 2 + t 2 ) = X − t 2 = 1 2 ( e t − e − t 2 − t ) = e t − e − t 4 − t 2 . \displaystyle
S(t)=\left(\frac{X}{2}+\frac{XY}{2}\right)-\left(\frac{XY}{2}+\frac{t}{2}\right)=\frac{X-t}{2}=\frac{1}{2}\left(\frac{e^{t}-e^{-t}}{2}-t\right)=\frac{e^{t}-e^{-t}}{4}-\frac{t}{2}. S ( t ) = ( 2 X + 2 X Y ) − ( 2 X Y + 2 t ) = 2 X − t = 2 1 ( 2 e t − e − t − t ) = 4 e t − e − t − 2 t . t = 0 t=0 t = 0 0 0 0 S ( 0 ) = 0 S(0)=0 S ( 0 ) = 0 A ( t ) = t 2 , S ( t ) = e t − e − t 4 − t 2 ( t ≧ 0 ) . \displaystyle
A(t)=\frac{t}{2},\qquad S(t)=\frac{e^{t}-e^{-t}}{4}-\frac{t}{2}\qquad(t\geqq0). A ( t ) = 2 t , S ( t ) = 4 e t − e − t − 2 t ( t ≧ 0 ) .
(3)
h ( t ) : = A ( t ) − S ( t ) = t 2 − ( e t − e − t 4 − t 2 ) = t − e t − e − t 4 ( t ≧ 0 ) . \displaystyle
h(t):=A(t)-S(t)=\frac{t}{2}-\left(\frac{e^{t}-e^{-t}}{4}-\frac{t}{2}\right)=t-\frac{e^{t}-e^{-t}}{4}\qquad(t\geqq0). h ( t ) := A ( t ) − S ( t ) = 2 t − ( 4 e t − e − t − 2 t ) = t − 4 e t − e − t ( t ≧ 0 ) . h ′ ( t ) = 1 − e t + e − t 4 = 1 − y ( t ) 2 . \displaystyle
h'(t)=1-\frac{e^{t}+e^{-t}}{4}=1-\frac{y(t)}{2}. h ′ ( t ) = 1 − 4 e t + e − t = 1 − 2 y ( t ) . y ( t ) = e t + e − t 2 \displaystyle y(t)=\dfrac{e^{t}+e^{-t}}{2} y ( t ) = 2 e t + e − t t ≧ 0 t\geqq0 t ≧ 0 y ′ ( t ) = x ( t ) ≧ 0 y'(t)=x(t)\geqq0 y ′ ( t ) = x ( t ) ≧ 0 y ( 0 ) = 1 y(0)=1 y ( 0 ) = 1 t → ∞ t\to\infty t → ∞ y ( t ) → ∞ y(t)\to\infty y ( t ) → ∞ h ′ ( t ) = 0 h'(t)=0 h ′ ( t ) = 0 y ( t ) = 2 y(t)=2 y ( t ) = 2 t t t h ′ ( t ) h'(t) h ′ ( t ) y ( t ) < 2 y(t)<2 y ( t ) < 2 h ′ ( t ) > 0 h'(t)>0 h ′ ( t ) > 0 y ( t ) > 2 y(t)>2 y ( t ) > 2 h ′ ( t ) < 0 h'(t)<0 h ′ ( t ) < 0 y ( t ) = 2 であるための必要十分条件は e t + e − t 2 = 2 、すなわち e 2 t − 4 e t + 1 = 0. \displaystyle
y(t)=2\text{であるための必要十分条件は}\frac{e^{t}+e^{-t}}{2}=2\text{、すなわち} e^{2t}-4e^{t}+1=0. y ( t ) = 2 であるための必要十分条件は 2 e t + e − t = 2 、すなわち e 2 t − 4 e t + 1 = 0. u = e t ( u > 0 ) u=e^{t}\,(u>0) u = e t ( u > 0 ) u 2 − 4 u + 1 = 0 u^{2}-4u+1=0 u 2 − 4 u + 1 = 0 u = 2 ± 3 u=2\pm\sqrt3 u = 2 ± 3 t ≧ 0 t\geqq0 t ≧ 0 u = e t ≧ 1 u=e^{t}\geqq1 u = e t ≧ 1 2 − 3 < 1 2-\sqrt3<1 2 − 3 < 1 2 + 3 > 1 2+\sqrt3>1 2 + 3 > 1 e t = 2 + 3 , t = log ( 2 + 3 ) . e^{t}=2+\sqrt3,\qquad t=\log(2+\sqrt3). e t = 2 + 3 , t = log ( 2 + 3 ) . t t t h ′ ( t ) h'(t) h ′ ( t ) h ( t ) = A ( t ) − S ( t ) h(t)=A(t)-S(t) h ( t ) = A ( t ) − S ( t ) t = log ( 2 + 3 ) t=\log(2+\sqrt3) t = log ( 2 + 3 ) t ≧ 0 t\geqq0 t ≧ 0 t = log ( 2 + 3 ) \boxed{t=\log(2+\sqrt3)} t = log ( 2 + 3 )
