第三章：懸鏈線的性質。
33 概述。34 懸鏈線的普遍方程。35 其切線的幾何結構。36 懸鏈線的幾何結構。37 當懸鏈線上的一點及其該點處切線的方向已知時，懸鏈線是唯一的。38 懸鏈線必須存在的界限。39 ,40 ,41 可以透過兩個固定點繪製的懸鏈線的數量。42 函式 f 1 ( m ) {\displaystyle f_{1}(m)} f 2 ( m ) {\displaystyle f_{2}(m)} 43 函式 f 1 ( m ) {\displaystyle f_{1}(m)} 44 函式 f 2 ( m ) {\displaystyle f_{2}(m)} 45 對超越方程根的近似幾何結構。46 ,47 ,48 函式 f 1 ( m ) {\displaystyle f_{1}(m)} f 2 ( m ) {\displaystyle f_{2}(m)} 49 ,50 出現的不同情況和相應的懸鏈線數量。50 ,51 ,52 每種情況下透過兩個固定點的切線交點的定位。53 ,54 對兩條懸鏈線的公共切線。55 具有相同引數且僅在一個點相交的懸鏈線。56 林德洛夫定理。57 對同一個定理的第二個證明。58 ,59 討論旋轉最小曲面的幾種情況。60 ,61 應用於肥皂泡。第 33 條
在介紹本章的主題時，遵循了施瓦茲教授在柏林的講座。託德亨特在他的變分法研究 中以略有不同的形式推導了這些結果，第 54 頁；另見戈德施密特的獲獎論文，皇家天文學會月刊，第 12 卷，第 84 頁；傑勒特，變分法，1850 年，第 145 頁；莫尼奧和林德洛夫，變分法，1861 年，第 204 頁；等等。
第 34 條 [ 1]
y = 1 2 m [ e ( x − x 0 ′ ) / m + e − ( x − x 0 ′ ) / m ] {\displaystyle y={\frac {1}{2}}m[e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}]} 由此立即得出
m d y d x = ± y 2 − m 2 = 1 2 m [ e ( x − x 0 ′ ) / m + e − ( x − x 0 ′ ) / m ] {\displaystyle m{\frac {{\text{d}}y}{{\text{d}}x}}=\pm {\sqrt {y^{2}-m^{2}}}={\frac {1}{2}}m[e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}]} 方程的右邊是一個單值函式，但左邊是一個雙值函式。因此有必要定義左邊，使其成為對應於右邊的單值函式。
如果我們使 x > x 0 ′ {\displaystyle x>x_{0}'}
e ( x − x 0 ′ ) / m > e − ( x − x 0 ′ ) / m {\displaystyle e^{(x-x_{0}')/m}>e^{-(x-x_{0}')/m}} 因此， y 2 − m 2 {\displaystyle {\sqrt {y^{2}-m^{2}}}} x < x 0 ′ {\displaystyle x<x_{0}'}
e ( x − x 0 ′ ) / m < e − ( x − x 0 ′ ) / m {\displaystyle e^{(x-x_{0}')/m}<e^{-(x-x_{0}')/m}} 然後 y 2 − m 2 {\displaystyle {\sqrt {y^{2}-m^{2}}}} d y d x = 0 {\displaystyle {\frac {{\text{d}}y}{{\text{d}}x}}=0} x = x 0 ′ {\displaystyle x=x_{0}'} y {\displaystyle y} m {\displaystyle m}
此值 m {\displaystyle m} y {\displaystyle y} d y d x = 0 {\displaystyle {\frac {{\text{d}}y}{{\text{d}}x}}=0} d 2 y d x 2 {\displaystyle {\frac {{\text{d}}^{2}y}{{\text{d}}x^{2}}}} x = x 0 ′ {\displaystyle x=x_{0}'} m {\displaystyle m} y {\displaystyle y} y 2 − m 2 {\displaystyle {\sqrt {y^{2}-m^{2}}}} y {\displaystyle y} x = x 0 {\displaystyle x=x_{0}} y = m {\displaystyle y=m} X {\displaystyle X} d y d x = 0 {\displaystyle {\frac {{\text{d}}y}{{\text{d}}x}}=0}
第 35 條
d y d x = tan ( τ ) = y 2 − m 2 m {\displaystyle {\frac {{\text{d}}y}{{\text{d}}x}}=\tan(\tau )={\frac {\sqrt {y^{2}-m^{2}}}{m}}} 因此，要在懸鏈線的任意點上作切線，例如在 P {\displaystyle P} P Q {\displaystyle PQ} P Q {\displaystyle PQ} m {\displaystyle m} Q {\displaystyle Q} R {\displaystyle R} R {\displaystyle R} P {\displaystyle P} R P {\displaystyle RP}
再有 d s 2 = d x 2 + d y 2 = ( 1 + y 2 − m 2 m 2 ) d x 2 = y 2 d x 2 m 2 {\displaystyle {\text{d}}s^{2}={\text{d}}x^{2}+{\text{d}}y^{2}=\left(1+{\frac {y^{2}-m^{2}}{m^{2}}}\right){\text{d}}x^{2}={\frac {y^{2}{\text{d}}x^{2}}{m^{2}}}}
因此
d s = y d x m = 1 2 [ e ( x − x 0 ′ ) / m + e − ( x − x 0 ′ ) / m ] d x {\displaystyle {\text{d}}s={\frac {y{\text{d}}x}{m}}={\frac {1}{2}}[e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}]{\text{d}}x} 並積分，
s − s 0 ′ = 1 2 [ e ( x − x 0 ′ ) / m + e − ( x − x 0 ′ ) / m ] = y 2 − m 2 {\displaystyle s-s_{0}'={\frac {1}{2}}[e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}]={\sqrt {y^{2}-m^{2}}}} 其中 s 0 ′ {\displaystyle s_{0}'}
點 R {\displaystyle R} m {\displaystyle m} Q R {\displaystyle QR}
這條軌跡還具有一個非凡的性質：如果我們將它繞 X {\displaystyle X}
此外， P N {\displaystyle PN}
= y sec ( τ ) = y 2 m {\displaystyle =y\sec(\tau )={\frac {y^{2}}{m}}} ρ = [ 1 + ( d y d x ) 2 ] 3 / 2 d 2 y d x 2 = ( d x d x ) 3 d 2 y d x 2 = ( y m ) 3 y m 2 = y 2 m {\displaystyle \rho ={\frac {\left[1+\left({\frac {{\text{d}}y}{{\text{d}}x}}\right)^{2}\right]^{3/2}}{\frac {{\text{d}}^{2}y}{{\text{d}}x^{2}}}}={\frac {\left({\frac {{\text{d}}x}{{\text{d}}x}}\right)^{3}}{\frac {{\text{d}}^{2}y}{{\text{d}}x^{2}}}}={\frac {\left({\frac {y}{m}}\right)^{3}}{\frac {y}{m^{2}}}}={\frac {y^{2}}{m}}} 或者
P N = P C {\displaystyle PN=PC} 其中 P C {\displaystyle PC}
第 36 條 懸鏈線的幾何構造。 取一個等於 2 m {\displaystyle 2m} P {\displaystyle P} P {\displaystyle P} m {\displaystyle m} P B {\displaystyle PB} A {\displaystyle A} A 1 {\displaystyle A_{1}} A {\displaystyle A} P A 1 B 1 {\displaystyle PA_{1}B_{1}} P 1 {\displaystyle P_{1}} P 1 A 1 = A 1 B 1 {\displaystyle P_{1}A_{1}=A_{1}B_{1}} P 1 A 1 {\displaystyle P_{1}A_{1}} A 2 {\displaystyle A_{2}} A 1 {\displaystyle A_{1}} P 1 A 2 B 2 {\displaystyle P_{1}A_{2}B_{2}} P 2 {\displaystyle P_{2}} P 2 A 2 = A 2 B 2 {\displaystyle P_{2}A_{2}=A_{2}B_{2}} A {\displaystyle A}
附圖大致顯示了懸鏈線、其漸開線和軌跡的相對位置。
第 37 條
設 x ¯ {\displaystyle {\bar {x}}} y ¯ {\displaystyle {\bar {y}}} X {\displaystyle X} k {\displaystyle k}
y ¯ = m 2 [ e ( x ¯ − x 0 ′ ) / m + e − ( x ¯ − x 0 ′ ) / m ] {\displaystyle {\bar {y}}={\frac {m}{2}}[e^{({\bar {x}}-x_{0}')/m}+e^{-({\bar {x}}-x_{0}')/m}]} k = y ¯ ′ = 1 2 [ e ( x ¯ − x 0 ′ ) / m − e − ( x ¯ − x 0 ′ ) / m ] {\displaystyle k={\bar {y}}'={\frac {1}{2}}[e^{({\bar {x}}-x_{0}')/m}-e^{-({\bar {x}}-x_{0}')/m}]} 為了簡潔，我們寫成 e ( x ¯ − x 0 ′ ) / m = z {\displaystyle e^{({\bar {x}}-x_{0}')/m}=z}
y ¯ = m 2 ( z + z − 1 ) , k = 1 2 ( z − z − 1 ) {\displaystyle {\bar {y}}={\frac {m}{2}}(z+z^{-1}),\qquad k={\frac {1}{2}}(z-z^{-1})} 因此，
z 2 − 2 k z − 1 = 0 {\displaystyle z^{2}-2kz-1=0} 所以
z = k ± 1 + k 2 {\displaystyle z=k\pm {\sqrt {1+k^{2}}}} 以及
z − 1 = − k ± 1 + k 2 {\displaystyle z^{-1}=-k\pm {\sqrt {1+k^{2}}}} 因此我們有
y ¯ = ± m 1 + k 2 {\displaystyle {\bar {y}}=\pm m{\sqrt {1+k^{2}}}} 由於 y ¯ {\displaystyle {\bar {y}}} m {\displaystyle m}
k = tan ( α ) {\displaystyle k=\tan(\alpha )} 我們有
z = tan ( α ) + 1 + tan 2 ( α ) = sin ( α ) + 1 cos ( α {\displaystyle z=\tan(\alpha )+{\sqrt {1+\tan ^{2}(\alpha )}}={\frac {\sin(\alpha )+1}{\cos(\alpha }}} − z − 1 = tan ( α ) − 1 + tan 2 ( α ) = sin ( α ) − 1 cos ( α {\displaystyle -z^{-1}=\tan(\alpha )-{\sqrt {1+\tan ^{2}(\alpha )}}={\frac {\sin(\alpha )-1}{\cos(\alpha }}} 以及
m = y ¯ 1 + tan 2 ( α ) = y ¯ cos ( α ) {\displaystyle m={\frac {\bar {y}}{\sqrt {1+\tan ^{2}(\alpha )}}}={\bar {y}}\cos(\alpha )} 此外，由於 log ( z ) {\displaystyle \log(z)} z {\displaystyle z} x 0 ′ {\displaystyle x_{0}'}
x ¯ − x 0 ′ m = log ( z ) = log sin ( α ) + 1 cos ( α ) {\displaystyle {\frac {{\bar {x}}-x_{0}'}{m}}=\log(z)=\log {\frac {\sin(\alpha )+1}{\cos(\alpha )}}} 並且量 x 0 ′ {\displaystyle x_{0}'} m {\displaystyle m} x ¯ {\displaystyle {\bar {x}}} y ¯ {\displaystyle {\bar {y}}}
第 38 條 P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} ( − a , b ) {\displaystyle (-a,b)} ( a , b ) {\displaystyle (a,b)}
由於 x 0 ′ {\displaystyle x_{0}'}
y = m 2 ( e x / m + e − x / m ) {\displaystyle y={\frac {m}{2}}(e^{x/m}+e^{-x/m})} 因此
b = m 2 ( e a / m + e − a / m ) = ϕ ( m ) , [1] {\displaystyle b={\frac {m}{2}}(e^{a/m}+e^{-a/m})=\phi (m){\text{,}}\qquad {\text{[1]}}} 這裡我們把 α {\displaystyle \alpha } m {\displaystyle m}
我們希望確定這個最後一個方程是否為 m {\displaystyle m} ϕ ( m ) {\displaystyle \phi (m)} m = 0 {\displaystyle m=0} m = ∞ {\displaystyle m=\infty }
此外
2 ϕ ′ ( m ) = e a / m − e − a / m − a m ( e a / m − e − a / m ) {\displaystyle 2\phi '(m)=e^{a/m}-e^{-a/m}-{\frac {a}{m}}(e^{a/m}-e^{-a/m})} 或者
ϕ ′ ( m ) = 1 − 1 2 a 2 m 2 − 3 4 ! a 4 m 4 − ⋯ − 2 n − 1 ( 2 n ) ! a 2 n m 2 n − ⋯ {\displaystyle \phi '(m)=1-{\frac {1}{2}}{\frac {a^{2}}{m^{2}}}-{\frac {3}{4!}}{\frac {a^{4}}{m^{4}}}-\cdots -{\frac {2n-1}{(2n)!}}{\frac {a^{2n}}{m^{2n}}}-\cdots } 因此，當 m {\displaystyle m} ϕ ′ ( m ) {\displaystyle \phi '(m)} m {\displaystyle m} m {\displaystyle m} ϕ ( m ) {\displaystyle \phi (m)} ϕ ′ ( m ) = 0 {\displaystyle \phi '(m)=0} m {\displaystyle m}
如果給定的 b {\displaystyle b} ϕ ( m ) {\displaystyle \phi (m)} m {\displaystyle m} b {\displaystyle b} ϕ ( m ) {\displaystyle \phi (m)} m {\displaystyle m} b {\displaystyle b} ϕ ( m ) {\displaystyle \phi (m)} m {\displaystyle m}
Moigno 和 Lindelöf 證明了滿足
e a / m + e − a / m − a m ( e a / m − e − a / m ) = 0 {\displaystyle e^{a/m}+e^{-a/m}-{\frac {a}{m}}(e^{a/m}-e^{-a/m})=0} 大約為 a m = 1.19968.... {\displaystyle {\frac {a}{m}}=1.19968....} b m = 1.81017... {\displaystyle {\frac {b}{m}}=1.81017...} b a = 1.50888... = tan ( 56 ∘ 28 ′ ) {\displaystyle {\frac {b}{a}}=1.50888...=\tan(56^{\circ }28')} b a {\displaystyle {\frac {b}{a}}}
如果我們寫 k = b a = tan ( 56 ∘ 28 ′ ) {\displaystyle k={\frac {b}{a}}=\tan(56^{\circ }28')} y = k x {\displaystyle y=kx} y = − k x {\displaystyle y=-kx}
由於比率 b a {\displaystyle {\frac {b}{a}}} m {\displaystyle m} y = m 2 ( e x / m + e − x / m ) {\displaystyle y={\frac {m}{2}}(e^{x/m}+e^{-x/m})} m {\displaystyle m} x = ± 1.19968.... m {\displaystyle x=\pm 1.19968....m} y = 1.181017.... m {\displaystyle y=1.181017....m}
第 39 條 y = m 2 [ e ( x − x 0 ′ ) / m + e − ( x 0 ′ ) / m ] {\displaystyle y={\frac {m}{2}}[e^{(x-x_{0}')/m}+e{-(x_{0}')/m}]}
I. 可以透過固定點繪製兩條懸鏈線；
II. 可以透過這些點繪製一條懸鏈線；
III. 無法透過這兩個點繪製任何懸鏈線。
我們可以假設 y 1 ≥ y 0 {\displaystyle y_{1}\geq y_{0}} x 1 > x 0 {\displaystyle x_{1}>x_{0}} X {\displaystyle X} P 0 {\displaystyle P_{0}} P 1 ′ {\displaystyle P_{1}'} P 1 ′ {\displaystyle P_{1}'} P 1 {\displaystyle P_{1}} y 0 {\displaystyle y_{0}}
第 40 條
y 0 = m 2 [ e ( x 0 − x 0 ′ ) / m + e − ( x 0 − x 0 ′ ) / m ] {\displaystyle y_{0}={\frac {m}{2}}[e^{(x_{0}-x_{0}')/m}+e^{-(x_{0}-x_{0}')/m}]} 以及
y 0 2 − m 2 = m 2 4 [ e ( x 0 − x 0 ′ ) / m + e − ( x 0 − x 0 ′ ) / m ] 2 {\displaystyle y_{0}^{2}-m^{2}={\frac {m^{2}}{4}}[e^{(x_{0}-x_{0}')/m}+e^{-(x_{0}-x_{0}')/m}]^{2}} 因此
y 0 2 − m 2 = ± m 2 [ e ( x 0 − x 0 ′ ) / m + e − ( x 0 − x 0 ′ ) / m ] ; [I] {\displaystyle {\sqrt {y_{0}^{2}-m^{2}}}=\pm {\frac {m}{2}}[e^{(x_{0}-x_{0}')/m}+e^{-(x_{0}-x_{0}')/m}]{\text{;}}\qquad {\text{[I]}}} 從這個關係可以看出 y 0 2 − m 2 {\displaystyle {\sqrt {y_{0}^{2}-m^{2}}}} x 0 − x 0 ′ ≷ 0 {\displaystyle x_{0}-x_{0}'\gtrless 0}
x 0 − x 0 ′ m = ± ln y 0 + y 0 2 m . [a] {\displaystyle {\frac {x_{0}-x_{0}'}{m}}=\pm \ln {\frac {y_{0}+{\sqrt {y_{0}^{2}}}}{m}}{\text{.}}\qquad {\text{[a]}}} 第 41 條 y 1 ≥ y 0 {\displaystyle y_{1}\geq y_{0}}
y 1 = m 2 [ e ( x 1 − x 0 ′ ) / m + e − ( x 1 − x 0 ′ ) / m ] {\displaystyle y_{1}={\frac {m}{2}}[e^{(x_{1}-x_{0}')/m}+e^{-(x_{1}-x_{0}')/m}]} 可以從以下事實得出 x 1 − x 0 ′ {\displaystyle x_{1}-x_{0}'} y 0 ′ = m {\displaystyle y_{0}'=m} x 0 ′ {\displaystyle x_{0}'} x 0 ′ > x 1 {\displaystyle x_{0}'>x_{1}} y 1 ≥ y 0 {\displaystyle y_{1}\geq y_{0}} m ≤ y 0 {\displaystyle m\leq y_{0}} m ≤ y 1 {\displaystyle m\leq y_{1}} x 0 {\displaystyle x_{0}} x 0 ′ {\displaystyle x_{0}'} x {\displaystyle x} y {\displaystyle y} y {\displaystyle y} [ 2]
因此，
y 1 2 − m 2 = + m 2 [ e ( x 1 − x 0 ′ ) / m − e − ( x 1 − x 0 ′ ) / m ] {\displaystyle {\sqrt {y_{1}^{2}-m^{2}}}=+{\frac {m}{2}}[e^{(x_{1}-x_{0}')/m}-e^{-(x_{1}-x_{0}')/m}]} 並且，與公式[I]不同，這裡不能有減號 ，因此，
x 1 − x 0 ′ m = + ln y 1 + y 1 2 − m 2 m . [b] {\displaystyle {\frac {x_{1}-x_{0}'}{m}}=+\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}{\text{.}}\qquad {\text{[b]}}} 第 42 條 x 0 ′ {\displaystyle x_{0}'} ± {\displaystyle \pm } m {\displaystyle m}
f 1 ( m ) = ln y 1 + y 1 2 − m 2 m − ln y 0 + y 0 2 − m 2 m − x 1 − x 0 m {\displaystyle f_{1}(m)=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}-\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}-{\frac {x_{1}-x_{0}}{m}}} 以及
f 1 ( m ) = ln y 1 + y 1 2 − m 2 m + ln y 0 + y 0 2 − m 2 m − x 1 − x 0 m {\displaystyle f_{1}(m)=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}+\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}-{\frac {x_{1}-x_{0}}{m}}} 我們將討論兩個超越性質的函式。我們需要檢查 f 1 ( m ) = 0 {\displaystyle f_{1}(m)=0} f 2 ( m ) = 0 {\displaystyle f_{2}(m)=0} m {\displaystyle m} m {\displaystyle m} f 1 ( m ) = 0 {\displaystyle f_{1}(m)=0} f 2 ( m ) = 0 {\displaystyle f_{2}(m)=0} m {\displaystyle m} x 0 ′ {\displaystyle x_{0}'}
f 1 ( m ) {\displaystyle f_{1}(m)}
f 1 ′ ( m ) = 1 m [ ] . [c] {\displaystyle f_{1}'(m)={\frac {1}{m}}\left[\right]{\text{.}}\qquad {\text{[c]}}} 在這個表示式中，右邊 1 m {\displaystyle {\frac {1}{m}}} x 1 − x 0 m {\displaystyle {\frac {x_{1}-x_{0}}{m}}}
1 1 − m 2 y 0 2 − 1 1 − m 2 y 1 2 {\displaystyle {\frac {1}{\sqrt {1-{\frac {m^{2}}{y_{0}^{2}}}}}}-{\frac {1}{\sqrt {1-{\frac {m^{2}}{y_{1}^{2}}}}}}} y 1 > y 0 {\displaystyle y_{1}>y_{0}} 因此，在區間 0 ⋯ y 0 {\displaystyle 0\cdots y_{0}} f 1 ′ ( m ) {\displaystyle f_{1}'(m)}
此外，
f 1 ( 0 ) = ln ( 2 y 1 ) − ln ( m = 0 ) − ln ( 2 y 0 ) + ln ( m = 0 ) − [ x 1 − x 0 m ] m = 0 = − ∞ {\displaystyle f_{1}(0)=\ln(2y_{1})-\ln(m=0)-\ln(2y_{0})+\ln(m=0)-\left[{\frac {x_{1}-x_{0}}{m}}\right]_{m=0}=-\infty } 第 43 條 f 1 ( m ) {\displaystyle f_{1}(m)} 0 ⋯ y 0 {\displaystyle 0\cdots y_{0}} − ∞ {\displaystyle -\infty } f 1 ( m ) {\displaystyle f_{1}(m)}
再者
f 1 ( y 0 ) = ln y 1 + y 1 2 − y 0 2 y 0 − x 1 − x 0 y 0 . [II] {\displaystyle f_{1}(y_{0})=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-y_{0}^{2}}}}{y_{0}}}-{\frac {x_{1}-x_{0}}{y_{0}}}{\text{.}}\qquad {\text{[II]}}} 那麼，如果
I f 1 ( y 0 ) < 0 {\displaystyle f_{1}(y_{0})<0} f 1 ( m ) {\displaystyle f_{1}(m)}
II. f 1 ( y 0 ) = 0 {\displaystyle f_{1}(y_{0})=0} f 1 ( m ) {\displaystyle f_{1}(m)} m = y 0 {\displaystyle m=y_{0}}
III. f 1 ( y 0 ) > 0 {\displaystyle f_{1}(y_{0})>0} f 1 ( m ) {\displaystyle f_{1}(m)} m < y 0 {\displaystyle m<y_{0}}
當
f 1 ( y 0 ) < 0 {\displaystyle f_{1}(y_{0})<0} P 1 {\displaystyle P_{1}} f 1 ( y 0 ) − 0 {\displaystyle f_{1}(y_{0})-0} P 1 {\displaystyle P_{1}} f 1 ( y 0 ) > 0 {\displaystyle f_{1}(y_{0})>0} P 1 {\displaystyle P_{1}} 這可以透過以下方法證明
y = y 0 2 [ e ( x − x 0 ) / y 0 + e − ( x − x 0 ) / y 0 ] {\displaystyle y={\frac {y_{0}}{2}}\left[e^{(x-x_{0})/y_{0}}+e^{-(x-x_{0})/y_{0}}\right]} 因為當 y = m {\displaystyle y=m} x = x 0 ′ {\displaystyle x=x_{0}'} y = y 0 = m {\displaystyle y=y_{0}=m} x = x 0 {\displaystyle x=x_{0}}
y 2 − y 0 2 = y 0 2 4 [ e ( x − x 0 ) / y 0 − e − ( x − x ) 0 ) / y 0 ] 2 {\displaystyle y_{2}-y_{0}^{2}={\frac {y_{0}^{2}}{4}}\left[e^{(x-x_{0})/y_{0}}-e^{-(x-x)_{0})/y_{0}}\right]^{2}} 因此
y 2 − y 0 2 = ± y 0 x [ e ( x − x 0 ) / y 0 − e − ( x − x 0 ) / y 0 ] {\displaystyle {\sqrt {y^{2}-y_{0}^{2}}}=\pm {\frac {y_{0}}{x}}\left[e^{(x-x_{0})/y_{0}}-e^{-(x-x_{0})/y_{0}}\right]} 其中，當 x > x 0 {\displaystyle x>x_{0}} x < x 0 {\displaystyle x<x_{0}}
我們還有 x − x 0 = ln y + y 2 − y 0 2 y 0 {\displaystyle x-x_{0}=\ln {\frac {y+{\sqrt {y^{2}-y_{0}^{2}}}}{y_{0}}}}
x 1 − x 0 = y 0 ln y 1 + y 1 2 y 0 {\displaystyle x_{1}-x_{0}=y_{0}\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}}}}{y_{0}}}} P 1 {\displaystyle P_{1}} x 1 − x 0 > y 0 ln y 1 + y 1 2 y 0 {\displaystyle x_{1}-x_{0}>y_{0}\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}}}}{y_{0}}}} P 1 {\displaystyle P_{1}} x 1 − x 0 < y 0 ln y 1 + y 1 2 y 0 {\displaystyle x_{1}-x_{0}<y_{0}\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}}}}{y_{0}}}} P 1 {\displaystyle P_{1}} 因此，當 f 1 ( y ) > 0 {\displaystyle f_{1}(y)>0} 0 ⋯ y 0 {\displaystyle 0\cdots y_{0}} P 1 {\displaystyle P_{1}} P 0 {\displaystyle P_{0}} < x 0 {\displaystyle <x_{0}}
第 44 條 討論 f 2 ( m ) {\displaystyle f_{2}(m)}
f 2 ( m ) = ln y 1 + y 1 2 − m 2 m + ln y 0 + y 0 2 − m 2 m − x 1 − x 0 m {\displaystyle f_{2}(m)=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}+\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}-{\frac {x_{1}-x_{0}}{m}}} 因此
f 2 ′ ( m ) = − 1 m 2 ( y 1 m y 1 2 − m 2 + y 0 m y 0 2 − m 2 − ( x 1 − x 0 ) ) {\displaystyle f_{2}'(m)=-{\frac {1}{m^{2}}}\left({\frac {y_{1}m}{\sqrt {y_{1}^{2}-m^{2}}}}+{\frac {y_{0}m}{\sqrt {y_{0}^{2}-m^{2}}}}-(x_{1}-x_{0})\right)} 當 m {\displaystyle m} 0 {\displaystyle 0} y 0 {\displaystyle y_{0}} y 0 2 m 2 − 1 {\displaystyle {\sqrt {\frac {y_{0}^{2}}{m^{2}-1}}}} y 0 y 0 2 m 2 − 1 {\displaystyle {\frac {y_{0}}{\frac {y_{0}^{2}}{m^{2}-1}}}} − m 2 f 2 ′ ( m ) {\displaystyle -m^{2}f_{2}'(m)} 0 {\displaystyle 0} 0 {\displaystyle 0} y 0 {\displaystyle y_{0}} 0 {\displaystyle 0} m = 0 {\displaystyle m=0} − m 2 f 2 ′ ( m ) = − ( x 1 − x 0 ) {\displaystyle -m^{2}f_{2}'(m)=-(x_{1}-x_{0})} x 1 − x 0 > 0 {\displaystyle x_{1}-x_{0}>0} − m 2 f 2 ′ ( y 0 ) {\displaystyle -m^{2}f_{2}'(y_{0})} m = y 0 {\displaystyle m=y_{0}} − m 2 f 2 ′ ( y 0 ) = + ∞ {\displaystyle -m^{2}f_{2}'(y_{0})=+\infty } m {\displaystyle m}
設 μ {\displaystyle \mu } m {\displaystyle m}
y 1 μ y 1 2 − μ 2 + y 0 μ y 0 2 − μ 2 − ( x 1 − x 0 ) = 0 , [A] {\displaystyle {\frac {y_{1}\mu }{\sqrt {y_{1}^{2}-\mu ^{2}}}}+{\frac {y_{0}\mu }{\sqrt {y_{0}^{2}-\mu ^{2}}}}-(x_{1}-x_{0})=0{\text{,}}\qquad {\text{[A]}}} 這是一個關於 μ {\displaystyle \mu } μ 2 {\displaystyle \mu ^{2}}
第 45 條 對於位於 0 {\displaystyle 0} y 0 {\displaystyle y_{0}} 圖中可以看到三角形 P 0 Q 0 A 0 {\displaystyle P_{0}Q_{0}A_{0}} P 0 Q 0 C 0 {\displaystyle P_{0}Q_{0}C_{0}} P 1 Q 1 A 1 {\displaystyle P_{1}Q_{1}A_{1}} P 1 Q 1 C 1 {\displaystyle P_{1}Q_{1}C_{1}} m {\displaystyle m} Q 0 C 0 = Q 1 C 1 {\displaystyle Q_{0}C_{0}=Q_{1}C_{1}}
Q 0 A 0 = y 0 m y 0 2 − m 2 {\displaystyle Q_{0}A_{0}={\frac {y_{0}m}{\sqrt {y_{0}^{2}-m^{2}}}}} 以及
Q 1 A 1 = y 1 m y 1 2 − m 2 {\displaystyle Q_{1}A_{1}={\frac {y_{1}m}{\sqrt {y_{1}^{2}-m^{2}}}}} 在兩個半圓上取等長線段 Q 0 C 0 = Q 1 C 1 {\displaystyle Q_{0}C_{0}=Q_{1}C_{1}} P 0 C 0 {\displaystyle P_{0}C_{0}} P 1 C 1 {\displaystyle P_{1}C_{1}} X {\displaystyle X} S {\displaystyle S}
Q 0 S + Q 1 S = Q 0 Q 1 = x 1 − x 0 {\displaystyle Q_{0}S+Q_{1}S=Q_{0}Q_{1}=x_{1}-x_{0}} 由此可知
y 0 ⋅ Q 0 B 0 y 0 2 + Q 0 B 0 ¯ 2 + y 1 ⋅ Q 1 B 1 y 1 2 − Q 1 B 1 ¯ 2 = x 1 − x 0 {\displaystyle {\frac {y_{0}\cdot Q_{0}B_{0}}{\sqrt {y_{0}^{2}+{\overline {Q_{0}B_{0}}}^{2}}}}+{\frac {y_{1}\cdot Q_{1}B_{1}}{\sqrt {y_{1}^{2}-{\overline {Q_{1}B_{1}}}^{2}}}}=x_{1}-x_{0}} 與上面的方程 [A] 對比可知
Q 0 B 0 = Q 1 B 1 = μ {\displaystyle Q_{0}B_{0}=Q_{1}B_{1}=\mu } 第 46 條 函式 f 1 ( m ) {\displaystyle f_{1}(m)} f 2 ( m ) {\displaystyle f_{2}(m)} 長度 m {\displaystyle m} X {\displaystyle X} f 1 ′ ( y 0 ) = ∞ {\displaystyle f_{1}'(y_{0})=\infty } y = f 1 ( x ) {\displaystyle y=f_{1}(x)} y 0 {\displaystyle y_{0}} y {\displaystyle y} f 1 ( 0 ) = − ∞ {\displaystyle f_{1}(0)=-\infty } y {\displaystyle y} y = f 1 ( x ) {\displaystyle y=f_{1}(x)} y = f 1 ( x ) {\displaystyle y=f_{1}(x)} x = 0 {\displaystyle x=0}
第 47 條 接下來考慮曲線 y = f 2 ( m ) {\displaystyle y=f_{2}(m)} f 1 ( y 0 ) = f 2 ( y 0 ) {\displaystyle f_{1}(y_{0})=f_{2}(y_{0})} f 2 ′ ( y 0 ) = − ∞ {\displaystyle f_{2}'(y_{0})=-\infty } [ 3] y {\displaystyle y} y {\displaystyle y} y = f 2 ( m ) {\displaystyle y=f_{2}(m)} m = 0 {\displaystyle m=0} m = 0 {\displaystyle m=0}
f 2 ( m ) = ln y 1 + y 1 2 − m 2 m + ln y 0 + y 0 2 − m 2 m − x 1 − x 0 m = − x 1 − x 0 m − 2 ln ( m ) + P ( m ) {\displaystyle f_{2}(m)=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}+\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}-{\frac {x_{1}-x_{0}}{m}}=-{\frac {x_{1}-x_{0}}{m}}-2\ln(m)+P(m)} 其中 P ( m ) {\displaystyle P(m)} m {\displaystyle m} m = 0 {\displaystyle m=0}
第 48 條
f 2 ′ ( m ) = − 1 m 2 ( y 1 m y 1 2 − m 2 + y 0 m y 0 2 − m 2 − ( x 1 − x 0 ) ) {\displaystyle f_{2}'(m)=-{\frac {1}{m^{2}}}\left({\frac {y_{1}m}{\sqrt {y_{1}^{2}-m^{2}}}}+{\frac {y_{0}m}{\sqrt {y_{0}^{2}-m^{2}}}}-(x_{1}-x_{0})\right)} 對於值 m = μ {\displaystyle m=\mu } m = 0 {\displaystyle m=0} − ( x 1 − x 0 ) {\displaystyle -(x_{1}-x_{0})} m = 0 {\displaystyle m=0} m = y 0 {\displaystyle m=y_{0}}
y 1 m y 1 2 m 2 + y 0 m y 0 2 m 2 − ( x 1 − x 0 ) {\displaystyle {\frac {y_{1}m}{\sqrt {y_{1}^{2}m^{2}}}}+{\frac {y_{0}m}{\sqrt {y_{0}^{2}m^{2}}}}-(x_{1}-x_{0})} 越來越大，因此在值 m = 0 {\displaystyle m=0} m = μ {\displaystyle m=\mu }
此外， f 2 ′ ( m ) {\displaystyle f_{2}'(m)} m = 0 {\displaystyle m=0} m = μ {\displaystyle m=\mu } m = μ {\displaystyle m=\mu } m = y 0 {\displaystyle m=y_{0}}
因此， f 2 ( m ) {\displaystyle f_{2}(m)} m = 0 {\displaystyle m=0} m = μ {\displaystyle m=\mu } m = μ {\displaystyle m=\mu } m = y 0 {\displaystyle m=y_{0}} f 2 ( μ ) {\displaystyle f_{2}(\mu )}
第 49 條 f 2 ( m ) {\displaystyle f_{2}(m)} m {\displaystyle m} P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}}
我們有
情況一。 f 2 ( μ ) < 0 {\displaystyle f_{2}(\mu )<0}
在這種情況下， f 2 ( m ) {\displaystyle f_{2}(m)} f 2 ( m ) {\displaystyle f_{2}(m)} f 1 ( m ) {\displaystyle f_{1}(m)} f 2 ( y 0 ) < 0 {\displaystyle f_{2}(y_{0})<0} f 2 ( y 0 ) = f 1 ( y 0 ) {\displaystyle f_{2}(y_{0})=f_{1}(y_{0})} f 1 ( y 0 ) {\displaystyle f_{1}(y_{0})}
情況二。 f 2 ( μ ) = 0 {\displaystyle f_{2}(\mu )=0}
除了 μ {\displaystyle \mu } m {\displaystyle m} f 2 ( m ) {\displaystyle f_{2}(m)} f 2 ( m ) = 0 {\displaystyle f_{2}(m)=0} f 1 ( m ) {\displaystyle f_{1}(m)} f 2 ( y 0 ) < 0 {\displaystyle f_{2}(y_{0})<0} f 1 ( y 0 ) = f 2 ( y 0 ) {\displaystyle f_{1}(y_{0})=f_{2}(y_{0})} f 1 ( y 0 ) < 0 {\displaystyle f_{1}(y_{0})<0}
情況三。 f 2 ( μ ) > 0 {\displaystyle f_{2}(\mu )>0}
這裡有兩個懸鏈線。一個根 f 2 ( m ) = m {\displaystyle f_{2}(m)=m} 0 {\displaystyle 0} μ {\displaystyle \mu } 通常 另一個根位於 μ {\displaystyle \mu } y 0 {\displaystyle y_{0}}
f 2 ( + 0 ) = − ∞ {\displaystyle f_{2}(+0)=-\infty } f 2 ( μ ) > 0 {\displaystyle f_{2}(\mu )>0} 由於 f 2 ( m ) {\displaystyle f_{2}(m)} + 0 … μ {\displaystyle +0\ldots \mu } 0 {\displaystyle 0}
在區間 μ … y 0 {\displaystyle \mu \ldots y_{0}} f 2 ( m ) {\displaystyle f_{2}(m)} f 2 ( y 0 ) > 0 {\displaystyle f_{2}(y_{0})>0} f 2 ( m ) = 0 {\displaystyle f_{2}(m)=0} f 2 ( y 0 ) ≤ 0 {\displaystyle f_{2}(y_{0})\leq 0}
接下來，我們必須考慮 f 1 ( m ) {\displaystyle f_{1}(m)} f 2 ( y 0 ) {\displaystyle f_{2}(y_{0})} f 1 ( y 0 ) < 0 {\displaystyle f_{1}(y_{0})<0} f 1 ( m ) = 0 {\displaystyle f_{1}(m)=0} f 2 ( y 0 ) = 0 {\displaystyle f_{2}(y_{0})=0} f 1 ( y 0 ) = 0 {\displaystyle f_{1}(y_{0})=0} f 1 ( m ) = 0 {\displaystyle f_{1}(m)=0} m = y 0 {\displaystyle m=y_{0}}
因此
A) 當 f 2 ( y 0 ) < 0 {\displaystyle f_{2}(y_{0})<0} f 2 ( m ) {\displaystyle f_{2}(m)} f 2 ( y 0 ) = 0 {\displaystyle f_{2}(y_{0})=0} f 2 ( m ) {\displaystyle f_{2}(m)} f 2 ( y 0 ) = f 1 ( y 0 ) {\displaystyle f_{2}(y_{0})=f_{1}(y_{0})}
B) 但當 f 2 ( y 0 ) > 0 {\displaystyle f_{2}(y_{0})>0} f 2 ( m ) = 0 {\displaystyle f_{2}(m)=0} 0 … μ {\displaystyle 0\ldots \mu } m 1 {\displaystyle m_{1}}
第 50 條 f 1 ( m ) {\displaystyle f_{1}(m)} f 2 ( m ) {\displaystyle f_{2}(m)}
f 2 ( m ) = f 1 ( m ) + 2 ln y 0 + y 0 2 − m 2 m {\displaystyle f_{2}(m)=f_{1}(m)+2\ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}} 我們考慮 m {\displaystyle m} 0 … y 0 {\displaystyle 0\ldots y_{0}} m = 0 {\displaystyle m=0} y 0 + y 0 2 m = ∞ {\displaystyle {\frac {y_{0}+{\sqrt {y_{0}^{2}}}}{m}}=\infty } m = y 0 {\displaystyle m=y_{0}} y 0 + y 0 2 − m 2 m = 1 {\displaystyle {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}=1} ln y 0 + y 0 2 − m 2 m {\displaystyle \ln {\frac {y_{0}+{\sqrt {y_{0}^{2}-m^{2}}}}{m}}} f 2 ( m ) > f 1 ( m ) {\displaystyle f_{2}(m)>f_{1}(m)} f 2 ( m 1 ) = 0 {\displaystyle f_{2}(m_{1})=0} f 1 ( m 1 ) < 0 {\displaystyle f_{1}(m_{1})<0}
另一方面， f 1 ( y 0 ) = f 2 ( y 0 ) {\displaystyle f_{1}(y_{0})=f_{2}(y_{0})} f 2 ( y 0 ) > 0 {\displaystyle f_{2}(y_{0})>0} f 1 ( y 0 ) > 0 {\displaystyle f_{1}(y_{0})>0} 0 … m 1 {\displaystyle 0\ldots m_{1}} f 1 ( m ) {\displaystyle f_{1}(m)} f 1 ( + 0 ) < 0 {\displaystyle f_{1}(+0)<0} 0 … m 1 {\displaystyle 0\ldots m_{1}} f 1 ( m ) {\displaystyle f_{1}(m)} m 1 … y 0 {\displaystyle m_{1}\ldots y_{0}}
因此，在 B) 下， f 2 ( m ) {\displaystyle f_{2}(m)} 0 … μ {\displaystyle 0\ldots \mu } m 1 {\displaystyle m_{1}} f 1 ( m ) {\displaystyle f_{1}(m)} m 1 {\displaystyle m_{1}} y 0 {\displaystyle y_{0}}
我們有以下總結
1 0 {\displaystyle 1^{0}} f 2 ( μ ) < 0 {\displaystyle f_{2}(\mu )<0} 2 0 {\displaystyle 2^{0}} f 2 ( μ ) = 0 {\displaystyle f_{2}(\mu )=0} 3 0 {\displaystyle 3^{0}} f 2 ( μ ) > 0 {\displaystyle f_{2}(\mu )>0} 第 51 條 關於在點 P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}}
情況 I. 如上所示，沒有懸鏈線，因此對切線的考慮沒有意義。
情況二。 f 2 ( μ ) = 0 {\displaystyle f_{2}(\mu )=0}
這裡，懸鏈線具有一個顯著的性質，即在點 P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} X {\displaystyle X} P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} B 0 {\displaystyle B_{0}} B 1 {\displaystyle B_{1}} P 0 B 0 Q 0 {\displaystyle P_{0}B_{0}Q_{0}} P 1 B 1 Q 1 {\displaystyle P_{1}B_{1}Q_{1}} Q 0 B 0 = Q 1 B 1 {\displaystyle Q_{0}B_{0}=Q_{1}B_{1}} m = μ {\displaystyle m=\mu } P 0 B 0 {\displaystyle P_{0}B_{0}} P 1 B ! {\displaystyle P_{1}B_{!}} X {\displaystyle X}
情況三。 f 2 ( μ ) > 0 {\displaystyle f_{2}(\mu )>0}
A) f 2 ( y 0 ) ≤ 0 {\displaystyle f_{2}(y_{0})\leq 0}
然後，正如已經展示的， f 2 ( m ) = 0 {\displaystyle f_{2}(m)=0} 0 {\displaystyle 0} μ {\displaystyle \mu } μ {\displaystyle \mu } y 0 {\displaystyle y_{0}} m 1 {\displaystyle m_{1}} m 2 {\displaystyle m_{2}} m 1 {\displaystyle m_{1}}
Q 0 T 0 = y 0 m 1 y 0 2 − m 1 2 {\displaystyle Q_{0}T_{0}={\frac {y_{0}m_{1}}{\sqrt {y_{0}^{2}-m_{1}^{2}}}}} Q 1 T 1 = y 1 m 1 y 1 2 − m 1 2 {\displaystyle Q_{1}T_{1}={\frac {y_{1}m_{1}}{\sqrt {y_{1}^{2}-m_{1}^{2}}}}} 我們斷言，這裡在 P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} X {\displaystyle X}
為了證明這一點，我們只需要證明
Q 0 T 0 + Q 1 T 1 < Q 0 Q 1 {\displaystyle Q{0}T_{0}+Q_{1}T_{1}<Q_{0}Q_{1}} 這是這樣看到的
f 2 ′ ( m 1 ) = − 1 m 2 ( y 1 m 1 y 1 2 − m 1 2 + y 0 m 1 y 0 2 − m 1 2 − ( x 1 − x 0 ) ) {\displaystyle f_{2}'(m_{1})=-{\frac {1}{m^{2}}}\left({\frac {y_{1}m_{1}}{\sqrt {y_{1}^{2}-m_{1}^{2}}}}+{\frac {y_{0}m_{1}}{\sqrt {y_{0}^{2}-m_{1}^{2}}}}-(x_{1}-x_{0})\right)} 現在，由於區間 0 … μ {\displaystyle 0\ldots \mu } f 2 ′ ( m ) {\displaystyle f_{2}'(m)} m 1 {\displaystyle m_{1}} f 2 ′ ( m 1 ) {\displaystyle f_{2}'(m_{1})} − m 1 2 f 2 ′ ( m 1 ) {\displaystyle -m_{1}^{2}f_{2}'(m_{1})} q 0 T 0 + Q 1 T 1 − Q 0 Q 1 {\displaystyle q_{0}T_{0}+Q_{1}T_{1}-Q_{0}Q_{1}}
注：在這個考慮中，整個解釋取決於根位於區間 0 … μ {\displaystyle 0\ldots \mu } f 2 ( y 0 ) > 0 {\displaystyle f_{2}(y_{0})>0} 0 … μ {\displaystyle 0\ldots \mu }
第 52 條 關於根 m 2 {\displaystyle m_{2}}
1 ∘ {\displaystyle 1^{\circ }} f 2 ( y 0 ) ≤ 0 {\displaystyle f_{2}(y_{0})\leq 0}
根位於區間 μ … y 0 {\displaystyle \mu \ldots y_{0}} f 2 ′ ( m ) {\displaystyle f_{2}'(m)} − m 2 f 2 ′ ( m ) {\displaystyle -m^{2}f_{2}'(m)}
y 1 m 2 y 1 2 m 2 2 + y 0 m 2 y 0 2 m 2 2 − ( x 1 − x 0 ) > 0 {\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}m_{2}^{2}}}}+{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}m_{2}^{2}}}}-(x_{1}-x_{0})>0} 所以
Q 0 T 0 + Q 1 T 1 > Q 0 Q 1 {\displaystyle Q_{0}T_{0}+Q_{1}T_{1}>Q_{0}Q_{1}} 為了使 T {\displaystyle T} X {\displaystyle X}
2 ∘ {\displaystyle 2^{\circ }} f 2 ( y 0 ) > 0 {\displaystyle f_{2}(y_{0})>0} m 2 {\displaystyle m_{2}} f 1 ( m ) = 0 {\displaystyle f_{1}(m)=0}
y 1 m 2 y 1 2 − m 2 2 + y 0 m 2 y 0 2 − m 2 2 − ( x 1 − x 0 ) {\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}+{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}-(x_{1}-x_{0})} 在區間 0 … y 0 {\displaystyle 0\ldots y_{0}}
我們已經證明，在這個區間內 f 1 ′ ( m ) {\displaystyle f_{1}'(m)}
f 1 ′ ( m 2 ) = − 1 m 2 ( y 1 m 2 y 1 2 − m 2 2 − y 0 m 2 y 0 2 − m 2 2 − ( x 1 − x 0 ) ) {\displaystyle f_{1}'(m_{2})=-{\frac {1}{m^{2}}}\left({\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}-{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}-(x_{1}-x_{0})\right)} 是正的，因此可以得出
y 1 m 2 y 1 2 − m 2 2 − y 0 m 2 y 0 2 − m 2 2 − ( x 1 − x 0 ) {\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}-{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}-(x_{1}-x_{0})} 是負的。因此
y 1 m 2 y 1 2 − m 2 2 − y 0 m 2 y 0 2 − m 2 2 < ( x 1 − x 0 ) {\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}-{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}<(x_{1}-x_{0})} 因此
y 0 m 2 y 0 2 − m 2 2 − y 1 m 2 y 1 2 − m 2 2 > ( x 1 − x 0 ) {\displaystyle {\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}-{\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}>(x_{1}-x_{0})} 由於 y 1 m 2 y 1 2 − m 2 2 {\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}}
y 1 m 2 y 1 2 − m 2 2 + y 0 m 2 y 0 2 − m 2 2 > ( x 1 − x 0 ) {\displaystyle {\frac {y_{1}m_{2}}{\sqrt {y_{1}^{2}-m_{2}^{2}}}}+{\frac {y_{0}m_{2}}{\sqrt {y_{0}^{2}-m_{2}^{2}}}}>(x_{1}-x_{0})} 並且交點位於 X {\displaystyle X}
第五十三條 P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} m 1 {\displaystyle m_{1}} m 2 {\displaystyle m_{2}} C 1 {\displaystyle C_{1}} C 2 {\displaystyle C_{2}} C 2 {\displaystyle C_{2}} C 1 {\displaystyle C_{1}} P {\displaystyle P} P T {\displaystyle PT} C 2 {\displaystyle C_{2}} X {\displaystyle X} C 2 {\displaystyle C_{2}} C 1 {\displaystyle C_{1}} C 2 {\displaystyle C_{2}} C 1 {\displaystyle C_{1}} C 2 {\displaystyle C_{2}}
第 54 條 A T 0 {\displaystyle AT_{0}} C 1 {\displaystyle C_{1}} A T 1 {\displaystyle AT_{1}} C 2 {\displaystyle C_{2}} A T 0 {\displaystyle AT_{0}} A T 0 {\displaystyle AT_{0}} A T 1 {\displaystyle AT_{1}}
我們還看到，點 P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} C 1 {\displaystyle C_{1}} C 2 {\displaystyle C_{2}} P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} C 1 {\displaystyle C_{1}} C 2 {\displaystyle C_{2}} P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} μ {\displaystyle \mu } m {\displaystyle m} m 2 > μ > m 1 {\displaystyle m_{2}>\mu >m_{1}}
第 55 條 m {\displaystyle m}
y = m 2 [ e ( x − x 0 ′ ) / m + e − ( x − x 0 ′ ) / m ] {\displaystyle y={\frac {m}{2}}[e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}]} y = m 2 [ e ( x − x 0 ″ ) / m + e − ( x − x 0 ″ ) / m ] {\displaystyle y={\frac {m}{2}}[e^{(x-x_{0}'')/m}+e^{-(x-x_{0}'')/m}]} 這兩條懸鏈線僅在一個點相交。因為我們有：
e ( x − x 0 ′ ) / m + e − ( x − x 0 ′ ) / m = e ( x − x 0 ″ ) / m + e − ( x − x 0 ″ ) / m {\displaystyle e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}=e^{(x-x_{0}'')/m}+e^{-(x-x_{0}'')/m}} 所以
e x / m [ e − x 0 ′ / m − e − x 0 ″ / m ] = e − x / m [ e x 0 ″ / m − e x 0 ′ / m ] {\displaystyle e^{x/m}[e^{-x_{0}'/m}-e^{-x_{0}''/m}]=e^{-x/m}[e^{x_{0}''/m}-e^{x_{0}'/m}]} 或者
e 2 x / m = e x 0 ″ / m − e x 0 ′ / m e − x 0 ′ / m − e − x 0 ″ / m = 1 e − ( x 0 ′ + x 0 ″ ) / m ( e x 0 ″ / m − e x 0 ′ / m − e x 0 ′ / m + e x 0 ″ / m ) {\displaystyle e^{2x/m}={\frac {e^{x_{0}''/m}-e^{x_{0}'/m}}{e^{-x_{0}'/m}-e^{-x_{0}''/m}}}={\frac {1}{e^{-(x_{0}'+x_{0}'')/m}}}\left({\frac {e^{x_{0}''/m}-e^{x_{0}'/m}}{-e^{x_{0}'/m}+e^{x_{0}''/m}}}\right)} 因此
e 2 x / m = e ( x 0 ′ + x 0 ″ ) / m {\displaystyle e^{2x/m}=e^{(x_{0}'+x_{0}'')/m}} 因此
x = x 0 ′ + x 0 ″ 2 y = m 2 [ e ( x 0 ″ − x 0 ′ ) / ( 2 m ) + e − ( x 0 ″ − x 0 ′ ) / ( 2 m ) ) ] {\displaystyle x={\frac {x_{0}'+x_{0}''}{2}}\qquad y={\frac {m}{2}}[e^{(x_{0}''-x_{0}')/(2m)}+e^{-(x_{0}''-x_{0}')/(2m))}]} 這些是 _唯一_ 一個點的座標。
第 56 條 林德洛夫定理 (1860).
如果我們假設懸鏈線繞著 X {\displaystyle X} P 0 T {\displaystyle P_{0}T} P 1 T {\displaystyle P_{1}T} X {\displaystyle X} P 0 T {\displaystyle P_{0}T} P 1 T {\displaystyle P_{1}T} X {\displaystyle X}
假設以 T {\displaystyle T} P 0 P 1 {\displaystyle P_{0}P_{1}} P 0 {\displaystyle P_{0}} P 0 ′ {\displaystyle P_{0}'} P 1 {\displaystyle P_{1}} P 1 ′ {\displaystyle P_{1}'} P 0 P 0 ′ {\displaystyle P_{0}P_{0}'} P 1 P 1 ′ {\displaystyle P_{1}P_{1}'}
那麼
P ) T : P 0 ′ T = 1 : 1 − α {\displaystyle P_{)}T~:~P_{0}'T=1~:~1-\alpha } 為了簡化，令
M 0 {\displaystyle M_{0}} P 0 T {\displaystyle P_{0}T} M 0 ′ {\displaystyle M_{0}'} P 0 ′ T {\displaystyle P_{0}'T} M 1 {\displaystyle M_{1}} P 1 T {\displaystyle P_{1}T} M 1 ′ {\displaystyle M_{1}'} P 1 ′ T {\displaystyle P_{1}'T} S {\displaystyle S} P 0 P 1 {\displaystyle P_{0}P_{1}} S ′ {\displaystyle S'} P 0 ′ P 1 ′ {\displaystyle P_{0}'P_{1}'}
從應變的性質來看，切線 P 0 T {\displaystyle P_{0}T} P 1 T {\displaystyle P_{1}T} P 0 ′ {\displaystyle P_{0}'} P 1 ′ {\displaystyle P_{1}'} P 0 P 0 ′ P 1 ′ P 1 {\displaystyle P_{0}P_{0}'P_{1}'P_{1}} P 0 P 1 {\displaystyle P_{0}P_{1}}
可以看出
S : S ′ = 1 : ( 1 − α ) 2 {\displaystyle S~:~S'=1~:~(1-\alpha )^{2}} M 0 : M 0 ′ = 1 : ( 1 − α ) 2 {\displaystyle M_{0}~:~M_{0}'=1~:~(1-\alpha )^{2}} M 1 : M 1 ′ = 1 : ( 1 − α ) 2 {\displaystyle M_{1}~:~M_{1}'=1~:~(1-\alpha )^{2}} 從圖中可以看出， P 0 P 0 ′ P 1 P 1 ′ {\displaystyle P_{0}P_{0}'P_{1}P_{1}'}
( M 0 − M 0 ′ ) + S ′ + ( M 1 − M 1 ′ ) + [ ( α ) 2 ] = S {\displaystyle (M_{0}-M_{0}')+S'+(M_{1}-M_{1}')+[(\alpha )^{2}]=S} 其中 [ ( α ) 2 ] {\displaystyle [(\alpha )^{2}]}
因此
S − S ′ = ( M 0 − M 0 ′ ) + ( M 1 − M 1 ′ ) + [ ( α ) 2 ] {\displaystyle S-S'=(M_{0}-M_{0}')+(M_{1}-M_{1}')+[(\alpha )^{2}]} 因此
S [ 1 − ( 1 − α ) 2 ] = M 0 [ 1 − ( 1 − α ) 2 ] + M 1 [ 1 − ( 1 − α ) 2 ] + [ ( α 2 ) ] {\displaystyle S[1-(1-\alpha )^{2}]=M_{0}[1-(1-\alpha )^{2}]+M_{1}[1-(1-\alpha )^{2}]+[(\alpha ^{2})]} 因此
2 α S = 2 α M 0 + 2 α M 1 + [ ( α 2 ) ] {\displaystyle 2\alpha S=2\alpha M_{0}+2\alpha M_{1}+[(\alpha ^{2})]} 或者最終
S = M 0 ′ + M 1 ′ {\displaystyle S=M_{0}'+M_{1}'} 這是一個關於一階微分的正確結果。
以類似的方式
S ′ = M 0 ′ + M 1 ′ {\displaystyle S'=M_{0}'+M_{1}'} 因此
S − S ′ = ( M 0 − M 0 ′ ) + ( M 1 − M 1 ′ ) {\displaystyle S-S'=(M_{0}-M_{0}')+(M_{1}-M_{1}')} 或者
S = ( M 0 − M 0 ′ ) + S ′ + ( M 1 − M 1 ′ ) {\displaystyle S=(M_{0}-M_{0}')+S'+(M_{1}-M_{1}')} 是一個絕對正確的表示式。
第 57 條 另一種證明。
我們已經看到
y 0 μ y 0 − μ 2 + y 1 μ y 1 − μ 2 − ( x 1 − x 0 ) = 0 {\displaystyle {\frac {y_{0}\mu }{\sqrt {y_{0}-\mu ^{2}}}}+{\frac {y_{1}\mu }{\sqrt {y_{1}-\mu ^{2}}}}-(x_{1}-x_{0})=0} 以及（參見第 45 條中的圖）
P 0 S = y 0 2 y 0 − μ 2 {\displaystyle P_{0}S={\frac {y_{0}^{2}}{\sqrt {y_{0}-\mu ^{2}}}}} P 1 S = y 1 2 y 1 − μ 2 {\displaystyle P_{1}S={\frac {y_{1}^{2}}{\sqrt {y_{1}-\mu ^{2}}}}} 因此，兩個圓錐的表面積分別為
y 0 ⋅ y 0 2 π y 0 − μ 2 {\displaystyle {\frac {y_{0}\cdot y_{0}^{2}\pi }{\sqrt {y_{0}-\mu ^{2}}}}} y 1 ⋅ y 1 2 π y 1 − μ 2 {\displaystyle {\frac {y_{1}\cdot y_{1}^{2}\pi }{\sqrt {y_{1}-\mu ^{2}}}}} 懸鏈線產生的表面積為
∫ x 0 x 1 2 y π d s {\displaystyle \int _{x_{0}}^{x_{1}}2y\pi ~{\text{d}}s} 在懸鏈線中 d s = y m d x {\displaystyle {\text{d}}s={\frac {y}{m}}~{\text{d}}x}
∫ x 0 x 1 2 y π d s = ∫ x 0 x 1 2 y 2 π d x m = 2 π ∫ x 0 x 1 m 2 4 [ e 2 ( x − x 0 ′ ) / m + 2 + e − 2 ( x − x 0 ′ ) / m ] d x m {\displaystyle \int _{x_{0}}^{x_{1}}2y\pi ~{\text{d}}s=\int _{x_{0}}^{x_{1}}{\frac {2y^{2}\pi ~{\text{d}}x}{m}}=2\pi \int _{x_{0}}^{x_{1}}{\frac {m^{2}}{4}}[e^{2(x-x_{0}')/m}+2+e^{-2(x-x_{0}')/m}]{\frac {{\text{d}}x}{m}}} = π m 2 4 [ e 2 ( x − x 0 ′ ) / m − e − 2 ( x − x 0 ′ ) / m + 4 x m ] x 0 x 1 {\displaystyle ={\frac {\pi m^{2}}{4}}\left[e^{2(x-x_{0}')/m}-e^{-2(x-x_{0}')/m}+{\frac {4x}{m}}\right]_{x_{0}}^{x_{1}}} = π [ m 2 ( e ( x − x 0 ′ ) / m + e − ( x − x 0 ′ ) / m ) ⋅ m 2 ( e ( x − x 0 ′ ) / m − e − ( x − x 0 ′ ) / m ) + m x ] x 0 x 1 [A] {\displaystyle =\pi \left[{\frac {m}{2}}(e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m})\cdot {\frac {m}{2}}(e^{(x-x_{0}')/m}-e^{-(x-x_{0}')/m})+mx\right]_{x_{0}}^{x_{1}}\qquad {\text{[A]}}} = π [ ± y y 2 − m 2 + m x ] x 0 x 1 {\displaystyle =\pi [\pm y{\sqrt {y^{2}-m^{2}}}+mx]_{x_{0}}^{x_{1}}} = π [ y 1 y 1 − m 2 + y 0 y 0 − m 2 ] + m ( x 1 − x 0 ) , [B] {\displaystyle =\pi [y_{1}{\sqrt {y_{1}-m^{2}}}+y_{0}{\sqrt {y_{0}-m^{2}}}]+m(x_{1}-x_{0}){\text{,}}\qquad {\text{[B]}}} 其中我們取了 + {\displaystyle +} y 0 y 0 2 − m 2 {\displaystyle y_{0}{\sqrt {y_{0}^{2}-m^{2}}}} x 0 − x 0 ′ {\displaystyle x_{0}-x_{0}'} e ( x − x 0 ′ ) / m − e − ( x − x 0 ′ ) / m {\displaystyle e^{(x-x_{0}')/m}-e^{-(x-x_{0}')/m}}
但從[1]
x 1 − x 0 = y 1 μ y 1 2 − μ 2 + y 0 μ y 0 2 − μ 2 {\displaystyle x_{1}-x_{0}={\frac {y_{1}\mu }{\sqrt {y_{1}^{2}-\mu ^{2}}}}+{\frac {y_{0}\mu }{\sqrt {y_{0}^{2}-\mu ^{2}}}}} 將[B]代入，我們有，在使 m = μ {\displaystyle m=\mu }
π [ y 1 y 1 − μ 2 + y 1 μ 2 y 1 2 − μ 2 + y 0 y 0 2 − μ 2 + y 0 μ 2 y 0 2 − μ 2 ] = π [ y 1 3 y 1 2 − μ 2 + y 0 3 y 0 2 − μ 2 ] {\displaystyle \pi \left[y_{1}{\sqrt {y_{1}-\mu ^{2}}}+{\frac {y_{1}\mu ^{2}}{\sqrt {y_{1}^{2}-\mu ^{2}}}}+y_{0}{\sqrt {y_{0}^{2}-\mu ^{2}}}+{\frac {y_{0}\mu ^{2}}{\sqrt {y_{0}^{2}-\mu ^{2}}}}\right]=\pi \left[{\frac {y_{1}^{3}}{\sqrt {y_{1}^{2}-\mu ^{2}}}}+{\frac {y_{0}^{3}}{\sqrt {y_{0}^{2}-\mu ^{2}}}}\right]} 如上所示，這是兩個圓錐體表面積的總和。
第 58 條 [ 4] 1 ∘ {\displaystyle 1^{\circ }} X {\displaystyle X} P 0 ′ P 1 ′ {\displaystyle P_{0}'P_{1}'} P 0 ′ T {\displaystyle P_{0}'T} P 1 ′ T {\displaystyle P_{1}'T} P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} 2 ∘ {\displaystyle 2^{\circ }} P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} X {\displaystyle X} P 0 ′ P 1 ′ {\displaystyle P_{0}'P_{1}'} P 0 ′ P 0 ″ P 1 ″ P 1 ′ {\displaystyle P_{0}'P_{0}''P_{1}''P_{1}'} P 0 ′ P 0 ″ {\displaystyle P_{0}'P_{0}''} P 1 ′ P 1 ″ {\displaystyle P_{1}'P_{1}''} ≷ 0 {\displaystyle \gtrless 0} P 0 ′ {\displaystyle P_{0}'} P 0 ″ {\displaystyle P_{0}''} P 1 ′ {\displaystyle P_{1}'} P 1 ″ {\displaystyle P_{1}''} P 0 ′ P 0 ″ {\displaystyle P_{0}'P_{0}''} P 1 ′ P 1 ″ {\displaystyle P_{1}'P_{1}''}
上述情況 II 被稱為過渡情況，即切線的交點從 X {\displaystyle X} α {\displaystyle \alpha }
第 59 條 m {\displaystyle m} m 1 {\displaystyle m_{1}} m 2 {\displaystyle m_{2}} m 2 > m 1 {\displaystyle m_{2}>m_{1}} m 1 {\displaystyle m_{1}}
y 1 m 1 y 1 2 − m 1 2 + y 0 m 1 y 0 2 − m 1 2 < x 1 − x 0 . [A] {\displaystyle {\frac {y_{1}m_{1}}{\sqrt {y_{1}^{2}-m_{1}^{2}}}}+{\frac {y_{0}m_{1}}{\sqrt {y_{0}^{2}-m_{1}^{2}}}}<x_{1}-x_{0}{\text{.}}\qquad {\text{[A]}}} 曲線的切線方程為
d y d x = y ′ − y x ′ − x {\displaystyle {\frac {{\text{d}}y}{{\text{d}}x}}={\frac {y'-y}{x'-x}}} 其中， x ′ {\displaystyle x'} y ′ {\displaystyle y'} X {\displaystyle X}
x ′ − x = − y d y d x {\displaystyle x'-x=-{\frac {y}{\frac {{\text{d}}y}{{\text{d}}x}}}} x ′ = x − y d y d x {\displaystyle x'=x-{\frac {y}{\frac {{\text{d}}y}{{\text{d}}x}}}} 即 ,
x ′ = x − m [ e ( x − x 0 ′ ) / m + e − ( x − x 0 ′ ) / m e ( x − x 0 ′ ) / m − e − ( x − x 0 ′ ) / m ] {\displaystyle x'=x-m\left[{\frac {e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}}{e^{(x-x_{0}')/m}-e^{-(x-x_{0}')/m}}}\right]} 因此，當 x = x 0 ′ {\displaystyle x=x_{0}'} x ′ = − ∞ {\displaystyle x'=-\infty } x = + ∞ {\displaystyle x=+\infty } x ′ = + ∞ {\displaystyle x'=+\infty }
另一方面， d x ′ d x {\displaystyle {\frac {{\text{d}}x'}{{\text{d}}x}}} x ′ {\displaystyle x'} x {\displaystyle x} − ∞ {\displaystyle -\infty } X {\displaystyle X} + ∞ {\displaystyle +\infty } P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} X {\displaystyle X}
當 m = m 1 {\displaystyle m=m_{1}} X {\displaystyle X}
第 60 條 應用。 假設我們有兩個大小相同的環，它們連線到同一個軸上，該軸垂直穿過它們的中心。如果這些環的邊緣透過一層自由的液體薄膜（肥皂液）連線，那麼薄膜會呈現什麼形狀？
根據物理學定律，薄膜有使自身面積最小化的趨勢。因此，只有作為最小表面，薄膜才處於平衡狀態。設 O {\displaystyle O} O ′ {\displaystyle O'} O ″ {\displaystyle O''} O O ″ {\displaystyle OO''} O L {\displaystyle OL} O O ″ {\displaystyle OO''} O L {\displaystyle OL} O P ″ {\displaystyle OP''} O P ′ {\displaystyle OP'} P ′ {\displaystyle P'} P ″ {\displaystyle P''}
第 61 條 tan − 1 ( 3 / 2 ) {\displaystyle \tan ^{-1}(3/2)} O P ′ {\displaystyle OP'} O P ″ {\displaystyle OP''} O T ′ {\displaystyle OT'} O T ″ {\displaystyle OT''} T ′ O T ″ {\displaystyle T'OT''} P 0 {\displaystyle P_{0}} O L {\displaystyle OL} P 1 {\displaystyle P_{1}} P 0 {\displaystyle P_{0}} O L {\displaystyle OL} P 0 P 1 {\displaystyle P_{0}P_{1}} P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}}
假設我們最初將兩個環接觸在一起，然後以相同的速度沿軸線向相反方向推開它們，從點 O {\displaystyle O} P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} T ′ O T ″ {\displaystyle T'OT''} P 0 O P 1 < T ′ O T ″ {\displaystyle P_{0}OP_{1}<T'OT''} P 0 {\displaystyle P_{0}} P 1 {\displaystyle P_{1}} X {\displaystyle X} P 0 O P 1 {\displaystyle P_{0}OP_{1}} T ′ O T ″ {\displaystyle T'OT''}
↑ 在整個討論中， X {\displaystyle X} ↑ 換句話說， y 1 {\displaystyle y_{1}} y 0 {\displaystyle y_{0}} x 0 ′ {\displaystyle x_{0}'} x 1 {\displaystyle x_{1}} ↑ 當然，距離 y 0 {\displaystyle y_{0}} X {\displaystyle X} ↑ 另見 Todhunter，變分法的研究，第 29 頁。 
![{\displaystyle y={\frac {1}{2}}m[e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/ffdf65dce67353eb049c24b38b347b5ac9ddde62)
![{\displaystyle m{\frac {{\text{d}}y}{{\text{d}}x}}=\pm {\sqrt {y^{2}-m^{2}}}={\frac {1}{2}}m[e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e48e66be77eb066590e2bf99188a033013358d6f)
![{\displaystyle {\text{d}}s={\frac {y{\text{d}}x}{m}}={\frac {1}{2}}[e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}]{\text{d}}x}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a3b142d64148d4a64a0bba753b2d534fe292025f)
![{\displaystyle s-s_{0}'={\frac {1}{2}}[e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}]={\sqrt {y^{2}-m^{2}}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/5d0b8ccffe7293a87225523b8deaace14779a4d8)
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![{\displaystyle {\bar {y}}={\frac {m}{2}}[e^{({\bar {x}}-x_{0}')/m}+e^{-({\bar {x}}-x_{0}')/m}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6981eb9e749ad73bf8f9c77f6279f4372e51c037)
![{\displaystyle k={\bar {y}}'={\frac {1}{2}}[e^{({\bar {x}}-x_{0}')/m}-e^{-({\bar {x}}-x_{0}')/m}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/83610cb7b173bb39a2d0832b9edcf5ebbbeff4f9)
![{\displaystyle b={\frac {m}{2}}(e^{a/m}+e^{-a/m})=\phi (m){\text{,}}\qquad {\text{[1]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6328af8ef5223b658faa773b42172270a5f8b7ac)
![{\displaystyle y={\frac {m}{2}}[e^{(x-x_{0}')/m}+e{-(x_{0}')/m}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/00f3ff494ef41be8a65b08f1a05bcd9aa1dee66c)
![{\displaystyle y_{0}={\frac {m}{2}}[e^{(x_{0}-x_{0}')/m}+e^{-(x_{0}-x_{0}')/m}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/15d58718ee4af33aac276d29fd0898c6c01cd095)
![{\displaystyle y_{0}^{2}-m^{2}={\frac {m^{2}}{4}}[e^{(x_{0}-x_{0}')/m}+e^{-(x_{0}-x_{0}')/m}]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/48f4023df3f734f2412f3e731b13f692f4d6bc6c)
![{\displaystyle {\sqrt {y_{0}^{2}-m^{2}}}=\pm {\frac {m}{2}}[e^{(x_{0}-x_{0}')/m}+e^{-(x_{0}-x_{0}')/m}]{\text{;}}\qquad {\text{[I]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/a232fc37ae87b818639fc41a841cc8aa2a994e90)
![{\displaystyle {\frac {x_{0}-x_{0}'}{m}}=\pm \ln {\frac {y_{0}+{\sqrt {y_{0}^{2}}}}{m}}{\text{.}}\qquad {\text{[a]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/c80a31fe677652f74f67cf27bd592cc5f5aeb293)
![{\displaystyle y_{1}={\frac {m}{2}}[e^{(x_{1}-x_{0}')/m}+e^{-(x_{1}-x_{0}')/m}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7d1a96cec29388d19c378b8764482dbdd04e8799)
![{\displaystyle {\sqrt {y_{1}^{2}-m^{2}}}=+{\frac {m}{2}}[e^{(x_{1}-x_{0}')/m}-e^{-(x_{1}-x_{0}')/m}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/6cc33d630af4f37347742c9be2e66aa32e0e43e6)
![{\displaystyle {\frac {x_{1}-x_{0}'}{m}}=+\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-m^{2}}}}{m}}{\text{.}}\qquad {\text{[b]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/91556f298341f3a30a54d431aa9a9de3d00bacf9)
![{\displaystyle f_{1}'(m)={\frac {1}{m}}\left[\right]{\text{.}}\qquad {\text{[c]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/cd679d0fd5349e682288a06219fc0632e1125a1b)
![{\displaystyle f_{1}(0)=\ln(2y_{1})-\ln(m=0)-\ln(2y_{0})+\ln(m=0)-\left[{\frac {x_{1}-x_{0}}{m}}\right]_{m=0}=-\infty }](https://wikimedia.org/api/rest_v1/media/math/render/svg/c978465358aba081b113cf9544bf7f94b2e5c480)
![{\displaystyle f_{1}(y_{0})=\ln {\frac {y_{1}+{\sqrt {y_{1}^{2}-y_{0}^{2}}}}{y_{0}}}-{\frac {x_{1}-x_{0}}{y_{0}}}{\text{.}}\qquad {\text{[II]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/78f6f1027ce40bfeb5c5ef1df06cee6c93f89f68)
![{\displaystyle y={\frac {y_{0}}{2}}\left[e^{(x-x_{0})/y_{0}}+e^{-(x-x_{0})/y_{0}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d3c67e1162854f4ddb4612dab5dc0dd26aaaa41f)
![{\displaystyle y_{2}-y_{0}^{2}={\frac {y_{0}^{2}}{4}}\left[e^{(x-x_{0})/y_{0}}-e^{-(x-x)_{0})/y_{0}}\right]^{2}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/d2b0129441702519a1079353da91793ea0b303ea)
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![{\displaystyle {\frac {y_{1}\mu }{\sqrt {y_{1}^{2}-\mu ^{2}}}}+{\frac {y_{0}\mu }{\sqrt {y_{0}^{2}-\mu ^{2}}}}-(x_{1}-x_{0})=0{\text{,}}\qquad {\text{[A]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/246245fa27f503b8a93c3249b8f717e6fe43a6b7)
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![{\displaystyle [(\alpha )^{2}]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/e4c45d9dc4a313a2c370b096934479d3708d77da)
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![{\displaystyle 2\alpha S=2\alpha M_{0}+2\alpha M_{1}+[(\alpha ^{2})]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/98e1d1bc702e6b433d7cae61c244f57ff141c918)
![{\displaystyle \int _{x_{0}}^{x_{1}}2y\pi ~{\text{d}}s=\int _{x_{0}}^{x_{1}}{\frac {2y^{2}\pi ~{\text{d}}x}{m}}=2\pi \int _{x_{0}}^{x_{1}}{\frac {m^{2}}{4}}[e^{2(x-x_{0}')/m}+2+e^{-2(x-x_{0}')/m}]{\frac {{\text{d}}x}{m}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/1cf30e0586fce260f1ef03c2e0dd2dd78081c572)
![{\displaystyle ={\frac {\pi m^{2}}{4}}\left[e^{2(x-x_{0}')/m}-e^{-2(x-x_{0}')/m}+{\frac {4x}{m}}\right]_{x_{0}}^{x_{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/02e8896da88536220e68eef888f27f0e77da475a)
![{\displaystyle =\pi \left[{\frac {m}{2}}(e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m})\cdot {\frac {m}{2}}(e^{(x-x_{0}')/m}-e^{-(x-x_{0}')/m})+mx\right]_{x_{0}}^{x_{1}}\qquad {\text{[A]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/3ef6a883d77073488a13d2925ca60be5e3a73375)
![{\displaystyle =\pi [\pm y{\sqrt {y^{2}-m^{2}}}+mx]_{x_{0}}^{x_{1}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/dc12902119b33440df76d814fa3f635728c42139)
![{\displaystyle =\pi [y_{1}{\sqrt {y_{1}-m^{2}}}+y_{0}{\sqrt {y_{0}-m^{2}}}]+m(x_{1}-x_{0}){\text{,}}\qquad {\text{[B]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9996347b80b980b2fd30339f86a6c57d1d208b5c)
![{\displaystyle \pi \left[y_{1}{\sqrt {y_{1}-\mu ^{2}}}+{\frac {y_{1}\mu ^{2}}{\sqrt {y_{1}^{2}-\mu ^{2}}}}+y_{0}{\sqrt {y_{0}^{2}-\mu ^{2}}}+{\frac {y_{0}\mu ^{2}}{\sqrt {y_{0}^{2}-\mu ^{2}}}}\right]=\pi \left[{\frac {y_{1}^{3}}{\sqrt {y_{1}^{2}-\mu ^{2}}}}+{\frac {y_{0}^{3}}{\sqrt {y_{0}^{2}-\mu ^{2}}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/9e30f00a3d699ef22c695d7d58f8fa059c363d17)
![{\displaystyle {\frac {y_{1}m_{1}}{\sqrt {y_{1}^{2}-m_{1}^{2}}}}+{\frac {y_{0}m_{1}}{\sqrt {y_{0}^{2}-m_{1}^{2}}}}<x_{1}-x_{0}{\text{.}}\qquad {\text{[A]}}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2a10938305aacbc0866f9098f9c2c194058c5875)
![{\displaystyle x'=x-m\left[{\frac {e^{(x-x_{0}')/m}+e^{-(x-x_{0}')/m}}{e^{(x-x_{0}')/m}-e^{-(x-x_{0}')/m}}}\right]}](https://wikimedia.org/api/rest_v1/media/math/render/svg/7aaa12b2eeb4477649aa4cc55a570bb3be78fe23)
