Известно, что гибкие провода провисают по цепной линии
![y=a\left(\mathop{\mathrm{ch}}\dfrac xa-1\right) y=a\left(\mathop{\mathrm{ch}}\dfrac xa-1\right)](https://tex.z-dn.net/?f=+y%3Da%5Cleft%28%5Cmathop%7B%5Cmathrm%7Bch%7D%7D%5Cdfrac+xa-1%5Cright%29+)
(ch x = (exp(x) + exp(-x))/2 – гиперболический косинус, sh x = (exp(x) - exp(-x))/2 – гиперболический синус)
Введем систему координат так, как показано на рисунке. Тогда
![y=a\left(\mathop{\mathrm{ch}}\dfrac xa-1\right)+20 y=a\left(\mathop{\mathrm{ch}}\dfrac xa-1\right)+20](https://tex.z-dn.net/?f=+y%3Da%5Cleft%28%5Cmathop%7B%5Cmathrm%7Bch%7D%7D%5Cdfrac+xa-1%5Cright%29%2B20+)
Вычислим длину провода:
![\displaystyle L=\int_{-d}^d\sqrt{1+(y')^2}\,dx=\int_{-d}^d\sqrt{1+\mathrm{sh}^2\dfrac xa}\,dx=\int_{-d}^d\mathop{\mathrm{ch}}\dfrac xa\,dx=2\mathop{\mathrm{sh}}\dfrac da \displaystyle L=\int_{-d}^d\sqrt{1+(y')^2}\,dx=\int_{-d}^d\sqrt{1+\mathrm{sh}^2\dfrac xa}\,dx=\int_{-d}^d\mathop{\mathrm{ch}}\dfrac xa\,dx=2\mathop{\mathrm{sh}}\dfrac da](https://tex.z-dn.net/?f=+%5Cdisplaystyle+L%3D%5Cint_%7B-d%7D%5Ed%5Csqrt%7B1%2B%28y%27%29%5E2%7D%5C%2Cdx%3D%5Cint_%7B-d%7D%5Ed%5Csqrt%7B1%2B%5Cmathrm%7Bsh%7D%5E2%5Cdfrac+xa%7D%5C%2Cdx%3D%5Cint_%7B-d%7D%5Ed%5Cmathop%7B%5Cmathrm%7Bch%7D%7D%5Cdfrac+xa%5C%2Cdx%3D2%5Cmathop%7B%5Cmathrm%7Bsh%7D%7D%5Cdfrac+da+)
Неизвестные величины a и d необходимо определить из условий L = 80, y(d) = y(-d) = 50. Составляем систему уравнений:
![\begin{cases}a\left(\mathop{\mathrm{ch}}\dfrac da-1\right)+20=50\\2a\mathop{\mathrm{sh}}\dfrac da=80\end{cases} \begin{cases}\mathop{\mathrm{ch}}\dfrac da=1+\dfrac{30}a\\\mathop{\mathrm{sh}}\,\dfrac da=\dfrac{40}a\end{cases} \begin{cases}a\left(\mathop{\mathrm{ch}}\dfrac da-1\right)+20=50\\2a\mathop{\mathrm{sh}}\dfrac da=80\end{cases} \begin{cases}\mathop{\mathrm{ch}}\dfrac da=1+\dfrac{30}a\\\mathop{\mathrm{sh}}\,\dfrac da=\dfrac{40}a\end{cases}](https://tex.z-dn.net/?f=+%5Cbegin%7Bcases%7Da%5Cleft%28%5Cmathop%7B%5Cmathrm%7Bch%7D%7D%5Cdfrac+da-1%5Cright%29%2B20%3D50%5C%5C2a%5Cmathop%7B%5Cmathrm%7Bsh%7D%7D%5Cdfrac+da%3D80%5Cend%7Bcases%7D+%5Cbegin%7Bcases%7D%5Cmathop%7B%5Cmathrm%7Bch%7D%7D%5Cdfrac+da%3D1%2B%5Cdfrac%7B30%7Da%5C%5C%5Cmathop%7B%5Cmathrm%7Bsh%7D%7D%5C%2C%5Cdfrac+da%3D%5Cdfrac%7B40%7Da%5Cend%7Bcases%7D+)
Находим a:
![1=\mathop{\mathrm{ch}}^2\dfrac da-\mathop{\mathrm{sh}}^2\dfrac da=\left(1+\dfrac{30}a\right)^2-\left(\dfrac{40}a\right)^2\\ a^2=(a+30)^2-40^2\\ a^2=a^2+60a+900-1600\\ a=\dfrac{35}3 1=\mathop{\mathrm{ch}}^2\dfrac da-\mathop{\mathrm{sh}}^2\dfrac da=\left(1+\dfrac{30}a\right)^2-\left(\dfrac{40}a\right)^2\\ a^2=(a+30)^2-40^2\\ a^2=a^2+60a+900-1600\\ a=\dfrac{35}3](https://tex.z-dn.net/?f=+1%3D%5Cmathop%7B%5Cmathrm%7Bch%7D%7D%5E2%5Cdfrac+da-%5Cmathop%7B%5Cmathrm%7Bsh%7D%7D%5E2%5Cdfrac+da%3D%5Cleft%281%2B%5Cdfrac%7B30%7Da%5Cright%29%5E2-%5Cleft%28%5Cdfrac%7B40%7Da%5Cright%29%5E2%5C%5C+a%5E2%3D%28a%2B30%29%5E2-40%5E2%5C%5C+a%5E2%3Da%5E2%2B60a%2B900-1600%5C%5C+a%3D%5Cdfrac%7B35%7D3+)
Тогда
![\mathop{\mathrm{sh}}\,\dfrac da=\dfrac{40}{35/3}=\dfrac{24}7\\ e^{d/a}-e^{-d/a}=\dfrac{48}7\\ 7(e^{d/a})^2-48e^{d/a}-7=0\\ e^{d/a}=7\\ d=a\ln 7=\dfrac{35\ln 7}3 \mathop{\mathrm{sh}}\,\dfrac da=\dfrac{40}{35/3}=\dfrac{24}7\\ e^{d/a}-e^{-d/a}=\dfrac{48}7\\ 7(e^{d/a})^2-48e^{d/a}-7=0\\ e^{d/a}=7\\ d=a\ln 7=\dfrac{35\ln 7}3](https://tex.z-dn.net/?f=+%5Cmathop%7B%5Cmathrm%7Bsh%7D%7D%5C%2C%5Cdfrac+da%3D%5Cdfrac%7B40%7D%7B35%2F3%7D%3D%5Cdfrac%7B24%7D7%5C%5C+e%5E%7Bd%2Fa%7D-e%5E%7B-d%2Fa%7D%3D%5Cdfrac%7B48%7D7%5C%5C+7%28e%5E%7Bd%2Fa%7D%29%5E2-48e%5E%7Bd%2Fa%7D-7%3D0%5C%5C+e%5E%7Bd%2Fa%7D%3D7%5C%5C+d%3Da%5Cln+7%3D%5Cdfrac%7B35%5Cln+7%7D3+)
Расстояние между столбами
![2d=\dfrac{70\ln 7}3\approx45.4\text{ m} 2d=\dfrac{70\ln 7}3\approx45.4\text{ m}](https://tex.z-dn.net/?f=+2d%3D%5Cdfrac%7B70%5Cln+7%7D3%5Capprox45.4%5Ctext%7B+m%7D+)