Question

Supposes a flexible cable is suspended between two towers that are 200 feet apart forms a curve whose equation is y=75(e^x/150 + e^-x/150). Calculate the length of the cable.

Answer 1

$$x_1=-100, x_2=100$$

$$y=75(e^{x/150}+e^{-x/150})$$

$$y'=\dfrac{1}{2}(e^{x/150}-e^{-x/150})$$

$$1+(y')^2=1+(\dfrac{1}{2}(e^{x/150}-e^{-x/150}))^2$$

$$=(\dfrac{1}{2}(e^{x/150}+e^{-x/150}))^2$$

$$L=\displaystyle\int_{-100}^{100}\sqrt{1+(y')^2}dx$$

$$=\displaystyle\int_{-100}^{100}\sqrt{(\dfrac{1}{2}(e^{x/150}+e^{-x/150}))^2}dx$$

$$=\displaystyle\int_{-100}^{100}\dfrac{1}{2}(e^{x/150}+e^{-x/150})dx$$

$$=2(\dfrac{1}{2})\displaystyle\int_{0}^{100}(e^{x/150}+e^{-x/150})dx$$

$$=150\big[e^{x/150}-e^{-x/150}\big]\begin{matrix} 100 \\ 0 \end{matrix}$$

$$=150(e^{100/150}-e^{-100/150}-(1-1))$$

$$=150(e^{2/3}-e^{-2/3})\approx215.148\ m$$