A cable is suspended between two poles of the same height that are $20{\,{\rm{m}}}$ apart, as shown in Figure 29(a). If the poles are placed at $(-10,0)$ and $(10,0)$, the equation that models the height of the cable is $y=10\cosh \dfrac{x}{10}+15$. Find the angle $\theta$ at which the cable meets a pole.

Solution

The slope of the tangent line to the catenary is given by $$y^\prime =\dfrac{d}{dx}\left( 10\cosh \dfrac{x}{10}+15\right) =10\cdot \dfrac{ 1}{10}\sinh \dfrac{x}{10}=\sinh \dfrac{x}{10}$$

At $x=10,$ the slope $m_{\tan }$ of the tangent line is $m_{\tan }=\sinh \dfrac{10}{10}=\sinh 1.$

The angle $\theta$ at which the cable meets the pole equals the angle between the tangent line and the pole. To find $\theta ,$ we form a right triangle using the tangent line and the pole, as shown in Figure 29(b).

Figure 29

From Figure 29(b), we find that the slope of the tangent line is $m_{\tan }=\dfrac{\Delta y}{\Delta x}=\sinh 1.$ Then $\tan \theta =\dfrac{\Delta x}{\Delta y}=\dfrac{1}{\sinh 1}.$ So, $\theta =\tan ^{-1}\left( \dfrac{1}{\sinh 1}\right) \approx 0.7050$ radians $\approx 404^{\circ}$.