Question

A telephone line hangs between two poles \(14\) m apart in the shape of the catenary \(y = 20\cosh \left( {\frac{x}{{20}}} \right) - 15\), where \(x\) and \(y\) are measured in meters.

(a) Find the slope of this curve where it meets the right-hand pole.

(b) Find the angle \(\theta \) between the line and the pole.

Short answer

(a) The slope of the curve is \(0.3572\).

(b) The required angle between line and pole is \(\theta \approx 70.34^\circ \).

Step-by-step solution

Step 1: Find the slope of the curve where it meets the right-hand pole

(a) Consider the function \(y = 20\cosh \left( {\frac{x}{{20}}} \right) - 15\). As the function itself continuous smoothly in the upward past \(x = - 7\) or \(x = 7\) , it does not bend sharply at \(x = \pm 7\) like the telephone line in the diagram.

Differentiate the function.

\(\begin{aligned}\frac{{dy}}{{dx}} & = \frac{d}{{dx}}\left( {20\cosh \left( {\frac{x}{{20}}} \right) - 15} \right)\\ & = 20\frac{d}{{dx}}\left( {\cosh \left( {\frac{x}{{20}}} \right)} \right) - \frac{d}{{dx}}\left( {15} \right)\\ & = 20\sinh \left( {\frac{x}{{20}}} \right) \cdot \frac{d}{{dx}}\left( {\frac{x}{{20}}} \right) - 0\\ & = 20\sinh \left( {\frac{x}{{20}}} \right)\left( {\frac{1}{{20}}} \right)\\ & = \sinh \left( {\frac{x}{{20}}} \right)\end{aligned}\)

Put \(x = 7\) into obtained result.

\(\begin{aligned}y' & = \sinh \left( {\frac{7}{{20}}} \right)\\ \approx 0.3572\end{aligned}\)

Thus, the slope of the curve is \(0.3572\).

Find the angle \(\theta \) between the line and the pole

From the part\(\left( a \right)\)the slope of the tangent at the right pole is\(m = 0.3572\).

Apply the trigonometric function.

\(\begin{aligned}\cot \theta & = \frac{{\sinh \frac{7}{{10}}}}{1}\\\cot \theta & = \sinh \frac{7}{{10}}\\\theta & = {\mathop{\rm arccot}\nolimits} \left( {\sinh \frac{7}{{10}}} \right)\\ \approx 70.34^\circ \end{aligned}\)

Thus, the required angle between line and pole is \(\theta \approx 70.34^\circ \).