Chapter 6.10, Problem 68E

Sag angle Imagine a climber clipping onto the rope described in Example 7 and pulling himself to the rope’s midpoint. Because the rope is supporting the weight of the climber, it no longer takes the shape of the catenary y = 200 cosh(x/200). Instead, the rope (nearly) forms two sides of an isosceles triangle. Compute the sag angle θ illustrated in the figure, assuming that the rope does not stretch when weighted. Recall from Example 7 that the length of the rope is 101 ft.

EXAMPLE 7 Length of a catenary A climber anchors a rope at two points of equal height, separated by a distance of 100 ft, in order to perform a Tyrolean traverse. The rope follows the catenary f(x) = 200 cosh(x/200) over the interval [−50, 50] (Figure 6.98). Find the length of the rope between the two anchor points.