Question

Ian has climbed a pinnacle that is detached from the main peak by roping down into the notch dividing them and then climbing the pinnacle. He pulled an extra rope behind him so that he could get back to the main peak by using a Tyrolean traverse, meaning that he would use the rope to go directly back to the peak instead of descending and then climbing on rock again. When he anchors the rope, it hangs in a catenary curve, with equation

$r\left(x\right)=125\cos h(0.008x-0.6528)$

The point x = 0 is where the rope attaches to the main peak, while x = 136 is where it attaches to the pinnacle. Heights are measured in feet above the notch. (This exercise involves hyperbolic functions.)

(a) How much higher is the main peak than the detached pinnacle?

(b) Where is the low point of the rope as it hangs loosely? How high is the rope above the notch at that point?

(c) What angle does the rope make with the horizontal where it attaches to the main peak?

Short answer

(a) The main peak is $15feet$higher than the pinnacle.

(b) The height of the rope is about $66feet$.

(c) the angle of the rope tat it makes with the horizontal is $60°$.

Step-by-step solution

Part (a) Step 1. Given Information.

The function:

$r\left(x\right)=125\cos h(0.008x-0.6528)$

Part (a) Step 2. Substitute x=0.

Substitute x=0in the given function, for the height of the main peak

$$\begin{gathered} r\left(x\right)=125\cos h(0.008x-0.6528) \\ =125\cos h(0.008(0)-0.6528) \\ =125\cos h(-0.6528) \\ =152feet \end{gathered}$$

Substitute x=136,for the height of the detached pinnacle

$$\begin{gathered} r\left(x\right)=125\cos h(0.008(136)-0.6528) \\ =125\cos h(1.088-0.6528) \\ =137feet \end{gathered}$$

Part (a) Step 3. Find the distance between the peak and pinnacle.

The distance between main peak and pinnacle is given by,

$152-137=15feet$

Part (b) Step 1. Find the height of the rope.

As the rope hung loosely, the low point of the rope is below the point where it attached to the pinnacle.

From the given, the height is about $66feet$

Part (c) Step 1. Find the angle from horizontal.

Substitute x=0in the given function, for the height of the main peak

$$\begin{gathered} r\left(x\right)=125\cos h(0.008x-0.6528) \\ =125\cos h(0.008(0)-0.6528) \\ =125\cos h(-0.6528) \\ =152feet \end{gathered}$$

From the given figure, the angle is about $60°$from horizontal.