Question

A flexible cable hangs in the shape of a catenary y = acosh where the number a is the height of the lowest point on the cable (the lowest point occurs at x = 0). Suppose that this cable is fastened to support heights of 16 ft and 28 ft. C. Determine the height a for which the supports are farthest apart.

Please show your work. Serious answers only.

Answer 1

Answer:

a = 16

Δx = 16 arcosh(1.75) ≈ 18.541

Step-by-step explanation:

The position of the 16ft support is:

16 = a cosh(x₁ / a)

The position of the 28ft support is:

28 = a cosh(x₂ / a)

Solving for x:

x₁ = a arcosh(16/a)

x₂ = a arcosh(28/a)

The distance between the supports is:

x = x₂ − x₁

Δx = a arcosh(28/a) − a arcosh(16/a)

The distance is a maximum when the derivative is 0 or undefined.

dΔx/da = a [1 / √((28/a)² − 1)] (-28/a²) + arcosh(28/a) − a [1 / √((16/a)² − 1)] (-16/a²) − arcosh(16/a)

dΔx/da = [(-28/a) / √((28/a)² − 1)] + arcosh(28/a) − [(-16/a) / √((16/a)² − 1)] − arcosh(16/a)

Graphing this, we see the derivative is undefined at a = 16, where dΔx/da increases to positive infinity.

Evaluating Δx:

Δx = 16 arcosh(28/16) − 16 arcosh(16/16)

Δx = 16 arcosh(1.75) − 0

Δx ≈ 18.541

Here's a graph: desmos.com/calculator/cjitlz56qh