A cable hangs between two poles of equal height and 30 feet apart. set up a coordinate system where the poles are placed at x=−15 and x=15, where x is measured in feet. the height (in feet) of the cable at position x i

A cable hangs between two poles of equal height and 30 feet apart. Set up a coordinate system where the poles are placed at x=−15 and x=15, where x is measured in feet. The height (in feet) of the cable at position x is

h(x)=14cosh(x/14)

cosh(x)=(e^x+e^−x)/2 i

I don't know how to explain this, I'm really bad at it!

Answer Link

MrRoyal MrRoyal · Answer 2 · 2022-02-15T16:27:07+01:00

The length of the cable is the number of units on it.

The cable is 35.26 feet long

How to determine the cable length

The function is given as:

[tex]h(x) = 15\cos(x/15)[/tex]

For all the lengths, we have the following differential equation

[tex]Length = \int\limits^{15}_{-15} {\sqrt{1 + (\frac{dy}{dx})^2} \, dx[/tex]

So, we have:

[tex]Length = \int\limits^{15}_{-15} {\sqrt{1 + (15 * \frac{1}{15} * \tanh^2(\frac{x}{15}))} \, dx[/tex]

Evaluate

[tex]Length = \int\limits^{15}_{-15} {\sqrt{1 + \tanh^2(\frac{x}{15})} \, dx[/tex]

This gives

[tex]Length = \int\limits^{15}_{-15} {\sqrt{cosh^2(\frac{x}{15})} \, dx[/tex]

Evaluate the exponents

[tex]Length = \int\limits^{15}_{-15} {cosh(\frac{x}{15}) \, dx[/tex]

The above function is an even function.

So, we have:

[tex]Length = 2\int\limits^{15}_{0} {cosh(\frac{x}{15}) \, dx[/tex]

Integrate

[tex]Length = 2 * [\frac{sinh(\frac{x}{15})}{1/15}]|\limits^{15}_{0}[/tex]

Simplify

[tex]Length = 30 * [\sinh(\frac{x}{15})]|\limits^{15}_{0}[/tex]

Expand

[tex]Length = 30 * [\sinh(\frac{15}{15}) - \sinh(\frac{0}{15})][/tex]

Solve

[tex]Length = 35.26[/tex]

Hence, the length of the cable is 35.26 feet