What is the width needed to make a 52 inch wide corrugated panel 4 inches thick?

Question

A manufacturer of corrugated metal roofing wants to produce panels that are 52 inches wide and 4 inches thick by processing flat sheets of metal into the shape shown above

The profile of the roofing takes the shape of a sine wave. If x = 0, y = 0 is the left endpoint of the profile shown above and x = 52, y = 0 is the right endpoint, what is the equation, y=f(x), of the profile?

That was part A and the answer was y=2*sin((2*Pi*x)/13)

Pat B asks : What is the width (in inches) of a flat metal sheet that is needed to make a 52 inch wide corrugated panel 4 inches thick?
You may approximate the answer within .01 inches.

Could someone please explain part B for me.
Thanks

Answer 1

You must calculate the length of the curve.

Suppose (x1,y1) and (x2,y2) are two points of the curve very close each other. The distance between them will be

L = √(x2-x1)²+(y2-y1)² = |x2-x1| √[1+((y2-y1)/(x2-x1))²]

When x2 and x1 are infinitely close each other, their distance will be

dL = dx √(1+y'²)

The length of the curve will be

L = ∫(x from a to b) √(1+y'²) dx

In your case, y = 2 sin(2π x / 13), y' = (4π / 13) cos(2π x / 13).

Then

L = ∫(x from 0 to 52) √[1+((4π / 13) cos(2π x / 13))²] dx

which seems to be an elliptical integral.

I calculated the length numerically, and obtained

L = 61.2658