We can now use this formula to find the required width of our flat sheet of iron. Remember, we're finding the width needed for one wave, then we'll multiply by the number of waves.

In our example,

y = 1.35 sin 0.589x, so

`(dy)/(dx)=0.795\ cos\ 0.589x`

For one period, the lower limit is x = 0 and the upper limit is x = 10.67. Substituting these into our formula gives us one wavelength of the curved material:

`r` `=int_0^10.67 sqrt(1+(0.795\ cos\ 0.589x)^2\) dx` `=12.196`

(You can see where this answer comes from in Wolfram|Alpha.)

The corrugated sheet has `(106.7)/(10.67)=10` complete wavelengths across its width, so to give a corrugated width of `106.7\ "cm"`, the original flat sheet will need to be:

`10 × 12.196 = 122.0\ "cm"` wide.

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