Arc Length of a Curve - Metal Roof Sheet

Corrugated iron and aluminum alloy are used extensively throughout the world as a versatile building material. Bending the material into a regular wave pattern gives it greater strength than if a flat sheet is used.
Another example of a light, thin and weak sheet that is made much stronger by having regular folds is corrugated cardboard, used for protecting goods in transit.

To make corrugated iron, you need to bend a wide flat sheet into waves. The resulting corrugated sheet is then narrower, of course.
The corrugations are commonly in the form of a sine curve.

This panel has a finished width of 106.7 cm, a period of 10.67 cm (distance from the top of each wave to the top of the next), and has amplitude 1.35 cm (height from the mid-point of the wave to the top of a crest).
How wide should the flat sheet be to give us a corrugated sheet of width 106.7 cm?

We model the corrugations using the curve

y = 1.35 sin 0.589x

This has the required amplitude 1.35 and period 10.67.
(Within the sine expression, we use 2π/10.67 = 0.589 for the coefficient of x. 
 We'll find the width needed for one wave, then multiply by the number of waves.

Approximate answer: Next, let's approximate the length of the curve so we've got a rough idea what our exact length should be. (It's always good practice to estimate your answer first, and in this topic, it helps us understand the concept better).
We plot the points O (0, 0), A (2.65, 1.35), B (5.33, 0), C (7.99, -1.35) and D (10.65, 0), which are key points on the curve (at the mid-points, maximum and minimum values), and join the line segments.

We then use Pythagoras' Theorem to find the length OA:

$\displaystyle{O}{A}=\sqrt{{{2.65}^{2}+{1.35}^{2}}}={2.97}$OA=√​2.65​2​​+1.35​2​​​​​=2.97

The distances AB, BC, CD are all equal, so we can say:

OA + AB + BC + CD = 4 × 2.97 = 11.88

So we expect the curved distance OD to be around 12 cm.

 

 

 

 

 

Exact value

We'll use calculus to find the 'exact' value. But first, some background.
We zoom in near the center of the segment OA and we see the curve is almost straight.
For this portion, the curve EF is getting quite close to the straight line segment EF.

For this zoomed-in section, we have:

curved length EF $\displaystyle={r}\approx{\int_{{a}}^{{b}}}\sqrt{{{1}^{2}+{0.57}^{2}}}={1.15}$

General Form of the Length of a Curve

If the horizontal distance is "dx" (or "a small change in x") and the vertical height of the triangle is "dy" (or "a small change in y") then the length of the curved arc "dr" is approximated as:

$\displaystyle{d}{r}\approx\sqrt{{{\left.{d}{x}\right.}^{2}+{\left.{d}{y}\right.}^{2}}}$dr≈√​dx​2​​+dy​2​​​​​

Now, if we move point E very close to point F, we will have a very good approximation for the length of the curve in that local region.

We need to add all those infinitesimally small lengths. We use integration, as it represents the sum of such infinitely small distances. We have for the distance between where $\displaystyle{x}={a}$x=a to $\displaystyle{x}={b}$x=b:

$\displaystyle\text{length}={r}={\int_{{a}}^{{b}}}\sqrt{{{\left({\left.{d}{x}\right.}\right)}^{2}+{\left({\left.{d}{y}\right.}\right)}^{2}}}$length=r=∫​a​b​​√​(dx)​2​​+(dy)​2​​​​​

By performing simple surd manipulation, we can express this in more familiar form as follows.
The arc length of the curve y = f(x) from x = a to x = b is given by:

$\displaystyle\text{length}={r}={\int_{{a}}^{{b}}}\sqrt{{{1}+{\left(\frac{{{\left.{d}{y}\right.}}}{{{\left.{d}{x}\right.}}}\right)}^{2}}}{\left.{d}{x}\right.}$length=r=∫​a​b​​√​1+(​dx​​dy​​)​2​​​​​dx

Of course, we are assuming the function $\displaystyle{y}= f{{\left({x}\right)}}$y=f(x) is continuous in the region $\displaystyle{x}={a}$x=a to $\displaystyle{x}={b}$x=b (otherwise, the formula won't work).