# 11. Arc Length of a Curve using Integration

by M. Bourne

### Example 1 - Corrugated iron sheeting

Corrugated iron roof. [Image source]

Corrugated iron is 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.

The flat sheet is rolled into corrugations, and will be narrower.

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.

We take a real example of a 4.2" Corrugated Metal Panel, which is a "high profile, wavy style corrugated panel that can be used in almost any roofing, siding, or decorative application." [Source]

Corrugated metal, in the shape of a sine wave [Image source]

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?

### Example 1 Solution

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. *For background on this, see Period of a sine curve.)

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:

`OA=sqrt(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`= r ≈ int_a^bsqrt(1^2+0.57^2)=1.15`

Of course, the real curved length is slightly more than 1.15.

Let's generalise this.

## 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:

`dr~~sqrt(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 `x = a` to `x = b`:

`"length"=r=int_a^bsqrt((dx)^2+(dy)^2`

By performing simple surd manipulation, we can express this in a more familiar form as follows.

The **arc length of the curve** *y* = *f*(*x*) from *x = a * to *x* = *b* is given by:

`"length"=r=int_a^bsqrt(1+((dy)/(dx))^2)dx`

Of course, we are assuming the function `y = f(x)` is continuous in the region `x = a` to `x = b` (otherwise, the formula won't work).

### Back to the Width of the Sheet example

Use the above formula to find the required width of the metal sheet in our example.

Answer

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.

### Example 2 - Parabolic dish application

The cross section of a parabolic dish is modelled by *y* = 0.04*x*^{2} (in meters), from *x* = −5 m to *x* = 5 m. We need to construct a steel support band which is to be placed around the outer edge of the dish. How long should we make the steel band?

We need to find the curved distance AB.

Answer

**Estimate: ** Looking at the graph (which has equal scaling along the *x*- and *y*-axes), we can see the final answer should be a little more than 10 m, somewhere between 10 m and 11 m.

Now to find the exact length:

`text[length]=r=int_a^b sqrt[1+((dy)/(dx))^2]\ dx`

In this example,

y= 0.04x^{2}, so`(dy)/(dx)=0.08x`

The lower limit is *x* = −5 and the upper limit is *x* = 5. Substituting these into our formula gives:

`r=int_-5^5 sqrt(1+(0.08x)^2)\ dx` `= 10.261`

[**Note: **I used a computer algebra system to find the above integral. Many arc length problems lead to impossible integrals. (This example does have a solution, but it is not straightforward.) Often the only way to solve arc length problems is to do them numerically, or using a computer. You can see the answer in Wolfram|Alpha.]

So the length of the steel supporting band should be 10.26 m.

This is consistent with our earlier estimate.

### Example 3 - Golden Gate Bridge cables

The central span of the Golden Gate Bridge in San Francisco, USA, is `1280\ "m"` long.

Central span of the Golden Gate Bridge. [Image source]

The height of the tower is `152\ "m"` from the roadway. What is the length of the main suspension cable between the 2 towers?

Answer

We first need to model the curve (find an equation that represents the curve accurately).

A freely hanging cable takes the form of a **catenary**. In the case of a suspension bridge, we need to also consider the mass of the roadway, which will give us a flattened catenary. The general form for a **flattened catenary** is the sum of 2 exponential functions, based on the hyperbolic function `y=cosh x`:

`y=(a(e^[bx]+e^[-bx]))/(2)`

The Golden Gate Bridge cable is almost a catenary and almost a parabola, but not quite either (because of the weight of the cables, the suspender ropes and the roadway). For the sake of this discussion, we'll assume it is a flattened catenary.

For convenience, we'll place the origin at the lowest point of the cable.

The required curve (after initial analysis and some guess and check) passing through (-640, 152), (0, 0) and (640, 152) is given by:

`y=1280((e^[x//1326]+e^[-x//1326])/(2)-1)`

Here is the graph of the above equation. We can see it passes through the required points.

The derivative of our function is

`(dy)/(dx)=640/663((e^[x//1326]+e^[-x//1326])/(2))`

Using the length of a curve formula, with start point *x* = -640 and end point *x* = 640, we have:

`int_-640^640 sqrt[1+(640/663(e^[x/1326]+e^[-x/1326])/(2))^2] dx` ` = 1326.956`

So the length of the central span of the main cable is `1327.0\ "m"`. (You can see the answer in Wolfram|Alpha here.)

Of course, the cable continues on both sides of the towers. The total length of each cable is `2,332\ "m"`.

Get the Daily Math Tweet!

IntMath on Twitter

**Interesting trivia:** The main cables of the Golden Gate Bridge are nearly one meter in diameter (actually, 0.91 m) and the "total length of galvanized steel wire used in both main cables is 129,000 km". [Source]

### Search IntMath, blog and Forum

### Online Calculus Solver

This calculus solver can solve a wide range of math problems.

Go to: Online algebra solver

### Calculus Lessons on DVD

Math videos by MathTutorDVD.com

Easy to understand calculus lessons on DVD. See samples before you commit.

More info: Calculus videos

### The IntMath Newsletter

Sign up for the free **IntMath Newsletter**. Get math study tips, information, news and updates each fortnight. Join thousands of satisfied students, teachers and parents!