Chapter Contents

• Applications of Integration
• 1. Applications of the Indefinite Integral
• 2. Area Under a Curve
• 3. Area Between 2 Curves
• 4. Volume of Solid of Revolution
• 5. Centroid of an Area
• 6. Moments of Inertia
• 7. Work by a Variable Force
• 8. Electric Charges
• 9. Average Value
• Head Injury Criterion
• a. Normal Braking
• b. Crash Tests
• c. Data
• d. Severity Index
• e. Head Injury Criterion
• f. Modelling the HIC
• g. Conclusions
• 10. Force by Liquid Pressure
• 11. Arc Length of a Curve
• 12. Arc Length of a Curve in Parametric or Polar Coordinates
• Applications of Integration Problem Solver

• Comments, Questions?

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11. Arc Length of a Curve

by M. Bourne

Example 1 - Corrugated iron sheeting

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.

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 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

Example 2 - Parabolic dish application

The cross section of a parabolic dish is modelled by y = 0.04x² (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

Example 3 - Golden Gate Bridge cables

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

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

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]