Blog Entry #102**The Arc Length of a Sine Curve**

The sine curve is one the most interesting curves in mathematics.

Let

y = a * sin x ,

Where a is the amplitude of the sine curve.

We can find the arc length of a curve between limits x1 and x2 by the integral:

x2

∫ √ (1 + (dy/dx)^2) dx

x1

For the sine curve:

y = a sin x

dy/dx = a cos x

(dy/dx)^2 = a^2 cos^x

And the arc length is:

x2

∫ √(1 + a^2 cos^x) dx

x1

There is no anti-derivative for √(1 + a^2 cos^x). Therefore, numerical methods must be used. **Finding an Approximate Curve**

Using a TI nSpire CX CAS, I used the Spreadsheet, curve fitting, and graphing features to determine an approximate polynomial. The arc length is from the origin (0,0) to (π, 0).

Note: x1 = 0, x2 = π

Here is a shot summary of what I did:

1. Created a spreadsheet with the following columns:

Column A: A sequence of numbers from 0.25 to 5 in increments of 0.25. The resulting list is named *amplist*.

Column B: Use the nSpire's arcLen function to get the arc length of the sine curve using *amplist* as the values for *a*. This list is named *arc1*.

2. Pressing the **menu** key allowed me to access the **Statistics** menu. Using the **Stat Calculations** option, I used different types of regression, including power and quartic regression. What I was looking for was which regression had the best coefficient of determination (R^2). In general, the closer R^2 is to 1, the better the fit.

I ended up choosing the quartic regression (4th degree polynomial) with R^2 ≈ 0.9999919.

The approximate polynomial is

y = .0081196317102889 x^4 - .11577326164517 x^3 + .63914882375794 x^2 + .2071162669684 x + 3.0881429428239

This polynomial was save to the function f1(x).

3. I created a graphs page and made two plots:

* Scatter plot where x = amplist, y = arc1. (The dots in red)

* The function f1(x) (see step 2). (The curve in blue)

Setting zoom to fit the data, the curve looks like a good fit.

You can create a similar graph with the Data & Statstics module, but I thought I would be different this time.

How good of a fit is the polynomial?

4. I went back to the spreadsheet and added two more columns.

Column C: est1 = f1(amplist). (estimate arc lengths)

Column D: err1 = abs(arc1 - est1)

By scrolling down Column D, the quartic polynomial was accurate in estimating the arc length of the sine curve from 0 to π to at least two decimal places. **Conclusion**

We have been looking to find the arc length of the curve y = a sin x from x = 0 to x = π.

The exact value is:

π

∫ √ (1 + a^2 cos^2 x ) dx

0

However, a good estimate can be found (to 2-3 decimal places) with the polynomial:

y = .0081196317102889 x^4 - .11577326164517 x^3 + .63914882375794 x^2 + .2071162669684 x + 3.0881429428239

Thanks as always and talk to you soon!

Eddie

This blog is property of Edward Shore. © 2012

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