Length of arc of circle or curve

[cliverlong]

Anyone,

I suppose if I sat and thought about this I should be able to work it out. Is there any way in Autograph to find the length of a path along a curve or at least the arc of a sector?

I'm trying to create a minimax of a fixed perimeter sector similar to the rectangle with fixed perimeter.

Clive

[Support]

Hi Clive,

There is no built in functionality to do this. I would construct a graph to calculate the arc length for me.

For the function y = f(x), the length s of the curve between x = a and x = b is the integral from a to b of √(1 + (f'(x))²) with respect to x.

In Autograph let's look at the example of f(x) = x³:

1. Plot y = x³
2. Calculate f'(x)

f'(x) = 3x²

3. Plot y = √(1 + (3x²)²)
4. Select this curve and add points at x = a and x = b.
5. Select both points and find the area under the curve using Simpson's rule.
6. Select the area and add a Text Box, change the word "Area: " to "Arc length, s = ".
7. Hide everything except the original curve.

I've attached a file that demonstrates this.

The arc length of a sector of a circle is a bit simpler. Make sure Autograph is in radians. Then the arc length of a sector with angle θ and radius r is s = rθ.

1. On a new 2D Graph Page add a point at the origin and on it place a circle with radius r.
2. Select the circle and add a point at t = 0 and another at t = θ.
3. Select the point at t = 0 and the origin and add a line segment.
4. Select the point at t = θ and the origin and add a line segment.
5. Select the point at t = 0, the point at the origin, then the point at t = θ and right-click and choose Angle.
6. Add a point with coordinates (0, rθ)
7. Select this point and the point at the origin and add a line segment. The length of this line segment is the arc length.
8. Select the line segment and add a Text Box. Change "Area: " to "Arc Length, s = ".
9. Hide the line segment and all points.
10. Instruct the user to edit the start and finish points using the Constant Controller.

Again I've attached a file which demonstrates this.

Simon