Owen Lawrence (in ) said:

| I'm not checking my work, but I'll give it a shot. First, let's
| simplify a bit by locating your vertex at the origin. Your
| parabola is then y = x^2/4a. The formula for arc length can be
| derived to be
|
| L = int( x1, x2 ) sqrt( 1 + (dy/dx)^2) dx, where int( x1, x2 ) is
| the definite integral from x1 to x2 on your parabola.
|
| dy/dx = x/2a
| (dy/dx)^2 = x^2/4a^2
|
| so L = int( x1, x2 ) sqrt( 1 + x^2/4a^2 ) dx
| = 1/2a int( x1, x2 ) sqrt( 4a^2 + x^2 ) dx
|
| I looked up the indefinite integral in my CRC math tables to be
|
| int() sqrt( x^2 + c^2) dx = 1/2[ x sqrt( x^2 + c^2 ) + c^2 ln( x +
| sqrt( x^2 + c^2 ) ]
|
| In our case c = 4a^2. Since you want your arc length to be from
| the vertex, we can let x1 = 0. So our final formula becomes:
|
| L = 1/4a[ x sqrt( x^2 + (2a)^4 ) + (2a)^4 ln( x + sqrt( x^2 +
| (2a)^4 ) ]
|
| Check the algebra. I haven't had lunch yet so maybe I'm not
| thinking clearly.

Owen...

Thank you. Even without lunch you did better than I managed. :-)

--
Morris Dovey
DeSoto Solar
DeSoto, Iowa USA
http://www.iedu.com/DeSoto/solar.html