Hi!

According to this site http://www.intmath.com/applications-integration/11-arc-length-curve.php the arc length of the curve y = f(x) from x=a to x=b is given by:

length_ab = Derivative_ab( Sqrt ( 1 + (dy/dx) ^ 2 ) dx)

So, we got a sine wave function which is

y = A * sin (F * x + P) from x=a to x=b

the length of this is

length_sine = Derivative_ab( Sqrt ( 1 + (A * F * cos (F * x)^2 dx)

Example of this is in first link or here:

http://www.wolframalpha.com/input/?i=tell+me+the+arc+length+of+y+%3D+1.35*sin%280.589x%29+from+x+%3D+0+to+10.67

Now, my question is what is the algorithm to compute length for specified x=a to x=b. For example, let's say:

step 1) set length = 0;

step 2) for x = a to x = b {

increase x by 0.01;

length = length + derivative(x);

}

step 3) return length;

I have implemented such a function, but it gives me the wrong length according to wolfram. Is it because of the increasing x or wrong derivative function?

Thanks in advance!

Regards!

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

 

length_ab = Derivative_ab( Sqrt ( 1 + (dy/dx) ^ 2 ) dx)

Well, the very first thing is that arclength is not equal to the derivative of (stuff), but is equal to the integral of (stuff).

According to this site http://www.intmath.c...ength-curve.php the arc length of the curve y = f(x) from x=a to x=b is given by:

length_ab = Derivative_ab( Sqrt ( 1 + (dy/dx) ^ 2 ) dx)

Bignose is objecting to your incorrectly copied expression. Your link there has the correct formula in a box labeled "arc length of a curve" - and a couple of pictures and explanatory comments, motivating it.

The "surd manipulation" they talk about is a multiplication by (one) under the square root sign - the square root of the numerator (differential x squared) pops out to the right, the denominator goes under the two terms, and you have the integral form in good order (that's the kind of cheese engineers use a lot).

Edited October 3, 201312 yr by overtone