Luk4 Posted October 2, 2013 Share Posted October 2, 2013 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! Link to comment Share on other sites More sharing options...
Bignose Posted October 3, 2013 Share Posted October 3, 2013 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). Link to comment Share on other sites More sharing options...
overtone Posted October 3, 2013 Share Posted October 3, 2013 (edited) 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, 2013 by overtone Link to comment Share on other sites More sharing options...
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