Arc Length Of A Curve<Text-field layout="Heading 1" style="Heading 1"><Font family="Lucida Bright">We approxmiate the length of the curve <Font executable="false" opaque="false">
</Font></Font><Equation style="Heading 1">NiMqJi0lJGNvc0c2IyUidEciIiIiIiMhIiI=</Equation><Font executable="false" family="Lucida Bright" opaque="false"> i + </Font><Equation style="Heading 1">NiMtJSRzaW5HNiMlInRH</Equation><Font executable="false" family="Lucida Bright" opaque="false"> j + </Font><Equation style="Heading 1">NiMqJiUidEciIiIiIzUhIiI=</Equation><Font executable="false" family="Lucida Bright" opaque="false"> k 1<t<3</Font><Font family="Lucida Bright">
by finding the length of a <Font bold="true" size="18" style="_cstyle256">piecewise linear curve</Font> close to the original curve. </Font></Text-field>The curve isrestart:
with(VectorCalculus):
r := t -> <cos(t)/2, sin(t), t/10> ;
a := 1; b := 3;Subdivide the interval 1<t<3 into N parts, h is the length of the subdivision.
The curve is approximated by a piecewise linear curve joining the points r(a), r(a + h), r(a+2h), ....r(a+Nh)=r(b) on the curve. The approximate length of the helix with N subdivisions ish :=(b-a)/N :
segi := r(a+(i+1)*h) - r(a+i*h):
approxlen := Sum( sqrt(segi.segi), i=1..'N' );Taking N=10, we get the approximate length of the helix isN := 100;
evalf(approxlen, 8);Change the value of N above to 100, then 1000, then 10000 etc to get better approximations.Compare this with the length of the curve obtained by the integral formular := <cos(t)/2, sin(t), t/10>;
v := diff(r,t):
speed := sqrt(v.v):
exactlen := Int( speed, t=a..b ):
evalf(exactlen, 8);