xt33t yt23 at the originSolution xt3tt3 yt3t2 for a parametr

Solution

x(t)=3t-t^3 y(t)=3t^2 for a parametric curve x(t) y(t) the arc length from t=a to b is given by integral sqrt( x\'(t)^2 + y\'(t)^2 ) dt for t=a to b first we need to find the limits of integration looking at a graph of the curve the loop begins and ends when x=0 and y>0 so x=0 3t-t^3=0 3t=t^3 t=0 by this makes y=0 so 3=t^2 t=+-sqrt(3) so our limits of integration are -sqrt(3) and sqrt(3) now we have x\'(t)=3-3t^2 y\'(t)=6t and we have integral sqrt( (3-3t^2)^2+36t^2) dt integral sqrt(9-18t^2+9t^4+36t^2) dt integral sqrt(9t^4+18t^2+9) dt integral sqrt((3t^2+3)^2) dt integral 3t^2+3 dt t^3+3t now evaluate this for t=-sqrt(3) to sqrt(3) 3*sqrt(3)+3sqrt(3)+3sqrt(3)+3sqrt(3)= 12*sqrt(3) so the arc length is 12*sqrt(3)