Find the length of the loop of the given curve. x = 9t - 3t^3 y = 9t^2 Solution

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ok dear x(t)=9t-3t^3 y(t)=9t^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 9t-3t^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)=9-9t^2 y'(t)=18t and we have integral sqrt( (9-9t^2)^2+324t^2) dt integral sqrt(81-162t^2+81t^4+324t^2) dt integral sqrt(81t^4+162t^2+81) dt integral sqrt((9t^2+9)^2) dt integral 9t^2+9 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)

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