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

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Explanation / Answer

Since x = 3t(3 - t^2) and y = 9t^2. Note that the beginning and the end of the loop occur when x = 0 (and t nonzero) ==> t = ± v3. Link of plot: http://www.wolframalpha.com/input/?i=plo… So, the arc length is given by ? v[(dx/dt)^2 + (dy/dt)^2] dt = ?(t = -v3 to v3) v[(9 - 9t^2)^2 + (18t)^2] dt = 2 ?(t = 0 to v3) v[(9 - 9t^2)^2 + (18t)^2] dt, by evenness of the integrand = 2 ?(t = 0 to v3) 9 v[(1 - t^2)^2 + (2t)^2] dt = 18 ?(t = 0 to v3) v(t^4 + 2t^2 + 1) dt = 18 ?(t = 0 to v3) v(t^2 + 1)^2 dt = 18 ?(t = 0 to v3) (t^2 + 1) dt = 18 (t^3/3 + t) {for t = 0 to v3} = 6(t^3 + 3t) {for t = 0 to v3} = 36v3. I hope this helps!

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