What is the arclength of f(t)=(3t^2-1, 3t^3-t) on t∈[-1/sqrt(3),1/sqrt(3)]?

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Answer 1 · 2018-06-23T02:18:37

${4sqrt3}/3$

For the given curve, we have

$x(t) = 3t^2-1,qquad y(t) = 3t^3-t$

Hence

$dx/dt = 6t,qquad dy/dt = 9t^2-1$

Thus, the infinitesimal arc length is

$ds = sqrt{(dx/dt)^2+(dy/dt)^2}dt$
$qquad = sqrt{36t^2+(9t^2-1)^2}dt$
$qquad = sqrt{(9t^2+1)^2}dt$
$qquad = (9t^2+1)dt$

Thus, the arc length required is

$int_{-1/sqrt3}^{+1/sqrt3}(9t^2+1)dt = (3t^3+t)_{-1/sqrt3}^{+1/sqrt3}$

$qquad qquad = {t(3t^2+1)}_{-1/sqrt3}^{+1/sqrt3}=4/sqrt3$