Calculus

Compute the curvature.

r(t)=(t^2,2t^3/3) t >0

The answer is 1/2t(1+t^2)^3/2


I have tried multiple times but i cannot arrive to this answer.

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  1. recall that for parametric curves, the curvature is

    x'y" - x"y'
    -----------------
    (x'^2 + y'^2)^(3/2)

    so, for your function, that is
    x' = 2t
    x" = 2
    y' = 2t^2
    y" = 4t

    (2t)(4t)-(2)(2t^2)
    ----------------------- = 4t^2/(8t^3(1+t^2)) = 1/(2t(1+t^2)^(3/2)
    (4t^2+4t^4)^(3/2)

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    posted by Steve

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