# Calculus

Compute the curvature.

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

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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