Question

Find the curvature x=t^2, y=t^3
Using k=|xÿ-yx|/(x^2+y^2)^3/2
First get the formula right:

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

Then just plug and chug:

x' = 2t
x" = 2
y' = 3t^2
y" = 6t

k = |2t*6t - 2*3t^2|/(4t^2+9t^4)^(3/2)
= 6t/(4+9t^2)^(3/2)

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