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)
Find the curvature x=t^2, y=t^3
Using k=|xÿ-yx|/(x^2+y^2)^3/2
1 answer