Find the curvature of the curve in the given parametrically by: x(t)=cos^3 (t), y(t)=sin^3 (t) at t=π/4
Recall that
k = (x'y" - x"y')/(x'^2 + y'^2)^(3/2)
Now, all we need is these derivatives:
x' = -3 sint cos^2t
x" = -3 cost cos2t
y' = 3 cost sin^2t
y" = 3 sint cos2t
You can see that things are going to get messy, but cancel out some.
So, what do you get? What does that mean? Where is that point on the curve?