x=3t^2 , y=t^3-t Determine the open t-intervals on which the curve is concave downward or concave upward.

Given parametric equations are:
x=3t^2,y=t^3-t
We need to find the second derivative, to determine the concavity of the curve.
dy/dx=(dy/dt)/(dx/dt)
Let's take the derivative of x and y with respect to t,
dx/dt=3*2t=6t
dy/dt=3t^2-1
dy/dx=(3t^2-1)/(6t)
dy/dx=(3t^2)/(6t)-1/(6t)
dy/dx=t/2-1/(6t)
(d^2y)/dx^2=d/dx[dy/dx]
=(d/dt[dy/dx])/(dx/dt)
=(d/dt(t/2-1/(6t)))/(6t)
=(1/2-1/6(-1)t^(-2))/(6t)
=(1/2+1/(6t^2))/(6t)
=((3t^2+1)/(6t^2))/(6t)
=(3t^2+1)/(6t^2(6t))
=(3t^2+1)/(36t^3)
Curve is concave upwards if second derivative is positive and concave downwards if it is negative,
So, the curve is concave upward for t>0
Curve is concave downward for t<0