Nature of concavity for parametric curve

For what intervals of t is the curve described by the given parametric equation concave up? Concave down?

x=t^2  ;      y=t^(3) + 3t

Student  X

The signage of the second order derivative( ie positive,negative or zero) would help to discover the nature of concavity for the curve at various intervals.

dy/dx =dy/dt * dt/dx = (3t^2 + 3)/ (2t) 

(d^2 y)/(dx^2)= d/dx (dy/dx) = d/dt (dy/dx) * ( dt/dx) 

                    =  [2t*(6t)-2(3t^2 + 3)]/(4t^2)  * 1/(2t) 

                   =  (6t^2 - 6)/(8t^3)

=  (3t^2 - 3)/(4t^3)     

(Note that since dy/dx is actually a function in t, implicit differentiation must be employed when seeking the expression for (d^2 y)/(dx^2) .)

When the curve is concave downwards, 

(d^2 y)/(dx^2)<0 =====> (3t^2 -3)/(4t^3) < 0

[(t-1)(t+1)]/ (t^3) < 0 

ie  t < -1   or    0 < t < 1  (shown)   

When the curve is concave upwards, 

(d^2 y)/(dx^2) =====> (3t^2 - 3)/(4t^3) > 0

[(t-1)(t+1)]/ (t^3) > 0 

ie  -1 < t < 0   or    t > 1  (shown) 

Hope this helps. Peace.          

Best Regards,

Mr Koh