Find dy/dx and d^2y/dx^2 and then determine for which t values the curve is concave up:

Question

x=t^3 - 18t y=t^2 - 1

Answer 1

dy/dt = 2t

dx/dt = 3t^2 - 18

dy/dx = (3t^2 - 18)/(2t)

d^2y/d^2x = (2t(6t) - 2(3t^2-18))/(4t^2) by quotient rule
> 0 for concave up
the denominator will always be positive for x≠0
so
12t^2 - 6t^2 + 36 >
6t^2 + 6 > 0
t^2 + 6 > 0

which is true for all values of t