The question is:
A curve is defined by the parametric equations
$$ x = t^2 + a $$
$$ y = t(t-a)^2 $$
Find the range of values for t in terms of a where the function is concave up?
What I have done
$$ y = t^3 - 2at^2 + a^2t $$
$$ dy/dt = 3t^2 - 4at + a^2 $$
$$ dx/dt = 2t $$
$$ dy/dx = dy/dt * dt/dx $$
$$ dy/dx = (3t^2 - 4at + a^2))/(2t) $$
$$ d^2y/dx^2 = d/dt(dy/dx) * dt/dx $$
$$ d^2y/dx^2 = [(12t^2 - 8at - 6t^2 + 8at - 2a^2)]/(4t^2) * 1/2t > 0 $$
$$ d^2y/dx^2 = 6t^2 - 2a^2 > 0 $$
$$ 2(3t^2 - a^2) > 0 $$
$$ t = ± \sqrt{a^2\over3} $$
So the answer should be
$$ t > \sqrt{a^2\over3} $$ $and$ $$ t < -\sqrt{a^2\over3}$$
However my book states only $t > \sqrt{a^2\over3}$
What is correct?
You properly found that $$\frac{dy}{dx}=\frac{3 t^2-4 a t+a^2}{2 t}$$ So, $$\frac d{dt}\left(\frac{dy}{dx}\right)=\frac{3t^2-a^2}{2t^2}$$ and then $$\frac{d^2y}{dx^2}=\frac{3t^2-a^2}{2t^2}\times \frac 1 {2t}=\frac{3t^2-a^2}{4t^3}$$ That is to say that $\frac{d^2y}{dx^2}$ has the same sign as $t(3t^2-a^2)$.
Is this helping you ?