Given $y=t^3-\frac{5}{2}t^2$ and $x=\sqrt t$, for $t>0$,
a) Use parametric differentiation to express $\frac{dy}{dx}$ in terms of $t$ in simplified form.
b) Show that $\frac{d^2y}{dx^2}=at^2+bt$, determining the constants $a$ and $b$.
c) Obtain an equation for the tangent to the curve which passes through the point of inflection.
Presumably you already have $$\frac {dy}{dx}=6t^{\frac 52}-10t^{\frac 32}$$ Then $$\frac{d^2y}{dx^2}=\frac{d}{dt}\left(\frac{dy}{dx}\right)\frac{dt}{dx}$$ $$=30t^2-30t$$