Answer
\[\frac{dy}{dx}=\frac{2t-1}{3t^2}\]
\[\frac{d^2y}{dx^2}=\frac{2(1-t)}{9t^5}\]
The Curve is concave down if $t\in(-\infty,0)\cup (1,\infty)$.
Work Step by Step
\[x=t^3+1\;\;,\;\;y=t^2-t\]
Differentiate $x$ with respect to $t$
\[\frac{dx}{dt}=3t^2\;\;\;\;\;\ldots (1)\]
Differentiate $y$ with respect to $t$
\[\frac{dy}{dt}=2t-1\]
\[\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t-1}{3t^2}\]
\[\frac{dy}{dx}=\frac{2t-1}{3t^2}\]
\[\Rightarrow \frac{dy}{dx}=\frac{2}{3t}-\frac{1}{3t^2}\]
Differentiate $\displaystyle\frac{dy}{dx}$ with respect to $x$ using chain rule:
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dx}(\frac{2}{3t}-\frac{1}{3t^2})\]
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dt}(\frac{2}{3t}-\frac{1}{3t^2})\cdot\frac{dt}{dx}\]
\[\Rightarrow \frac{d^2y}{dx^2}=(\frac{-2}{3t^2}+\frac{2}{3t^3})\cdot\frac{dt}{dx}\]
Using (1)
\[\Rightarrow \frac{d^2y}{dx^2}=(\frac{-2}{3t^2}+\frac{2}{3t^3})\cdot\frac{1}{3t^2}\]
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{2(1-t)}{9t^5}\]
and $\displaystyle\frac{d^2y}{dx^2}<0$ if $t<0$ and $t>1$ so curve is concave down if $t\in(-\infty,0)\cup (1,\infty)$.
