Answer
\[\frac{dy}{dx}=\frac{2t+1}{2t}\]
\[\frac{d^2y}{dx^2}=\frac{-1}{4t^3}\]
The Curve is concave down if $t<0$.
Work Step by Step
\[x=t^2+1\;\;,\;\;y=t^2+t\]
Differentiate $x$ with respect to $t$
\[\frac{dx}{dt}=2t\;\;\;\;\;\ldots (1)\]
Differentiate $y$ with respect to $t$
\[\frac{dy}{dt}=2t+1\]
\[\Rightarrow \frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}=\frac{2t+1}{2t}\]
\[\frac{dy}{dx}=\frac{2t+1}{2t}\]
\[\Rightarrow \frac{dy}{dx}=1+\frac{1}{2t}\]
Differentiate $\displaystyle\frac{dy}{dx}$ with respect to $x$ using chain rule:
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dx}(1+\frac{1}{2t})\]
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{d}{dt}(1+\frac{1}{2t})\cdot\frac{dt}{dx}\]
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{-1}{2t^2}\cdot\frac{dt}{dx}\]
Using (1)
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{-1}{2t^2}\cdot\frac{1}{2t}\]
\[\Rightarrow \frac{d^2y}{dx^2}=\frac{-1}{4t^3}\]
and $\displaystyle\frac{d^2y}{dx^2}<0$ if $t>0$ so curve is concave down if $t<0$.
