Answer
$\frac{dy}{dx} = \frac{2t-1}{3t^2}$
$\frac{d^2y}{dx^2} = \frac{2-2t}{9t^5}$
The curve is concave upward when $0 \lt t \lt 1$
Work Step by Step
$x = t^3+1$
$y = t^2-t$
We can find $\frac{dy}{dx}$:
$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{2t-1}{3t^2}$
We can find $\frac{d^2y}{dx^2}$:
$\frac{d^2y}{dx^2} = \frac{\frac{d}{dt}(\frac{dy}{dx})}{\frac{dx}{dt}} = \frac{\frac{6t^2-6t(2t-1)}{9t^4}}{3t^2} = \frac{-6t^2+6t}{27t^6} = \frac{2-2t}{9t^5}$
The curve is concave upward when $\frac{d^2y}{dx^2} \gt 0$
$2-2t \gt 0$ when $t \lt 1$
$9t^5 \gt 0$ when $t \gt 0$
$\frac{d^2y}{dx^2} \gt 0$ when $0 \lt t \lt 1$
$2-2t \lt 0$ when $t \gt 1$
$9t^5 \lt 0$ when $t \lt 0$
There are no values of $t$ such that both $2-2t$ and $9t^5$ are negative.
The curve is concave upward when $0 \lt t \lt 1$
