Answer
For the interval (0, 1), the curve is concave up.
Work Step by Step
$x=t^{3}+1$, $y=t^{2}-t$
Step 1
Find $\frac{dy}{dx}$
$\frac{dy}{dx}=\frac{\frac{dy}{dt}}{\frac{dx}{dt}}= \frac{2t-1}{3t^{2}}=\frac{2}{3t}- \frac{1}{3t^{2}}= \frac{2}{3}t^{-1}-\frac{1}{3}t^{-2}$
Step 2
Find the 2nd derivative
$\frac{d^{2}y}{dx^{2}}=\frac{\frac{d}{dt}(\frac{2}{3}t^{-1}-\frac{1}{3}t^{-2})}{3t^{2}}=\frac{x}{y}=\frac{-\frac{2}{3}t^{-2}+\frac{2}{3}t^{-3}}{3t^{2}} \frac{3t^{3}}{3t^{3}}$
$=\frac{2-2t}{9t^{5}}$
Step 3
Find where the second derivative is 0
$2-2t=0$
$2(1-t)=0$
$t=1$
The 2nd derivative is undefined when t=0.
Interval $(-\infty, 0), t=-1 $
$\frac{2-2t}{9t^{5}}=\frac{2-2(-1)}{9(-1)^{5}}=-\frac{4}{9} \lt 0$- concvace down
Interval $(0,1), t=0,5$
$\frac{2-2(0,5)}{9(0,5)^{5}}= 3.6 \gt 0$- concave up
Interval $(1, \infty), t=2$
$\frac{2-2(2)}{9(2)^{5}}=-\frac{1}{144} \lt 0$- concave down
