Sketch the plane curve r(t)=<t^3,t^2>, find r'(1) and r''(1) , then sketch the position vector r(1), r'(1), r''(1) at the graph of the curve r(t)
To get a general idea of what the curve looks like eliminate the variable t.
$\displaystyle x=t^3 \quad y=t^2$ this gives
$\displaystyle y=x^{2/3}=\sqrt[3]{x^2}$
Recall that
$\displaystyle \frac{d}{dt}\mathbf{r}(t)=<\frac{dx}{dt},\frac{dy} {dt}>$
Give this a shot and try back if you are still stuck.