Answer
$c\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right) = \left( {3t,9{t^2}} \right)$
Work Step by Step
We may write $x\left( t \right) = 3t$, so that $y\left( t \right) = 9{t^2}$. So, the parametrization is
$c\left( t \right) = \left( {x\left( t \right),y\left( t \right)} \right) = \left( {3t,9{t^2}} \right)$
When $t=0$, we have $c\left( 0 \right) = \left( {0,0} \right)$. When $t=1$, we have $c\left( 1 \right) = \left( {3,9} \right)$.
Thus, the path $c\left(t\right)$ traces the parabolic arc $y=x^2$ from $\left(0,0\right)$ to $\left(3,9\right)$ for $0 \le t \le 1$.
