The position $\vec r$ of a particle moving in an xy plane is given by $\vec r$ = $(1.00t^3 − 3.00t)i + (8.00 − 9.00t^4)j$, with $\vec r$ in meters and $t$ in seconds. In unit-vector notation, calculate the following for $t = 2.10 s$.
I found $\vec r$ by subbing in $2.1$ into the equation.
I differentiated the equation, then subbed $2.1$ to get velocity. The answer I have is $(3t^2 - 3)i + (0-9t^3)$, then $10.23 - 83.35$. It's not accepting that, though.
If $\vec r(t) = \hat i(t^3-3t)+\hat j(8-9t^4)$, then we find that
$$\vec r'(t) =\hat i (3t^2-3)+\hat j(-36t^3)$$