We first need the magnitude of the derivative of the vector function. This is,

\[\vec r'\left( t \right) = \left\langle {2t,6{t^2}, - 3{t^2}} \right\rangle \] \[\left\| {\vec r'\left( t \right)} \right\| = \sqrt {4{t^2} + 36{t^4} + 9{t^4}} = \sqrt {{t^2}\left( {4 + 45{t^2}} \right)} = \sqrt {{t^2}} \sqrt {4 + 45{t^2}} = \left| t \right|\sqrt {4 + 45{t^2}} = t\sqrt {4 + 45{t^2}} \]

For these kinds of problems make sure to simplify the magnitude as much as you can. It can mean the difference between a really simple problem and an incredibly difficult problem.

Note as well that because we are assuming that we are starting at \(t = 0\) for this kind of problem it is safe to assume that \(t \ge 0\) and so \(\sqrt {{t^2}} = \left| t \right| = t\).

Show Step 2