Answer
By taking the derivative of ${\bf{w}}\left( t \right)$ twice, we obtain
${\bf{w}}{\rm{''}}\left( t \right) = - 9{\bf{w}}\left( t \right)$
Work Step by Step
We have ${\bf{w}}\left( t \right) = \left( {\sin \left( {3t + 4} \right),\sin \left( {3t - 2} \right),\cos 3t} \right)$.
Taking the derivative of ${\bf{w}}\left( t \right)$ gives
${\bf{w}}'\left( t \right) = \left( {3\cos \left( {3t + 4} \right),3\cos \left( {3t - 2} \right), - 3\sin 3t} \right)$
Another derivative gives
${\bf{w}}{\rm{''}}\left( t \right) = \left( { - 9\sin \left( {3t + 4} \right), - 9\sin \left( {3t - 2} \right), - 9\cos 3t} \right)$
${\bf{w}}{\rm{''}}\left( t \right) = - 9\left( {\sin \left( {3t + 4} \right),\sin \left( {3t - 2} \right),\cos 3t} \right)$
Hence, ${\bf{w}}{\rm{''}}\left( t \right) = - 9{\bf{w}}\left( t \right)$.
