Answer
\(\vec{r}'(t)=\langle\sin t + t \cos t, e^t (\cos t -\sin t), \cos^2t - \sin^2t\rangle \)
Work Step by Step
\(\vec{r}(t)=\langle t \sin t, e^t \cos t, \sin t \cos t \rangle\)
In order to get \(\vec{r}(t)\) we simply take the derivate of each component.
\(\vec{r}'(t)=\langle\sin t + t \cos t, e^t (\cos t -\sin t), \cos^2t - \sin^2t\rangle \)
