## Calculus 8th Edition

$r(t) \times r'(t)= \omega a \times b$
As we are given that $r(t)=a \cos \omega t+b \sin \omega t$, Use cross product rule. $r'(t)=u'(t) \times v(t)+u(t) \times v'(t)$ $r(t)=a \cos \omega t+b \sin \omega t$ and $r'(t)= -\omega a \sin \omega t+ \omega b \cos \omega t$ Now, $r'(t)=\begin{vmatrix}i&j&k \\\cos \omega t&\sin \omega t&0\\-\omega \sin \omega t&\omega \cos \omega t&0\end{vmatrix}$ This gives: $=\omega c \cos^2 \omega t+ \omega c \sin^2 \omega t$ Now, $\omega a \times b=\begin{vmatrix}a&b&c \\ \omega&0&0\\0&1&1 \end{vmatrix}$ $\omega a \times b=\omega c$ Hence, the result. $r(t) \times r'(t)= \omega a \times b$