VC1.8

From Exampleproblems

Solution If $a=\sin \theta i+\cos \theta j+\theta k,b=\cos \theta i-\sin \theta j-3k,c=2i+3j-k\,$ .Find $\frac{d}{d\theta}[a\times(b\times c)]\,$ at $\theta=0\,$

$b\times c=\begin{vmatrix} i & j & k \\ \cos \theta & -\sin \theta & -3 \\ 2 & 3 & -1 \end{vmatrix}\,$

$b\times c=(\sin \theta+9)i+(\cos \theta-6)j+(3\cos \theta+2\sin \theta)k\,$

Therefore, $a\times(b\times c)=\begin{vmatrix} i & j & k \\ \sin \theta & \cos \theta & \theta \\ \sin \theta + 9 & \cos \theta -9 & 3\cos \theta + 2\sin \theta \end{vmatrix}\,$

or $a\times(b\times c)=(3\cos^2 \theta+2\sin \theta \cos \theta-\theta \cos \theta+6\theta)i-3(\sin \theta \cos theta+2\sin^2 \theta-\theta\sin \theta-9\theta)j+(\sin \theta \cos \theta-6\sin \theta-\sin \theta \cos \theta-9\cos \theta)k\,$

Therefore, $\frac{d}{d\theta}(a\times(b\times c)=\frac{d}{d\theta}(3\cos^2 \theta+2\sin \theta \cos \theta-\theta \cos \theta+6\theta)i-\frac{d}{d\theta}[\frac{3}{2}\sin 2\theta+2\sin^2 \theta-\theta \sin \theta-9\theta]j-\frac{d}{d\theta}(6\sin \theta+9\cos \theta)k\,$

$[-6\cos \theta \sin \theta+2\cos 2\theta-(\cos \theta-\theta \sin \theta)+6]i-[3\cos 2\theta+4\sin \theta \cos \theta-(\sin \theta+\theta \cos \theta)-9]j-(6\cos \theta-9\sin \theta)k\,$

When $\theta =0\,$ , $\frac{d}{d\theta}(a\times(b\times c)=(2-1+6)i-(3-9)j-6k=7i+6j-6k\,$

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