Question

If \({\rm{u(t)}}\)and \({\rm{v(t)}}\) are differentiable vector functions, then

\(\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{(u(t) \times v(t)) = }}{{\rm{u}}^'}{\rm{(t) \times }}{{\rm{v}}^'}{\rm{(t)}}{\rm{.}}\)

Short answer

The statement is false.

Step-by-step solution

Let's take some value for\({\rm{u(t)}}\)and\({\rm{v(t)}}\),

Let,\({\rm{u(t) = ti}}\)and \({\rm{v(t) = tj}}\)

Then,

\({\rm{u(t) \times v(t) = }}{{\rm{t}}^{\rm{2}}}{\rm{k}}\)

Differentiate the equation,

\(\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{\{ u(t) \times v(t)\} = }}\frac{{\rm{d}}}{{{\rm{dt}}}}\left\{ {{{\rm{t}}^{\rm{2}}}} \right\}{\rm{k}}\)

\({\rm{ = 2tk}}\)

Differentiate \({\rm{u(t)}}\)and\({\rm{v(t)}}\),

\({{\rm{u}}^'}{\rm{(t) = i}}\)

\({{\rm{v}}^'}{\rm{(t) = j}}\)

\(\begin{aligned}{c}{{\rm{u}}^'}{\rm{(t) \times }}{{\rm{v}}^'}{\rm{(t) = i \times j}}\\{\rm{ = k}}\end{aligned}\)

Therefore, \(\frac{{\rm{d}}}{{{\rm{dt}}}}{\rm{\{ u(t) \times v(t)\} }} \ne {{\rm{u}}^'}{\rm{(t) \times }}{{\rm{v}}^'}{\rm{(t)}}\).

Therefore, the statement is false.