Question

 If A, B are differentiable vector point function of scalar variable f over domain S, then prove that d/dt( AxB) = (dA/dt x B) + (A x dB/dt) 

Answer 1

Given A, B are differentiable vector point functions of scalar variable f over domain S.

then we have to prove that: $\frac{d(A \times B)}{dt}$ = $\frac{dA}{dt} \times B$ + $A \times \frac{dB}{dt}$

Let A = $<$ x₁, y₁, z₁ $>$ and B = $<$ x₂, y₂, z₂ $>$

$\implies$ $\frac{dA}{dt}$ = $<$ x₁', y₁', z₁' $>$ and $\frac{dB}{dt}$ = $<$ x₂', y₂', z₂' $>$

We have,

A $\times$ B = $\begin{vmatrix} i & j & k \\ x_1 & y_1 & z_1\\ x_2 & y_2 & z_2 \end{vmatrix}$ = = $<$ y₁z₂ − z₁y₂, z₁x₂ − x₁z₂, x₁y₂ − y₁x₂ $>$

Taking derivatives using the product rule from single variable calculus, we get a lot of terms, which we can group to prove the vector formula.

$\frac{d(A \times B)}{dt}$ = $<$ y₁'z₂ + y₁z₂' − z₁'y₂ − z₁y₂', z₁'x₂ + z₁x₂' − x₁'z₂ − x₁z₂', x₁'y₂ + x₁ y₂' − y₁'x₂ − y₁x₂' $>$

= $<$ (y₁'z₂ − z₁'y₂)+ (y₁z₂'− z₁y₂'), (z₁'x₂ − x₁'z₂) + (z₁x₂'− x₁z₂'), (x₁'y₂ − y₁'x₂) + (x₁y₂'− y₁x₂') $>$

= $<$ x₁', y₁', z₁' $>$ $\times$ $<$ x₂, y₂, z₂ $>$ + $<$ x₁, y₁, z₁ $>$ $\times$ $<$ x₂', y₂', z₂' $>$

= $\frac{dA}{dt} \times B$ + $A \times \frac{dB}{dt}$

Hence proved,

$\frac{d(A \times B)}{dt}$ = $\frac{dA}{dt} \times B$ + $A \times \frac{dB}{dt}$