Differentiating cross product

By Anonymous on Tuesday, February 27, 2001 - 09:48 pm :

If a and b are vectors which depend on time, prove that

d/dt(a x b ) = (d/dt a ) x b + a x (d/dt b )

I have been given a hint to differentiate components in (a x b ) assuming base vectors are constants

By Kerwin Hui (Kwkh2) on Tuesday, February 27, 2001 - 10:22 pm :

Suppose

a =a₁ i +a₂ j +a₃ k
b =b₁ i +b₂ j +b₃ k

then for simplicity, consider the i component:

(a x b )₁ = (a₂ b₃ -a₃ b₂ )

so d/dt(a x b )₁ = d/dt(a₂ )b₃ +a₂ d/dt(b₃ )-d/dt(a₃ )b₂ -a₃ d/dt(b₂ )
=[d/dt(a₂ )b₃ -d/dt(a₃ )b₂ ]+[a₂ d/dt(b₃ )-a₃ d/dt(b₂ )]
=[d/dt(a ) x b + a x d/dt(b )]₁

Similar for other components.

Kerwin

By Anonymous on Wednesday, February 28, 2001 - 01:46 pm :

Isn't this just a case of the product rule of differentiation?

By Barkley Bellinger (Bb246) on Wednesday, February 28, 2001 - 09:14 pm :

Yes, but you may need to use suffix notation and the summation convention to be convinced.

By Michael Doré (Md285) on Thursday, March 1, 2001 - 12:55 pm :

Well we can use an alternative (but equivalent) definition of differentiation of vectors:

Define:

$a' (t) = (a (t + δ t) - a (t)) / δ t$

with $δ t$ tending to 0.

From here you get:

$d (a \times b) / dt = (a (t + δ t) \times b (t + δ t) - a (t) \times b (t)) / δ t$

with $δ t$ tending to 0. You can write this as:

$(a (t + δ t) \times b (t + δ t) - a (t + δ t) \times b (t) + a (t + δ t) \times b (t) - a (t) \times b (t)) / δ t$

This can then be split into two terms:

$(a (t + δ t) \times b (t + δ t) - a (t + δ t) \times b (t)) / δ t$

and

$(a (t + δ t) \times b (t) - a (t) \times b (t)) / δ t$

Using the fact that the cross product is distibutive over addition, these two terms can be written:

$a (t + δ t) \times (b (t + δ t) - b (t)) / δ t$

and

$[(a (t + δ t) - a (t)) / δ t] \times b (t)$

These converge to $a (t) \times b' (t)$ and $a' (t) \times b (t)$ so $d (a \times b) / dt$ converges to:

$a (t) \times b' (t) + a' (t) \times b (t)$

as required.