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 + dt) - a (t))/dt
with dt tending to 0.
From here you get:
d(a ×b )/dt = (a (t+dt) × b (t+dt) - a (t) ×b (t))/dt
with dt tending to 0. You can write this as:
(a (t+dt) ×b (t+dt) - a (t+d t) ×b (t) + a (t+dt) ×b (t) - a (t) ×b (t))/dt
This can then be split into two terms:
(a (t+dt) ×b (t+dt) - a (t+ dt) ×b (t))/dt
and
(a (t+dt) ×b (t) - a (t) × b (t))/dt
Using the fact that the cross product is distibutive over addition, these two
terms can be written:
a (t+dt) ×(b (t+dt) - b (t))/ dt
and
[(a (t+dt) - a (t))/dt] ×b (t)
These converge to a (t) ×b ' (t) and
a ' (t) ×b (t) so d(a ×b )/dt converges to:
a (t) ×b ' (t) + a ' (t) ×b(t)
as required.