Vectors differentiation formulas for Dot and Box Product how?

[abrowaqas]

Let A ,B and C represent vectors.
we have

1) d/dt (A . B) = A. dB/dt + dA/dt .B

2) d/dt [ A . (BxC) ] = A . (Bx dC/dt) + A . ( dB/dt x C) + dA/dt . (B xC)

now the problem in these formulas is that
we know that Dot product between two vectors and Scalar triple product of vectors is always a scalar. now if we find their derivative it results always Zero.

then how these formulas has been defined since the derivative remains zero always for constant hence it always yield zero result.

please explain

 

[tiny-tim]

hi abrowaqas!

we know that Dot product between two vectors and Scalar triple product of vectors is always a scalar. now if we find their derivative it results always Zero.

no … the derivative of a scalar is zero only if the scalar is constant

 

[Char. Limit]

Let A ,B and C represent vectors.
we have

1) d/dt (A . B) = A. dB/dt + dA/dt .B

2) d/dt [ A . (BxC) ] = A . (Bx dC/dt) + A . ( dB/dt x C) + dA/dt . (B xC)

now the problem in these formulas is that
we know that Dot product between two vectors and Scalar triple product of vectors is always a scalar. now if we find their derivative it results always Zero.

then how these formulas has been defined since the derivative remains zero always for constant hence it always yield zero result.

please explain

A vector doesn't have to be constant.

 

[wisvuze]

I think you are just confused ( or I'm confused about what you are asking ):

if f is a function that is a "constant" function , so that f ( x ) = a fixed c for all x's, then Df = 0

You can see that the dot product of two vectors v( t ) , w( t ) can be a non constant function, even though the result is a scalar. Since the dot product of two different pairs of vectors can give you different results

 

[abrowaqas]

Thanks Wisvuse..
I got it but can you explain it by giving one example..

 

[HallsofIvy]

Let u be the vector <t, 0, 3t>, v be the vector <t^2, t- 1, 2t>. Then the dot product of u and v is t^3+ 6t^2 and the derivative of that is 3t^2+ 12t.

The derivative of u is <1, 0, 3> and the derivative of v is <2t, 1, 2>. The "product rule gives the derivative of u dot v as u'v+ uv'= <1, 0, 3>.<t^2, t-1, 2t>+ <t, 0, 3t>.<2t, 1, 2>= (t^2+ 6t)+ (2t^2+ 6t)= 3t^2+ 12t as before.