Scalar Triple Product Derivative

SPhy
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Homework Statement



Find an expression equivalent for the derivative of the scalar triple product

a(t) . (b(t) x c(t))

The Attempt at a Solution



Initially I figured since whatever comes out of B X C is being dotted with A, I can use the derivative rules of a dot product:

(a(t)' . (b(t) x c(t))) + ( a(t) . (b(t) x c(t))' )

However, the solution just gives an expression for the scalar triple product in a 3x3 matrix form, that is,a(t) . (b(t) x c(t)) = [ a1 a2 a3
b1 b2 b3
c1 c2 c3 ]

What gives?
 
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SPhy said:

Homework Statement



Find an expression equivalent for the derivative of the scalar triple product

a(t) . (b(t) x c(t))


The Attempt at a Solution



Initially I figured since whatever comes out of B X C is being dotted with A, I can use the derivative rules of a dot product:

(a(t)' . (b(t) x c(t))) + ( a(t) . (b(t) x c(t))' )

However, the solution just gives an expression for the scalar triple product in a 3x3 matrix form, that is,


a(t) . (b(t) x c(t)) = [ a1 a2 a3
b1 b2 b3
c1 c2 c3 ]

What gives?

That would be helpful if you know how to differentiate a determinant whose entries are functions. But I like your original idea. What is the formula for ##(\vec b(t) \times \vec c(t))'##? Is there a similar formula as you used for the dot product? Just keep going...
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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