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Find an expression for the derivative (d/dt) of [r(t)·r'(t)×r''(t)] in simplified form.

I am attempting to use properties for both the dot product and cross product but I cannot seem to find the ones that will work to logically simplify the expression. Any clues?

aort01
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1 Answers1

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It's a triple product:

$$r(t) \cdot (r'(t) \times r''(t))=\det(r(t),r'(t),r''(t))$$

The derivative of a determinant can be obtained by summing the determinants of the matrices obtained by differentiating the different columns, one at a time (see here), a process underlined by the use of red color for the "primes" below:

$$\underbrace{\det(r\color{red}{'}(t),r'(t),r''(t))}_{D_1}+\underbrace{\det(r(t),r'\color{red}{'}(t),r''(t))}_{D_2}+\underbrace{\det(r(t),r'(t),r''\color{red}{'}(t))}_{D_3}$$

$D_1=0$ and $D_2=0$ (because of $2$ identical columns).

As a consequence, the final result is $D_3$ that can be written under the form:

$$r(t) \cdot (r'(t) \times r'''(t))$$

Jean Marie
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  • Just to add as a note, this problem could also have been solved by taking the expansion of the expression while deriving (where V is r' and A is r'') ; to give us V(VXA) + r(VXA'). We know from the definition of the cross product that VVXA will be orthogonal (i.e. = 0) so we are therefore left with as you stated above (and making V and A back into elements of r ), r(r' X r'''). – aort01 Nov 10 '21 at 16:54
  • That's right. Frankly, I hadn't thought to this way. – Jean Marie Nov 10 '21 at 17:23