Show that u'(t) = r(t).(r'(t)Xr'''(t))

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Homework Statement


Let u(t) be a vector valued function, where

u(t) = r(t).(r'(t)Xr''(t))

where r(t) is a vector valued function, and (r'(t)Xr''(t)) the cross product of the first and second derivative of r(t). Show that

u'(t) = r(t).(r'(t)Xr'''(t))

where r'''(t) is the 3rd derivative of r(t).

Homework Equations





The Attempt at a Solution



I got this question on an exam and did not know how to solve it. I started out by computing the cross product of (r'(t)Xr''(t)) by using determinants, and then took the dot product of r(t).(r'(t)Xr''(t)), which gave me a really ugly vector. I ran out of time while computing r(t).(r'(t)Xr'''(t)). Is there some other technique I could've applied to prove this?

Thanks,
M
 
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u(t) is a scalar, not vector-valued. You can see that many terms in the derivative of u cancel or vanish by using the identities

\vec{a}\times \vec{b} = - \vec{b} \times {a}

\vec{a}\cdot (\vec{b}\times \vec{c}) = \vec{c} \cdot( \vec{a}\times{b}).
 
I'm using ' for derivative & without t in parantheses.:--

u=r.(r'Xr"), differentiating both sides, use these two formulas: (a.b)'=a'.b + a.b'
(aXb)'=a'Xb + aXb'

u' = r'.(r'Xr") + r.(r'Xr")'
= 0 + r.(r"Xr" + r'Xr''') = r(r'Xr''') since a.(aXb )= 0 and aXa = 0
proved
 
let r=a\vec{i}+b\vec{j}+c\vec{k}

<br /> \begin{bmatrix}<br /> a&#039;b&#039;c&#039;&#039;\ -a&#039;b&#039;&#039;c&#039;\ +b&#039;c&#039;a&#039;&#039;\ -b&#039;c&#039;&#039;a&#039;\ +c&#039;a&#039;b&#039;&#039;\ -c&#039;a&#039;&#039;b&#039; \\ <br /> ab&#039;&#039;c&#039;&#039;\ -ab&#039;&#039;&#039;c&#039;\ +bc&#039;&#039;a&#039;&#039;\ -bc&#039;&#039;&#039;a&#039;\ +ca&#039;&#039;b&#039;&#039;\ -ca&#039;&#039;&#039;b&#039; \\ <br /> ab&#039;c&#039;&#039;&#039;\ -ab&#039;&#039;c&#039;&#039;\ +bc&#039;a&#039;&#039;&#039;\ -bc&#039;&#039;a&#039;&#039;\ +ca&#039;b&#039;&#039;&#039;\ -ca&#039;&#039;b&#039;&#039; <br /> \end{bmatrix}<br /> \\ <br />
It's not a matrix, you should read it as one line addition.yep, it' spretty ugly if you're under stress and hurry.

Row 1 completely disappear 1with 4, 2-5, 3-6
Then
r2c1 with r3c2
r2c3 with r3c4
r2c5 with r3c6

Remain 6 terms, the other equation.
 
Last edited:
Thanks everyone. I understand it now. I should have studied the properties of cross product more closely.

M
 
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