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

## 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).

## 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

fzero
Homework Helper
Gold Member
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 $$r=a\vec{i}+b\vec{j}+c\vec{k}$$

$$\begin{bmatrix} a'b'c''\ -a'b''c'\ +b'c'a''\ -b'c''a'\ +c'a'b''\ -c'a''b' \\ ab''c''\ -ab'''c'\ +bc''a''\ -bc'''a'\ +ca''b''\ -ca'''b' \\ ab'c'''\ -ab''c''\ +bc'a'''\ -bc''a''\ +ca'b'''\ -ca''b'' \end{bmatrix} \\$$
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