# Vector Proof Question for position of particle

1. Feb 5, 2013

### Momentous

1. The problem statement, all variables and given/known data

If ~r, ~v and ~a denote the position, velocity and acceleration of a particle, prove that

2. Relevant equations

(d/dt)[ a * ( v x r)] = (da/dt) * (v x r)

"*" is a dot product
"x" is a cross product

3. The attempt at a solution

I didn't really get that far. The only thing I could think of was to make v = dr/dt, and a = d^2r, dt^2

2. Feb 5, 2013

### kevinferreira

What are the properties of the cross product that you know about?

3. Feb 5, 2013

### Simon Bridge

If ~r, ~v and ~a denote the position, velocity and acceleration of a particle, prove that
(d/dt)[ a * ( v x r)] = (da/dt) * (v x r)

OK - let me just tidy that up a bit:

$\vec{r}$, $\vec{v}$, $\vec{a}$, are the vectors ... the relation to prove is:$$\frac{d}{dt}\left [ \vec{a}\cdot(\vec{v}\times\vec{r}) \right ]=\frac{d\vec{a}}{dt}\cdot(\vec{v}\times\vec{r})$$... you correctly realize that $$\vec{v}=\frac{d\vec{r}}{dt}\; \text{&}\; \vec{a}=\frac{d\vec{r}}{dt}$$ will probably be important.

You will also need to review the rules for dot and cross products (how do they combine?) and you have to figure out how the product rule works with them.
These should be in your notes. If you have trouble locating them there, you'll find them online - just google for dot and cross products.

4. Feb 5, 2013

### tiny-tim

Welcome to PF!

Hi Momentous! Welcome to PF!

You don't seem to be aware that the product rule applies …

(a.bxc)' = a'.bxc + a.b'xc + a.bxc'

5. Feb 5, 2013

### Momentous

Thank you guys for the welcoming! I really appreciate the help. My knowledge of vectors is definitely lacking. I totally missed the product rule there.

So by following that, it seems like I would get

(a.(v x r))' = a'.(v x r) + a.(a x r) + a.(v x v)

with v x v = 0, I would have

(a.(v x r))' = a'.(v x r) + a.(a x r)

That seems a little bit closer, but I know I'm still off. Would rewriting the last piece as a.(dv/dt x r) help out there? I'm wondering if that would be some type of vector identity.

6. Feb 5, 2013

### Simon Bridge

If u,v, and w are all vectors, expand u.(vxw)
hint: compare what you have with what you need - what must be true in order to get there?

7. Feb 5, 2013

### Momentous

I don't know how to expand that. It's been 2 years since I've taken calc 3, and I haven't really used it since.

8. Feb 5, 2013

### Simon Bridge

so do the comparison...

(if only there were some easily searchable database of information at our fingertips where stuff like the properties of cross and dot products could be looked up...)

9. Feb 5, 2013

### Momentous

Alright, so with the information you guys have given me, I think this is the closest I can get.

(a.(v x r))' = a'.(v x r) + a.(a x r) + a.(v x v) by product rule

Rewritten

(a.(v x r))' = a'.(v x r) + a.(a x r)

Rewrite as

(a.(v x r))' = a'.(v x r) + r.(a x a)

from the link that Simon Bridge gave me

Rewrite as

(a.(v x r))' = a'.(v x r)

And that's it? I think.
Thanks for the help. I was really lost

10. Feb 5, 2013

### Simon Bridge

Which is the relation you needed. Well done.
It was just a matter of knowing a bit of algebra.

You should probably brush up on these operations.

11. Feb 5, 2013

### vela

Staff Emeritus
You could also get to the result you needed by applying some basic information you should know about cross and dot products. In particular, what do you know about the direction of $\vec{a}$ compared to that of $\vec{a}\times\vec{r}$?

12. Feb 5, 2013

### Momentous

I don't want to sound like I'm making excuses, but my Vector Calc skills are minimal at best. Most of my physics classes haven't made much use of applying calculus. We usually don't go into it that much. I know these are things that I should know, but most of it I don't.

Vela, I imagine that you are getting at the fact that a is orthogonal to a x r. That would make a . (a x r) the dot product of two orthogonal vectors, which would be zero. Now that you point that out, it's easy to see that I don't need to switch around anything with a triple product. But still, these are things I'm not going to see until the courses I'm in right now really use Vector Calc on a regular basis.

Again, thank you guys so much for the help. I plan on using these forums a lot when I need the assistance. You guys are all genius's.

13. Feb 5, 2013

### Simon Bridge

Ahhh - we've just been doing this long enough to make all those mistakes already ;)

14. Feb 6, 2013

### tiny-tim

(just got up :zzz:)

another formula you need to learn …

a.(bxc) = c.(axb) = b.(cxa)

(and they're all the volume of a parallepiped, ie a squashed box, of sides with vectors a b and c)