Deriving Cross Vectors: Understanding d/dt[a * (v x r)] = a (v x r)

[happyparticle]

Homework Statement

derivate a cross vectors product

Relevant Equations

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

Hi,
I need to prove that d/dt[a * (v x r)] = a (v x r) if r,v and a denote the position, velocity and the acceleration of a particle.

I see someone else posted the same question, but I didn't understand the answer.

Actually, I don't know how to derivate a cross vectors product. I'm not even sure where to begin.

This is what I did.
d/dt[a * (v x r)] = da/dt * (v x r) + a (d/dt[(v x r)])

I'm not sure at all about what I did, but anyway I'm stuck here. I'm wondering if you guys can tell me if I'm wrong and some tips about d/dt[(v x r)].

 

[archaic]

$$\frac{d}{dt}(\mathbf u\times\mathbf v)=\frac{d\mathbf u}{dt}\times\mathbf v+\mathbf u\times\frac{d\mathbf v}{dt}$$

 

[happyparticle]

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

If the final answer is da/dt * (v x r) then a (dv/dt x r + v x dr/dt) should be = 0, right?

 

[PeroK]

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

If the final answer is da/dt * (v x r) then a (dv/dt x r + v x dr/dt) should be = 0, right?

Are you using $*$ for the scalar product? If so, it shouldn't be disappearing from your expressions.

 

[archaic]

Along with what PeroK has said, bear in mind that $\mathbf u\times\mathbf u=\mathbf 0$, and that $\frac{d\mathbf r}{dt}=\mathbf v$.

 

[etotheipi]

I need to prove that d/dt[a * (v x r)] = a (v x r) if r,v and a denote the position, velocity and the acceleration of a particle.

I second PeroK's question about whether '$*$' denotes the scalar product, however I thought I would also point out that in any case, that result is dimensionally inconsistent, so it can't be right.

Are you sure the result isn't $\dot{\vec{a}} \cdot (\vec{v} \times \vec{r})$, or something?

 

[happyparticle]

Yes, the result should be a ⋅ (v x r), sorry.

I keep working on it. Thanks for you help.

 

[etotheipi]

Yes, the result should be a ⋅ (v x r), sorry.

Surely the question cannot be to prove that $\frac{d}{dt} \left( \vec{a} \cdot (\vec{v} \times \vec{r}) \right) = \vec{a} \cdot (\vec{v} \times \vec{r})$... because that is not true! Does the $\vec{a}$ have a dot on it, $\dot{\vec{a}}$?

 

[happyparticle]

I made few mistakes by typing my question. a (point above) like da/dx ⋅ ( a x r).
Is that make more sense?

 

[etotheipi]

I made few mistakes by typing my question. a (point above) like da/dx ⋅ ( a x r).
Is that make more sense?

Hang on, what's $x$ got to do with it now? And why has the $\vec{v}$ in the cross product changed to $\vec{r}$? I'm inclined to think the question actually asks you to show$$\frac{d}{dt} (\vec{a} \cdot (\vec{v} \times \vec{r})) = \frac{d\vec{a}}{dt} \cdot (\vec{v} \times \vec{r})$$Is that what's written?

 

[happyparticle]

With all the examples and try and error I switched the vectors, but yes it is what you are written. Sorry for the mistakes.

 

[happyparticle]

Alright, I did few more steps.
da/dt ⋅ (v x r) + a(dv/dt x r + v x dr/dt) = da/dt ⋅ (v x r) + a(dv/dt x r + v x v) = da/dt ⋅ (v x r) + a(dv/dt x r) = da/dt ⋅ (v x r) + a(a x r).Am l on the right way?

 

[etotheipi]

Edit: Posted in response to #11; response to #12 is below.

OK, good. It's always a good idea to figure out what we're meant to prove, before we start to prove it . You want to evaluate $$u = \frac{d}{dt} (\vec{a} \cdot (\vec{v} \times \vec{r}))$$Firstly, notice that the derivative of a dot product is$$\frac{d}{dt} (\vec{\alpha} \cdot \vec{\beta}) = \frac{d\vec{\alpha}}{dt}\cdot \vec{\beta} + \vec{\alpha} \cdot \frac{d\vec{\beta}}{dt}$$Can you do this step for the expression in the question, e.g. using that $\vec{\alpha} = \vec{a}$ and $\vec{\beta} = \vec{v} \times \vec{r}$?

Once you've done this, the second term will be a derivative of a cross product. You can evaluate this using the formula @archaic wrote in #2. Finally, you will need to use the fact that the vector triple product is unchanged under cyclic shifts of the arguments, and also that $\vec{\gamma} \times \vec{\gamma} = \vec{0}$ for any vector $\vec{\gamma}$. But we can deal with that when we get there!

 

[etotheipi]

Alright, I did few more steps.
da/dt ⋅ (v x r) + a(dv/dt x r + v x dr/dt) = da/dt ⋅ (v x r) + a(dv/dt x r + v x v) = da/dt ⋅ (v x r) + a(dv/dt x r) = da/dt ⋅ (v x r) + a(a x r).Am l on the right way?

Yes, that looks right, except you're forgetting to write in the scalar product between $\vec{a}$ and $(\vec{a} \times \vec{r})$ in the second term. Insert the dot, and it's all good.

So now consider the term on the right. The triple product is unchanged under cyclic shifts of the operators or the arguments, so $\vec{a} \cdot (\vec{a} \times \vec{r}) = (\vec{a} \times \vec{a}) \cdot \vec{r}$...

 

[happyparticle]

Alright, so basically, the last steps are da/dt ⋅ (v x r) + a ⋅ ( a x r) = da/dt ⋅ (v x r) + (a x a) ⋅ r = da/dt ⋅ (v x r) + 0 ⋅ r = da/dt ⋅ (v x r).

 

[etotheipi]

Yeah, that looks fine . You can also notice that $\vec{a} \times \vec{r}$ is a vector perpendicular to $\vec{a}$, so when we take the dot product with $\vec{a}$ we'll get zero.

N.B. It might be a good idea in the future to learn to typeset in latex, since that will help avoid some of the confusion that occurred earlier on!

 

[happyparticle]

Alright, thanks a lot. Yeah, I'll learn it right now.