Identity involving the vectors for position, velocity and aceleration

[fer Mnaj]

Homework Statement

If r and v are both explicit functions of time, show that:
d/dt [r × (v × r)] = r´^2a + (r · v)v − (v^2 + r · a)r (All of them are vectors)

Relevant Equations

v=dr/dt a=dv/dt

any tip to start?

 

[kuruman]

Expand the triple cross product. Look up the relevant identity for that and start taking derivatives.

 

[fer Mnaj]

What happens when I've got the derivative of aceleration? and how can I plug in the dot product?? what did you mean when you say relevant identity?So far

 

[kuruman]

What happens when I've got the derivative of aceleration? and how can I plug in the dot product?? what did you mean when you say relevant identity?

It is not possible to get the time derivative of the acceleration when you take the first derivative of
r x (v x r).
Do some web research on "vector triple product". There is an identity that allows you write what you have in terms of scalar products. It's otherwise known as the BAC-CAB rule. That's what I mean by relevant identity.

 

[fer Mnaj]

i checked it,thanks, but don't know if i did wrong when I canceled r· v because they were orthogonals, I just got the first part of the proof

 

[ehild]

i checked it,thanks, but don't know if i did wrong when I canceled r· v because they were orthogonals,

$\vec r$ and $\vec v$ are not orthogonal in general.

 

[PeroK]

Homework Statement:: If r and v are both explicit functions of time, show that:
d/dt [r × (v × r)] = r´^2a + (r · v)v − (v^2 + r · a)r (All of them are vectors)
Relevant Equations:: v=dr/dt a=dv/dt

any tip to start?

Another way to tackle questions with the cross product is do each component separately. In this case:
$$(\vec r \times (\vec v \times \vec r))_x = y(\vec v \times \vec r)_z - z(\vec v \times \vec r)_y$$
Then you can expand the RHS again, take its derivative and show that it is equal to the $x$ component of that complicated expression involving $\vec a, \vec v$ and $\vec r$.

In general, if the identity holds for the x-component then, by symmetry, it must hold for the y and z components as well. So, generally, you only need to show the identify for the x-component and you are done.

 

[fer Mnaj]

so far, now it looks more like the expresion I want to get. But what can I do with (r· v)r. Have I done something wrong? At the end it should be (v2+r· a)r, instead of what I've got (v2+r· a)v

 

[PeroK]

You need to check your differentiation of $r^2 = \vec r \cdot \vec r$.

 

[fer Mnaj]

whats wrong? the derivative of r2 =2r+dr/dt

 

[kuruman]

The derivative of r² is 2r(dr/dt).

 

[PeroK]

whats wrong? the derivative of r2 =2r+dr/dt

I'm not sure what that is, but it's simpler to treat it as a dot product.

 

[PeroK]

The derivative of r² is 2r(dr/dt).

Yes, but $\frac{dr}{dt}$ is more complicated than $\frac{dr^2}{dt}$ was in the first place!

 

[fer Mnaj]

ók... now what should I do?

 

[PeroK]

ók... now what should I do?

Fix the differentiation of $r^2$.

 

[fer Mnaj]

besides that

 

[kuruman]

Yes, but $\frac{dr}{dt}$ is more complicated than $\frac{dr^2}{dt}$ was in the first place!

I guess in this case going the $\vec{r} \cdot \vec r$ route is better.

 

[PeroK]

besides that

Just keep things expressed as dot products and differentiate them all.

 

[fer Mnaj]

how can I differentiate d/dt ((v2+r⋅ a)r)? I did it as a product but that gets the derivative of aceleration and I don't feel its that way

 

[kuruman]

Part of $\dfrac{d}{dt} \left[ (\vec r \cdot \vec v)\vec r \right]$ is $(\vec r \cdot \vec a)\vec r$, There are no more derivatives with respect to time for this part. Why do you feel you need to take the derivative of the acceleration? I begin to suspect that you are unsure about how to take derivatives when you have a product.

 

[fer Mnaj]

d/dt(r2v) its a product, isn't it? in that case, the derivative would be 2rv+r2a, wouldn't it??

and I referred to d/dt ((v2+r⋅ a)r), which I did it as a the derivative of a product and that's my biggest doubt. Did I do something wrong before?

 

[kuruman]

d/dt(r2v) its a product, isn't it? in that case, the derivative would be 2rv+r2a, wouldn't it??

Yes it would if you want to write that way, but it is not recommended for this problem. Anyway, if you choose to write this way and get 2rv+r2a, you are done taking derivatives of this part. Why are asking

how can I differentiate d/dt ((v2+r⋅ a)r)?

?

 

[fer Mnaj]

Part of $\dfrac{d}{dt} \left[ (\vec r \cdot \vec v)\vec r \right]$ is $(\vec r \cdot \vec a)\vec r$

How did you get that?

 

[fer Mnaj]

I saw it as a triple product, probably there I did something wrong

 

[fer Mnaj]

Now my biggest doubt is: How can I derivate a dot product that's being multiplied by a vector?? take for example (r⋅ r)v

 

[kuruman]

Now my biggest doubt is: How can I derivate a dot product that's being multiplied by a vector?? take for example (r⋅ r)v

You apply the rule of multiplication as always: you distribute the time derivative operator,
$$\frac{d}{dt}[(\vec r \cdot \vec r)\vec v]=\left(\frac{d\vec r}{dt} \cdot \vec r\right)\vec v+ \left(\vec r \cdot \frac{d\vec r}{dt}\right)\vec v+(\vec r \cdot \vec r)\dfrac{d \vec v}{dt}.$$

 

[PeroK]

You apply the rule of multiplication as always: you distribute the time derivative operator,
$$\frac{d}{dt}[(\vec r \cdot \vec r)\vec v]=\left(\frac{d\vec r}{dt} \cdot \vec r\right)\vec v+ \left(\vec r \cdot \frac{d\vec r}{dt}\right)\vec v+(\vec r \cdot \vec r)\dfrac{d \vec v}{dt}.$$

... which is a special case of:
$$\frac{d}{dt}(a \vec b) = \frac{da}{dt}\vec b + a\frac{d \vec b}{dt}$$
and
$$\frac{d}{dt}(\vec b \cdot \vec c) = \frac{d \vec b}{dt} \cdot\vec c + \vec b \cdot \frac{d \vec c}{dt}$$
For any vectors $\vec b, \vec c$ and scalar $a$.