Derivative dot product cross product

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Discussion Overview

The discussion revolves around the differentiation of the dot product involving a cross product, specifically the expression $$\frac{d}{dt}[\mathbf{a}\cdot (\mathbf{v}\times\mathbf{r})]$$. Participants explore the implications of applying the derivative to this expression and the relationships between acceleration, velocity, and position vectors in the context of physics.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants question why the derivative does not appear to affect the cross product in the expression $$\frac{d}{dt}[\mathbf{a}\cdot (\mathbf{v}\times\mathbf{r})]$$.
  • Others provide a more complete formula for the derivative, suggesting it should include terms for the derivatives of both the velocity and position vectors: $$\dot{\mathbf{a}}\cdot (\mathbf{v}\times\mathbf{r})+{\mathbf{a}}\cdot (\dot{\mathbf{v}}\times\mathbf{r})+\mathbf{a}\cdot (\mathbf{v}\times\dot{\mathbf{r}})$$.
  • One participant notes that if $$\dot{\mathbf{v}}=\mathbf{a}$$ and $$\dot{\mathbf{r}}=\mathbf{v}$$, this leads to further implications in the context of the discussion.
  • There is a humorous remark about the interpretation of symbols, suggesting that context is crucial for understanding the equations presented.
  • Participants discuss the properties of cross products, specifically that $$\mathbf{v} \times \mathbf{v}$$ is zero, and question why $$\mathbf{a} \times \mathbf{r}$$ would also be considered zero.
  • One participant clarifies that $$\mathbf{a} \times \mathbf{r}$$ is not zero, but the result of the cross product is perpendicular to $$\mathbf{a}$$, which affects the outcome of the dot product.

Areas of Agreement / Disagreement

Participants express differing views on the application of the derivative to the cross product and the interpretation of the variables involved. There is no consensus on the implications of the derivative or the conditions under which the equations hold true.

Contextual Notes

Participants acknowledge that the definitions of the variables and the context of the problem are important, but these aspects remain somewhat ambiguous and unresolved in the discussion.

Dustinsfl
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$$
\frac{d}{dt}[\mathbf{a}\cdot (\mathbf{v}\times\mathbf{r})] = \dot{\mathbf{a}}\cdot (\mathbf{v}\times\mathbf{r}).
$$
How is this true? Shouldn't the derivative affect the cross product as well?
 
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dwsmith said:
$$
\frac{d}{dt}[\mathbf{a}\cdot (\mathbf{v}\times\mathbf{r})] = \dot{\mathbf{a}}\cdot (\mathbf{v}\times\mathbf{r}).
$$
How is this true? Shouldn't the derivative affect the cross product as well?

The formula is

$
\dfrac{d}{dt}[\mathbf{a}\cdot (\mathbf{v}\times\mathbf{r})] = \dot{\mathbf{a}}\cdot (\mathbf{v}\times\mathbf{r})+{\mathbf{a}}\cdot (\dot{\mathbf{v}}\times\mathbf{r})+\mathbf{a}\cdot (\mathbf{v}\times\dot{\mathbf{r}})
$
 
Fernando Revilla said:
The formula is

$
\dfrac{d}{dt}[\mathbf{a}\cdot (\mathbf{v}\times\mathbf{r})] = \dot{\mathbf{a}}\cdot (\mathbf{v}\times\mathbf{r})+{\mathbf{a}}\cdot (\dot{\mathbf{v}}\times\mathbf{r})+\mathbf{a}\cdot (\mathbf{v}\times\dot{\mathbf{r}})
$

But since $\dot{\mathbf{v}}=\mathbf{a}$ and $\dot{\mathbf{r}}=\mathbf{v}$, you get...
 
Ackbach said:
But since $\dot{\mathbf{v}}=\mathbf{a}$ and $\dot{\mathbf{r}}=\mathbf{v}$, you get...

Well, I don't suppose $\mathbf{a}$ means acceleration, etc in the same way that I don't suppose that a matrix $U\in\mathbb{C}^{n\times n}$ is a unitary matrix. :)
 
Fernando Revilla said:
Well, I don't suppose $\mathbf{a}$ means acceleration, etc in the same way that I don't suppose that a matrix $U\in\mathbb{C}^{n\times n}$ is a unitary matrix. :)

Well, certainly not a priori. I'm guessing that there's some context for this problem, and that in that context, the equations I wrote before are true. It was a guess. Think of it as thinking more like a physicist than a mathematician. (Evilgrin)
 
a is acceleration and v is velocity.
 
v cross v is zero but why is a cross r zero?
 
dwsmith said:
v cross v is zero but why is a cross r zero?

It isn't. But the result is perpendicular to $\mathbf{a}$, so when you do the outer dot product...
 

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