MHB Derivative dot product cross product

[Dustinsfl]

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

 

[Fernando Revilla]

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

 

[Ackbach]

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...

 

[Fernando Revilla]

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. :)

 

[Ackbach]

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)

 

[Dustinsfl]

a is acceleration and v is velocity.

 

[Dustinsfl]

v cross v is zero but why is a cross r zero?

 

[Ackbach]

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...