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[SOLVED] derivative dot product cross product

dwsmith

Well-known member
Feb 1, 2012
1,673
$$
\frac{d}{dt}[\mathbf{a}\cdot (\mathbf{v}\times\mathbf{r})] = \dot{\mathbf{a}}\cdot (\mathbf{v}\times\mathbf{r}).
$$
How is this true? Shouldn't the derivative affect the cross product as well?
 

Fernando Revilla

Well-known member
MHB Math Helper
Jan 29, 2012
661
$$
\frac{d}{dt}[\mathbf{a}\cdot (\mathbf{v}\times\mathbf{r})] = \dot{\mathbf{a}}\cdot (\mathbf{v}\times\mathbf{r}).
$$
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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

Well-known member
MHB Math Helper
Jan 29, 2012
661
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

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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)
 

dwsmith

Well-known member
Feb 1, 2012
1,673
a is acceleration and v is velocity.
 

dwsmith

Well-known member
Feb 1, 2012
1,673
v cross v is zero but why is a cross r zero?
 

Ackbach

Indicium Physicus
Staff member
Jan 26, 2012
4,192
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...