# [SOLVED]derivative dot product cross product

#### dwsmith

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

#### Fernando Revilla

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

#### dwsmith

##### Well-known member
a is acceleration and v is velocity.

#### dwsmith

##### Well-known member
v cross v is zero but why is a cross r zero?

#### Ackbach

##### Indicium Physicus
Staff member
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...