The discussion focuses on the derivative of the dot product involving a vector and the cross product of two other vectors, specifically the equation $$\frac{d}{dt}[\mathbf{a}\cdot (\mathbf{v}\times\mathbf{r})] = \dot{\mathbf{a}}\cdot (\mathbf{v}\times\mathbf{r}) + {\mathbf{a}}\cdot (\dot{\mathbf{v}}\times\mathbf{r}) + \mathbf{a}\cdot (\mathbf{v}\times\dot{\mathbf{r}})$$. Participants clarify that while the derivative does not directly affect the cross product, it does influence the overall expression through the product rule. The discussion emphasizes the importance of context in interpreting the variables, particularly that $\dot{\mathbf{v}}$ represents acceleration and $\dot{\mathbf{r}}$ represents velocity.
PREREQUISITESStudents and professionals in physics, mathematics, and engineering who are working with vector calculus and need to understand the differentiation of vector products.
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?
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}})
$
Ackbach said:But since $\dot{\mathbf{v}}=\mathbf{a}$ and $\dot{\mathbf{r}}=\mathbf{v}$, you get...
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. :)
dwsmith said:v cross v is zero but why is a cross r zero?