Having trouble proving a result from Taylor's Classical Mechanics

Question

I'm working through John Taylor's Classical Mechanics. Question 1.19 asks for proof that

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

Where $\vec{r}$, $\vec{v}$, and $\vec{a}$ denote a particle's position, velocity, and acceleration. I applied the product rule to the LHS, and one of the resulting terms was also $\frac{d\vec{a}}{dt} \cdot (\vec{v} \times \vec{r})$, so I subtracted that from both sides yielding $\vec{a} \cdot \frac{d}{dt} (\vec{v} \times \vec{r}) = 0$. But this is clearly false. Where have I gone wrong? Very confused. Any help appreciated.

Answer 1

To derive this, we need the product rule for cross products: https://en.wikipedia.org/wiki/Cross_product. We also need the product rule for dot products (which you already used as you mentioned)
I'll denote vectors with boldface letters.

So we have: \begin{align}\label{eq:1}\tag{eq:1} \frac{d}{dt} \left[\mathbf{a} \cdot (\mathbf{v} \times \mathbf{r}) \right]= \frac{d\mathbf{a}}{dt}\cdot (\mathbf{v} \times \mathbf{r}) + \underbrace{\frac{d (\mathbf{v} \times \mathbf{r})}{dt}\cdot \mathbf{a}}_{\Large{(*)}} \end{align} Now apply the product rule for cross-products to $(*)$: \begin{align}\label{eq:2} \tag{*} \frac{d}{dt}[\mathbf{v} \times \mathbf{r}] &= \frac{d \mathbf{v}}{dt} \times\mathbf{r} \:\:+ \:\:\mathbf{v} \times \frac{d \mathbf{r}}{dt}\\ &= \mathbf{a} \times \mathbf{r} + \mathbf{v} \times \mathbf{v} \end{align} as per definition of velocity and acceleration. Now by the properties of the cross product $\mathbf{v} \times \mathbf{v} = \mathbf{0}$.

plugging the result of (\ref{eq:2}) into \ref{eq:1} we get: \begin{align*} \frac{d}{dt} \left[\mathbf{a} \cdot (\mathbf{v} \times \mathbf{r}) \right] &= \frac{d \mathbf{a}}{dt} \cdot (\mathbf{v} \times \mathbf{r}) + (\mathbf{a} \times \mathbf{r}) \cdot \mathbf{a} \end{align*} Now recall that the cross product is a vector perpendicular to both vectors that are in the cross product and that the dot product is equal to $0$ when two vectors are perpendicular, meaning that $(\mathbf{a} \times \mathbf{r}) \cdot \mathbf{a} = 0$ resulting in the equation: \begin{align} \frac{d}{dt}[\mathbf{a} \cdot (\mathbf{v} \times \mathbf{r})] = \frac{d \mathbf{a}}{dt} \cdot (\mathbf{v} \times \mathbf{r}) \end{align} I hope this helped!

Answer 2

$\vec{a} \cdot \frac{d}{dt} (\vec{v} \times \vec{r}) = 0$

This is in fact true.

Expanding $\frac{d}{dt} (\vec{v} \times \vec{r})$ gives $(\frac{d\vec{v}}{dt} \times \vec{r}) + (\vec{v} \times \frac{d\vec{r}}{dt})$.

But $\frac{d\vec{v}}{dt} = \vec{a}$ and $\frac{d\vec{r}}{dt} = \vec{v}$.

So the full expansion is $\vec{a} \cdot (\vec{a} \times \vec{r} + \vec{v} \times \vec{v})$. The first term in the parentheses is perpendicular to $\vec{a}$, and the second is zero, as $\vec{v}$ is parallel to itself. So the expression as a whole is zero.