The given problem asks to evaluate $\mathbf{r} \times \frac{d\mathbf{r}}{dt}$, where:

$\mathbf{r} = \cos(\omega t) \mathbf{a} + \sin(\omega t) \mathbf{b},$

and $\mathbf{a}$ and $\mathbf{b}$ are constant vectors.

Step 1: Calculate $\frac{d\mathbf{r}}{dt}$

$\frac{d\mathbf{r}}{dt} = \frac{d}{dt} \left[\cos(\omega t) \mathbf{a} + \sin(\omega t) \mathbf{b}\right].$

Using the derivatives of $\cos(\omega t)$ and $\sin(\omega t)$: $\frac{d\mathbf{r}}{dt} = -\omega \sin(\omega t) \mathbf{a} + \omega \cos(\omega t) \mathbf{b}.$

Step 2: Compute $\mathbf{r} \times \frac{d\mathbf{r}}{dt}$

Now substitute $\mathbf{r}$ and $\frac{d\mathbf{r}}{dt}$ into the cross product formula:

[ \mathbf{r} \times \frac{d\mathbf{r}}{dt} = \left[\cos(\omega t) \mathbf{a} + \sin(\omega t) \mathbf{b}\right] \times \left[-\omega \sin(\omega t) \mathbf{a} + \omega \cos(\omega t) \mathbf{b}\right]. ]

Using the distributive property of the cross product: [ \mathbf{r} \times \frac{d\mathbf{r}}{dt} = \cos(\omega t) \left[-\omega \sin(\omega t) (\mathbf{a} \times \mathbf{a}) + \omega \cos(\omega t) (\mathbf{a} \times \mathbf{b})\right]

• \sin(\omega t) \left[-\omega \sin(\omega t) (\mathbf{b} \times \mathbf{a}) + \omega \cos(\omega t) (\mathbf{b} \times \mathbf{b})\right]. ]

Step 3: Simplify using vector properties

1. The cross product of a vector with itself is zero: $\mathbf{a} \times \mathbf{a} = \mathbf{0}, \quad \mathbf{b} \times \mathbf{b} = \mathbf{0}.$

2. The cross product is anti-commutative: $\mathbf{b} \times \mathbf{a} = - (\mathbf{a} \times \mathbf{b}).$

Substituting these into the expression: