To show that $\mathbf{r} \times \mathbf{v} = r^2 \frac{d\theta}{dt} \hat{z}$, where $\mathbf{r}(t)$ and $\mathbf{v}(t)$ are defined in the xy-plane, let's proceed with the given definitions:

1. Position Vector:
   $\mathbf{r}(t) = (x(t), y(t), 0) = (r(t)\cos\theta(t), r(t)\sin\theta(t), 0)$.

2. Velocity Vector:
   $\mathbf{v}(t) = \frac{d\mathbf{r}}{dt} = \left( \frac{dx}{dt}, \frac{dy}{dt}, 0 \right)$.

We start by finding $\mathbf{v}(t)$.

1. Calculate the Velocity Vector $\mathbf{v}(t)$:

$\mathbf{v}(t) = \frac{d}{dt}(r(t)\cos\theta(t), r(t)\sin\theta(t), 0).$ Using the product rule for differentiation:

• $\frac{dx}{dt} = \frac{d}{dt} (r(t)\cos\theta(t)) = \frac{dr}{dt} \cos\theta - r \sin\theta \frac{d\theta}{dt}$.

• $\frac{dy}{dt} = \frac{d}{dt} (r(t)\sin\theta(t)) = \frac{dr}{dt} \sin\theta + r \cos\theta \frac{d\theta}{dt}$.

Thus, the velocity vector is: $\mathbf{v}(t) = \left( \frac{dr}{dt} \cos\theta - r \sin\theta \frac{d\theta}{dt}, \frac{dr}{dt} \sin\theta + r \cos\theta \frac{d\theta}{dt}, 0 \right).$

2. Calculate the Cross Product $\mathbf{r} \times \mathbf{v}$:

The cross product of $\mathbf{r}(t)$ and $\mathbf{v}(t)$ is given by:

\begin{vmatrix} \hat{i} & \hat{j} & \hat{k} \\ r\cos\theta & r\sin\theta & 0 \\ \frac{dr}{dt} \cos\theta - r \sin\theta \frac{d\theta}{dt} & \frac{dr}{dt} \sin\theta + r \cos\theta \frac{d\theta}{dt} & 0 \end{vmatrix}.$$ Evaluating this determinant, we get: $$\mathbf{r} \times \mathbf{v} = \hat{k} \left[ r\cos\theta \left( \frac{dr}{dt} \sin\theta + r \cos\theta \frac{d\theta}{dt} \right) - r\sin\theta \left( \frac{dr}{dt} \cos\theta - r \sin\theta \frac{d\theta}{dt} \right) \right].$$ Simplifying the expression inside the brackets: $$\mathbf{r} \times \mathbf{v} = \hat{k} \left[ r\frac{dr}{dt} \cos\theta \sin\theta + r^2 \cos^2 \theta \frac{d\theta}{dt} - r\frac{dr}{dt} \cos\theta \sin\theta + r^2 \sin^2 \theta \frac{d\theta}{dt} \right].$$ Further simplification leads to: $$\mathbf{r} \times \mathbf{v} = \hat{k} \left[ r^2 \cos^2 \theta \frac{d\theta}{dt} + r^2 \sin^2 \theta \frac{d\theta}{dt} \right].$$ Since $$\cos^2\theta + \sin^2\theta = 1$$, we obtain: $$\mathbf{r} \times \mathbf{v} = \hat{k} \left[ r^2 \frac{d\theta}{dt} \right].$$ ### Final Result: $$\mathbf{r} \times \mathbf{v} = r^2 \frac{d\theta}{dt} \hat{z}.$$ This shows the desired result. ### Do you have any questions or need further clarification? Here are 5 related questions: 1. What does the magnitude of $$\mathbf{r} \times \mathbf{v}$$ represent in orbital mechanics? 2. How does the cross product $$\mathbf{r} \times \mathbf{v}$$ relate to angular momentum? 3. How would the expression change if the motion were not restricted to the xy-plane? 4. What is the physical significance of the term $$r^2 \frac{d\theta}{dt}$$? 5. How do we interpret the derivative $$\frac{d\theta}{dt}$$ in terms of orbital motion? **Tip:** Understanding cross products in vector calculus is crucial for solving problems involving angular momentum and rotational dynamics.