We now let ( ) ( ), ( ),0 ( ) cos ( ), ( )sin ( ),0r t x t y t r t t r t t

  , so that the orbital plane is the xy-plane.
Expand the cross product to show that
2
r v d
r dt


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:

\[
\mathbf{r} \times \mathbf{v} = 
\begin{vmatrix}
\hat{i} & \hat{j} & \hat{k} \\
r\cos\theta & r\sin\theta & 0 \\
\frac{dr}{dt} \cos\theta - r \sin\theta \frac{d\theta}{dt} & \frac{dr}{dt} \sin\theta + r \cos\theta \frac{d\theta}{dt} & 0
\end{vmatrix}.
\]

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.

We now let r(t) = (x(t), y(t), 0) = (r(t)cos(θ(t)), r(t)sin(θ(t)), 0), so that the orbital plane is the xy-plane. Expand the cross product to show that r × v = r^2 (dθ/dt) ẑ.

Explore how to expand the cross product of r(t) and v(t) in the xy-plane for orbital motion. This problem uses vector calculus and key trigonometric identities to show that r × v = r^2 (dθ/dt) ẑ. A detailed step-by-step solution is provided, suitable for undergraduate students.

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