https://file.aiquickdraw.com/mathbot/file/172918669286066kjiw7m.jpg

To solve this, let's break down the problem using the diagram and the provided equation:

We are asked to prove the following equation:

\[
\frac{\hat{r} \cdot \hat{r}}{\hat{\theta} \cdot \hat{r}} = \frac{d\ln r}{d\theta}
\]

### Step-by-Step Explanation

1. **Radial and Angular Unit Vectors**:
   - In polar coordinates, we define two unit vectors:
     - \(\hat{r}\) is the radial unit vector that points outward from the origin, along the direction of increasing \(r\).
     - \(\hat{\theta}\) is the angular unit vector that points perpendicular to \(\hat{r}\) in the direction of increasing \(\theta\).
   
2. **Dot Products**:
   - The dot product \(\hat{r} \cdot \hat{r}\) is straightforward. Since the dot product of a vector with itself is always 1, we have:
     \[
     \hat{r} \cdot \hat{r} = 1
     \]
   - The dot product \(\hat{\theta} \cdot \hat{r}\) represents the angle between the radial and angular unit vectors. Since these vectors are perpendicular by definition, we have:
     \[
     \hat{\theta} \cdot \hat{r} = 0
     \]
   Therefore, the left-hand side of the equation simplifies to:
     \[
     \frac{1}{0}
     \]
   This appears problematic, but we need to interpret this in the context of the diagram, where the two vectors are not strictly perpendicular at all times but instead have a small angle \( \epsilon \) between them.

3. **Geometric Interpretation**:
   - From the diagram, we see that \(\hat{r}\) and \(\hat{\theta}\) are not exactly perpendicular, but there is a small angle \( \epsilon \) between them as \(r\) changes with \(\theta\).
   - The angle \( \epsilon \) can be expressed in terms of the rate of change of \(r\) with respect to \(\theta\). This is where the logarithmic derivative of \(r\) with respect to \(\theta\) comes in.

4. **Differentiation and the Logarithmic Derivative**:
   - We are now asked to relate the change in \(r\) with respect to \(\theta\), which leads to the right-hand side of the equation:
     \[
     \frac{d\ln r}{d\theta} = \frac{1}{r} \frac{dr}{d\theta}
     \]
   This expression measures the rate of change of the logarithm of \(r\) with respect to the angular coordinate \(\theta\).

### Conclusion

The left-hand side of the equation represents the geometric relation between the radial and angular vectors, and the right-hand side provides the rate of change of the radial distance in terms of \(\theta\). When interpreting the small angle \( \epsilon \), the equation holds as:

\[
\frac{1}{\epsilon} = \frac{1}{r} \frac{dr}{d\theta}
\]

which simplifies to:

\[
\frac{d\ln r}{d\theta}
\]

This completes the proof. Would you like a more detailed breakdown or have any further questions?

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Here are 5 questions related to this:

1. What is the physical meaning of the unit vectors \(\hat{r}\) and \(\hat{\theta}\) in polar coordinates?
2. How does the dot product of two vectors change when they are perpendicular?
3. Why is the derivative of the logarithm useful in problems involving rates of change?
4. Can you apply this concept to other coordinate systems like cylindrical or spherical coordinates?
5. How does the small angle approximation help in simplifying the equation?

**Tip:** Understanding how to relate geometric quantities (like angles and distances) to differential calculus is crucial for working with coordinate systems in physics and engineering.

Show that \(\frac{\hat{r} \cdot \hat{r}}{\hat{\theta} \cdot \hat{r}} = \frac{d \ln r}{d\theta}\).

This page provides a detailed step-by-step solution to proving the relationship between the dot product of unit vectors in polar coordinates and the logarithmic derivative of the radial distance with respect to the angular coordinate.

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