To solve this problem, let's analyze the given mathematical context.

Given:

1. Unit vector $\hat{r}$ is in the direction of vector $\mathbf{r}$.
2. The problem is asking for $\left| \hat{r} \times \frac{d\hat{r}}{dt} \right|$.

Conceptual Understanding:

1. A unit vector $\hat{r}$ is defined as: $\hat{r} = \frac{\mathbf{r}}{|\mathbf{r}|},$ where $|\mathbf{r}|$ is the magnitude of $\mathbf{r}$.

2. Differentiating $\hat{r}$ with respect to time $t$ gives: $\frac{d\hat{r}}{dt}.$

3. The vector cross product $\hat{r} \times \frac{d\hat{r}}{dt}$ is evaluated:

   • Since $\hat{r} \cdot \hat{r} = 1$, differentiating both sides gives: $\hat{r} \cdot \frac{d\hat{r}}{dt} = 0.$ This indicates that $\frac{d\hat{r}}{dt}$ is perpendicular to $\hat{r}$.

4. The magnitude of $\hat{r} \times \frac{d\hat{r}}{dt}$ is thus equal to the magnitude of $\frac{d\hat{r}}{dt}$, which means: $\left| \hat{r} \times \frac{d\hat{r}}{dt} \right| = \left| \frac{d\hat{r}}{dt} \right|.$

Answer:

The correct option is: $\boxed{\text{(b)} \ \left| \frac{d\hat{r}}{dt} \right|}$

Let me know if you'd like further clarification or a more detailed breakdown! 😊

Here are 5 related questions:

1. What is the derivative of a unit vector in general?
2. How does the cross product of two vectors relate to their magnitudes and the sine of the angle between them?
3. Why is the derivative of $\hat{r}$ perpendicular to $\hat{r}$?
4. What happens if $\mathbf{r}$ is a constant vector?
5. How does this concept relate to circular motion?

Tip: For a unit vector, always remember $\hat{r} \cdot \hat{r} = 1$, and differentiating this identity helps solve many problems!