91Ó°ÊÓ

When dealing with calculus problems, derivatives are a fundamental concept used to understand how a quantity changes in relation to another. In this exercise, we're focused on how the force, denoted by \( F \), between two atoms changes as the distance \( r \) between them changes. This requires finding the derivative of \( F \) with respect to \( r \), which helps us understand the rate of change of force as the distance changes.

To compute this derivative, we need to apply basic rules of differentiation. The force equation given is \( F = -\frac{A}{r^2} + \frac{B}{r^3} \). Each term is differentiated individually:

• The term \( -\frac{A}{r^2} \) differentiates to \( \frac{2A}{r^3} \).
• The term \( \frac{B}{r^3} \) differentiates to \( -\frac{3B}{r^4} \).

Combining these, the derivative \( \frac{dF}{dr} = \frac{2A}{r^3} - \frac{3B}{r^4} \) gives us the rate at which force changes with respect to distance.

The rate of change is crucial for understanding how one quantity changes over another, which is precisely what the derivative tells us. In this problem, part (a) sought the rate of change of force as a function of distance, while part (b) asked how force changes with time given that distance itself is changing.

In part (b), we use the chain rule to find the rate of change of force with respect to time. This is necessary because when distance \( r \) changes over time, the force \( F \) also changes. Applying the chain rule involves multiplying the derivative of force with respect to distance \( \frac{dF}{dr} \) by the rate at which distance changes over time \( \frac{dr}{dt} \):

• Since \( \frac{dr}{dt} = k \), the rate of distance change is \( k \),
• The rate of force change over time is now \( \frac{dF}{dt} = \left( \frac{2A}{r^3} - \frac{3B}{r^4} \right) \cdot k \).

Understanding the rate of change helps in modeling how forces act, especially when dynamic processes are involved.

When working with calculus in physical contexts, understanding the units of the result is just as critical as performing the math. In this problem, the importance of unit analysis helps to make sense of results in real-world applications.

For part (a), the derivative \( \frac{dF}{dr} \) has units that reflect the change in force per unit distance. Force is generally measured in newtons (N), and distance in meters (m), resulting in units of newtons per meter (N/m). These units represent how many newtons of force change for every meter of distance change.

In part (b), the derivative \( \frac{dF}{dt} \) incorporates time. It has units derived from multiplying the units of \( \frac{dF}{dr} \) (N/m) by the units of \( \frac{dr}{dt} \) (m/s). Thus, \( \frac{dF}{dt} \) will have units of newtons per second (N/s). This indicates how much force changes per second over time, integrating both spatial and temporal understanding into your calculations. Keeping track of these units helps ensure consistency and correctness when applying calculus to physics problems, offering insights into force dynamics across varying states.