To solve this problem, we'll follow these steps:

Step 1: Derive the rate of change formula

The rate of change of $V$ with respect to $t$ is given by the derivative of $V(t)$: $V(t) = 7,000e^{-0.6233t}$ The derivative, $V'(t)$, is computed as: $V'(t) = \frac{d}{dt} \left[ 7,000e^{-0.6233t} \right] = 7,000 \cdot (-0.6233) \cdot e^{-0.6233t}$ $V'(t) = -4,363.1 \cdot e^{-0.6233t}$

Step 2: Compute $V'(t)$ at $t = 4$ and $t = 6$

Substitute $t = 4$ and $t = 6$ into $V'(t)$.

1. At $t = 4$: $V'(4) = -4,363.1 \cdot e^{-0.6233 \cdot 4}$

2. At $t = 6$: $V'(6) = -4,363.1 \cdot e^{-0.6233 \cdot 6}$

I'll calculate these values. math

Constants

coefficient = -4363.1

Compute V'(t) at t = 4 and t = 6

t4_rate = coefficient * math.exp(-0.6233 * 4) t6_rate = coefficient * math.exp(-0.6233 * 6)

t4_rate, t6_rateThe rates of change of $V$ with respect to $t$ are as follows:

1. At $t = 4$: $V'(4) \approx -360.59$
2. At $t = 6$: $V'(6) \approx -103.66$

These values indicate the rate at which the value $V$ is decreasing at those times.

Would you like a detailed explanation of the steps, or assistance graphing this function? Here are some related questions to expand on:

1. What does the negative sign in the rate of change indicate in this context?
2. How does the rate of change relate to the graph of the function $V(t)$?
3. Can the function $V(t)$ ever reach a value of zero? Why or why not?
4. How would the rate of change differ if the decay constant ($-0.6233$) was smaller or larger?
5. What are some real-world applications of exponential decay models like this one?

Tip: Always interpret the derivative in context to better understand how the function behaves over time.