To find the rate at which the population was changing on January 1, 1981, we need to determine the derivative of the population function $P = 22000(0.91)^t$ with respect to $t$ and evaluate it at $t = 21$ (since 1981 is 21 years after 1960).

Step 1: Differentiate the Population Function

The function given is: $P = 22000(0.91)^t$ To find the rate of change, we differentiate $P$ with respect to $t$.

Let: $P = 22000 \cdot e^{t \ln(0.91)}$ Using the chain rule, the derivative $\frac{dP}{dt}$ is: $\frac{dP}{dt} = 22000 \cdot e^{t \ln(0.91)} \cdot \ln(0.91)$ Simplifying, we can substitute back: $\frac{dP}{dt} = 22000 \cdot (0.91)^t \cdot \ln(0.91)$

Step 2: Substitute $t = 21$

Now, plug $t = 21$ into the derivative to find the rate of change at that time.

$\frac{dP}{dt} \Big|_{t=21} = 22000 \cdot (0.91)^{21} \cdot \ln(0.91)$

Step 3: Calculate the Value

1. Calculate $(0.91)^{21}$.
2. Calculate $\ln(0.91)$ (approximately -0.0943).
3. Multiply these values together with 22000.

Let me calculate this.The population of Slim Chance was changing at a rate of approximately $-286.32$ people per year on January 1, 1981. This negative rate indicates that the population was decreasing at that time.

Would you like further explanation on any part of this solution? Here are some additional questions you might find interesting:

1. How would the rate of change differ for other years, such as 1970 or 1990?
2. How is the rate of change affected by different growth factors?
3. Can we determine when the population would theoretically reach zero?
4. How would we interpret the result if the growth factor were greater than 1?
5. How does the natural logarithm affect growth or decay rates in exponential models?

Tip: In exponential decay models like this, a negative rate of change over time indicates that the quantity (population, in this case) is decreasing as expected for each subsequent time period.