The population of a town is increasing at a rate of 1.4% annually. The current population is 2000. We need to calculate the instantaneous rate of change of the population 10 years from now.

1. Problem Description

The population of a town is increasing at a rate of 1.4% annually. The current population is

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0. We need to calculate the instantaneous rate of change of the population 10 years from now.

2. Solution Steps

First, we need to find the population after 10 years. The population growth follows the formula:

$P(t) = P_0 * (1 + r)^t$

where $P(t)$ is the population at time $t$, $P_0$ is the initial population, $r$ is the growth rate, and $t$ is the time in years.

In this case, $P_0 = 2000$, $r = 1.4\% = 0.014$, and $t = 10$.

$P(10) = 2000 * (1 + 0.014)^{10} = 2000 * (1.014)^{10}$

$P(10) = 2000 * 1.149435 \approx 2298.87$

Next, we need to find the instantaneous rate of change at $t=10$. The instantaneous rate of change is the derivative of the population function with respect to time.

Since $P(t) = P_0 (1+r)^t$, the derivative is:

$\frac{dP}{dt} = P_0 (1+r)^t \ln(1+r)$

At $t = 10$, we have:

$\frac{dP}{dt}|_{t=10} = 2000 (1.014)^{10} \ln(1.014)$

$\frac{dP}{dt}|_{t=10} = 2000 * 1.149435 * \ln(1.014)$

$\ln(1.014) \approx 0.01390$

$\frac{dP}{dt}|_{t=10} \approx 2000 * 1.149435 * 0.01390 \approx 31.95$

Alternatively, we can approximate the instantaneous rate of change at $t=10$ by multiplying the population at $t=10$ by the annual growth rate:

Rate of change = $P(10) * r = 2298.87 * 0.014 \approx 32.18$

The closest answer choice is 31.

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3. Final Answer

31.95