3. Population Growth The population of a suburb grows at a rate proportional to the population. Suppose the population doubles in size from 4000 to 8000 in an 18 -month period and continues at the current rate of growth. (a) Write a differential equation that models the population \( P \) at time \( t \) in months. (b) Find the general solution to the differential equation. (c) Find the particular solution to the differential equation with the initial condition \( P(0)=4000 \). (d) What will the population be in 4 years \( [t=48] \) ?

The growth of a population can be described using the classic exponential growth model, where the rate of change of the population is proportional to its current size. This is often seen in real-world situations like bacteria growth or urban expansion. To model this mathematically, we can write the differential equation as \( \frac{dP}{dt} = kP \), where \( k \) is the growth constant. Given that the population doubles from 4000 to 8000 in 18 months, we can derive \( k \) and solve the equation to find the specific growth function for the suburb. Now, let’s tackle the specifics of the problem step by step. ### (a) The differential equation modeling the population \( P \) at time \( t \) in months is: \[ \frac{dP}{dt} = kP \] ### (b) The general solution of this differential equation is: \[ P(t) = Ce^{kt} \] where \( C \) is a constant determined by initial conditions. ### (c) To find the particular solution using the initial condition \( P(0) = 4000 \): 1. Substitute \( t = 0 \) into the general solution: \[ P(0) = Ce^{k \cdot 0} = C = 4000 \] 2. The particular solution then becomes: \[ P(t) = 4000e^{kt} \] To determine \( k \), we use the information about the population doubling: - At \( t = 18 \): \[ 8000 = 4000e^{18k} \] Dividing both sides by 4000: \[ 2 = e^{18k} \] Taking the natural logarithm gives: \[ 18k = \ln(2) \quad \Rightarrow \quad k = \frac{\ln(2)}{18} \] So, the particular solution is: \[ P(t) = 4000e^{\frac{\ln(2)}{18}t} \] ### (d) To find the population in 4 years (48 months), we evaluate \( P(48) \): \[ P(48) = 4000e^{\frac{\ln(2)}{18}(48)} \] Calculating \( \frac{48}{18} = \frac{8}{3} \): \[ P(48) = 4000e^{\ln(2) \cdot \frac{8}{3}} = 4000 \cdot 2^{\frac{8}{3}} \approx 4000 \cdot 10.0794 \approx 40317.58 \] So the estimated population after 4 years will be approximately **40,318**. It’s astounding to see how quickly populations can grow when rates are exponential!