Step 1 :Given the rates of change in population for two cities are \(P^{\prime}(t)=42\) for Alphaville and \(Q^{\prime}(t)=102 e^{0.05 t}\) for Betaburgh, where \(t\) is the number of years since 1990, and \(P^{\prime}\) and \(Q^{\prime}\) are measured in people per year. In 1990, Alphaville had a population of 6000, and Betaburgh had a population of 3500.
Step 2 :a) To determine the population models for both cities, we integrate the rates of change. For Alphaville, \(P(t) = \int P^{\prime}(t) dt = \int 42 dt = 42t + C_1\). Given that the population in 1990 (when \(t=0\)) was 6000, we find \(C_1 = 6000\). So, the population model for Alphaville is \(P(t) = 42t + 6000\).
Step 3 :For Betaburgh, \(Q(t) = \int Q^{\prime}(t) dt = \int 102 e^{0.05 t} dt = 2040 e^{0.05 t} + C_2\). Given that the population in 1990 (when \(t=0\)) was 3500, we find \(C_2 = 3500 - 2040 = 1460\). So, the population model for Betaburgh is \(Q(t) = 2040 e^{0.05 t} + 1460\).
Step 4 :b) To find the populations of Alphaville and Betaburgh in 2000, we substitute \(t = 2000 - 1990 = 10\) into the population models. For Alphaville, \(P(10) = 42*10 + 6000 = 6420\), which rounds to 6400 to the nearest hundred. For Betaburgh, \(Q(10) = 2040 e^{0.05 * 10} + 1460 \approx 4796.42\), which rounds to 4800 to the nearest hundred.
Step 5 :c) To sketch the graph of each city's population model and estimate the year in which the two cities have the same population, we set \(P(t) = Q(t)\) and solve for \(t\). This requires numerical methods or graphing software, and is not included in this solution.