The rates of change in population for two cities are $P^{\prime }(t)=42$ for Alphaville and $Q^{\prime }(t)=102e^{0.05t}$ 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 . Answer parts a) through c).
a) Determine the population models for both cities.
The population model for Alphaville is $P(t)=42t+6000$.
The population model for Betaburgh is $Q(t)=2040e^{0.05t}+1460$.
b) What were the populations of Alphaville and Betaburgh, to the nearest hundred, in 2000 ?
The population of Alphaville in 2000 was 6400 people.
(Round to the nearest hundred as needed.)
The population of Betaburgh in 2000 was 4800 people.
(Round to the nearest hundred as needed.)
c) Sketch the graph of each city's population model and estimate the year in which the two cities have the same population. Choose the correct graph below.
A.
B.
c.
D.

Step 1 ：Given the rates of change in population for two cities are $P^{\prime }(t)=42$ for Alphaville and $Q^{\prime }(t)=102e^{0.05t}$ 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, $P(t)=\int P^{\prime }(t)dt=\int 42dt=42t+C_1$. 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, $Q(t)=\int Q^{\prime }(t)dt=\int 102e^{0.05t}dt=2040e^{0.05t}+C_2$. 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 $Q(t)=2040e^{0.05t}+1460$.

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, $P(10)=42\ast 10+6000=6420$, which rounds to 6400 to the nearest hundred. For Betaburgh, $Q(10)=2040e^{0.05\ast 10}+1460\approx 4796.42$, 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.