Question

the rates of change in population for two cities are ( p(t)=45 ) for alphaville and ( q(t)=102e^{0.05t} ) for betaburgh, where ( t ) is the number of years since 1990, and ( p ) and ( q ) are measured in people per year in 1990, alphaville had a population of 5000, 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)=square )

Explanation:

Step1: Find the population model for Alphaville

We know that if \(P^{\prime}(t)\) is the rate of change of the population, then \(P(t)=\int P^{\prime}(t)dt + C\).
Since \(P^{\prime}(t) = 45\), then \(\int P^{\prime}(t)dt=\int45dt\).
Using the power - rule of integration \(\int kdt=kt + C\) (where \(k = 45\)), we have \(\int45dt=45t + C\).
In 1990 (\(t = 0\)), \(P(0)=5000\). Substitute \(t = 0\) and \(P(0)=5000\) into \(P(t)=45t + C\).
We get \(P(0)=45\times0 + C\), so \(C = 5000\).

Step2: Find the population model for Betaburgh

We know that \(Q(t)=\int Q^{\prime}(t)dt + C\). Since \(Q^{\prime}(t)=102e^{0.05t}\), then \(\int Q^{\prime}(t)dt=\int102e^{0.05t}dt\).
Using the formula \(\int ae^{bt}dt=\frac{a}{b}e^{bt}+C\) (where \(a = 102\) and \(b=0.05\)), we have \(\int102e^{0.05t}dt=\frac{102}{0.05}e^{0.05t}+C=2040e^{0.05t}+C\).
In 1990 (\(t = 0\)), \(Q(0) = 3500\). Substitute \(t = 0\) and \(Q(0)=3500\) into \(Q(t)=2040e^{0.05t}+C\).
We get \(Q(0)=2040e^{0}+C\), so \(3500=2040 + C\), and \(C=3500 - 2040=1460\).

Answer:

The population model for Alphaville is \(P(t)=45t + 5000\).
The population model for Betaburgh is \(Q(t)=2040e^{0.05t}+1460\).