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).
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 5500 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 citys population model and estimate the year in which the two cities have the same population. choose the correct graph below.
the two cities have the same population in the year ( square ). (round to the nearest year as needed.)

Explanation:

Step1: Find the population model for Alphaville

Since \(P^{\prime}(t) = 45\), integrating with respect to \(t\) gives \(P(t)=\int45dt = 45t + C\). Using the initial condition \(P(0)=5000\) (when \(t = 0\) in 1990), we substitute \(t = 0\) and \(P(0)=5000\) into \(P(t)=45t + C\). Then \(5000=45\times0 + C\), so \(C = 5000\). Thus, \(P(t)=45t+5000\).

Step2: Set \(P(t)=Q(t)\)

We know \(P(t)=45t + 5000\) and \(Q(t)=2040e^{0.05t}+1460\). Set \(45t + 5000=2040e^{0.05t}+1460\), which simplifies to \(45t+3540 = 2040e^{0.05t}\).

We can use a numerical method (such as the Newton - Raphson method or trial - and - error). Let's try some values of \(t\):

When \(t = 10\):
\(P(10)=45\times10 + 5000=5450\)
\(Q(10)=2040e^{0.05\times10}+1460=2040e^{0.5}+1460\approx2040\times1.6487+1460\approx3363+1460 = 4823\)

When \(t = 15\):
\(P(15)=45\times15+5000=675 + 5000=5675\)
\(Q(15)=2040e^{0.05\times15}+1460=2040e^{0.75}+1460\approx2040\times2.117+1460\approx4319+1460=5779\)

When \(t = 14\):
\(P(14)=45\times14 + 5000=630+5000 = 5630\)
\(Q(14)=2040e^{0.05\times14}+1460=2040e^{0.7}+1460\approx2040\times2.0138+1460\approx4108+1460 = 5568\)

When \(t = 13\):
\(P(13)=45\times13+5000=585 + 5000=5585\)
\(Q(13)=2040e^{0.05\times13}+1460=2040e^{0.65}+1460\approx2040\times1.9155+1460\approx3908+1460=5368\)

Step3: Use a more accurate method (Newton - Raphson)

Let \(f(t)=2040e^{0.05t}-45t - 3540\)
\(f^{\prime}(t)=2040\times0.05e^{0.05t}-45=102e^{0.05t}-45\)

Starting with an initial guess \(t_0 = 14\)
\(f(14)=2040e^{0.7}-45\times14-3540\approx2040\times2.0138-630 - 3540\approx4108-630 - 3540=-62\)
\(f^{\prime}(14)=102e^{0.7}-45\approx102\times2.0138-45\approx205.41 - 45=160.41\)

\(t_1=t_0-\frac{f(t_0)}{f^{\prime}(t_0)}=14+\frac{62}{160.41}\approx14.39\)

\(f(14.39)=2040e^{0.05\times14.39}-45\times14.39-3540\)
\(=2040e^{0.7195}-647.55 - 3540\approx2040\times2.053-647.55 - 3540\approx4188 - 647.55 - 3540\approx0.45\)
\(f^{\prime}(14.39)=102e^{0.7195}-45\approx102\times2.053-45\approx209.41 - 45 = 164.41\)

\(t_2=t_1-\frac{f(t_1)}{f^{\prime}(t_1)}=14.39-\frac{0.45}{164.41}\approx14.39\)

Answer:

The two cities have the same population in the year \(1990 + 14=2004\)