Answer :
The population of the city in 2010 was approximately 311883, assuming that the population increases at a rate proportional to the number of inhabitants present at any time.
Let P(t) be the population at time t. We know that the rate of change of the population is proportional to the population itself. This can be expressed mathematically as:
dP/dt = kP
where k is the constant of proportionality.
We can solve this differential equation by separation of variables:
dP/P = k dt
Integrating both sides, we get:
ln(P) = kt + C
where C is the constant of integration. To find the value of C, we use the initial condition that the population was 200000 in 1990:
ln(200000) = k(1990) + C
To find the value of k, we use the second initial condition that the population was 250000 in 2000:
ln(250000) = k(2000) + C
Subtracting the first equation from the second, we get:
ln(250000/200000) = k(2000-1990)
ln(5/4) = 10k
k = ln(5/4) / 10
Now we can use the equation for P(t) to find the population in 2010:
ln(P) = (ln(5/4)/10)t + C
ln(P) = (1/10)ln(5/4)t + C
ln(P) = (1/10)ln(5/4)(2010-1990) + C
ln(P) = (1/10)ln(5/4)20 + C
ln(P) = 0.0182 + C
P = e^(0.0182 + C)
To find the value of C, we use the initial condition that the population was 200000 in 1990:
200000 = e^(0.0182 + C)
C = ln(200000) - 0.0182
C = 12.209
Therefore, the population in 2010 is:
P = e^(0.0182 + 12.209) ≈ 311882.7
Therefore, the population in 2010 was approximately 311883.
Learn more about calculus here: brainly.com/question/6581270
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