Question

Assume that the population of a certain city increases at a rate proportional to the number of inhabitants at any time. If the population doubles in 40 years, in how many years will it triple?

Short answer

The population will triple in approximately \(t' \approx 62.33\) years.

Step-by-step solution

Understand the Variables

Let P(t) be the population of the city at time t (in years). The growth rate, dP/dt, is proportional to the population, P(t).

Formulate the Proportional Growth Differential Equation

We can represent the proportionality between the growth rate and the population as follows: \[ \frac{dP}{dt} = k P(t) \] where k is the proportionality constant.

Separate Variables and Integrate

Now we need to solve this differential equation. Separate the variables and integrate both sides: \[\int \frac{1}{P} dP = \int k dt\] When integrating both sides we get: \[\ln(P(t)) = kt+C\]

Solve for P(t)

To solve for P(t), we take the exponential of both sides: \[P(t) = e^{kt+C}\] Now, let P₀ = P(0) be the initial population: \[P₀ = e^{k\cdot0+C}\] \[P₀ = e^C\] So our equation for P(t) now becomes: \[P(t) = P₀e^{kt}\]

Find k value using the given population doubling

It's given that the population doubles in 40 years. We can use this information to find the value of k: \[2P₀ = P₀e^{40k}\] Divide both sides by P₀: \[2 = e^{40k}\] Now solve for k: \[k = \frac{\ln(2)}{40}\]

Find the time for population to triple

We need to find the time, t', when the population triples: \[3P₀ = P₀e^{t'\cdot \frac{\ln(2)}{40}}\] Divide both sides by P₀: \[3 = e^{t'\cdot \frac{\ln(2)}{40}}\] Take the natural logarithm of both sides: \[\ln(3) = t' \cdot \frac{\ln(2)}{40}\] Finally, solve for t': \[t' = \frac{40\ln(3)}{\ln(2)}\] Calculating this, we get approximately: \[t' \approx 62.33 \text{ years}\] So, it will take approximately 62.33 years for the population to triple.