Problem 52 - Applied Mathematics: For the Managerial, Life, and Social Sciences

Question

In an endeavor to curb population growth in a Southeast Asian island state, the government has decided to launch an extensive propaganda campaign. Without curbs, the government expects the rate of population growth to have been $$ 60 e^{0.02 t} $$ thousand people/year, \(t\) yr from now, over the next 5 yr. However, successful implementation of the proposed campaign is expected to result in a population growth rate of $$ -t^{2}+60 $$ thousand people/year, \(t\) yr from now, over the next 5 yr. Assuming that the campaign is mounted, how many fewer people will there be in that country 5 yr from now than there would have been if no curbs had been imposed?

Step-by-Step Solution

Answer

The difference in population growth after 5 years due to the imposed curbs is approximately 314,970 fewer people.

Step 1: Integrate population growth rate without curbs

First, find the total population growth without curbs by integrating the given function: \( \displaystyle\int_{0}^{5} 60e^{0.02t} dt \) Here, we use the integral of exponential function, which is: \( \displaystyle\int e^{at} dt = \frac{1}{a} e^{at} + C \) Hence, \( \displaystyle\int_{0}^{5} 60e^{0.02t} dt = 60 \int_{0}^{5} e^{0.02t} dt = 60\cdot\left(\frac{1}{0.02}\right)\left(e^{0.02\cdot 5}-e^{0.02\cdot 0}\right) = 3000\left(e^{0.1}-1\right) \)

Step 2: Integrate population growth rate with curbs

Next, find the total population growth with curbs by integrating the given function: \( \displaystyle\int_{0}^{5} (-t^{2}+60) dt \) Here, we integrate using the power rule: \( \displaystyle\int (-t^{2}+60) dt = -\frac{1}{3}t^{3}+60t+C \) Hence, \( \displaystyle\int_{0}^{5} (-t^{2}+60) dt = \left(-\frac{1}{3}(5)^{3}+60(5)\right) - \left(-\frac{1}{3}(0)^{3}+60(0)\right) = -\frac{125}{3}+300 \)

Step 3: Calculate the difference in population growth

Finally, find the difference between population growth without curbs and population growth with curbs: Difference = Total population growth without curbs - Total population growth with curbs Difference = \( 3000\left(e^{0.1}-1\right) - \left(-\frac{125}{3}+300\right) \) Calculating the difference numerically, we get: Difference ≈ 314.97 thousand people Therefore, there will be approximately 314,970 fewer people in the country, 5 years from now, if the curbs are imposed.

Key Concepts

Exponential Growth

Exponential growth refers to a type of growth where the population increases at a rate proportional to its current size. This type of growth is common in populations with ample resources and no constraints. In mathematical terms, an exponential growth model is generally expressed as \( P(t) = P_0 e^{rt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( r \) is the growth rate, and \( e \) is Euler's number. This model shows that as time (\( t\)) increases, the population grows rapidly.

In the context of our exercise, the rate of population increase without interventions was given by \( 60 e^{0.02t} \), where 60 represents thousands of people per year, and 0.02 is the growth rate. This means that initially, the population would grow rapidly, and over time, the increase becomes significantly larger as the years progress. One important aspect to note is that exponential growth does not take into account limitations like resources, space, or other natural constraints, which is why it can't continue indefinitely in real-world scenarios.

Integration Techniques

Integration is a fundamental technique used in calculus to find the accumulated value, such as area under a curve, or total change given a rate of change. It's like the "opposite" of differentiation.

1. **Exponential Functions:** When integrating exponential functions, you often encounter formulas like \( \int e^{at} dt = \frac{1}{a} e^{at} + C \). In our problem, this technique was applied to find the total population growth without curbs.

2. **Power Rule of Integration:** This rule is used when you have polynomials in the form \( -t^2 + 60 \). The power rule states that \( \int x^n dx = \frac{1}{n+1} x^{n+1} + C \). For the polynomial component, each term is integrated separately, so \( -t^2 \) becomes \( -\frac{1}{3}t^3 \), and the constant 60 becomes \( 60t \). This approach helps compute the total population growth with curbs implemented.

Learning integration techniques enhances the ability to solve various problems pertaining to areas, populations, or any quantity that accumulates over time.

Definite Integrals

Definite integrals are used to calculate the exact area under a curve between two limits, which corresponds to a total change. Unlike indefinite integrals, which include a constant \( C \), definite integrals are evaluated over a specific interval and result in a single value.

In the exercise provided, definite integrals are employed to determine total population growth over a 5-year period. The calculations included endpoints: from \( t = 0 \) to \( t = 5 \). This means finding the "net change" over this interval. The expression \( \int_{0}^{5} \) sets the specific bounds (from \( t = 0 \) to \( t = 5 \)).

The result of these integrations gives us two quantities: one without curbs (using the exponential model) and one with curbs (using a polynomial). By comparing these, the impact of curbing measures becomes clear. Notably, the subtraction of these results depicts the difference in population growth attributable to the proposed campaign. Understanding definite integrals is key to evaluating real-world situations where the exact sum or total is required.