Problem 1 - Advanced Engineering Mathematics

Question

Let \(P=P(t)\) be the population at time \(t,\) and \(P_{0}\) the initial population. From \(d P / d t=k P\) we obtain \(P=P_{0} e^{k t}\) \(\operatorname{Using} P(5)=2 P_{0}\) we find \(k=\frac{1}{5} \ln 2\) and \(P=P_{0} e^{(\ln 2) t / 5} .\) Setting \(P(t)=3 P_{0}\) we have \(3=e^{(\ln 2) t / 5},\) so \(\ln 3=\frac{(\ln 2) t}{5} \quad\) and \(\quad t=\frac{5 \ln 3}{\ln 2} \approx 7.9\) years. Setting \(P(t)=4 P_{0}\) we have \(4=e^{(\ln 2) t / 5},\) so \(\ln 4=\frac{(\ln 2) t}{5} \quad\) and \(\quad t \approx 10\) years

Step-by-Step Solution

Answer

It takes approximately 7.9 years for the population to triple and 10 years to quadruple.

Step 1: Understand the Population Growth Equation

We start with the differential equation \( \frac{dP}{dt} = kP \), which represents the exponential growth model of a population. Solving this yields the formula \( P = P_0 e^{kt} \), where \( P_0 \) is the initial population.

Step 2: Use Given Condition to Find k

We use the condition \( P(5) = 2P_0 \) to find the constant \( k \). Substituting into the population formula gives:\[ 2P_0 = P_0 e^{5k} \]Canceling \( P_0 \) from both sides gives \( 2 = e^{5k} \). Taking the natural log of both sides, we get:\[ \ln 2 = 5k \]Solving for \( k \), we find:\[ k = \frac{1}{5} \ln 2 \]

Step 3: Find General Population Formula with k

Substitute \( k = \frac{1}{5} \ln 2 \) back into the equation \( P = P_0 e^{kt} \):\[ P = P_0 e^{(\ln 2)t/5} \]

Step 4: Solve for Time when Population is Triple

To find when the population triples, set \( P(t) = 3P_0 \):\[ 3 = e^{(\ln 2)t/5} \]Taking the natural log of both sides gives:\[ \ln 3 = \frac{(\ln 2)t}{5} \]Solving for \( t \), we get:\[ t = \frac{5 \ln 3}{\ln 2} \approx 7.9 \text{ years} \]

Step 5: Solve for Time when Population is Quadruple

To find when the population quadruples, set \( P(t) = 4P_0 \):\[ 4 = e^{(\ln 2)t/5} \]Taking the natural log of both sides gives:\[ \ln 4 = \frac{(\ln 2)t}{5} \]Solving for \( t \), we find:\[ t = \frac{5 \ln 4}{\ln 2} \approx 10 \text{ years} \]

Key Concepts

Population Growth

Population growth is a key concept in understanding how the size of a population changes over time. At its simplest, population growth occurs when the birth rate exceeds the death rate. This difference leads to an increase in population size. Exponential growth is a specific type of population growth where the rate of growth is proportional to the current population. This means that as the population gets larger, it grows faster. In mathematical terms, this is represented by the equation: \[ \frac{dP}{dt} = kP \] Here, \( P \) represents the population size at time \( t \), and \( k \) is a constant growth rate. The solution to this differential equation gives us the exponential growth formula: \[ P = P_0 e^{kt} \] where \( P_0 \) is the initial population size. In real-life scenarios, exponential growth can occur in populations with abundant resources and no limiting factors. However, it's important to note that in nature, growth eventually slows or stops due to constraints such as limited resources.

Differential Equations

A differential equation is a mathematical equation that involves one or more functions and their derivatives. They are used to describe a wide variety of phenomena in physics, engineering, biology, and economics. In the context of population growth, the differential equation \( \frac{dP}{dt} = kP \) is a simple yet powerful model that represents the rate of change of the population. - **Solving Differential Equations:** Solving a differential equation involves finding the function \( P(t) \) that satisfies the equation for a given initial condition. In our example, this gives us the equation \( P = P_0 e^{kt} \), which describes how the population evolves over time. - **Initial Conditions:** These conditions are necessary to find a particular solution to a differential equation. For population growth, an initial population size \( P_0 \) at time \( t = 0 \) is used to determine the constant of integration.Differential equations are fundamental in modeling dynamic systems, where variables change with respect to time.

Natural Logarithms

Natural logarithms, denoted as \( \ln \), are logarithms with the base \( e \), where \( e \approx 2.71828 \). The natural logarithm of a number is its logarithm to the base \( e \). - **Why Natural Logarithms?** They provide a way to deal with exponential relationships, which are common in growth processes. In population growth, natural logarithms can be used to solve equations like \( e^{kt} = a \) by converting them into \( \ln a = kt \). This makes it easier to isolate variables and understand relationships.For example, take \( e^{5k} = 2 \). By applying the natural logarithm, you get: \[ \ln 2 = 5k \]This transformation simplifies solving for \( k \) and illustrates how logarithms help solve exponential equations in population growth models. - **Properties of Natural Logarithms:** They have properties that are useful in simplifying complex growth equations, such as \( \ln(ab) = \ln a + \ln b \) and \( \ln(a^b) = b \cdot \ln a \). These properties make calculations more manageable when working with exponential growth equations.