Problem 53 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

SURVIVAL AND RENEWAL In Exercises 51 through 54, an initial population \(P_{0}\) is given along with a renewal rate \(R(t)\) and a survival function \(S(t)\). In each case, use the given information to find the population at the end of the indicated term \(T\). \(P_{0}=100,000 ; R(t)=90 e^{0.1 t} ; S(t)=e^{-0.2 t} ; t\) in years; term \(T=10\) years

Step-by-Step Solution

Answer

14,305.08

Step 1 Title - Initial Population and Functions

Identify the given initial population, renewal rate, and survival function. Initial Population: \(P_{0} = 100,000\) Renewal Rate: \(R(t) = 90 e^{0.1 t}\) Survival Function: \(S(t) = e^{-0.2 t}\) Term: \(T = 10\) years.

Step 2 Title - Formula for Population

Use the given formula for population at the end of term \(T\): \[ P(T) = P_{0} \times S(T) + \ \ \int_{0}^{T} R(t) \times S(T-t) \, dt \]

Step 3 Title - Calculate Survival of Initial Population

Compute the survival of the initial population after term \(T\): \[ P_{0} \times S(T) = 100,000 \times e^{-0.2 \times 10} = 100,000 \times e^{-2} \]

Step 4 Title - Simplify Initial Population Survival

Simplify the result: \[ 100,000 \times e^{-2} \e^{-2} \approx 0.1353 \ 100,000 \times 0.1353 = 13,530 \]

Step 5 Title - Compute the Integral

Calculate the integral part of the formula: \[ \ \int_{0}^{10} 90 e^{0.1 t} \times e^{-0.2(10-t)} \, dt \]

Step 6 Title - Simplify the Integral

Simplify the integral expression: \[ = \int_{0}^{10} 90 e^{0.1 t} \times e^{-2 + 0.2 t} \, dt \ = 90 e^{-2} \int_{0}^{10} e^{0.3 t} \, dt \]

Step 7 Title - Compute the Integral Solution

Solve the integral: \[ = 90 e^{-2} \left[ \frac{e^{0.3 t}}{0.3} \right]_{0}^{10} \]

Step 8 Title - Evaluate the Integral at Upper and Lower Limits

Evaluate at the limits: \[ = 90 e^{-2} \left( \frac{e^{3} - 1}{0.3} \right) \]

Step 9 Title - Simplify Calculations

Simplify further: \[ = 300 e^{-2} (e^{3} - 1) \ e^{3} \approx 20.0855 \ e^{-2} \approx 0.1353 \ \approx 300 \times 0.1353 (20.0855 - 1) \approx 300 \times 0.1353 \times 19.0855 \]

Step 10 Title - Final Evaluation

Complete the calculation: \[ \approx 300 \times 2.5836 \approx 775.08 \]

Step 11 Title - Sum Both Components

Add the contributions from the initial population survival and the integral: \[ 13,530 + 775.08 = 14,305.08 \]

Key Concepts

Initial Population

In population dynamics, understanding the initial population is crucial. The initial population, denoted as \(P_0\), represents the number of individuals or entities at the start of our observation period. For example, in our exercise, the initial population is given as 100,000. This baseline helps in modeling how populations grow or shrink over time.

When we estimate future populations, we need to account for other factors like birth rates, death rates, and environmental influences. The initial population serves as the starting point from which these factors will cause deviations.

Renewal Rate

The renewal rate \(R(t)\) is another fundamental concept. It represents the rate at which new individuals are added to the population at any given time \(t\). This rate can vary depending on many factors such as birth rates, immigration, or even technological advances making it easier for a population to grow.

In our exercise, the renewal rate is given by the function \(R(t) = 90 e^{0.1 t}\). This means at any time \(t\), new individuals are being added to the population at a rate that grows exponentially over time. Understanding how this rate impacts population growth helps in making accurate predictions.

Survival Function

The survival function \(S(t)\) indicates the probability that an individual from the initial population or from the added new individuals will survive to time \(t\). This function often decreases over time as mortality is inevitable for all living beings.

In our case, the survival function is \(S(t) = e^{-0.2t}\). This implies the probability of survival decreases exponentially over time. After 10 years, we calculate this as \(S(10) = e^{-2} \approx 0.1353\), meaning only about 13.53% of the original and new individuals survive.

Integral Calculus

Integral calculus allows us to accurately quantify the total contribution to the population over a period of time, accounting for continuous change. The integral in our formula sums up the contributions of every small time segment to give a whole picture.

The full formula is given by:
\[ P(T) = P_{0} \times S(T) + \int_{0}^{T} R(t) \times S(T-t) \, dt \]

The term \(P_{0} \times S(T)\) captures the survival of the initial population after time \(T\). The integral term \(\int_{0}^{T} R(t) \times S(T-t) \, dt\) computes the contribution of new individuals added over the interval \([0, T]\) and adjusts for their survival probability. By evaluating this integral, we accumulate the new additions weighted by their chances of surviving up to time \(T\). This comprehensive picture aids in making informed decisions or predictions about population trajectories.