Problem 4 - Biology Concepts and Applications

Question

A population of worms is growing exponentially in a compost heap. Thirty days ago there were 400 worms and now there are \(800 .\) How many worms will there be thirty days from now, assuming conditions remain constant? a. 1,200 b. 1,600 c. 3,200 d. 6,400

Step-by-Step Solution

Answer

The population will be 1,600 worms in 30 days more.

Step 1: Understanding Exponential Growth

Exponential growth means that the population is increasing by a consistent percentage over equal time intervals. This can be described by the equation for exponential growth: \[ P(t) = P_0 e^{kt} \] where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( k \) is the growth rate, and \( t \) is the time.

Step 2: Determine the Growth Rate

Given that the population of worms increased from 400 to 800 in 30 days, we can use this equation:\[ 800 = 400 e^{30k} \]Solving for \( k \), divide both sides by 400:\[ 2 = e^{30k} \]Take the natural logarithm of both sides:\[ \ln(2) = 30k \]Solve for \( k \):\[ k = \frac{\ln(2)}{30} \].

Step 3: Calculate the Population in 60 Days

Using \( k = \frac{\ln(2)}{30} \) and plugging into the exponential growth equation:\[ P(60) = 400 e^{60 \times \frac{\ln(2)}{30}} \]This simplifies to:\[ P(60) = 400 e^{2 \ln(2)} \]Using the property \( e^{\ln(a^b)} = a^b \):\[ P(60) = 400 \times 2^2 \]Simplifying gives:\[ P(60) = 400 \times 4 = 1600 \].

Step 4: Verify With Initial Conditions

Check the accuracy with initial data: Initially: \( P(0) = 400 \), \( P(30) = 400 \times 2 = 800 \), verifies since \( 30 \) days yield double.Thus in 60 days, population further doubles: \( 1600 \).

Key Concepts

Population Dynamics

Population dynamics is the study of how and why the number of individuals in a population changes over time. It considers various factors that affect these changes, such as birth rates, death rates, immigration, and emigration. In the context of the worm population in the compost heap, the population dynamics involve examining how the worm population size changes when conditions such as resources and environment remain constant. The worms breed continuously and, under ideal conditions, experience exponential growth. This growth pattern is typical for populations with abundant resources, where the population size increases rapidly without constraints. Understanding population dynamics is crucial for predicting future population sizes and for making decisions regarding resource management and conservation. In exponential growth scenarios, we see a swift increase in numbers, showcasing a unique pattern where the rate of growth of the population is proportional to its current size. This insight allows us to accurately model and predict future population sizes given stable environmental conditions.

Growth Rate Calculation

Calculating the growth rate is an essential part of understanding exponential growth. In our exercise, the growth rate helps determine how quickly the worm population is doubling over a specific time frame. The growth rate is denoted by the variable "\(k\)" in the exponential growth formula \( P(t) = P_0 e^{kt} \). To determine \(k\), we use known population sizes at specific times. In the example, the population of worms increased from 400 to 800 in 30 days.
By setting up the equation:

• \(800 = 400 e^{30k}\)

We solve for \(k\) using logarithmic rearrangement:

• Divide both sides by 400: \(2 = e^{30k}\)
• Take the natural logarithm: \(\ln(2) = 30k\)
• Solve for \(k\): \(k = \frac{\ln(2)}{30}\)

This calculation gives us a specific numerical value for the growth rate, which is crucial for predicting future population scenarios under consistent conditions.

Exponential Growth Equation

The exponential growth equation is a powerful tool for predicting the future size of a population. It captures the essence of populations growing at a constant rate over time. The general form of the equation is:

• \[ P(t) = P_0 e^{kt} \]

Where:

• \(P(t)\) is the population size at time \(t\)
• \(P_0\) is the initial population size
• \(k\) is the growth rate
• \(t\) is the time period

In our example, with an initial population of 400 worms, the equation helps predict how many worms there will be 60 days later:

• First, determine the growth rate \(k\) which we've calculated as \(\frac{\ln(2)}{30}\)
• Then, use this in the equation: \(P(60) = 400 e^{60 \times \frac{\ln(2)}{30}}\)
• Simplify by recognizing the property \(e^{2 \ln(2)} = 2^2\), so: \(P(60) = 400 \times 4 = 1600\)

This demonstrates how the population doubles every 30 days, and forecasts a population of 1,600 worms at the 60-day mark. The exponential growth formula is invaluable for quickly calculating population sizes over time when growth rates and initial sizes are known.