Problem 39 - College Algebra

Question

Use this scenario: A biologist recorded a count of 360 bacteria present in a culture after 5 minutes and 1,000 bacteria present after 20 minutes. Rounding to six significant digits, write an exponential equation representing this situation. To the nearest minute, how long did it take the population to double?

Step-by-Step Solution

Answer

The exponential equation is \( P(t) = 267.699 e^{0.0663076t} \). The population doubles in approximately 10 minutes.

Step 1: Understand the Problem

We are given two data points of bacteria population counts: 360 bacteria after 5 minutes and 1000 bacteria after 20 minutes. We need to find an exponential equation that models this growth and determine the time it takes for the population to double.

Step 2: Formulate the Exponential Growth Model

The general form of an exponential growth equation is \( P(t) = P_0 e^{kt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, \( k \) is the growth rate, and \( t \) is time in minutes. We need to use the given data to find values for \( P_0 \) and \( k \).

Step 3: Solve for the Initial Population \( P_0 \)

Using \( P(5) = 360 \), substitute \( t = 5 \) into the equation: \( 360 = P_0 e^{5k} \). Similarly, use \( P(20) = 1000 \): \( 1000 = P_0 e^{20k} \).

Step 4: Eliminate \( P_0 \) by Dividing the Equations

Divide the second equation by the first to eliminate \( P_0 \): \( \frac{1000}{360} = \frac{P_0 e^{20k}}{P_0 e^{5k}} \). Simplify to obtain \( \frac{1000}{360} = e^{15k} \).

Step 5: Solve for the Growth Rate \( k \)

To solve for \( k \), take the natural logarithm of both sides: \( \ln(\frac{1000}{360}) = 15k \). Calculate \( k = \frac{\ln(\frac{1000}{360})}{15} \).

Step 6: Calculate the Growth Rate Numerically

First compute \( \frac{1000}{360} \approx 2.77778 \). Then, find \( k = \frac{\ln(2.77778)}{15} \approx 0.0663076 \).

Step 7: Find the Initial Population \( P_0 \)

Substitute \( k \) back into one of the equations to find \( P_0 \). Using \( 360 = P_0 e^{5 \times 0.0663076} \), solve for \( P_0 \): \( P_0 \approx 360 / e^{0.331538} \approx 267.699 \.\).

Step 8: Write the Exponential Model

The exponential growth equation is \( P(t) = 267.699 e^{0.0663076t} \), rounded to six significant digits.

Step 9: Determine the Doubling Time

To find the doubling time, solve \( 2P_0 = P_0 e^{kt_d} \). Simplifying yields \( 2 = e^{kt_d} \) or \( \ln(2) = kt_d \). Solve \( t_d = \frac{\ln(2)}{k} \).

Step 10: Calculate the Doubling Time

Using \( k = 0.0663076 \), \( t_d = \frac{\ln(2)}{0.0663076} \approx 10.4558 \) minutes. Round to the nearest minute: \( t_d \approx 10 \) minutes.

Key Concepts

Growth Rate

In the context of exponential growth, the *growth rate* is a crucial parameter that tells us how quickly a population increases over time. It is often denoted by the symbol \( k \) in the exponential equation. Growth rate is typically determined by observing the population at two different times and using these observations to calculate how fast the population is changing.

To calculate the growth rate, one common method is to use the exponential growth model:

• Given: \( P(t) = P_0 e^{kt} \), where \( P(t) \) is the population at time \( t \), \( P_0 \) is the initial population, and \( k \) is the growth rate.
• Set up equations based on known values of the population at different times, like \( P(5) = 360 \) and \( P(20) = 1000 \).
• Use logarithms to isolate and solve for \( k \).

In this exercise, by setting up and solving the equation with given data on bacteria counts, the growth rate \( k \) is found to be approximately \( 0.0663076 \). Having an accurate growth rate allows predictions and assessments of the future population size.

Doubling Time

The concept of *doubling time* is related to how long it takes for a quantity experiencing exponential growth to double in size. It's a helpful measure for quickly understanding how rapid the growth is. Doubling time is dependent on the growth rate \( k \), and can be calculated mathematically.

To determine doubling time, use the equation:

• Starting with the equation for doubling: \( 2P_0 = P_0 e^{kt_d} \).
• This simplifies to \( 2 = e^{kt_d} \).
• Taking the natural log gives \( \ln(2) = kt_d \).
• Finally, solve for \( t_d \) using \( t_d = \frac{\ln(2)}{k} \).

In this exercise, with \( k = 0.0663076 \), the bacteria culture takes approximately 10.46 minutes to double in size when rounded to the nearest minute, this results in a doubling time of approximately 10 minutes.

Exponential Equation

An *exponential equation* effectively models scenarios involving rapid increases or decreases over time. In biological contexts, such as the growth of bacteria, exponential equations help predict population changes. The standard form of an exponential growth equation is \( P(t) = P_0 e^{kt} \), capturing key parameters: initial population size \( P_0 \) and growth rate \( k \).

Developing an accurate exponential equation involves several steps:

• Start by interpreting the conditions given, such as two known population values after different time intervals.
• Use these conditions to create equations based on \( P(t) = P_0 e^{kt} \).
• Manipulate these equations to solve for unknowns, often requiring logarithmic transformations.
• Combine results to form a complete exponential equation.

In the provided example, the equation derived is \( P(t) = 267.699 e^{0.0663076t} \). This model can predict the number of bacteria at any given time, given the initial conditions described.