Question

A culture of bacteria has a population of 150 cells when it is first observed. The population doubles every 12 hr, which means its population is governed by the function \(p(t)=150 \cdot 2^{t / 12},\) where \(t\) is the number of hours after the first observation. a. Verify that \(p(0)=150,\) as claimed. b. Show that the population doubles every \(12 \mathrm{hr}\), as claimed. c. What is the population 4 days after the first observation? d. How long does it take the population to triple in size? e. How long does it take the population to reach \(10,000 ?\)

Short answer

Answer: 150 Part b: Does the population double every 12 hours? Answer: Yes Part c: What is the population of the bacteria culture after 4 days? Answer: 38,400 Part d: How long does it take for the population to triple? Answer: Approximately 18.95 hours Part e: How long does it take for the population to reach 10,000? Answer: Approximately 59.79 hours

Step-by-step solution

Part a: Verify initial population

At the time when the culture is first observed (t=0), the population is claimed to be 150. We can verify this by substituting t=0 into the function and see if it matches the given population: \(p(0)=150 \cdot 2^{0 / 12}=150 \cdot 2^0=150 \cdot 1=150\) So, the claim is correct that \(p(0)=150\).

Part b: Show the population doubles every 12 hours

We need to show that if we substitute t = 12 in the function, the population equals to twice the initial population (which is 150). So, \(p(12)=150 \cdot 2^{12 / 12}=150 \cdot 2^1=150 \cdot 2=300\) Since 300 is twice 150, the claim is correct that the population doubles every 12 hours.

Part c: Population after 4 days

We need to find the population 4 days after the first observation. There are 24 hours in a day, so after 4 days, the time elapsed is 4 * 24 = 96 hours. Substitute t = 96 into the function: \(p(96)=150 \cdot 2^{96 / 12}=150 \cdot 2^8=150 \cdot 256=38400\) The population after 4 days is 38,400.

Part d: Time for population to triple

We need to find the time it takes for the population to triple, which is \(150*3=450\). Let's solve for t in the function \(p(t)=450\): \(450=150 \cdot 2^{t / 12}\) Divide both sides by 150, we get: \(3=2^{t / 12}\) Now, take the logarithm of both sides with base 2: \(\log_{2}(3)=\frac{t}{12}\) Now, multiply both sides by 12 to isolate t, and we get: \(t=12\log_{2}(3) \approx 18.95 \mathrm{hr}\) So, it takes approximately 18.95 hours for the population to triple.

Part e: Time for population to reach 10,000

We need to find the time it takes for the population to reach 10,000. Let's solve for t in the function \(p(t) = 10000\): \(10000=150\cdot 2^{\frac{t}{12}}\) Divide both sides by 150, we get: \(\frac{10000}{150}=2^{\frac{t}{12}}\) Now, take the logarithm of both sides with base 2: \(\log_{2}\left(\frac{10000}{150}\right)=\frac{t}{12}\) Now, multiply both sides by 12 to isolate t, and we get: \(t=12\log_{2}\left(\frac{10000}{150}\right) \approx 59.79 \mathrm{hr}\) So, it takes approximately 59.79 hours for the population to reach 10,000.

Key concepts

Population Growth Model

Understanding population growth is crucial for many fields, especially biology and environmental science. A typical population growth model represents how populations change over time under ideal conditions. One classic example is the exponential growth model, used to describe a scenario where resources are unlimited, and the population can grow without restraint. In this model, the population size at any time is given by the formula

\( p(t) = p_0 \cdot b^{\frac{t}{d}} \)

where

• \( p(t) \) is the population at time \( t \),
• \( p_0 \) is the initial population size,
• \( b \) is the base of the exponential function, often \( e \) (Euler's number) or 2 for doubling,
• and \( d \) is the time taken for the population to grow by a certain factor (e.g., doubling time).

The exercise shows bacteria growth modeled by exponential increase where the population doubles every 12 hours, accurately reflecting how some bacteria colonies rapidly expand in optimal conditions.

Doubling Time

The concept of doubling time is integral to understanding exponential growth. It refers to the amount of time required for a quantity to double in size or value at a constant growth rate. In the example of our bacterial culture, the doubling time is 12 hours. Mathematical expressions for doubling time arise from the exponential growth model with

\( 2 = b^{\frac{t_d}{d}} \)

solving for \( t_d \) (the doubling time), where \( b \) is the base of the growth (here, b=2).

Calculating Doubling Time Using Logarithms

To find the doubling time for any exponential growth, we can rearrange the equation and apply logarithmic functions:

\( t_d = d \cdot \frac{\log{2}}{\log{b}} \)

Notice that when \( b \) is 2, then the doubling time \( t_d \) is simply \( d \), as \( \log{2} \) of base 2 is 1.

Logarithmic Functions

Expanding on the previous concepts, logarithmic functions are the inverses of exponential functions. They are indispensable when working with exponential growth models, especially when determining the time it takes for a population to reach a certain size. Logarithms allow us to solve for the exponent in an equation, as seen in the exercise when calculating the time for the bacteria population to triple and to hit 10,000.

Applying Logarithms to Exponential Equations

When faced with an equation like

\( b^{x} = y \)

taking the logarithm of both sides (preferably with a like base of \( b \)) yields the solution for the unknown exponent:

\( x = \log_{b}(y) \)

In our practice problem, we used \( \log_2 \) to solve for the time \( t \), demonstrating the practical application of logarithms in biology-related calculations.

Calculus in Biology

Calculus may not be the first thing that comes to mind when thinking about biology, but it's incredibly useful in modeling dynamic life processes, like population growth. Calculus in biology covers a range of applications, from determining rates of change to optimizing biological functions.

Exponential Growth and Derivatives

In the context of exponential growth, understanding the rate at which a population grows over time is a derivative concept from calculus. For our bacteria population described by

\( p(t) = 150 \cdot 2^{t / 12} \)

the derivative with respect to time, \( p'(t) \), tells us the growth rate at any moment. This expression of calculus aligns deeply with the study of life, aiding in predicting outcomes of population dynamics, and informing environmental policies and healthcare practices.