Question

The number of a certain type of bacteria increases continuously at a rate proportional to the number present. There are 150 present at a given time and 450 present 5 hours later. (a) How many will there be 10 hours after the initial time? (b) How long will it take for the population to double? (c) Does the answer to part (b) depend on the starting time? Explain your reasoning.

Short answer

(a) There will be 1350 bacteria 10 hours after the initial time. (b) It will take approximately \(\frac{5}{\ln(3)} \ln(2)\) hours for the population to double. (c) The doubling time does not depend on the starting time, because the growth rate is constant.

Step-by-step solution

Find the growth rate k

Firstly, use the information that we have 150 bacteria present at a given time and 450 present 5 hours later. substitute these values into the exponential growth formula. So, we get the equation \(450 = 150e^{5k}\). Solve for \(k\). After rearranging, we get \( e^{5k} = 3\), and by taking a natural logarithm on both sides, we get \(k = \frac{1}{5} \ln(3)\).

Calculate the number of bacteria after 10 hours

Substitute the initial amount \(N_0 = 150\), the time \(t = 10\) hours, and the growth rate \(k = \frac{1}{5} \ln(3)\) into the formula \(N = N_0 e^{kt}\). This gives us \(N = 150 e^{10 \cdot \frac{1}{5} \ln(3)} = 150 e^{2\ln(3)} = 150 \cdot 3^2 = 1350\). So, there will be 1350 bacteria after 10 hours.

Calculate how long it will take for the population to double

Setting \(N = 2N_0\) in the formula and solve for \(t\). We get \(2 = e^{kt}\), so \(t = \frac{1}{k} \ln(2)\). Substituting the calculated \(k\), we get \(t = \frac{5}{\ln(3)} \ln(2)\).

Discuss whether the doubling time depends on the starting time

The answer to part (b) does not depend on the starting time because the growth rate is constant. No matter when we start, it will take the same amount of time for the bacteria population to double.

Key concepts

Bacteria Population

Bacteria populations often grow exponentially, which means they increase rapidly over time. Imagine you start with a small number of bacteria, like the 150 bacteria mentioned in the original exercise. Bacteria reproduce by dividing, which means in optimal conditions, each bacterium can become two. In our example, after five hours, the 150 bacteria grow to become 450. This demonstrates the typical exponential growth pattern, where the rate of growth is proportional to the current population size.

Exponential growth is quite common in nature, especially with bacteria due to their rapid division. Such growth can be modeled using the equation \( N = N_0 e^{kt} \), where \( N \) is the population size at time \( t \), \( N_0 \) is the initial population size, \( k \) is the growth rate, and \( e \) is the mathematical constant approximately equal to 2.718. It illustrates how swiftly bacteria can multiply, particularly in environments with ample resources and favorable conditions.

In the given problem, we started with an initial count of 150 bacteria, and by using the exponential formula, the number projected at future times can be calculated easily. Understanding this concept helps predict how swiftly organisms like bacteria can multiply, which is crucial for various fields like microbiology, medicine, and ecology.

Growth Rate

The growth rate is a crucial parameter in understanding how fast a population is increasing. It is represented by \( k \) in the exponential growth equation. In our bacteria example, the growth rate stays constant no matter when you start observing the population.

To calculate the growth rate, we used the information that the bacteria population tripled from 150 to 450 over 5 hours. By substituting these values into the formula \( 450 = 150 e^{5k} \), we solve for \( k \). After some algebra, and using the natural logarithm, we find \( k = \frac{1}{5} \ln(3) \).

• A constant growth rate implies a steady percentage increase per unit time.
• It's independent of the initial population count, meaning no matter how many bacteria you start with, if conditions remain unchanged, the growth rate remains the same.

This constant measure is essential for predicting population size and understanding how entities like bacteria react over time under stable conditions.

Doubling Time

Doubling time refers to the duration it takes for a population to double in size. It's a helpful measure, especially for populations that grow exponentially, like bacteria. In our example, we calculated the doubling time using the equation \( 2 = e^{kt} \).

By rearranging, this gives us \( t = \frac{1}{k} \ln(2) \). With the growth rate \( k \) from the previous section, substituting values gives us the exact time it takes for the bacteria population to double. For our exercise, this time wasn't dependent on when we started counting; it was consistent at any initiation point because the growth rate was constant.

Key points to remember about doubling time include:

• It provides a simple way to understand exponential growth by expressing it in terms of time.
• In practice, helps with planning and resource allocation as population sizes increase.
• Can guide decision-making in contexts like health and environmental policies.

By knowing how quickly a population can double, researchers, policymakers, and business leaders can make informed decisions based on anticipated future conditions.