Problem 45 - A Graphical Approach to College Algebra

Question

Escherichia coli is a strain of bacteria that occurs naturally in many organisms. Under certain conditions, the number of bacteria present in a colony is approximated by $$A(t)=A_{0} e^{0.023 t}$$ where \(t\) is in minutes. If \(A_{0}=2,400,000,\) find the number of bacteria at each time. Round to the nearest hundred thousand. (a) 5 minutes (b) 10 minutes (c) 60 minutes

Step-by-Step Solution

Answer

(a) 2,700,000 (b) 3,000,000 (c) 9,200,000

Step 1: Understand the Problem

The problem requires calculating the bacterial population at different time intervals using the formula \(A(t) = A_0 e^{0.023 t}\). Here, \(A_0 = 2,400,000\) is the initial number of bacteria. We need to find \(A(t)\) for 5, 10, and 60 minutes and round the results to the nearest hundred thousand.

Step 2: Calculate Bacteria at 5 Minutes

Substitute \(t = 5\) into the formula: \[A(5) = 2,400,000 \times e^{0.023 \times 5}\]. Calculate \(0.023 \times 5 = 0.115\). Then compute the exponential: \(e^{0.115}\). Now, multiply by the initial amount: \(2,400,000 \times e^{0.115}\). Finally, round the result to the nearest hundred thousand.

Step 3: Calculate Bacteria at 10 Minutes

Substitute \(t = 10\) into the formula: \[A(10) = 2,400,000 \times e^{0.023 \times 10}\]. Calculate \(0.023 \times 10 = 0.23\). Then compute the exponential: \(e^{0.23}\). Multiply by \(2,400,000\). Finally, round the result to the nearest hundred thousand.

Step 4: Calculate Bacteria at 60 Minutes

Substitute \(t = 60\) into the formula: \[A(60) = 2,400,000 \times e^{0.023 \times 60}\]. Calculate \(0.023 \times 60 = 1.38\). Then compute the exponential: \(e^{1.38}\). Multiply by \(2,400,000\). Finally, round the result to the nearest hundred thousand.

Key Concepts

Bacterial Population

Bacteria like Escherichia coli can multiply rapidly under the right conditions. Their growth is often modeled to help understand how quickly they multiply in certain environments. In our exercise, we are looking at the population growth over time. Under these conditions, bacteria often double quickly, leading to exponential growth. The initial amount, given as 2,400,000, shows how many bacteria we start with. As time progresses, this number becomes significantly larger. This characteristic growth pattern can also be seen in other natural phenomena, like populations of animals or the spread of a virus. In the context of our exercise, we calculate the number of bacteria at different times to observe this exponential behavior.

Mathematical Modeling

Mathematical modeling is a way to represent real-world phenomena using equations. It helps predict outcomes based on certain variables. In our exercise, we use the formula \(A(t) = A_0 e^{0.023 t}\) to model the growth of a bacterial population. Here, \(A(t)\) represents the number of bacteria at time \(t\), and \(A_0\) is the initial number of bacteria. The value \(0.023\) is an example of a growth rate, showing how quickly the population expands per minute. By substituting different time values (5, 10, and 60 minutes), we can see how the population evolves. Mathematical models like this enable scientists and researchers to simulate complex behaviors and predict future trends with accuracy.

Natural Exponential Function

The natural exponential function is a key part of modeling exponential growth. It's represented by the function \(e^x\), where \(e\) is a mathematical constant approximately equal to 2.71828. In exponential growth problems, like with bacterial populations, we use this function because it describes any process where the rate of change is proportional to the current value. In our formula, \(e^{0.023 t}\) captures the growth rate over time. The exponential function's unique properties make it ideal for calculating continuous growth, offering a smooth curve that represents how quantities like populations increase. Understanding the role of the natural exponential function is crucial for interpreting behaviors in natural and social sciences.