Problem 80 - A Graphical Approach to Precalculus with Limits

Question

Growth of \(\mathbf{E}\) coli Bacteria \(\mathbf{A}\) type of bacteria that inhabits the intestines of animals is named \(E\). coli (Escherichia coli). These bacteria are capable of rapid growth and can be dangerous to humans- especially children. In one study, \(E\) coli bacteria were capable of doubling in number every 49.5 minutes. Their number after \(x\) minutes can be modeled by the function $$ N(x)=N_{0} e^{0.014 x} $$ (Source: Stent, G. S... Molecular Biology of Bacterial Viruses, W. H. Freeman.) Suppose \(N_{0}=500,000\) is the initial number of bacteria per milliliter. (a) Make a conjecture about the number of bacteria per milliliter after 99 minutes. Verify your conjecture. (b) Estimate graphically the time elapsed until there were 25 million bacteria per milliliter.

Step-by-Step Solution

Answer

(a) About 1,999,500 bacteria after 99 minutes. (b) 279 minutes to reach 25 million bacteria.

Step 1: Understand the Doubling Formula

Given the growth model for the E. coli bacteria, the formula is \(N(x) = N_0 e^{0.014x}\). To find the number of bacteria after 99 minutes, substitute \(x = 99\) and \(N_0 = 500,000\) into the formula.

Step 2: Calculate Bacteria Count for 99 Minutes

Substitute the values into the formula: \[ N(99) = 500,000 \times e^{0.014 \times 99} \]Calculate the exponent:\[ 0.014 \times 99 = 1.386 \]Calculate \(e^{1.386}\) using a calculator to get approximately 3.999. Therefore:\[ N(99) = 500,000 \times 3.999 \approx 1,999,500 \]So, the number of bacteria per milliliter after 99 minutes is 1,999,500.

Step 3: Graphically Estimate Time for 25 Million Bacteria

We need to find \(x\) such that \(N(x) = 25,000,000\). Rearrange the equation to solve for \(x\):\[ 25,000,000 = 500,000 \times e^{0.014x} \]Divide both sides by 500,000:\[ 50 = e^{0.014x} \]Take the natural logarithm (ln) of both sides:\[ \ln(50) = 0.014x \]Calculate \(\ln(50)\) using a calculator:\[ \ln(50) \approx 3.912 \]Solve for \(x\):\[ x = \frac{3.912}{0.014} \approx 279.43 \]So, it would take approximately 279 minutes. Estimate graphically by plotting \(N(x) = 500,000 e^{0.014x}\) and finding when it crosses 25 million.

Step 4: Verification through Doubling Time

To verify part (a), calculate the doubling time:\[ 2N_0 = N_0 e^{0.014(T)} \] simplifies to \[ 2 = e^{0.014(49.5)} \].Calculating using a calculator: \(e^{0.6945} \approx 2\), confirming that doubling time is about 49.5 minutes.For part (b), note that after approximately 279 minutes, the bacteria count reaches 25 million, reaffirmed by computation steps.

Key Concepts

Bacteria Growth

Bacteria, like E. coli, exhibit a fascinating phenomenon known as exponential growth. This means that under ideal conditions, their population can double at a consistent rate. In the exercise, the E. coli bacteria began with an initial population, denoted as \(N_0\), of 500,000 bacteria per milliliter. Their growth over time is governed by the equation \(N(x) = N_0 e^{0.014x}\). This equation implies that the population not only grows rapidly but accelerates in growth as time proceeds, an intrinsic characteristic of exponential growth. Understanding this growth pattern is crucial because it helps in predicting how quickly a population of bacteria can become dangerous if unchecked. For instance, in clinical and laboratory settings, understanding such growth curves is vital for developing effective interventions.

Doubling Time

Doubling time is a concept that provides insight into how long it takes for a population to double in size. For E. coli bacteria, this time period is crucial to understanding their rapid proliferation in optimal conditions. In our case, E. coli doubles every 49.5 minutes. This means if you start with a certain number of bacteria, it only takes 49.5 minutes for that number to double. To verify the doubling time mathematically, you can use the formula \[2 = e^{0.014T}\], where \(T\) is the doubling time. Upon solving, you find that indeed \(T = 49.5\). This provides not just an estimate but a calculated confirmation of the growth behavior, emphasizing the need to monitor and control bacterial populations effectively, especially in environments like hospitals where infection risk is high.

Logarithmic Functions

Logarithmic functions play a pivotal role when estimating how long it takes a population to reach a certain size. In the exercise, we needed to determine the time required for the E. coli population to grow to 25 million. This involves solving the equation \(e^{0.014x} = 50\), where \(x\) represents time in minutes. To solve this equation, the natural logarithm function, denoted as \(\ln\), is used. This mathematical tool helps to "undo" the exponential effect so we can solve for \(x\). Using \(\ln\):

• Take the natural logarithm of both sides, resulting in \(\ln(50) = 0.014x\).
• Calculate \(\ln(50)\); it is approximately 3.912.
• Solve for \(x\): \(x = \frac{3.912}{0.014} \approx 279.43\).

Logarithmic functions thus offer a mathematical approach to unravel exponential scenarios, which can translate to real-world applications in population dynamics.

Graphical Estimation

Graphical estimation provides a visual approach to solving exponential growth problems like estimating when a bacterial population will reach a significant number. By plotting the bacteria growth function \(N(x) = 500,000 e^{0.014x}\), you can physically see how the numbers soar over time. In this exercise, plotting allows us to estimate when \(N(x)\) reaches 25 million. By observing the graph, you look for the point where the curve crosses 25 million on the y-axis. This intersection helps affirm calculations found by other means, such as logarithmic functions, offering a second layer of verification. Graphical estimation is helpful as it offers a quick approximation that is invaluable in situations where decisions need to be made swiftly. It's also accessible for those who learn better visually, bridging gaps left by purely numerical analysis.