Problem 2 - Essential Calculus Early Transcendentals

Question

A common inhabitant of human intestines is the bacterium Escherichia coli. A cell of this bacterium in a nutrient-broth medium divides into two cells every 20 minutes. The initial population of a culture is 60 cells. (a) Find the relative growth rate. (b) Find an expression for the number of cells after \(t\) hours. (c) Find the number of cells after 8 hours. (d) Find the rate of growth after 8 hours. (e) When will the population reach \(20,000\) cells?

Step-by-Step Solution

Answer

(a) The growth rate is \(3\ln(2)\). (b) Cells after \(t\) hours: \(60 \cdot 2^{3t}\). (c) After 8 hours: 1,006,632,960 cells. (d) Rate at 8 hours: \(180 \ln(2) \times 16,777,216\). (e) 20,000 cells in about 3.24 hours.

Step 1: Understand the Doubling Time

Each Escherichia coli cell divides into two cells every 20 minutes. This implies a doubling time of 20 minutes, which is essential for calculating the relative growth rate.

Step 2: Calculate the Relative Growth Rate

The relative growth rate can be found using the formula:\[ r = \frac{\ln(2)}{T} \]where \(T\) is the doubling time in the same time unit. Convert 20 minutes into hours: 20 minutes = \(\frac{1}{3}\) hours. Then,\[ r = \frac{\ln(2)}{\frac{1}{3}} = 3\ln(2)\].

Step 3: Expression for Number of Cells After Time t

The expression for exponential growth is:\[ N(t) = N_0 e^{rt}\]where \(N(t)\) is the number of cells at time \(t\), \(N_0\) is the initial cell number (60), and \(r\) is the growth rate. Plugging the values, we get:\[ N(t) = 60 e^{3\ln(2)t} = 60 \cdot 2^{3t}\].

Step 4: Number of Cells After 8 Hours

Substitute \(t = 8\) into the expression:\[ N(8) = 60 \cdot 2^{3 \times 8} = 60 \cdot 2^{24}\].Calculate: \(2^{24} = 16,777,216\) hence,\[ N(8) = 60 \times 16,777,216 = 1,006,632,960\].

Step 5: Rate of Growth After 8 Hours

The rate of growth is the derivative of \(N(t)\) with respect to \(t\):\[ N'(t) = \frac{d}{dt}[60 \cdot 2^{3t}] = 60 \cdot 3 \ln(2) \cdot 2^{3t} = 180 \ln(2) \cdot 2^{3t}\].At \(t = 8\):\[ N'(8) = 180 \ln(2) \cdot 2^{24}\].Using \(2^{24} = 16,777,216\),\[ N'(8) = 180 \ln(2) \times 16,777,216\].

Step 6: Find Time to Reach 20,000 Cells

Set \(N(t) = 20,000\) and solve for \(t\):\[ 20,000 = 60 \cdot 2^{3t}\]\[ \frac{20,000}{60} = 2^{3t} \Rightarrow 333.33 = 2^{3t}\]\[ \ln(333.33) = 3t \cdot \ln(2)\]\[ t = \frac{\ln(333.33)}{3\ln(2)}\]Evaluating this gives \(t \approx 3.24\) hours.

Key Concepts

Relative Growth Rate

The relative growth rate is a measure of how quickly a population grows, relative to its size. In exponential growth scenarios, such as with bacteria like *Escherichia coli*, understanding this rate is crucial.
This rate can be calculated using the formula:

• \[ r = \frac{\ln(2)}{T} \]

where \( T \) is the doubling time. Doubling time is the period it takes for a quantity to double in size or value and is often represented in consistent time units, like hours. In our exercise, since the bacteria double every 20 minutes, this needs to be converted into hours: 20 minutes is equivalent to \( \frac{1}{3} \) hours. Using this conversion, we find:

• \[ r = \frac{\ln(2)}{\frac{1}{3}} = 3\ln(2) \]

This calculation indicates that for every unit of time (in this case, an hour), the growth rate is proportional to \( 3\ln(2) \). Understanding this concept helps gauge the rapidity of bacterial population expansion.

Doubling Time

Doubling time is a key predictor in understanding exponential growth patterns. It refers to the time it takes for a population to double. For bacterial growth such as with *Escherichia coli*, which doubles every 20 minutes, it becomes essential for calculating other growth parameters.
The conversion to consistent time units is necessary for accurate calculations. Since our problem deals with hours, 20 minutes is converted into \( \frac{1}{3} \) hours. This information allows one to derive related formulas, such as the relative growth rate.Doubling time is integral in the context of:

• Estimating how quickly a population grows
• Deriving exponential growth formulas
• Applying consistent units for accurate calculations

By knowing the doubling time, you can assess how infectious a bacteria might become over a specific period, thereby providing insights into potential outcomes over hours and days.

Derivatives

Derivatives are a central concept in calculus and are used to find the rate of change of a quantity. In the context of population growth, the derivative of a function representing the number of cells gives the rate at which the population is increasing at any given time.
For our exponential growth situation, the number of bacteria over time is represented by the function:

• \[ N(t) = 60 \cdot 2^{3t} \]

The derivative, \( N'(t) \), represents the growth rate at time \( t \). To find this, we differentiate the expression concerning \( t \):

• \[ N'(t) = \frac{d}{dt}[60 \cdot 2^{3t}] = 180 \ln(2) \cdot 2^{3t} \]

Calculating \( N'(8) \) determines how quickly the population is growing at 8 hours:

• \[ N'(8) = 180 \ln(2) \times 16,777,216 \]

Derivatives provide a snapshot of growth dynamics over small intervals of time, helping predict how populations evolve.

Exponential Functions

Exponential functions are mathematical expressions where the independent variable, often time, appears in the exponent. This function type is significant in modeling growth processes such as bacterial reproduction.
An exponential function has the general form:

• \[ N(t) = N_0 e^{rt} \]

In the case of our exercise, *Escherichia coli* cells double in consistent intervals, possessing an exponential growth factor determined by their relative growth rate. We use \( 2 \) to characterize the doubling nature:

• \[ N(t) = 60 \cdot 2^{3t} \]

This equation allows one to compute the number of cells at any point in time \( t \) by using the calculated exponential.
Exponential functions model situations involving:

• Rapid population growth
• Continuous compounding processes
• Situations where rates are proportional to current value

Understanding how exponential functions operate is vital in various fields, including biology, finance, and technology, wherever growth or decay processes are rapid and proportional.