Problem 30 - Calculus Early Transcendentals

Question

A cell of the bacterium \(E\) coli divides into two cells every 20 minutes when placed in a nutrient culture. Let \(y=y(t)\) be the number of cells that are present \(t\) minutes after a single cell is placed in the culture. Assume that the growth of the bacteria is approximated by an exponential growth model. (a) Find an initial-value problem whose solution is \(y(t)\) (b) Find a formula for \(y(t) .\) (c) How many cells are present after 2 hours? (d) How long does it take for the number of cells to reach \(1,000,000 ?\)

Step-by-Step Solution

Answer

(a) \( \frac{dy}{dt} = ky, y(0)=1 \) (b) \( y(t) = e^{\left(\frac{\ln(2)}{20}\right)t} \) (c) 64 cells (d) \( t = \frac{20 \cdot \ln(1,000,000)}{\ln(2)} \) minutes

Step 1 - Define the problem (initial-value problem)

The bacterium divides into two cells every 20 minutes, indicating exponential growth. We start with one cell at time zero: \( y(0) = 1 \). The doubling time is 20 minutes. The general form of the exponential growth model is \( y(t) = y_0e^{kt} \), where \( y_0 \) is the initial number of cells and \( k \) is the growth constant. The initial-value problem can be represented as a differential equation: \( \frac{dy}{dt} = ky \), with the initial condition \( y(0) = 1 \).

Step 2 - Determine the growth constant k

The bacteria double every 20 minutes, so \( y(20) = 2 \). Applying the exponential growth model \( y(t) = y_0e^{kt} \), substitute \( y(20) = 2 \) and \( y_0 = 1 \): \( 2 = 1 \cdot e^{20k} \). Solve for \( k \) by taking the natural logarithm: \( 20k = \ln(2) \), hence \( k = \frac{\ln(2)}{20} \).

Step 3 - Construct the formula for y(t)

Now that we have \( k = \frac{\ln(2)}{20} \), substitute back into the exponential model with \( y_0 = 1 \): \[ y(t) = e^{\left(\frac{\ln(2)}{20}\right)t} \]. This is the formula for the number of bacteria after \( t \) minutes.

Step 4 - Calculate the number of cells after 2 hours

Convert 2 hours into minutes, giving \( t = 120 \). Substitute \( t = 120 \) into the formula: \[ y(120) = e^{\left(\frac{\ln(2)}{20}\right)\cdot120} = e^{6\ln(2)} = 2^6 = 64 \]. Thus, there are 64 cells after 2 hours.

Step 5 - Determine time to reach 1,000,000 cells

We use the formula and set \( y(t) = 1,000,000 \):\[ 1,000,000 = e^{\left(\frac{\ln(2)}{20}\right)t} \]. Taking the natural log gives:\[ \ln(1,000,000) = \frac{\ln(2)}{20}t \]. Solving for \( t \) gives:\[ t = \frac{20 \cdot \ln(1,000,000)}{\ln(2)} \]. Calculate \( t \) to find the time it takes to reach 1,000,000 cells.

Key Concepts

Differential Equation

A differential equation is a mathematical equation that involves derivatives, expressing how a particular quantity changes over time. In the case of exponential growth, like our bacterium problem, the rate of change of the number of cells is proportional to the number itself. This is expressed by the differential equation \( \frac{dy}{dt} = ky \), where \( dy/dt \) is the derivative of \( y \) with respect to time \( t \), and \( k \) is a constant. This form dictates that the growth rate increases as the number of cells increases—characteristic of exponential processes.

The beauty of using a differential equation in this context is its ability to model real-world growth scenarios. It provides a mathematical framework that predicts future population sizes, based on current sizes and known growth rates. In solving these problems, we often seek the particular solution to the differential equation under given initial conditions.

Growth Constant

The growth constant \( k \) is a crucial parameter in the context of exponential growth models. It determines the rate at which the population grows. Mathematical models often use the term \( e^{kt} \) where \( k \) affects the curve of growth exponentially over time.

In our bacterial culture problem, since the bacteria double every 20 minutes, we derived \( k \) using the relationship \( y(20) = 2y_0 \). This allowed us to solve for \( k \) using natural logarithms: \( k = \frac{\ln(2)}{20} \).

Understanding the growth constant is fundamental because it adapts our model to different scenarios, whether we're studying populations, investments, or radioactive decay, by reflecting how quickly these entities change.

Doubling Time

Doubling time is the time it takes for a quantity to double in size or value, pivotal in understanding exponential growth. In our scenario, the doubling time is observed to be 20 minutes. It fits perfectly into the equation \( y(t) = y_0e^{kt} \) by ensuring that when \( t \) equals the doubling time, \( y(t) = 2y_0 \).

Mathematically, we can express doubling time using \( k \):

• Start with \( y(t) = 2y_0 \)
• Substitute \( t = T \) (doubling time): \( 2y_0 = y_0e^{kT} \)
• Solving for \( T \), we find \( T = \frac{\ln(2)}{k} \)

In practical terms, knowing the doubling time helps determine how quickly a population or investment grows, directly affecting strategic decision-making processes.

Initial-Value Problem

An initial-value problem involves finding a function that satisfies a differential equation and meets a specific condition at the outset. Here, the condition is \( y(0) = 1 \), indicating that our culture starts with a single bacterium at time zero.

To solve initial-value problems like ours:

• Identify the differential equation: \( \frac{dy}{dt} = ky \)
• Apply initial conditions to find constants, here \( y(0) = 1 \)
• Reconvene with a precise model for \( y(t) \)

Addressing an initial-value problem equips one to predict outcomes from specific beginnings, essential for planning and evaluation purposes in scientific and engineering challenges.