Problem 29 - Calculus Early Transcendentals

Question

Suppose that an initial population of \(10,000\) bacteria grows exponentially at a rate of \(2 \%\) per hour and that \(y=y(t)\) is the number of bacteria present \(t\) hours later. (a) Find an initial-value problem whose solution is \(y(t)\) (b) Find a formula for \(y(t) .\) (c) How long does it take for the initial population of bacteria to double? (d) How long does it take for the population of bacteria to reach \(45,000 ?\)

Step-by-Step Solution

Answer

(a) \( y(0) = 10,000 \), \( \frac{dy}{dt} = 0.02y \). (b) \( y(t) = 10,000 e^{0.02t} \). (c) \( t \approx 34.65 \) hours. (d) \( t \approx 74.96 \) hours.

Step 1: Define Initial-Value Problem

The initial-value problem can be defined using the formula for exponential growth, which is \( y(t) = y_0 e^{kt} \), where \( y_0 = 10,000 \) is the initial number of bacteria, and \( k \) is the growth rate per unit time. Since the growth rate is \( 2\% \) per hour, \( k = 0.02 \). Therefore, the initial-value problem is \( y(0) = 10,000 \), with \( \frac{dy}{dt} = 0.02y \).

Step 2: Find the Formula for y(t)

Use the solution form of the differential equation: \( y(t) = y_0 e^{kt} \). Substituting the given values, \( y_0 = 10,000 \) and \( k = 0.02 \), gives \( y(t) = 10,000 e^{0.02t} \). This is the formula for the population at any time \( t \).

Step 3: Calculate Time to Double the Population

To find the time \( t \) at which the population doubles, set \( y(t) = 2y_0 = 20,000 \). Using the formula \( 20,000 = 10,000 e^{0.02t} \), solve for \( t \):\[ e^{0.02t} = 2 \]\[ 0.02t = \ln(2) \]\[ t = \frac{\ln(2)}{0.02} \]. Calculate \( t \approx 34.65 \) hours.

Step 4: Calculate Time to Reach 45,000

For the population to reach \( 45,000 \), solve \( y(t) = 45,000 \) using the formula. Start with \( 45,000 = 10,000 e^{0.02t} \):\[ e^{0.02t} = 4.5 \]\[ 0.02t = \ln(4.5) \]\[ t = \frac{\ln(4.5)}{0.02} \]. Calculate \( t \approx 74.96 \) hours.

Key Concepts

Differential Equations

Imagine a differential equation as a powerful tool to describe how things change over time. In our bacterial growth example, it helps us model how the population of bacteria increases as time progresses. A differential equation involves derivatives, which tell us the rate at which one quantity changes compared to another.

In the case of exponential growth, this rate of change is proportional to the current size, expressed with:

• \( \frac{dy}{dt} = ky \)

Here, \( y \) represents the population size, \( t \) is time, and \( k \) is a constant called the growth rate.

This equation means that the larger the population, the faster it grows. By solving this equation, we obtain a formula for \( y(t) \), allowing us to predict future population sizes based on current conditions.

Initial Value Problem

An initial value problem is a special kind of differential equation problem where we know the starting point. It’s like having the starting point on a map, helping you find your way to where you want to go.

In our bacteria example, we start with an initial number of bacteria, say \( 10,000 \), and use this information to solve the differential equation:

• The equation was \( \frac{dy}{dt} = 0.02y \)
• The initial condition was \( y(0) = 10,000 \)

By including the initial condition, we gain specific insights into the behavior of the population over time. This transforms the equation into a complete problem with a unique solution, expressed as \( y(t) = 10,000 e^{0.02t} \). This solution helps us evaluate the size of the population at any chosen time.

Doubling Time

Doubling time is a handy concept telling us how long it takes for a quantity to double in size. For exponential growth, the time it takes for the population to double is constant, regardless of the initial population size. This is possible because of the nature of exponential growth.

In our bacterial growth, we want to find when the population grows from \( 10,000 \) to \( 20,000 \). We set the equation \( 10,000 e^{0.02t} = 20,000 \), and by solving it, we find:

• \( t = \frac{\ln(2)}{0.02} \approx 34.65 \; \text{hours} \)

This means it takes approximately 34.65 hours for the bacterial population to double.

Population Growth Rate

The population growth rate is a measure of how quickly a population increases over time. It is presented as a percentage and influences how fast the size of a population grows.

In our exercise, the bacteria grow at a rate of \( 2\% \) per hour. This rate connects directly to the constant \( k \) in the differential equation \( \frac{dy}{dt} = ky \). With \( k = 0.02 \), it reflects a steady, predictable growth pattern.

Understanding the growth rate lets us make informed predictions about future population size. By inserting this growth rate into our exponential model, we can quickly find the population at any future point, such as when it reaches \( 45,000 \) individuals. By using the formula:

• Set \( 10,000 e^{0.02t} = 45,000 \)
• Solve to find \( t \approx 74.96 \; \text{hours} \)

showing us the time needed to hit this new population milestone.