Problem 25 - University Calculus: Early Transcendentals

Question

When a bactericide was added to a nutrient broth in which bacteria were growing, the bacterium population continued to grow for a while, but then stopped growing and began to decline. The size of the population at time \(t\) (hours) was \(b=10^{6}+10^{4} t-10^{3} t^{2}\). Find the growth rates at a. \(t=0\) hours. b. \(t=5\) hours. c. \(t=10\) hours.

Step-by-Step Solution

Answer

At t=0 hours, growth rate is 10,000; at t=5 hours, it is 0; at t=10 hours, it is -10,000.

Step 1: Differentiate the population function

To find the growth rate, we need the derivative of the population function with respect to time, \(t\). The function is given as \(b = 10^{6} + 10^{4}t - 10^{3}t^{2}\). We will differentiate this function.The derivative, denoted \(b'(t)\), is:\[b'(t) = \frac{d}{dt}(10^{6} + 10^{4}t - 10^{3}t^{2})\]Applying basic differentiation rules, we get:\[b'(t) = 0 + 10^{4} - 2 \times 10^{3}t = 10^{4} - 2 \times 10^{3}t\]

Step 2: Calculate Growth Rate at t=0

Now that we have the derivative \(b'(t) = 10^{4} - 2 \times 10^{3}t\), we can calculate the growth rate at \(t=0\) hours.Substitute \(t=0\) into \(b'(t)\):\[b'(0) = 10^{4} - 2 \times 10^{3} \times 0 = 10^{4}\]Thus, the growth rate at \(t=0\) hours is \(10^{4}\).

Step 3: Calculate Growth Rate at t=5

Next, find the growth rate at \(t=5\) hours using the derivative \(b'(t)\).Substitute \(t=5\) into \(b'(t)\):\[b'(5) = 10^{4} - 2 \times 10^{3} \times 5\]Calculating:\[b'(5) = 10^{4} - 10 \times 10^{3} = 10^{4} - 10^{4} = 0\]The growth rate at \(t=5\) hours is \(0\). This indicates that the population stops growing at this point.

Step 4: Calculate Growth Rate at t=10

Finally, calculate the growth rate at \(t=10\) hours.Substitute \(t=10\) into \(b'(t)\):\[b'(10) = 10^{4} - 2 \times 10^{3} \times 10\]Calculate the result:\[b'(10) = 10^{4} - 20 \times 10^{3} = 10^{4} - 2 \times 10^{4} = -10^{4}\]The growth rate at \(t=10\) hours is \(-10^{4}\), indicating the population is declining.

Key Concepts

Bacterial Growth

Bacterial growth is a fascinating process where bacteria multiply and increase in numbers, usually in a predictable manner. When bacteria are placed in a nutrient-rich environment, they can replicate rapidly. However, various factors, such as limited nutrients or the addition of bactericides, can affect this growth.
Bacteria typically reproduce through a process called binary fission, where one cell splits into two identical cells. This results in exponential growth, meaning the population doubles over regular intervals.
However, not all growth follows this perfect pattern. Influences like temperature, pH levels, and other external conditions can cause fluctuations in the growth rate, leading to different phases in the bacteria's growth curve:

• Lag Phase: Adaptation period where bacteria are getting accustomed to their environment.
• Log Phase: Rapid exponential growth phase.
• Stationary Phase: Growth slows as resources become limited.
• Decline Phase: Bacteria begin to die off as resources are depleted or inhibitors, like bactericides, take effect.

Understanding bacterial growth helps in various applications like infection control, fermentation processes, and antibiotic effectiveness.

Population Function

The population function is a mathematical expression that describes how the size of a population changes over time. In our exercise, the population function, expressed as \( b = 10^{6} + 10^{4}t - 10^{3}t^{2} \), models the bacterial population in a specific environment.
This function is a quadratic equation, which means its graph will be a parabola. The terms signify different aspects:

• The constant \(10^{6}\) denotes the initial population when time \(t=0\).
• The linear term \(10^{4}t\) represents a component of growth that increases with time.
• The quadratic term \(-10^{3}t^{2}\) implies a deceleration in population growth, eventually leading to a decline.

In analyzing this quadratic function, we can infer that initially, as \(t\) increases, the bacteria population grows, but after a certain point, the \(-10^{3}t^{2}\) term dominates, indicating a population decrease. Understanding this function provides insights into when interventions may be needed to control growth or stimulate it.

Growth Rate Calculation

Growth rate calculation involves finding out how fast the population changes at any point in time. To calculate growth rates, we use calculus, specifically the derivative of the population function concerning time. This helps determine the rate at which the population size is changing.
For our given function, the derivative is \( b'(t) = 10^{4} - 2 \times 10^{3}t \). This expression allows us to assess the growth or decline over time:

• At \(t = 0\), substitute in the derivative: \(b'(0) = 10^{4}\). This positive rate shows that the population is growing initially.
• At \(t = 5\), \(b'(5) = 0\) indicates the population has stopped growing, reaching a peak or plateau.
• At \(t = 10\), \(b'(10) = -10^{4}\) shows the population is declining, perhaps due to depleted resources or other inhibiting factors.

Calculating growth rates is crucial for predicting population trends, understanding growth dynamics, and effectively managing ecological, biological, or resource-based systems.