Problem 6 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Suppose a bacterium Vibrio natriegens is growing in a beaker and cell concentration \(C\) at time \(t\) in minutes is given by $$C(t)=0.87 \times 1.02^{t} \quad \text { million cells per ml }$$ a. Approximate \(C(t)\) and \(C^{\prime}(t)\) for \(t=0,10,20,30,\) and 40 minutes. b. Plot a graph of \(C^{\prime}(t)\) vs \(C(t)\) using the five pairs of values you just computed.

Step-by-Step Solution

Answer

Evaluate \( C(t) \) and \( C'(t) \) at those times, then plot \( C'(t) \) vs \( C(t) \).

Step 1: Calculate the Cell Concentration

We need to evaluate the cell concentration \( C(t) \) for the given times \( t = 0, 10, 20, 30, \) and \( 40 \) minutes. The formula given is \( C(t) = 0.87 \times 1.02^t \). Calculate each: - For \( t = 0 \): \( C(0) = 0.87 \times 1.02^0 = 0.87 \) million cells per ml. - For \( t = 10 \): \( C(10) = 0.87 \times 1.02^{10} \approx 1.06 \) million cells per ml. - For \( t = 20 \): \( C(20) = 0.87 \times 1.02^{20} \approx 1.30 \) million cells per ml. - For \( t = 30 \): \( C(30) = 0.87 \times 1.02^{30} \approx 1.59 \) million cells per ml. - For \( t = 40 \): \( C(40) = 0.87 \times 1.02^{40} \approx 1.94 \) million cells per ml.

Step 2: Derive the Rate of Change of Cell Concentration

To find \( C'(t) \), differentiate \( C(t) = 0.87 \times 1.02^t \). Using the chain rule, \( C'(t) = 0.87 \times ln(1.02) \times 1.02^t \). Specifically, \( ln(1.02) \approx 0.0198 \), so \( C'(t) = 0.87 \times 0.0198 \times 1.02^t \).

Step 3: Calculate the Rate Function Values

Now use the derived formula to find \( C'(t) \) for \( t = 0, 10, 20, 30, 40 \): - For \( t = 0 \): \( C'(0) = 0.87 \times 0.0198 \times 1.02^0 \approx 0.0172 \) million cells per ml per minute. - For \( t = 10 \): \( C'(10) = 0.87 \times 0.0198 \times 1.02^{10} \approx 0.0210 \) million cells per ml per minute. - For \( t = 20 \): \( C'(20) = 0.87 \times 0.0198 \times 1.02^{20} \approx 0.0257 \) million cells per ml per minute. - For \( t = 30 \): \( C'(30) = 0.87 \times 0.0198 \times 1.02^{30} \approx 0.0313 \) million cells per ml per minute. - For \( t = 40 \): \( C'(40) = 0.87 \times 0.0198 \times 1.02^{40} \approx 0.0381 \) million cells per ml per minute.

Step 4: Prepare the Graph Data

Using the values computed, plot \( C'(t) \) on the y-axis and \( C(t) \) on the x-axis. The pairs you will plot are: - \( (0.87, 0.0172) \)- \( (1.06, 0.0210) \)- \( (1.30, 0.0257) \)- \( (1.59, 0.0313) \)- \( (1.94, 0.0381) \).

Step 5: Plot the Graph

Draw a graph with \( C(t) \) on the x-axis and \( C'(t) \) on the y-axis. Place the points calculated in Step 4 and connect them to observe how the rate of change of cell concentration varies as concentration increases.

Key Concepts

Calculus

Calculus is a branch of mathematics that studies the rate at which quantities change. Originating from the Latin word "calculus," which means "small stone," it captures the dynamics of moving systems in both science and engineering contexts.
It primarily involves two main branches:

• Differential Calculus: This focuses on finding the rate of change or the slope of curves. It deals with the concept of a derivative, which provides tools to understand and predict changes within a given function.
• Integral Calculus: This handles the accumulation of quantities, such as areas under curves, and involves the concept of integrals.

In this context, exponential growth problems like the growth of bacteria over time often serve as practical examples to illustrate how calculus can be applied to determine instantaneous rates and predict future growth. By applying calculus principles, we can determine rates such as the speed of bacterial growth and analyze how quickly changes happen at any given time.

Differentiation

Differentiation is a fundamental concept in calculus that involves computing the derivative of a function. A derivative represents the rate of change of a function with respect to one of its variables. This can be thought of as calculating how a small change in the input of a function results in a change in the output.
For example, in the problem with Vibrio natriegens, we calculated the derivative to find out how rapidly the concentration of bacterial cells changes over time.

• The formula given was: \( C(t) = 0.87 \times 1.02^t \).
• Using differentiation, we derived \( C'(t) = 0.87 \times \ln(1.02) \times 1.02^t \), where \( \ln(1.02) \approx 0.0198 \).
• This indicates how fast the cell concentration changes at any time \( t \).

Differentiation is invaluable for analyzing real-world scenarios, especially those involving growth or decay, like biology, finance, and physics, where understanding how things evolve over time is crucial.

Graphing Functions

Graphing functions is an essential skill in both calculus and algebra. It involves plotting points on a coordinate plane to visualize how a function behaves. By doing so, it becomes easier to understand the relationship between variables, identify trends, and predict future behavior.
For the bacteria growth problem, graphing allows us to see how the rate of change \( C'(t) \) relates to the concentration \( C(t) \) across different time intervals.

• The graph plots these pairs \((C(t), C'(t))\) such as \((0.87, 0.0172)\), \((1.06, 0.0210)\), etc.
• On the graph, \( C(t) \) typically forms the x-axis while \( C'(t) \) forms the y-axis.
• This creates a visual representation of exponential growth and gives insights into how quickly concentrations increase.

By effectively graphing these functions, students can visually interpret the real impact of mathematical computations, turning abstract numbers into tangible patterns and behaviors.