Problem 7 - Calculus for the Life Sciences: A Modeling Approach Volume I

Question

Suppose penicillin concentration in the serum of a patient \(t\) minutes after a bolus injection of \(2 \mathrm{~g}\) is given by $$P(t)=200 \times 0.96^{t} \quad \mu \mathrm{g} / \mathrm{ml}$$ a. Approximate \(P(t)\) and \(P^{\prime}(t)\) for \(t=0,5,10,15,\) and 20 minutes. b. Plot a graph of \(P^{\prime}(t)\) vs \(P(t)\) using the five pairs of values you just computed.

Step-by-Step Solution

Answer

Evaluate P(t) and P'(t), then plot P'(t) vs P(t).

Step 1: Evaluate P(t) at given time points

To find the concentration of penicillin at time points \(t = 0, 5, 10, 15, 20\), we'll substitute these values into the function \(P(t) = 200 \times 0.96^t\).- For \(t = 0\), \(P(0) = 200 \times 0.96^0 = 200 \, \mu g/ml\)- For \(t = 5\), \(P(5) = 200 \times 0.96^5 \approx 167.78 \, \mu g/ml\)- For \(t = 10\), \(P(10) = 200 \times 0.96^{10} \approx 140.76 \, \mu g/ml\)- For \(t = 15\), \(P(15) = 200 \times 0.96^{15} \approx 118.13 \, \mu g/ml\)- For \(t = 20\), \(P(20) = 200 \times 0.96^{20} \approx 99.11 \, \mu g/ml\)

Step 2: Differentiate P(t) to find P'(t)

The derivative \(P'(t)\) represents the rate of change of the concentration.Use the chain rule for differentiation on \(P(t) = 200 \times 0.96^t\):\[ P'(t) = 200 \times \ln(0.96) \times 0.96^t \]The natural log of 0.96 is approximately -0.0408.

Step 3: Evaluate P'(t) at given time points

Now, evaluate \(P'(t)\) at \(t = 0, 5, 10, 15, 20\) using the derived formula:- For \(t = 0\), \(P'(0) = 200 \times (-0.0408) \times 0.96^0 \approx -8.16 \, \mu g/ml\cdot/min\)- For \(t = 5\), \(P'(5) = 200 \times (-0.0408) \times 0.96^5 \approx -6.83 \, \mu g/ml\cdot/min\)- For \(t = 10\), \(P'(10) = 200 \times (-0.0408) \times 0.96^{10} \approx -5.73 \, \mu g/ml\cdot/min\)- For \(t = 15\), \(P'(15) = 200 \times (-0.0408) \times 0.96^{15} \approx -4.80 \, \mu g/ml\cdot/min\)- For \(t = 20\), \(P'(20) = 200 \times (-0.0408) \times 0.96^{20} \approx -4.03 \, \mu g/ml\cdot/min\)

Step 4: Plot P'(t) vs P(t)

Create a plot with \(P(t)\) on the x-axis and \(P'(t)\) on the y-axis using the computed values:- When \(P(t) = 200 \, \mu g/ml\), \(P'(t) = -8.16 \, \mu g/ml\cdot/min\)- When \(P(t) = 167.78 \, \mu g/ml\), \(P'(t) = -6.83 \, \mu g/ml\cdot/min\)- When \(P(t) = 140.76 \, \mu g/ml\), \(P'(t) = -5.73 \, \mu g/ml\cdot/min\)- When \(P(t) = 118.13 \, \mu g/ml\), \(P'(t) = -4.80 \, \mu g/ml\cdot/min\)- When \(P(t) = 99.11 \, \mu g/ml\), \(P'(t) = -4.03 \, \mu g/ml\cdot/min\)This plot will show that as the concentration decreases, the rate of change (derivative) also becomes less negative, indicating slowing decline.

Key Concepts

Derivative Calculation

In the context of analyzing penicillin concentration over time, calculating the derivative is crucial to understanding how the concentration changes at any given moment. Here, we began with the function given as \( P(t) = 200 \times 0.96^t \), representing how the concentration decreases exponentially over time. Derivatives help in identifying the rate at which the concentration diminishes.

To calculate the derivative \( P'(t) \), we apply the chain rule. The chain rule is a fundamental technique in calculus used to differentiate composite functions. In our case, the function involves an exponential term, so we use the property that the derivative of \( a^t \) is \( a^t \ln(a) \), where \( a \) is a constant.

Thus, for our problem, we differentiate using the formula:

• \( P'(t) = 200 \times \ln(0.96) \times 0.96^t \)

The natural logarithm of 0.96 is approximately -0.0408, which indicates that the rate of change in concentration is negative, showing a decay over time.

Rate of Change

In mathematics, the rate of change is often associated with the derivative of a function. For the penicillin concentration in the patient, the rate of change \( P'(t) \) informs us how rapidly the concentration is decreasing at any given moment.

By computing \( P'(t) \) for different time values \( t = 0, 5, 10, 15, \) and 20 minutes, we discover how the concentration evolves:

• At \( t = 0 \), the rate is \( -8.16 \, \mu g/ml \cdot/min \)
• At \( t = 5 \), it slows to \( -6.83 \, \mu g/ml \cdot/min \)
• At \( t = 10 \), further slows to \( -5.73 \, \mu g/ml \cdot/min \)
• These values indicate that as time progresses, the concentration decreases at a slower rate, reflecting its nature in exponential decay.

The slow reduction in the absolute value of \( P'(t) \) across time demonstrates a decelerating decline in drug concentration, also highlighting how derivatives provide insights into the behavior of a real-world process.

Plotting Graphs

Graphical representation is a powerful tool to visualize mathematical relationships. In our problem, plotting \( P'(t) \) versus \( P(t) \) illustrates the relationship between the concentration of penicillin and its rate of reduction.

By assigning \( P(t) \) to the x-axis and \( P'(t) \) to the y-axis, we clearly see:

• Initially high concentration values correspond to more negative rates of change.
• As concentration decreases, the rate of change becomes less negative, indicating a slowing decline.

This plot offers a clear visual representation, showing an unmistakable trend that as the initial concentration is high, the decay rate is steep. Over time, the concentration approaches lower values, the slope of the plot softens, depicting the decelerating rate of decay in the drug concentration. Analyzing graphs helps in drawing conclusions about the system's dynamics, fostering a deeper understanding of the interaction between different quantities.