Problem 25 - Biocalculus Calculus for the Life Sciences

Question

Drug pharmacokinetics The plasma drug concentration of a new drug was modeled by the function \(C(t)=23 t e^{-2 t}\) where \(t\) is measured in hours and \(C\) in \(\mu \mathrm{g} / \mathrm{mL}\) . (a) What is the maximum drug concentration and when did it occur? (b) Calculate \(\int_{0}^{\infty} C(t) d t\) and explain its significance.

Step-by-Step Solution

Answer

(a) Maximum concentration: \( \frac{23}{2e} \) at \( t = \frac{1}{2} \) hours. (b) Total drug exposure: \( \frac{23}{4} \) \( \mu \mathrm{g} / \mathrm{mL} \cdot \mathrm{hr} \).

Step 1: Differentiate the Function

First, differentiate the function \( C(t) = 23t e^{-2t} \) to find the critical points where the maximum concentration might occur. Use the product rule: if \( u(t) = 23t \) and \( v(t) = e^{-2t} \), then \( C'(t) = u'(t)v(t) + u(t)v'(t) \).

Step 2: Solve for Critical Points

Calculate the derivatives: \( u'(t) = 23 \) and \( v'(t) = -2e^{-2t} \). Thus, the derivative of \( C(t) \) is \( C'(t) = 23e^{-2t} - 46te^{-2t} = 23e^{-2t}(1-2t) \). Solve \( C'(t) = 0 \) to find the critical points: \( 23e^{-2t}(1-2t) = 0 \). The only solution is \( t = \frac{1}{2} \) because \( e^{-2t} \) is never zero.

Step 3: Verify the Maximum Point

To confirm that \( t = \frac{1}{2} \) is a maximum, check the second derivative \( C''(t) \) or evaluate the sign of \( C'(t) \) around this point. Alternatively, compute \( C''(t) \), which is negative at \( t = \frac{1}{2} \), confirming a local maximum.

Step 4: Calculate Maximum Concentration

Substitute \( t = \frac{1}{2} \) into the original function to find the maximum concentration: \( C\left(\frac{1}{2}\right) = 23 \times \frac{1}{2} \times e^{-1} = \frac{23}{2e} \).

Step 5: Integrate to Calculate Total Drug Exposure

Calculate the integral \( \int_{0}^{\infty} C(t) \, dt = \int_{0}^{\infty} 23t e^{-2t} \, dt \). Use integration by parts with \( u = 23t \), \( dv = e^{-2t} \, dt \), \( du = 23 \, dt \), and \( v = -\frac{1}{2}e^{-2t} \).

Step 6: Evaluate the Integral

Calculate the integration by parts expression: Evaluate \( -\frac{23}{2}te^{-2t} \bigg|_0^\infty + \frac{23}{2} \int_0^\infty e^{-2t} \, dt \). The first term evaluates to 0 at both bounds due to \( te^{-2t} \). The second term, through standard integration, gives \( \frac{23}{2} \times \frac{1}{2} \). Thus, the integral evaluates to \( \frac{23}{4} \).

Step 7: Discuss the Significance of the Integral

The integral \( \int_{0}^{\infty} C(t) \, dt \) represents the total drug exposure over time, as it calculates the total amount of drug present in the plasma.

Key Concepts

Drug Concentration

In pharmacokinetics, drug concentration refers to the amount of drug present in a given volume of plasma or blood, often expressed in micrograms per milliliter (\( \mu \text{g/mL} \)). It is crucial for determining the drug's efficacy and safety. Understanding how drug concentration changes over time helps in tailoring proper dosing regimens. Analyzing concentration-time curves, such as the function \( C(t) = 23t e^{-2t} \), is key. This function describes how the concentration (\( C \)) of a drug varies with time (\( t \)) after administration. Studying these variations can assist in identifying peak concentrations and duration of therapeutic effects, ensuring efficient and safe use of drugs. It's essential for healthcare providers and pharmacists to be familiar with these curves for optimal dosing.

Calculus in Biology

Calculus is a powerful mathematical tool used in biology to interpret and predict biological processes. In the context of pharmacokinetics, calculus helps analyze how drug concentrations change over time. By doing this, we can quantitatively understand drug absorption, distribution, metabolism, and excretion. Calculus plays a significant role in determining key pharmacokinetic parameters, such as:

• Peak drug concentration and time to reach it (\( t_{max} \))
• Total drug exposure over time, measured through an area under the curve (AUC)
• Rate of change of drug concentration

By integrating these principles, healthcare professionals can better design effective and safe drug therapies for patients.

Integration by Parts

Integration by parts is a technique used in calculus to solve integrals where direct integration is challenging. It transforms the integral of a product of functions into simpler integrals. The formula is given by:\[ \int u \, dv = uv - \int v \, du \]In the context of the problem, integration by parts was used to evaluate the integral \( \int_{0}^{\infty} 23t e^{-2t} \, dt \). Here, the choice of functions was:

• \( u = 23t \)
• \( dv = e^{-2t} \, dt \)

This integration helps find total drug exposure by determining the area under the concentration-time curve (AUC), a vital pharmacokinetic parameter that indicates how much drug is present in the body over time.

Maximum Concentration

Maximum drug concentration, or \( C_{max} \), refers to the peak plasma concentration that a drug achieves after administration and before elimination occurs. Identifying \( C_{max} \) is important as it can influence both the efficacy and toxicity of a drug. A higher concentration can improve therapeutic effects but might increase the risk of adverse reactions. To find when \( C_{max} \) occurs, differentiation is used to identify critical points. For the given function \( C(t) = 23t e^{-2t} \), the maximum concentration is at \( t = \frac{1}{2} \). At this time, substituting into the original function reveals the peak concentration is \( \frac{23}{2e} \). This provides valuable information for determining proper dosing.

Differentiation in Calculus

Differentiation is a fundamental concept in calculus used to find the rate at which a quantity changes. In pharmacokinetics, it is essential for identifying when a drug reaches its maximum concentration (\( C_{max} \)). The process involves finding the derivative of the concentration-time function to locate critical points—where maximum and minimum values occur. For example, differentiating \( C(t) = 23t e^{-2t} \) using the product rule helps discover when the drug concentration is at a peak. The derivative, found to be \( C'(t) = 23e^{-2t}(1-2t) \), is set to zero to solve for \( t \), revealing the time at which \( C_{max} \) occurs. This helps ensure that drugs remain within therapeutic windows, maximizing efficacy and minimizing risks.