Problem 75 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

CONCENTRATION OF DRUG Suppose that \(t\) hours after an antibiotic is administered orally, its concentration in the patient's bloodstream is given by a surge function of the form \(C(t)=A t e^{-k t}\), where \(A\) and \(k\) are positive constants and \(C\) is measured in micrograms per milliliter of blood. Blood samples are taken periodically, and it is determined that the maximum concentration of drug occurs 2 hours after it is administered and is 10 micrograms per milliliter. a. Use this information to determine \(A\) and \(k\). b. A new dose will be administered when the concentration falls to 1 microgram per milliliter. When does this occur?

Step-by-Step Solution

Answer

A \(\approx 13.59\), k \(=\frac{1}{2}\), and the concentration falls to 1 microgram per milliliter approximately at \( t \) \( \approx 6.8 \) hours.

Step 1: Set Up the Given Conditions for Maximum Concentration

Given that the maximum concentration occurs at 2 hours and is 10 micrograms per milliliter, use this information to find equations involving A and k. The conditions are: 1. \(C(2) = 10\) micrograms per milliliter. 2. The derivative \( C'(t) \) zero at \( t = 2 \) since it is a maximum.

Step 2: Differentiate the Concentration Function

Differentiate \( C(t) = A t e^{-k t} \) with respect to \( t \) to find the critical points. \[ C'(t) = A e^{-k t} (1 - k t) \].

Step 3: Find the Value of k

Set the derivative equal to zero and solve for \( k \). \[ A e^{-k t} (1 - k t) = 0 \ 1 - k \cdot 2 = 0 \ k = \frac{1}{2} \]

Step 4: Find the Constant A Using the Maximum Concentration

Plug \( t = 2 \) and \( k = \frac{1}{2} \) into the function to find A. \[ C(2) = A \cdot 2 \cdot e^{-\frac{1}{2} \cdot 2} = 10 \ 2A e^{-1} = 10 \ A = 5e \ \approx 13.59 \ \text{(since } e \approx 2.718 \text{)} \]

Step 5: Determine When the Concentration Falls to 1 Microgram per Milliliter

Set \( C(t) \) to 1 microgram per milliliter and solve for t. \[ 1 = 13.59 t e^{-\frac{1}{2} t} \]

Step 6: Solve for t Using Iterative or Graphical Methods

Solving analytically, \( t \) is non-trivial and may require numerical methods or graphing: Use numerical methods such as Newton-Raphson to find the solution. The approximate solution can be found around \( t \approx 6.8 \) hours.

Key Concepts

surge function

In this exercise, the concentration of a drug in a patient’s bloodstream is modeled using a surge function. This type of function is typically represented in the form: \( C(t) = A \, t \, e^{-kt} \) where:

* \( C(t) \) is the drug concentration at time \( t \)
* \( A \) is a positive constant
* \( k \) is another positive constant
* \( t \) is the time in hours after administration

This equation combines a linear term, \( A t \), with an exponential decay term, \( e^{-kt} \), leading to an initial rise followed by a fall in the concentration over time. The surge function is useful for modeling scenarios where a quantity rapidly increases to a peak before gradually decreasing.

derivative

To solve parts of this problem, we need the derivative of the surge function. The derivative, \( C'(t) \), helps us find critical points where the concentration is at a maximum or minimum. For the function: \( C(t) = A t e^{-kt} \) The derivative is found using the product rule: \( (uv)' = u'v + uv' \). In this case:

* \( u = A t \)
* \( v = e^{-kt} \)
* \( u' = A \)
* \( v' = -k e^{-kt} \)

So the derivative is: \( C'(t) = A e^{-kt} + A t (-k e^{-kt}) \)

Units simplify to: \( C'(t) = A e^{-kt} (1 - kt) \) We set this equal to zero to find critical points as: \( A e^{-kt}(1 - kt) = 0 \) Given \( A \, e^{-kt} eq 0 \), we solve for \( k \) when \( t = 2 \).

maximum concentration

To find maximum concentration, we know from the problem that the drug’s concentration peaks at 2 hours and reaches 10 micrograms per milliliter. From the derivative, setting \( t = 2 \) leads to: \( 1 - 2k = 0 \) So, \( k = \frac{1}{2} \). Next, using the concentration function at the peak: \( C(2) = 10 \) Plugging in \( t = 2 \) and \( k = \frac{1}{2} \): \( 10 = A \cdot 2 \cdot e^{-\frac{1}{2} \cdot 2} \) This becomes: \( 10 = 2A e^{-1} \) So: \( 2A \approx 10 \, e \)
\( A \approx 5 e \approx 13.59 \)

numerical methods

Finally, we determine when the drug concentration falls to 1 microgram per milliliter. So, we set: \( 1 = 13.59 t e^{-\frac{1}{2} t} \) This equation is nonlinear and tricky to solve analytically. Therefore, we use numerical methods, such as the Newton-Raphson method, to approximate \( t \). By iterating, we find the value of \( t \) where the concentration falls to 1 microgram per milliliter.In practice, numerical methods involve:

• Starting with an initial guess for \( t \)
• Using the derivative function to improve the guess
• Repeating until the solution converges

For our specific case, calculations show that \( t \approx 6.8 \) hours is a close approximation to when the concentration dips to 1 microgram per milliliter.