Problem 68 - Single Variable Calculus

Question

Suppose that a dose of \(C\) units of a certain drug is administered to a patient and that the fraction of the dose remaining in the patient's bloodstream \(t\) hr after the dose is administered is given by \(\mathrm{Ce}^{-k t}\), where \(k\) is a positive constant. a. Show that the residual concentration of the drug in the bloodstream after extended treatment when a dose of \(C\) units is administered at intervals of \(t\) hr is given by $$ R=\frac{C e^{-k t}}{1-e^{-k t}} $$ b. If the highest concentration of this particular drug that is considered safe is \(S\) units, find the minimal time that must exist between doses. Hint: \(C+R \leq S\)

Step-by-Step Solution

Answer

The residual concentration formula for the drug after extended treatment when a dose of \(C\) units is administered at intervals of \(t\) hours is given by: \[ R = \frac{C e^{-k t}}{1-e^{-k t}} \] To maintain a safe drug concentration, the minimum time between doses \(t\) must be: \[t \geq \frac{1}{k} \ln \left( \frac{1}{1 - \frac{C}{S}} \right)\]

Step 1: Part (a): Residual Concentration Formula

We can find the residual concentration by considering the sum of the present dose and all prior doses, considering their corresponding residual concentrations. Let's denote the \(n\)-th residual concentration by \(R_n\). Suppose the patient has already received \(n\) doses, each separated by time interval \(t\). Then: 1. The first dose contributes a concentration of \(C e^{-knt}\) at time \(nt\). 2. The second dose contributes a concentration of \(Ce^{-k(nt- t)}\) at time \(nt\). 3. The third dose contributes a concentration of \(Ce^{-k(nt-2t)}\) at time \(nt\). 4. ... 5. The (n-1)-th dose contributes a concentration of \(Ce^{-k(nt-(n-2)t)}\) at time \(nt\). 6. The n-th dose contributes a concentration of \(Ce^{-kt}\) at time \(nt\). Now we want to find the residual concentration, which is the sum of the concentrations contributed by each dose: \[ R_{n} = C \left(e^{-k n t} + e^{-k (n - 1) t} + \cdots + e^{-k t}\right) \] We can notice that this is a geometric sum with common ratio \(e^{-kt}\). So, \[ R_{n} = \frac{C(e^{-kt} - e^{-knt})}{1 - e^{-kt}} \] As the treatment continues, n becomes infinitely large and the residual concentration formula converges to: \[R = \lim_{n \to \infty} R_{n} = \frac{C e^{-k t}}{1-e^{-k t}}\]

Step 2: Part (b): Minimum Time Between Doses

Now we need to find the minimum time between doses to keep the drug concentration below the safe level. According to the hint, we have: \[ C + R \leq S\] Substitute the expression for R from part (a) into the inequality: \[ C + \frac{C e^{-k t}}{1-e^{-k t}} \leq S\] To find the minimum value of \(t\), we need to solve this inequality for \(t\). First, let's simplify the inequality: \[ \frac{C(1-e^{-kt} + e^{-kt})}{1-e^{-kt}} \leq S\] \[ \frac{C}{1-e^{-kt}} \leq S\] Now we can solve for \(t\): \[ 1 - e^{-kt} \geq \frac{C}{S}\] \[ e^{-kt} \leq 1 - \frac{C}{S}\] Now, we need to take the logarithm on both sides. Remember that \(k\) is a positive constant: \[ -kt \leq \ln \left(1 - \frac{C}{S}\right)\] The minimum time between doses \(t\) is therefore given by: \[t \geq \frac{1}{k} \ln \left( \frac{1}{1 - \frac{C}{S}} \right)\]

Key Concepts

Residual Concentration Formula

When a drug is administered repeatedly, understanding its residual concentration in the body is crucial. This formula helps us find the remaining concentration from previous doses. Imagine you receive a dose of a drug, and over time, only a fraction remains in your bloodstream. This leftover fraction can be calculated using the expression \( R = \frac{C e^{-kt}}{1-e^{-kt}} \), where:

• \(C\): Initial dose amount.
• \(e^{-kt}\): The fraction left after time \(t\).
• \(k\): A constant rate of decay.

As more doses are given, their collective residual amount accumulates, forming a series. The formula accounts for these accumulations, resulting in a final simplified expression of the total residual concentration.

Geometric Series

When you deal with repeated doses of a drug, they form a geometric series in terms of their residue. Each subsequent dose contributes a smaller fraction to the total concentration because of exponential decay. Let's break it down:

• The nth dose contributes \(Ce^{-knt}\), the (n-1)-th dose contributes \(Ce^{-k(n-1)t}\), and so forth.
• These contributions form a geometric progression with the common ratio \(e^{-kt}\).

A geometric series has a specific form that allows us to sum these terms easily:\( R_{n} = C(e^{-kt} + e^{-2kt} + \,\ldots\, + e^{-knt}) \).As \(n\) becomes very large (∞), this series typically converges to a simpler form, represented by our residual concentration formula. This convergence is what lets us understand the long-term accumulation of the drug in the bloodstream.

Inequality Solving

Inequalities help us determine safe conditions for drug administration, like ensuring drug concentration remains below a safe threshold \(S\). Here’s a step-by-step breakdown:

• The inequality \( C + R \leq S \) ensures that both the new and residual concentration stay safe.
• You substitute the residual concentration formula to solve: \( C + \frac{C e^{-kt}}{1-e^{-kt}} \leq S \).
• Simplifying gives: \( \frac{C}{1-e^{-kt}} \leq S \).

To find \(t\), rearrange and solve for \(e^{-kt}\), leading to:\( e^{-kt} \leq 1 - \frac{C}{S} \).By taking logarithms, you find the time gap needed to be:\( t \geq \frac{1}{k} \ln \left( \frac{1}{1 - \frac{C}{S}} \right) \).This solution helps figure out how long to wait between doses to keep patients safe.

Drug Safety

Drug safety ensures medications remain beneficial without causing harm. In our context, it means keeping blood drug levels under a safe upper limit \(S\). This involves:

• Understanding the body's rate of breaking down and excreting the drug, characterized by the decay constant \(k\).
• Making sure that the combined effect of the initial and residual drug concentration does not exceed \(S\).
• Using the formula \( C + R \leq S \) to calculate safe intervals between doses.

This knowledge empowers physicians to optimize dose timing, maintaining drug efficiency while reducing risks of toxicity. Ultimately, balancing therapeutic effects and safety is crucial for patient well-being.