Problem 32 - Applied Calculus

Question

At a time \(t\) hours after taking a tablet, the rate at which a drug is being eliminated is $$ r(t)=50\left(e^{-0.1 t}-e^{-0.2 t}\right) \mathrm{mg} / \mathrm{hr} $$ Assuming that all the drug is eventually eliminated, calculate the original dose.

Step-by-Step Solution

Answer

The original dose of the drug is 250 mg.

Step 1: Understanding the Problem

We need to find the original dose of the drug. The given function \( r(t) \) represents the rate of elimination of the drug at time \( t \), measured in mg/hr. The original dose can be found by integrating this rate over time from 0 to infinity.

Step 2: Set Up the Integral

The original dose of the drug is the total amount of drug eliminated over time. Therefore, we need to evaluate the integral of \( r(t) \) from 0 to infinity:\[\text{Dose} = \int_0^{\infty} 50\left(e^{-0.1 t} - e^{-0.2 t}\right) \, dt\]

Step 3: Evaluate the Integral

Calculate the integral by splitting it into two parts:\[\int_0^{\infty} 50e^{-0.1t} \, dt - \int_0^{\infty} 50e^{-0.2t} \, dt\]

Step 4: Compute Each Integral Separately

Calculate the first integral:\[\int_0^{\infty} 50e^{-0.1t} \, dt = 50 \left[ \frac{-1}{0.1} e^{-0.1t} \right]_0^{\infty} = 500\]and the second integral:\[\int_0^{\infty} 50e^{-0.2t} \, dt = 50 \left[ \frac{-1}{0.2} e^{-0.2t} \right]_0^{\infty} = 250\]

Step 5: Calculate the Original Dose

Subtract the second integral from the first to find the total dose eliminated:\[\text{Dose} = 500 - 250 = 250 \, \text{mg}\]

Key Concepts

Drug Elimination

Drug elimination is a critical process in pharmacology, describing how a drug is removed from the body. When a drug is administered, your body begins to metabolize and eliminate it through processes like excretion via urine or breakdown in the liver. Understanding this rate of elimination is vital for drug dosing and safety. In our exercise, the aim is to determine how much of a drug was originally taken by understanding how quickly and efficiently it leaves the system over time.

In practical terms, drug elimination is often modeled with mathematical equations to predict how quickly a drug's concentration decreases. This helps in deciding dosing schedules and assessing drug interactions. The function given in the exercise is a typical way to represent the rate of elimination. It uses exponential decay to show how the drug concentration diminishes over time.

Exponential Functions

Exponential functions play a significant role in modeling real-world phenomena, such as growth processes and decay, like our drug elimination example. An exponential function involves a constant base raised to a variable exponent. In the exercise, we see this in the form of two terms: \( e^{-0.1t} \) and \( e^{-0.2t} \).

Here, the base \( e \) (Euler's number, approximately 2.718) is a constant that makes exponential functions particularly suitable for continuous growth or decay scenarios. When the exponent is negative, as in this exercise, it describes a decay process, reflecting how the drug reduces in concentration over time. Understanding these functions is essential for biology and pharmacology as they allow for accurate simulations of variable processes over time.

Definite Integrals

Definite integrals provide a powerful tool for calculating the total accumulation of quantities, such as the total amount of a drug eliminated over time in our exercise. Simply put, the definite integral of a function over an interval gives you a number that represents the total sum or net area under the curve of a graph. This is exactly what we use it for in this scenario.

The mathematical expression used involved calculating the integral of the rate function from 0 to infinity. This represents the total drug that was initially taken. For an exponential decay function like the one in the exercise, computing this integral is feasible because these functions approach zero as time goes to infinity, making them converge. By solving these integrals, we attain the original dose administered. Mastering the use of definite integrals is pivotal for understanding how various quantities accumulate and interact over time in scientific calculations.