Problem 22
Question
The rate at which a drug leaves the bloodstream and passes into the urine is proportional to the quantity of the drug in the blood at that time. If an initial dose of \(Q_{0}\) is injected directly into the blood, \(20 \%\) is left in the blood after 3 hours. (a) Write and solve a differential equation for the quantity, \(Q,\) of the drug in the blood after \(t\) hours. (b) How much of this drug is in a patient's body after 6 hours if the patient is given 100 mg initially?
Step-by-Step Solution
Verified Answer
After 6 hours, 4 mg of the drug remains in the bloodstream.
1Step 1: Understand the Problem
We are given that the rate at which the drug leaves the bloodstream is proportional to its current quantity in the blood. This implies a first-order differential equation. We also know that 20% of the initial amount remains in the blood after 3 hours.
2Step 2: Formulate the Differential Equation
Since the rate of change of the drug quantity, \( Q \), is proportional to \( Q \), we can express this relationship as \( \frac{dQ}{dt} = -kQ \), where \( k \) is the proportionality constant.
3Step 3: Solve the Differential Equation
To solve the differential equation \( \frac{dQ}{dt} = -kQ \), separate variables and integrate: \( \int \frac{1}{Q} \ dQ = -\int k \ dt \). This gives \( \ln |Q| = -kt + C \), or \( Q = Ce^{-kt} \) after exponentiating both sides.
4Step 4: Apply Initial Conditions
Given \( Q(0) = Q_0 \), the initial condition, substitute into the solution to find \( C \): \( Q_0 = Ce^0 \rightarrow C = Q_0 \). Thus, \( Q(t) = Q_0 e^{-kt} \).
5Step 5: Determine the Proportionality Constant
We know that after 3 hours (\( t = 3 \)), 20% of the drug is left, so \( Q(3) = 0.2 Q_0 \). Substitute into the equation: \( 0.2 Q_0 = Q_0 e^{-3k} \), which simplifies to \( e^{-3k} = 0.2 \). Solving for \( k \), we get \( k = -\frac{1}{3} \ln(0.2) \).
6Step 6: Solve for Quantity After 6 Hours
Substitute \( t = 6 \) and \( Q_0 = 100 \) mg into the equation \( Q(t) = Q_0 e^{-kt} \): \( Q(6) = 100 e^{-6\left(-\frac{1}{3} \ln(0.2)\right)} \). Compute \( Q(6) = 100 \cdot (0.2)^{2} \).
7Step 7: Final Calculation
Calculate \( Q(6) = 100 \cdot 0.04 = 4 \) mg. Thus, 4 mg of the drug remains in the bloodstream after 6 hours.
Key Concepts
Exponential DecayFirst-order Differential EquationProportionality Constant
Exponential Decay
Exponential decay is a mathematical concept used to describe processes that decrease rapidly over time. It is often found in natural phenomena, such as radioactive decay or the cooling of a hot object. In the context of the exercise, we are looking at how a drug leaves the bloodstream, diminishing in a similar exponential manner.
This means that the rate of decrease of the drug's quantity is proportional to its current amount in the bloodstream. If you think of it like this: the more drug there is, the faster it leaves, but as the quantity lessens, the rate of decay slows down.
For example, in the original problem, 20% of the drug remains after 3 hours. This information is crucial because it allows us to find the exact rate at which the drug is disappearing using our exponential decay formula. Mathematically, this is typically expressed by the equation:
This means that the rate of decrease of the drug's quantity is proportional to its current amount in the bloodstream. If you think of it like this: the more drug there is, the faster it leaves, but as the quantity lessens, the rate of decay slows down.
For example, in the original problem, 20% of the drug remains after 3 hours. This information is crucial because it allows us to find the exact rate at which the drug is disappearing using our exponential decay formula. Mathematically, this is typically expressed by the equation:
- \( Q(t) = Q_0 e^{-kt} \)
First-order Differential Equation
First-order differential equations are ordinary differential equations that involve the first derivative of an unknown function. This is the simplest type of differential equation and forms the backbone of modeling various natural phenomena, such as the drug decay issue at hand.
The equation given in the solution, \( \frac{dQ}{dt} = -kQ \), is a textbook example of a first-order linear differential equation. Here, the derivative \( \frac{dQ}{dt} \) denotes the rate of change of the drug quantity \( Q \) with respect to time \( t \), which is directly proportional to the quantity \( Q \) itself.
This concept is pivotal in understanding how certain quantities, such as the level of drug in the bloodstream, change over time. Through integration of such equations, we derive the general solution \( Q(t) = Q_0 e^{-kt} \), showing how \( Q \) decreases over time.
The equation given in the solution, \( \frac{dQ}{dt} = -kQ \), is a textbook example of a first-order linear differential equation. Here, the derivative \( \frac{dQ}{dt} \) denotes the rate of change of the drug quantity \( Q \) with respect to time \( t \), which is directly proportional to the quantity \( Q \) itself.
This concept is pivotal in understanding how certain quantities, such as the level of drug in the bloodstream, change over time. Through integration of such equations, we derive the general solution \( Q(t) = Q_0 e^{-kt} \), showing how \( Q \) decreases over time.
Proportionality Constant
The proportionality constant, often denoted as \( k \), plays a crucial role in differential equations. It tells us how strongly one quantity depends on another. In our drug example, \( k \) indicates how quickly the drug is cleared from the bloodstream relative to its current level.
To find \( k \), we use given conditions from the problem. For instance, when 20% of the drug remains after 3 hours, we substitute these values into our differential equation solution:\( Q(3) = 0.2 Q_0 = Q_0 e^{-3k} \). Solving this gives us \( e^{-3k} = 0.2 \), allowing us to find \( k \) as \( k = -\frac{1}{3} \ln(0.2) \). This precisely quantifies the rate of decay of the drug.
Understanding \( k \) helps us apply differential equations to real-world problems, allowing accurate predictions and informed decision-making.
To find \( k \), we use given conditions from the problem. For instance, when 20% of the drug remains after 3 hours, we substitute these values into our differential equation solution:\( Q(3) = 0.2 Q_0 = Q_0 e^{-3k} \). Solving this gives us \( e^{-3k} = 0.2 \), allowing us to find \( k \) as \( k = -\frac{1}{3} \ln(0.2) \). This precisely quantifies the rate of decay of the drug.
Understanding \( k \) helps us apply differential equations to real-world problems, allowing accurate predictions and informed decision-making.
Other exercises in this chapter
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