Problem 12 - Calculus

Question

Caffeine is metabolized and excreted at a continuous rate of about \(17 \%\) per hour. A person with no caffeine in the body starts drinking coffee, containing 130 mg of caffeine per cup, at 7 am. The person drinks coffee continuously all day at the rate of one cup an hour. Write a differential equation for \(A\), the amount of caffeine in the body \(t\) hours after 7 am and give the particular solution to this differential equation. How much caffeine is in the person's body at \(5 \mathrm{pm} ?\)

Step-by-Step Solution

Answer

There is approximately 765.5 mg of caffeine in the person's body at 5 PM.

Step 1: Understand the Metabolism Rate

The caffeine is metabolized and excreted at a continuous rate of 17% per hour. This means that the amount of caffeine decreases at a rate proportional to its current amount.

Step 2: Determine the Caffeine Inflow

A person drinks coffee continuously at a rate of one cup per hour, with each cup containing 130 mg of caffeine. Thus, the rate of caffeine intake is constant at 130 mg/hour.

Step 3: Set Up the Differential Equation

The rate of change of the amount of caffeine, \(A\), in the body can be modeled by the differential equation:\[\frac{dA}{dt} = 130 - 0.17A\]The equation considers both the inflow of caffeine and the metabolism rate, where the rate of intake is 130 mg/hour and the outflow rate is 17% of the current amount of caffeine.

Step 4: Solve the Differential Equation

To solve \(\frac{dA}{dt} = 130 - 0.17A\), we separate variables:\[\frac{dA}{130 - 0.17A} = dt\]Integrate both sides to find an expression for \(A\):\[\int \frac{1}{130 - 0.17A} dA = \int dt\]The solution is:\[A(t) = \frac{130}{0.17} + Ce^{-0.17t}\]

Step 5: Determine Constant C

Since the person starts with no caffeine (\(A(0) = 0\)), substitute into the equation:\[0 = \frac{130}{0.17} + Ce^0\]Solving gives:\[C = -\frac{130}{0.17}\]

Step 6: Formulate the Particular Solution

Plug \(C\) back into the general solution:\[A(t) = \frac{130}{0.17} - \frac{130}{0.17}e^{-0.17t}\]

Step 7: Calculate Caffeine at 5 PM

5 PM is 10 hours after 7 AM. Substitute \(t = 10\) into the particular solution:\[A(10) = \frac{130}{0.17} - \frac{130}{0.17}e^{-0.17 \times 10}\]Calculate the value:\[A(10) \approx \frac{130}{0.17}(1 - e^{-1.7}) \approx 765.5 \text{ mg}\]

Key Concepts

Caffeine Metabolism

Caffeine metabolism is the biological process by which caffeine is broken down in the body.
In this exercise, the metabolism rate is key. It is indicated that caffeine is metabolized and excreted at a continuous rate of \(17\%\) per hour.
This implies that the caffeine level in the body decreases over time, as some of it gets processed and eliminated.

• The metabolism rate (\(17\%\)) means each hour, \(17\%\) of the caffeine in the body is metabolized.
• Understanding this percentage is crucial as it affects how much caffeine remains in the body at any given time.

Caffeine metabolism, like many biological processes, is modeled mathematically using a rate of change, combined with other factors such as caffeine intake, to form a comprehensive differential equation. Understanding this helps grasp how caffeine builds up and dissipates in the system.

Rate of Change

Rate of change refers to how quickly a quantity, like caffeine level, changes over time.
In the case of this problem, caffeine in the body changes due to two factors: intake from drinking coffee and elimination by metabolism. These changes are addressed by the differential equation, \(\frac{dA}{dt} = 130 - 0.17A\).

• \(\frac{dA}{dt}\) is the mathematical expression for the rate of change of caffeine in the body, \(A\).
• The first part of the equation, \(130\), represents a constant intake (one cup of coffee per hour).
• The second part, \(-0.17A\), represents the rate of caffeine being metabolized, where \(A\) is the current amount of caffeine in the body.

This rate of change equation gives a dynamic picture of how caffeine influences the body throughout the day. As time progresses, these dynamics shift based on consumption patterns and the body's natural metabolic rate.

Particular Solution

A particular solution is a specific answer to a differential equation that satisfies some initial conditions.
In this scenario, the initial condition is when no caffeine is in the body at the starting point, \(t = 0\).

• The general solution to our equation is \(A(t) = \frac{130}{0.17} + Ce^{-0.17t}\).
• By applying \(A(0) = 0\) (the initial condition), we determine that \(C = -\frac{130}{0.17}\).
• Substituting \(C\) back gives us the particular solution: \[A(t) = \frac{130}{0.17} - \frac{130}{0.17}e^{-0.17t}\].

This solution allows us to predict how much caffeine is in the body at any future point, such as at 5 PM (10 hours after initial consumption begins). Calculations show \(A(10) \approx 765.5\) mg of caffeine.