Problem 16 - Applied Calculus

Question

A chain smoker smokes five cigarettes every hour. From each cigarette, 0.4 mg of nicotine is absorbed into the person's bloodstream. Nicotine leaves the body at a rate proportional to the amount present, with constant of proportionality -0.346 if \(t\) is in hours. (a) Write a differential equation for the level of nicotine in the body, \(N,\) in \(\mathrm{mg},\) as a function of time, \(t,\) in hours. (b) Solve the differential equation from part (a). Initially there is no nicotine in the blood. (c) The person wakes up at 7 am and begins smoking. How much nicotine is in the blood when the person goes to sleep at 11 pm ( 16 hours later)?

Step-by-Step Solution

Answer

The differential equation is \( \frac{dN}{dt} = 2 - 0.346N \). After 16 hours, approximately 5.78 mg of nicotine is in the blood.

Step 1: Formulate the Differential Equation

We are given that nicotine is absorbed into the bloodstream at a rate of 0.4 mg per cigarette, and the smoker smokes 5 cigarettes every hour. Thus, the nicotine entering the bloodstream per hour is \(5 \times 0.4 = 2\) mg. The elimination of nicotine from the bloodstream is proportional to the current amount present, which gives us a rate of \(-0.346N\). Therefore, the differential equation governing the nicotine level \(N(t)\) is: \[ \frac{dN}{dt} = 2 - 0.346N \]

Step 2: Solve the Differential Equation

To solve the differential equation \( \frac{dN}{dt} = 2 - 0.346N \), we rearrange it to the standard form: \( \frac{dN}{dt} + 0.346N = 2 \). This is a linear first-order differential equation and can be solved using an integrating factor. The integrating factor is \( \mu(t) = e^{\int 0.346 \, dt} = e^{0.346t} \). Multiply through by the integrating factor: \[ e^{0.346t} \frac{dN}{dt} + 0.346e^{0.346t}N = 2e^{0.346t} \] \[ \frac{d}{dt}(e^{0.346t}N) = 2e^{0.346t} \] Integrate both sides with respect to \(t\): \[ e^{0.346t}N = \int 2e^{0.346t} \, dt \] \[ e^{0.346t}N = \frac{2}{0.346}e^{0.346t} + C \] Solving for \(N(t)\), we have: \[ N(t) = \frac{2}{0.346} + Ce^{-0.346t} \]

Step 3: Apply Initial Condition

The initial condition is that initially, there is no nicotine in the blood: \(N(0) = 0\). Substitute \(t = 0\) into the solution to find \(C\): \[ 0 = \frac{2}{0.346} + Ce^{0} \] \[ C = -\frac{2}{0.346} \] Substitute \(C\) back into the equation: \[ N(t) = \frac{2}{0.346} - \frac{2}{0.346}e^{-0.346t} \] Simplify to obtain: \[ N(t) = \frac{2}{0.346}(1 - e^{-0.346t}) \]

Step 4: Calculate Nicotine Level After 16 Hours

The person smokes from 7 am to 11 pm, which is a total of 16 hours. Substitute \(t = 16\) into \(N(t)\): \[ N(16) = \frac{2}{0.346}(1 - e^{-0.346 \times 16}) \] Calculate the numerical value: \[ N(16) \approx 5.78 \text{ mg} \]

Key Concepts

Nicotine Absorption

When a person smokes a cigarette, nicotine is introduced into the bloodstream. In the case of the exercise, the chain smoker consumes 5 cigarettes per hour, with each cigarette contributing 0.4 mg of nicotine.
This consistent intake results in 2 mg of nicotine being absorbed into the bloodstream every hour. This information is crucial because it establishes the constant rate at which nicotine is being introduced into the body.
Understanding nicotine absorption is important in creating a mathematical model of its presence in the body. The specific value of 2 mg per hour enters the differential equation as a term that accounts for the increase of nicotine, balancing against the constant elimination rate.

Rate of Change

The concept of rate of change is central to understanding how systems evolve over time. In this exercise, we describe how the level of nicotine changes in the bloodstream.
Differential equations are mathematical tools that model these changes. The rate of change of nicotine, as given in \( \frac{dN}{dt} = 2 - 0.346N \), combines both the absorption and elimination of nicotine.
The term 2 represents the constant addition of nicotine, while the term \(-0.346N\) indicates the rate at which nicotine is expelled from the body. It's a descriptive way of how nicotine levels increase and decrease simultaneously. By solving this equation, one can predict nicotine levels over time.

Initial Condition

An initial condition is a starting point used to solve differential equations, providing specific data needed to find a unique solution.
In this context, the initial condition is that at time \(t = 0\), \(N(0) = 0\), meaning that initially, there is no nicotine in the bloodstream.
This condition is essential for finding the integration constant, \(C\), in the process of solving the differential equation \(N(t) = \frac{2}{0.346} + Ce^{-0.346t}\). By substituting \(N(0) = 0\) into this equation, we can calculate \(C\).
This step ensures that the solution fits the real-world scenario described in the problem.

Exponential Decay

The concept of exponential decay characterizes processes where quantities decrease at a rate proportional to their current value. In the context of this exercise, nicotine leaves the body according to an exponential decay model.
The decay is expressed by the term \(-0.346N\) in the differential equation, which shows that as the nicotine level \(N\) increases, the rate of its decrease also increases. This is typical of exponential decay, where the quantity diminishes faster as it grows larger.
In the final solution, \(N(t) = \frac{2}{0.346}(1 - e^{-0.346t})\), the exponential portion \(e^{-0.346t}\) represents this decay. This term approaches zero as time goes to infinity, indicating that, eventually, nicotine will leave the system if no new nicotine is added. The solution reveals the dynamic balance of absorption and elimination over time.