Problem 16 - Applied Calculus

Question

A chain smoker smokes five cigarettes every hour. From cach cigarette, \(0.4 \mathrm{mg}\) 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 \(11 \mathrm{pm}(16\) hours later \() ?\)

Step-by-Step Solution

Answer

(a) \( \frac{dN}{dt} = 2 - 0.346N \); (b) \( N(t) = \frac{2}{0.346} (1 - e^{-0.346t}) \); (c) Nicotine at 11 pm is approximately 5.78 mg.

Step 1: Understand the Problem

The problem requires us to set up a differential equation to describe how the nicotine level changes over time due to smoking and natural removal from the bloodstream. We need to find the amount of nicotine in the blood at a later time when starting from no nicotine.

Step 2: Formulate the Differential Equation (Part a)

We know 5 cigarettes are smoked every hour, with each cigarette contributing 0.4 mg of nicotine. Thus, nicotine inputs are constant at 2 mg/hour. Nicotine is removed at a rate proportional to its current level, where the constant of proportionality is -0.346. So, the differential equation is \( \frac{dN}{dt} = 2 - 0.346N \).

Step 3: Identify the Initial Condition

The problem states that initially there is no nicotine in the blood. Therefore, the initial condition is \( N(0) = 0 \).

Step 4: Solve the Differential Equation (Part b)

The differential equation \( \frac{dN}{dt} = 2 - 0.346N \) is a linear first-order differential equation. It can be solved using the integrating factor method. The integrating factor is \( e^{0.346t} \). Multiply the whole differential equation by this factor and solve for \( N(t) \). After integrating and applying the initial condition \( N(0) = 0 \), we find \( N(t) = \frac{2}{0.346} (1 - e^{-0.346t}) \).

Step 5: Calculate Nicotine at 11 pm (Part c)

At 7 am, \( t = 0 \), and at 11 pm, \( t = 16 \) hours. We use \( N(t) = \frac{2}{0.346} (1 - e^{-0.346 \times 16}) \) to find \( N(16) \). Calculate the expression to find the nicotine level at 11 pm.

Key Concepts

Nicotine Elimination

Nicotine elimination from the body is an important aspect to consider, especially when analyzing its accumulation due to smoking habits. The rate of nicotine removal is not constant but depends on the current level of nicotine in the bloodstream. This is because the body metabolizes nicotine at a rate proportional to how much is present. In this case, this relationship is captured by a negative proportionality constant, -0.346, indicating a decreasing rate of nicotine as it gets eliminated. The negative sign in the proportionality constant signifies that as time progresses, the nicotine level decreases. This type of relationship, where the rate of change of a substance is proportional to its current amount, is common in natural processes and is represented mathematically by a differential equation. Understanding this concept helps us grasp how substances like nicotine are processed in the body over time.

Rate of Change

The rate of change in the level of nicotine in the body can be mathematically expressed using a differential equation. The foundational equation given in the exercise is \( \frac{dN}{dt} = 2 - 0.346N \), where \( N \) represents the nicotine concentration, and \( t \) is time in hours.

• The first term, 2 mg, represents the constant nicotine input from regular smoking.
• The second term, \( -0.346N \), represents the removal of nicotine at a rate proportional to its current level.

The constant of proportionality, -0.346, reflects the body's efficiency in processing nicotine. Solving this differential equation provides us with the nicotine level in the body as a function of time. By examining the rate of change, we gain insights into how habits or treatments might affect nicotine levels and overall health over time. Understanding these dynamics is crucial in fields like pharmacology and health sciences.

Initial Condition

In solving differential equations, the initial condition refers to the state of the system at the beginning of the process. It is a vital component since it allows us to find a particular solution that matches the specific scenario under consideration. In this exercise, the initial condition is straightforward: at 7 am, when the smoker begins their day, there is no nicotine in the bloodstream. Mathematically, this is stated as \( N(0) = 0 \). This initial state is crucial as it anchors the solution, enabling us to incorporate how changes occur over time from that starting point.Using the initial condition, we solve for the particular solution \( N(t) = \frac{2}{0.346} (1 - e^{-0.346t}) \). By ensuring that the initial condition is factored correctly into our calculations, the solution embodies both the input of nicotine and its elimination from the bloodstream over a given period, providing a realistic depiction of how nicotine levels evolve through the day.