Problem 15 - Calculus

Question

(a) Define the variables. (b) Write a differential equation to describe the relationship. (c) Solve the differential equation. Nicotine leaves the body at a rate proportional to the amount present, with constant of proportionality 0.347 if the amount of nicotine is in \(\mathrm{mg}\) and time is in hours. The amount of nicotine in the body immediately after smoking a cigarette is \(0.4 \mathrm{mg}\).

Step-by-Step Solution

Answer

The solution is \( N(t) = 0.4e^{-0.347t} \).

Step 1: Define the Variables

Let's define the variables needed to solve this problem. Let \( N(t) \) be the amount of nicotine in the body at any time \( t \), where \( t \) is measured in hours.

Step 2: Establish the Differential Equation

The problem states that the rate at which nicotine leaves the body is proportional to the amount present. This can be formulated as a differential equation: \( \frac{dN}{dt} = -kN \), where \( k \) is the constant of proportionality and is given as 0.347.

Step 3: Apply the Initial Condition

We are given that immediately after smoking a cigarette, the amount of nicotine in the body is \(0.4 \mathrm{mg}\). This gives us the initial condition \( N(0) = 0.4 \).

Step 4: Solve the Differential Equation

Since \( \frac{dN}{dt} = -0.347N \) is a separable differential equation, we separate and integrate: \[ \int \frac{1}{N} \, dN = -0.347 \int \, dt \]. This results in \( \ln |N| = -0.347t + C \). Solve for \( N \) by exponentiating both sides to get \( N(t) = Ce^{-0.347t} \).

Step 5: Use the Initial Condition to Find Constant

With \( N(0) = 0.4 \), we substitute into the equation to find \( C \): \( 0.4 = Ce^{-0.347 \cdot 0} \). Simplifying, \( C = 0.4 \). Thus, the solution is \( N(t) = 0.4e^{-0.347t} \).

Key Concepts

Separable Differential Equation

A separable differential equation is one where the variables can be separated onto different sides of the equation. This makes it easier to solve by integration. In our nicotine example, we have \( \frac{dN}{dt} = -0.347N \). Here, we can rearrange this equation to separate the variables \( N \) and \( t \).
By dividing both sides by \( N \) and multiplying both sides by \( dt \), we obtain:

• \( \frac{1}{N} \, dN = -0.347 \, dt \).

Next, we integrate both sides. The left side with respect to \( N \) and the right side with respect to \( t \). This results in:

• \( \int \frac{1}{N} \, dN = \ln |N| \),
• \( -0.347 \int \, dt = -0.347t \).

These steps highlight the power of separation in solving differential equations.

Rate of Change

The 'rate of change' describes how fast a quantity changes over time. In our nicotine scenario, the rate of change of nicotine amount \( N \) in the body is proportional to the amount present. This is expressed by the differential equation \( \frac{dN}{dt} = -kN \), where \( k \) is the rate constant.
The negative sign indicates that the nicotine amount is decreasing over time. A higher \( k \) value would imply a faster rate of decay. For our case, \( k = 0.347 \) means that each hour, the rate at which nicotine leaves is proportional to how much remains.

Exponential Decay

Exponential decay refers to a decrease that happens at a consistent percentage rate per unit of time. For nicotine, once we solve the differential equation, we get the expression \( N(t) = Ce^{-0.347t} \).
Here, \( e \) is the base of the natural logarithm (approximately 2.718), \( -0.347 \) is the decay rate, and \( t \) is time in hours. The characteristic feature of exponential decay is that the amount of nicotine decreases rapidly at first and then the rate of decrease slows down over time.
This type of decay is common in many natural processes, including radioactive decay and cooling of hot objects.

Initial Condition

The initial condition helps us determine specific solutions of differential equations where general solutions arise from integration. In this nicotine exercise, the initial condition is \( N(0) = 0.4 \), meaning 0.4 mg of nicotine is present right after smoking a cigarette.
To find the specific solution, we substitute the initial condition into the integrated equation \( N(t) = Ce^{-0.347t} \). Evaluating at \( t = 0 \), we have:

• \( N(0) = Ce^{0} = C \).

Solving gives \( C = 0.4 \).
Now we know \( N(t) = 0.4e^{-0.347t} \), fully incorporating the given initial condition into our solution. This step ensures the solution is tailor-made to reflect the real-world situation described in the problem.