Problem 48 - Calculus Single Variable

Question

Suppose that during wartime, a particular type of weapon is produced at a constant rate \(\mu\) and destroyed in battle at a constant relative rate \(\delta .\) If there were \(N_{0}\) of these weapons at the outbreak of the war \((t=0),\) then how many were there at time \(t ?\) If the war is a lengthy one, about how many of these weapons will be on hand at time \(t\) when \(t\) is large?

Step-by-Step Solution

Answer

At time \(t\), the number of weapons is \( N(t) = (N_0 - \frac{\mu}{\delta})e^{-\delta t} + \frac{\mu}{\delta} \). For a lengthy war, \( N(t) \to \frac{\mu}{\delta} \).

Step 1: Define the Differential Equation

The number of weapons at time \(t\) is \(N(t)\). These weapons are produced at a constant rate \(\mu\) and destroyed at a relative rate \(\delta\), so we can write the rate of change as \( \frac{dN}{dt} = \mu - \delta N \).

Step 2: Solve the Homogeneous Equation

First, let's consider the homogeneous part of the differential equation \( \frac{dN}{dt} = -\delta N \). The solution to this is of the form \( N_h(t) = Ce^{-\delta t} \), where \(C\) is a constant.

Step 3: Find a Particular Solution

Assume a particular solution of the form \( N_p(t) = \frac{\mu}{\delta} \). This satisfies the equation \( \frac{dN_p}{dt} = 0 = \mu - \delta \frac{\mu}{\delta} \), which holds true. Hence, \( N_p(t) = \frac{\mu}{\delta} \) is indeed a particular solution.

Step 4: General Solution

The general solution to the differential equation \( \frac{dN}{dt} = \mu - \delta N \) is the sum of the homogeneous and particular solutions: \( N(t) = Ce^{-\delta t} + \frac{\mu}{\delta} \).

Step 5: Apply Initial Condition

At time \( t = 0 \), there are \( N_0 \) weapons, so \( N(0) = C + \frac{\mu}{\delta} = N_0 \). Solving for \(C\), we get \( C = N_0 - \frac{\mu}{\delta} \).

Step 6: Substitute Back to Find N(t)

Substitute \( C = N_0 - \frac{\mu}{\delta} \) back into the general solution to get \( N(t) = \left(N_0 - \frac{\mu}{\delta}\right)e^{-\delta t} + \frac{\mu}{\delta} \).

Step 7: Find Long-Term Behavior

As \( t \to \infty \), the term \( (N_0 - \frac{\mu}{\delta})e^{-\delta t} \to 0 \) because of the exponential decay. Hence, for large \(t\), the number of weapons approximates to \( \frac{\mu}{\delta} \).

Key Concepts

Understanding Exponential Decay

Exponential decay is a key concept in differential equations, often describing how quantities diminish over time. In our exercise, the destruction of weapons over time is an example of this process. We express it mathematically with the term \( e^{-\delta t} \), which signifies that as time \( t \) increases, this expression decreases rapidly.

• The base of the exponential function is \( e \), approximately equal to 2.71828, and is the natural logarithm base.
• \( \delta \) is the decay constant - it dictates the rate at which the weapons are destroyed.
• As \( t \) becomes larger, \( e^{-\delta t} \) approaches zero.

This exponential decay indicates the diminishing influence of the homogeneous solution component over time, emphasizing the steady state or long-term solution derived from the particular solution.

Exploring Rate of Change

The rate of change in differential equations provides the speed at which quantities increase or decrease over time. In our situation, the rate of change of the weapon count is characterized by the differential equation \( \frac{dN}{dt} = \mu - \delta N \).

• \( \mu \) represents the constant rate of production of weapons.
• \( \delta \) represents the relative rate of destruction.

This equation captures the balance between weapon production and destruction. As production and destruction occur simultaneously, the net change in weapon count over time is determined by both rates.

Importance of Initial Condition

The initial condition in a differential equation provides the starting point from which solutions evolve. In this context, at the outset of war \((t=0)\), there are \(N_0\) weapons available.

• The general solution \( N(t) = Ce^{-\delta t} + \frac{\mu}{\delta} \) depends on this initial condition.
• We determine \( C \) by substituting the initial condition, leading to \( C = N_0 - \frac{\mu}{\delta} \).

This piece of information allows us to solve the differential equation completely, tailoring the solution to fit the real-world scenario described. The specific value of \( C \) ensures that the solution reflects the actual number of weapons available at the start.

Determining Long-Term Behavior

The long-term behavior of a system described by a differential equation shows the state the system settles into after a long time. In our example, we're interested in the amount of weapons as time becomes large \( (t \to \infty) \).

• As time progresses, the effect of the initial condition \((N_0 - \frac{\mu}{\delta})e^{-\delta t} \) diminishes to zero due to exponential decay.
• Thus, the weapon count approaches the stable value of \( \frac{\mu}{\delta} \).

This represents a balanced state where weapon production compensates perfectly for destruction, indicating a steady situation or equilibrium in the context of weapon availability in the wartime scenario.