Problem 58 - Calculus and its applications

Question

The reaction \(R\) in pleasure units by a consumer receiving \(S\) units of a product can be modeled by the differential equation \(\frac{d R}{d S}=\frac{k}{S+1}\) where \(k\) is a positive constant. a) Use separation of variables to solve the differential equation. b) Rewrite the solution in terms of the initial condition \(R(0)=0\) c) Explain why the condition \(R(0)=0\) is reasonable.

Step-by-Step Solution

Answer

The solution is \( R = k \ln(S+1) \). The condition \( R(0)=0 \) is reasonable because no product received results in no pleasure.

Step 1: Separation of Variables

To use separation of variables, we start with the given differential equation \( \frac{dR}{dS} = \frac{k}{S+1} \). We separate the variables to get \( dR = \frac{k}{S+1} \cdot dS \).

Step 2: Integrate Both Sides

Next, integrate both sides with respect to their respective variables: \[ \int dR = \int \frac{k}{S+1} \cdot dS \] This gives \[ R = k \ln|S+1| + C \] where \( C \) is the constant of integration.

Step 3: Apply the Initial Condition

We use the initial condition \( R(0) = 0 \) to find \( C \). Substitute \( S = 0 \) and \( R = 0 \) into the equation: \[ 0 = k \ln|0+1| + C \] This simplifies to \( C = 0 \).

Step 4: Rewrite the Solution

With \( C = 0 \), the solution to the differential equation is \[ R = k \ln(S+1) \].

Step 5: Justify the Initial Condition

The initial condition \( R(0) = 0 \) is reasonable because it implies that if the consumer has not received any units of the product (\( S = 0 \)), their reaction in pleasure units (\( R \)) should naturally be zero, since no interaction with the product has occurred.

Key Concepts

Differential Equations

Differential equations are mathematical equations that relate a function with its derivatives. In simpler terms, these equations involve rates of change and describe how one variable changes with respect to another.
They are essential in modeling real-world phenomena because many processes naturally change over time. For example, they can model population growth, heat transfer, or, as in our exercise, a consumer's reaction to receiving a product.
In our case, the differential equation is \( \frac{dR}{dS} = \frac{k}{S+1} \), where \( R \) is the reaction of a consumer and \( S \) is the number of product units. The left side of the equation, \( \frac{dR}{dS} \), represents how quickly the reaction \( R \) changes as more product units are received.

Integration

Integration is a fundamental concept in calculus that helps us find functions from their rate of change (derivatives). Think of it as the reverse operation of differentiation.
When solving differential equations, we often use integration to find the general solution. After separating variables, the next step is to integrate both sides of the equation.

• For the equation \( \frac{dR}{dS} = \frac{k}{S+1} \), separating variables gives us \( dR = \frac{k}{S+1} \cdot dS \).
• Integrating the left side, \( \int dR \), gives us \( R \).
• Integrating the right side, \( \int \frac{k}{S+1} \cdot dS \), results in \( k \ln|S+1| + C \), where \( C \) is the constant of integration.

The general solution tells us how \( R \) changes with \( S \), up to the unknown constant \( C \).

Initial Conditions

Initial conditions help us find unique solutions for differential equations. They provide specific values at a particular point, allowing us to determine any constants in the general solution.
In our exercise, the initial condition is \( R(0) = 0 \), meaning the reaction is zero when no product units have been received. This condition is crucial to find the specific value of \( C \).
Using the condition \( R(0) = 0 \):

• Substitute \( S = 0 \) and \( R = 0 \) into \( R = k \ln|S+1| + C \).
• This gives \( 0 = k \ln|0+1| + C \).
• Since \( \ln 1 = 0 \), it simplifies to \( C = 0 \).

Thus, the initial condition allows us to determine that the specific solution is \( R = k \ln(S+1) \). It logically makes sense because if no units are received, there should be no reaction in pleasure.