Problem 27 - Calculus Early Transcendentals

Question

(a) Suppose that a quantity \(y=y(t)\) increases at a rate that is proportional to the square of the amount present, and suppose that at time \(t=0,\) the amount present is \(y_{0} .\) Find an initial-value problem whose solution is \(y(t) .\) (b) Suppose that a quantity \(y=y(t)\) decreases at a rate that is proportional to the square of the amount present, and suppose that at a time \(t=0,\) the amount present is \(y_{0} .\) Find an initial-value problem whose solution is \(y(t) .\)

Step-by-Step Solution

Answer

(a) \( \frac{dy}{dt} = k y^2, y(0) = y_0 \); (b) \( \frac{dy}{dt} = -k y^2, y(0) = y_0 \).

Step 1: Understand the Problem

In this problem, we need to form differential equations based on the descriptions given for the rate of change of a quantity \( y(t) \), which either increases or decreases proportionally to the square of the amount present.

Step 2: Identify the Rate in Case A

For part (a) where \( y(t) \) increases, the rate of change of \( y \) can be expressed by \( \frac{dy}{dt} = k y^2 \), where \( k \) is a positive constant of proportionality. This is because the rate of increase is proportional to the square of the current amount \( y \).

Step 3: Formulate the Initial Condition for Case A

The initial condition is given as \( y(0) = y_0 \) where \( y_0 \) is the initial amount present when \( t = 0 \). Thus, the initial value problem for part (a) is: \[ \begin{cases} \frac{dy}{dt} = k y^2 \ y(0) = y_0 \end{cases} \]

Step 4: Identify the Rate in Case B

For part (b) where \( y(t) \) decreases, the rate of change of \( y \) can be expressed by \( \frac{dy}{dt} = -k y^2 \) where \( k \) is a positive constant. This is because the rate of change is negative due to the decrease and is proportional to the square of the current amount \( y \).

Step 5: Formulate the Initial Condition for Case B

The initial condition is given as \( y(0) = y_0 \) where \( y_0 \) is the initial amount present when \( t = 0 \). Thus, the initial value problem for part (b) is:\[ \begin{cases} \frac{dy}{dt} = -k y^2 \ y(0) = y_0 \end{cases} \]

Key Concepts

Initial Value Problems

An initial value problem is a type of differential equation that begins with a given initial condition. This condition specifies the value of the solution at a particular point, usually at the starting time, which is noted as time zero. In this exercise, the initial value is given as:

• For part (a): The initial value is stated as \(y(0) = y_0\), indicating the amount at time zero when the quantity is increasing.
• For part (b): The initial value is similarly \(y(0) = y_0\), yet here it relates to the situation where the quantity decreases.

The purpose of an initial value is important because it allows us to find a specific solution when solving the differential equation. Without this condition, we would obtain a family of solutions rather than a unique one that fits the physical or real-world context described by the problem.

Proportionality in Rates of Change

Proportionality in rates of change refers to how the rate at which a quantity changes is directly related to some function of the quantity itself. In this case, it's related to the square of the quantity:

• For an increasing quantity, the rate of change is described by \( \frac{dy}{dt} = k y^2 \). This means that as the amount of \( y \) grows, its rate of increase becomes faster as it is being squared.
• For a decreasing quantity, the rate is given by \( \frac{dy}{dt} = -k y^2 \). Here, the negative sign indicates that the quantity is declining, and again, the rate of decrease accelerates as \( y \) is squared.

The constant \( k \) is known as the proportionality constant, and its positive value ensures that the relationships maintain their stated direction - increasing or decreasing. Whether a rate is growing or diminishing, proportionality gives us a clear mathematical way to model these changes over time.

Differential Equation Formulation

Differential equations are powerful mathematical tools used to model situations where rates of change are dependent on one variable. In this exercise, you learn how such equations can be formulated from a description of a physical process:

• Part (a) required formulating an equation when the change was increasing: \( \frac{dy}{dt} = k y^2 \). This indicates growth at a rate proportional to \( y^2 \).
• Part (b) required formulating an equation for a decreasing scenario: \( \frac{dy}{dt} = -k y^2 \). This reveals a decreasing rate proportional to the square of \( y \).

The equations show how dependent variables like \( y \) evolve over time, given their rate of change properties specified by the problem. All differential equations formulated here are initial value problems, which allow you to solve them comprehensively using their initial conditions to ensure that they precisely describe the real-world processes involved.