Problem 18 - Calculus

Question

Suppose \(Q=C e^{k t}\) satisfies the differential equation $$\frac{d Q}{d t}=-0.03 Q.$$ What (if anything) does this tell you about the values of \(C\) and \(k ?\)

Step-by-Step Solution

Answer

\( k = -0.03 \); \( C \) is undetermined.

Step 1: Understand the Differential Equation

The given differential equation is \( \frac{dQ}{dt} = -0.03 Q \). This is a first-order linear differential equation, which typically describes exponential growth or decay scenarios. Here, the equation suggests that \( Q(t) \) decreases exponentially over time as the rate of change \( \frac{dQ}{dt} \) is proportional to \( Q(t) \) itself, with a proportionality constant of \(-0.03\).

Step 2: Express the Exponential Model

We are given that \( Q = Ce^{kt} \). This represents an exponential function where \( C \) is the initial quantity when \( t = 0 \), and \( k \) is a constant that affects the rate of growth or decay of \( Q \).

Step 3: Differentiate the Exponential Function

Differentiate \( Q = Ce^{kt} \) with respect to \( t \).The derivative is: \[ \frac{dQ}{dt} = Cke^{kt} \]

Step 4: Set the Differential Equation Equal to the Derivative

Substitute the expression for \( \frac{dQ}{dt} \) from Step 3 into the given differential equation:\[ \frac{dQ}{dt} = -0.03Q \]Expanding the right-hand side gives:\[ -0.03Ce^{kt} \]Set these equal:\[ Cke^{kt} = -0.03Ce^{kt} \]

Step 5: Solve for Constants

Since \( Ce^{kt} \) is common to both sides, it cancels out provided \( C eq 0 \). This yields:\[ k = -0.03 \]

Step 6: Interpret the Values of \( C \) and \( k \)

The solution indicates that \( k = -0.03 \), showing that the process described is an exponential decay. Meanwhile the value of \( C \) is arbitrary as it acts as an initial condition and is not determined by the differential equation.

Key Concepts

Exponential Decay

Exponential decay refers to a process where a quantity decreases at a rate proportional to its current value. It is often modeled by the mathematical expression \( Q = Ce^{kt} \), where \( Q \) denotes the amount at time \( t \), \( C \) is the initial amount at time \( t = 0 \), and \( k \) is a constant that indicates the rate of decay.

• If \( k \) is negative, as in our equation, it indicates a declining or decaying process.
• This decrease is not linear; instead, it follows a curve that gradually flattens.

In the problem, the differential equation \( \frac{dQ}{dt} = -0.03Q \) illustrates exponential decay, since the rate of change, \( \frac{dQ}{dt} \), is negative and proportional to \( Q \).

First-order Linear Differential Equation

A first-order linear differential equation is an equation that involves the first derivative of a function. It can be written in the standard form \( \frac{dy}{dt} + P(t)y = Q(t) \). If \( Q(t) = 0 \), it is called homogeneous. These equations are known for their ability to model systems with proportionality and constant rates.

• In our case, \( \frac{dQ}{dt} = -0.03Q \) is already in a simplified form.
• The solution to this type of equation usually involves an exponential function, reflecting how the rate of change of the system is proportional to its current state.

This characteristic makes them ideal for representing real-world phenomena such as population growth or radioactive decay, wherein the rate of change is inherently connected to the current amount.

Proportionality Constant

The proportionality constant is a significant feature of differential equations that tells us how the rate of change of a quantity relates to the quantity itself. In this exercise, the proportionality constant is \(-0.03\).

• This constant determines the direction and rate of change: since it is negative, it leads to a decay process.
• Its magnitude \( |-0.03| \) informs us about the steepness of the decay curve; a larger magnitude would indicate a faster decay, whereas a smaller magnitude suggests a gentler slope.

Here, this constant is essential in understanding how quickly \( Q \) decreases over time, impacting both the speed and form of the exponentially decaying curve.