Problem 19 - Applied 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

The differential equation implies that \( k = -0.03 \). Nothing specific is determined about \( C \).

Step 1: Understand the form of the solution

The function \( Q = C e^{k t} \) models the amount \( Q \) with respect to time \( t \), where \( C \) is a constant and \( k \) is the growth rate.

Step 2: Differentiate the given function

Differentiate \( Q = C e^{k t} \) with respect to \( t \). The derivative is \( \frac{dQ}{dt} = C k e^{k t} \).

Step 3: Set the derivative equal to the differential equation

According to the problem, \( \frac{dQ}{dt} = -0.03 Q \). Substitute \( Q = C e^{k t} \) into this equation to get \( C k e^{k t} = -0.03 C e^{k t} \).

Step 4: Solve for k

Divide both sides of the equation \( C k e^{k t} = -0.03 C e^{k t} \) by \( C e^{k t} \) (assuming \( C eq 0 \) and \( e^{k t} eq 0 \)) to find \( k = -0.03 \).

Step 5: Determine the value of C

Since \( C \) does not appear in the resulting expression for \( k \), the differential equation does not give any specific information about the value of \( C \).

Key Concepts

Exponential Growth and Decay

Exponential growth and decay describe processes that increase or decrease at a rate proportional to their current value. This kind of model is very common in natural phenomena, such as population growth, radioactive decay, or even interest calculations. For an exponential function expressed as \( Q = C e^{kt} \):

• \( C \) represents the initial value or amount at time \( t = 0 \).
• \( k \) is the constant growth rate if \( k > 0 \) or decay rate if \( k < 0 \). This rate determines whether the function grows or shrinks over time.

In the context of the problem, we have \( k = -0.03 \), indicating an exponential decay. This suggests that the quantity \( Q \) decreases by 3% per unit time. Understanding exponential decay allows you to predict how the quantity diminishes over time, using the simple relationship where the rate of change is proportional to\( Q \). It's a crucial concept for analyzing any situation where something decreases steadily at a consistent rate.

Derivative

The derivative is a fundamental concept in calculus that describes the rate at which a function changes. In the context of this exercise, the derivative \( \frac{dQ}{dt} \) tells us how the quantity \( Q \) changes with respect to time \( t \). When you differentiate a function like \( Q = C e^{kt} \), the process involves:

• Applying the chain rule, which in this case gives us \( k \) as a factor, because it derives from \( e^{kt} \).
• The result, \( Cke^{kt} \), incorporates both constant multiplication and exponent rules.

The derivative equates to the right side of the differential equation, \( -0.03Q \), after substituting the expression for \( Q \). This entirely describes the rate of change of \( Q \) with precise mathematical representation.In summary, differential equations like the one in the problem give us powerful insight into how a function behaves as we change its variables, thanks largely to the properties of derivatives.

Constant of Integration

In the solution of differential equations, the constant of integration is an important concept that arises when integrating a function. It accounts for any arbitrary constant that may be added to an antiderivative.In this specific scenario, the constant \( C \) in \( Q = C e^{kt} \) represents an initial condition or point of reference at \( t = 0 \). This is because when integrating a differential equation, you get a family of solutions that differ by a constant.

• \( C \) dictates the initial value of the quantity \( Q \) before any changes due to \( t \) occur.
• It ensures the uniqueness of the solution depending on specific initial conditions of the problem.

The differential equation used in the problem does not provide explicit information about \( C \), meaning it remains an unknown quantity unless additional information, like an initial condition, is provided. Understanding the constant of integration helps in solving real-world problems by ensuring that the solution not only satisfies the differential equation but also adheres to initial conditions provided in a given context.