Question

\mathrm{\\{} E X P L ~ B e s i d e s ~ p r o v i d i n g ~ a n ~ e a s y ~ w a y ~ t o ~ d i f f e r e n t i a t e ~ p r o d u c t s , ~ l o g . ~ arithmic differentiation also provides a measure of the relative or fractional rate of change, defined as $y^{\prime} / y Prove that if the relative rate of change is a negative constant then the function must represent exponential decay.

Short answer

If the relative rate of change is a negative constant, then \( y = Ae^{-kx} \) shows exponential decay.

Step-by-step solution

Understand the Relative Rate of Change

The relative rate of change of a function \( y \) is given by \( \frac{y'}{y} \), where \( y' \) is the derivative of \( y \). This represents how fast a function is changing at a given point relative to its value.

Set Up Equation for Negative Constant Rate

Given that the relative rate of change is a negative constant, set \( \frac{y'}{y} = -k \), where \( k \) is a positive constant (since the rate is negative).

Solve the Differential Equation

Rewrite the equation \( \frac{y'}{y} = -k \) as \( y' = -ky \). This is a separable differential equation. Separate variables to get \( \frac{dy}{y} = -k \, dx \).

Integrate Both Sides

Integrate both sides: \( \int \frac{dy}{y} = \int -k \, dx \). The left side yields \( \ln |y| \), and the right side yields \(-kx + C\), where \( C \) is the constant of integration. So, \( \ln |y| = -kx + C \).

Solve for y

Exponentiate both sides to solve for \( y \): \( e^{\ln |y|} = e^{-kx + C} \) simplifies to \( |y| = e^C e^{-kx} \). Let \( e^C = A \), a positive constant, then \( |y| = Ae^{-kx} \). Since \( y \) must be positive, \( y = Ae^{-kx} \) is the solution, representing exponential decay.

Key concepts

Relative Rate of Change

The concept of relative rate of change helps us understand how quickly a function changes compared to its current value. When we have a function, often denoted as \( y \), its rate of change is described by its derivative, \( y' \). The relative rate of change is simply the ratio \( \frac{y'}{y} \). This measurement gives us insight into the function's growth or decay in a relative sense.

For example, if \( y' = 2y \), the relative rate of change \( \frac{y'}{y} = 2 \), meaning the function grows at a rate proportional to its size. But when this relative rate is negative, it describes a situation where the function is decaying, or decreasing, over time. This has real-life implications like radioactive decay or depreciation of assets, where substances or values continuously decrease proportionally to their remaining quantity.

• Positive \( \frac{y'}{y} \) = growth
• Zero \( \frac{y'}{y} \) = constant
• Negative \( \frac{y'}{y} \) = decay

Understanding this can help predict how a function behaves in a changing environment.

Differential Equation

A differential equation is essentially an equation that relates a function with its derivatives. In our exercise, when the relative rate of change is a constant \( -k \), \( \frac{y'}{y} = -k \), transforming into \( y' = -ky \) creates a differential equation to solve. This particular form is called a separable differential equation { because it can be rearranged so that each side depends on only one variable.To solve this, we can separate the variables: \( \frac{dy}{y} = -k \, dx \). This allows us to integrate both sides separately.

• Left Side: Integrate \( \frac{dy}{y} \) -> \( \ln |y| \)
• Right Side: Integrate \( -k \, dx \) -> \( -kx + C \)

Now, you'll have the equation \( \ln |y| = -kx + C \). This is a standard form, indicating that our solution should revolve around exponential functions. The importance of differential equations lies in their wide application across various scientific fields, serving as the backbone for modeling real-world processes.

Logarithmic Differentiation

Logarithmic differentiation can be a powerful technique, especially when dealing with functions that multiply or divide multiple expressions involving variables. By taking the logarithm of a function, the properties of logarithms can simplify differentiation.In our current context, it aids in solving for \( y \) when the differential equation \( y' = -ky \) leads us to integrate \( \frac{dy}{y} \). Integrating this gives \( \ln |y| \), which when equated to \( -kx + C \), shows the usefulness of logarithmic properties.

• Utilizes properties like \( \ln(ab) = \ln a + \ln b \)
• Simplifies exponentiation post-integration into manageable forms
• Transforms power functions, easing differentiation

After transforming the logarithmic integration back into a regular function by solving \( e^{\ln |y|} = e^{-kx + C} \), we find \( y = Ae^{-kx} \). This representation is a form of exponential decay.Logarithmic differentiation helps streamline complex calculations by exploiting the simplifications that logarithmic properties offer, providing insight into exponential growth and decay behaviors.