Problem 53 - Precalculus

Question

Consider the models of exponential decay \(f(t)=(2+c) e^{-k_{i} t}\) and \(g(t)=c e^{-k_{2} t} .\) Suppose that \(f(1)=g(1),\) what is the relationship between \(k_{1}\) and \(k_{2} ?\)

Step-by-Step Solution

Answer

The relationship is \( k_1 - k_2 = \ln(1 + \frac{2}{c}) \).

Step 1: Set up the given equality

We are given that \( f(1) = g(1) \). Substitute 1 into both functions to establish an equation. This yields: \\( (2+c)e^{-k_1 \cdot 1} = ce^{-k_2 \cdot 1} \) or equivalently \( (2+c)e^{-k_1} = ce^{-k_2} \).

Step 2: Isolate the exponential terms

Divide both sides of the equation from Step 1 by \(c\), assuming \(c eq 0\). This results in: \\( \frac{2+c}{c} e^{-k_1} = e^{-k_2} \). \Simplify to \( \left(1 + \frac{2}{c}\right) e^{-k_1} = e^{-k_2} \).

Step 3: Solve for the ratio of exponential decay rates

Take the natural logarithm of both sides to solve for the relationship between \(k_1\) and \(k_2\): \\[ \ln\left(1 + \frac{2}{c}\right) - k_1 = -k_2 \]. \Rearrange to find \( \ln\left(1 + \frac{2}{c}\right) = k_1 - k_2 \).

Step 4: Express the final relationship

The relationship between \(k_1\) and \(k_2\) is given by: \\[ k_1 - k_2 = \ln\left(1 + \frac{2}{c}\right) \].

Key Concepts

Understanding the Natural Logarithm

The natural logarithm is a crucial mathematical concept often used to simplify complex exponential equations. Represented as \( \ln x \), it is the logarithm to the base \( e \), where \( e \approx 2.71828 \). The natural logarithm allows us to determine the power to which the base \( e \) must be raised to obtain a given number. In the context of exponential decay problems, the natural logarithm can help us linearize an equation by converting the exponential terms into a more manageable algebraic form.

For example, if we have an equation such as \( e^{y} = x \), taking the natural logarithm of both sides gives us \( y = \ln x \). This transformation is handy when looking for relationships between variables in equations involving exponential functions. In our exercise, taking the natural logarithm of both sides allows us to isolate and compare the decay constants \( k_1 \) and \( k_2 \). Remember, by applying the natural logarithm, we simplify the comparison and create a clearer path to our solution.

Exponential Functions and Their Behavior

Exponential functions are mathematical functions in the form of \( a e^{bx} \), where \( a \) and \( b \) are constants, and \( e \) is the base of natural logarithms. They are characterized by their rapid growth or decay, making them ideal for modeling real-world processes such as population growth, radioactive decay, and interest calculations.

Exponential growth occurs when the constant \( b \) is positive, while exponential decay occurs when \( b \) is negative. The rate of growth or decay is determined by the value of this constant. The larger the absolute value of \( b \), the steeper the curve of the function, illustrating faster growth or decay.

In our given functions \( f(t) = (2+c)e^{-k_1 t} \) and \( g(t) = ce^{-k_2 t} \), both exhibit the property of exponential decay, as indicated by the negative sign in their exponents. Evaluating the functions at a specific point, such as \( t=1 \), offers an equation that helps to explore their respective decay rates \( k_1 \) and \( k_2 \). With understanding exponential functions, we recognize how changes in these rates affect the values and behavior of the functions over time.

Equality of Functions in Context

The equality of functions means that two functions yield the same value for all inputs within their domain; however, it’s more common to check equality at specific points of interest. In this problem, we are given that \( f(1) = g(1) \) which means, for the given time \( t=1 \), the outputs of the functions \( f(t) \) and \( g(t) \) are equal. This provides a basis to explore the relationship between functions specifically at that point.

Setting up the equation \( (2+c)e^{-k_1} = ce^{-k_2} \) is the first step to explore the equality for an input \( t=1 \). Dividing through by a common factor helps to isolate the exponential terms, leading to the step where we can logarithmically compare \( k_1 \) and \( k_2 \). By examining the equation \( \ln \left(1 + \frac{2}{c}\right) = k_1 - k_2 \), we determine their relationship.

This process illustrates how we can use specific points of equality to investigate the parameters of different functions. Understanding and proving the equality at such points give us powerful insight into proportional relationships between decay rates in exponential functions.