Problem 35 - Precalculus : Building Concepts and Connections

Question

The purchase price of a 2006 Ford \(\mathrm{F} 150\) longbed pickup truck is \(\$ 23,024 .\) After 1 year, the price of the Ford \(\mathrm{F} 150\) is \(\$ 17,160 .\) (Source: Kelley Blue Book) (a) Find an exponential function, \(P(t)=C e^{k t},\) that models the price of the truck, where \(t\) is the number of years since 2006 (b) What will be the value of the Ford \(\mathrm{F} 150\) in the year \(2009 ?\)

Step-by-Step Solution

Answer

The predicted price of the Ford F 150 in 2009 according to this model is approximately \$10,673.

Step 1: Determine the initial price and depreciation

Given, the purchase price of the Ford F 150 in 2006 was \$23,024 and the price after 1 year (2007) it was reduced to \$17,160.\nSo the equation becomes \(17,160 = 23,024 * e^{k*1}\). Solving this equation, we get the value of \(k\).

Step 2: Solve for \(k\)

Rearrange the equation to isolate \(e^{k}\). Dividing both sides by \$23,024, we have \(e^{k} = 17,160/23,024\).\nTaking the natural log of both sides of the equation, we have \(k = ln(17,160/23,024)\). Solving this we get \(k = -0.2889\) (approximately)

Step 3: Formulate the Exponential Function

Replace the \(C\) in the general formula with the initial price and \(k\) with the value we just determined. This will give us the function \(P(t) = 23,024 * e^{-0.2889t}\).

Step 4: Predict the Future Value

In the year 2009, 3 years would have passed since the initial purchase. Therefore, substituting \(t = 3\) in the equation, we can estimate the future price of the truck.\nThe price in 2009 will be \( P(3) = 23,024* e^{-0.2889*3}\). Solving this equation, the price can be calculated.

Key Concepts

Depreciation

Depreciation is a financial term used to describe the reduction in the value of an asset over time. It occurs due to several factors, such as wear and tear, market conditions, and technological obsolescence. In the context of the exercise, we saw the depreciation of a pickup truck's value from its initial purchase price. There are primarily two types of depreciation:

• Straight-line depreciation: This method spreads the loss of value evenly across the useful life of an asset. If a truck costs $20,000 and has a life span of 10 years, it uniformly loses $2,000 in value yearly.
• Exponential depreciation: This model assumes that an asset loses value rapidly in the initial years, slowing down over time. The mathematical model used in this exercise is an example of exponential depreciation.

In real-life scenarios, determining the correct depreciation method is crucial for accurate financial analysis, helping people understand the asset's real value over time. The exponential decay equation allows businesses to calculate an asset's value at any point within its useful life.

Exponential Function

An exponential function is a mathematical expression where a constant base is raised to a variable exponent. It takes the form of \(f(x) = a \cdot e^{bx}\), where \(a\) is a constant, \(e\) is Euler's number (approximately 2.718), and \(b\) represents the rate of decay or growth. Exponential functions can model various natural processes and financial patterns, including depreciation.The essential features of exponential functions include:

• A constant percentage rate of increase or decrease per unit time or event.
• Exponential growth, where the function rapidly increases due to a positive exponent.
• Exponential decay, where values decrease swiftly with time as seen in asset depreciation, represented by a negative exponent.

In our exercise, the function \(P(t) = 23,024 \cdot e^{-0.2889t}\) was formulated to describe the price depreciation of the Ford F150 pickup truck over time. By substituting various values for \(t\), we can determine the vehicle's value as the years progress.

Natural Logarithm

The natural logarithm is a logarithm with the base \(e\), an irrational constant approximately equal to 2.718. Represented as \(\ln(x)\), it is the inverse function of the exponential function. The natural logarithm plays a critical role in solving equations involving exponential growth and decay, such as the depreciation problem provided in this exercise.To solve the depreciation equation given by the function \(P(t) = C \cdot e^{kt}\), we often need to rearrange and take the natural logarithm to isolate and solve for the variable, as we did to determine the decay constant \(k\).The key aspects of natural logarithms include:

• \(\ln(1) = 0\), because \(e^0 = 1\).
• They allow for converting between exponential forms and more manageable linear forms.
• They are particularly useful in continuous growth or decay scenarios, providing clearer insight into exponential relationships.

Using the natural logarithm in our exercise allowed us to compute \(k\), representing the annual depreciation rate, which was crucial in modeling the truck's decreasing value.