Problem 84 - Precalculus : Building Concepts and Connections

Question

A new car that costs $$\$ 25,000$$ depreciates to $$80 \%$$ of its value in 3 years. (a) Assume the depreciation is linear. What is the linear function that models the value of this car \(t\) years after purchase? (b) Assume the value of the car is given by an exponential function \(y=A e^{h t},\) where \(A\) is the initial price of the car. Find the value of the constant \(k\) and the exponential function. (c) Using the linear model found in part (a), find the value of the car 5 years after purchase. Do the same using the exponential model found in part (b). (d) Graph both models using a graphing utility. Which model do you think is more realistic, and why?

Step-by-Step Solution

Answer

The linear depreciation function is \(y = -1666.67t + 25000\), and the exponential depreciation function is \(y = 25000 * e^{-0.074t}\). The value of the car 5 years after the purchase according to the linear model is \$15,833.33, while for the exponential model it's \$16,213.15. The exponential model is more realistic.

Step 1: Calculation of Linear Depreciation

In the given exercise, it is mentioned that the car depreciates to 80\% of its value in three years. The present value of the car is \$25,000.Based on the linear depreciation, the car loses value in a straight line fashion over the period. After 3 years, the value of the car would be \(80\%\) of the original value, which is \(0.80 * \$25,000 = \$20,000\).So if the linear depreciation is represented as \(y = mt + c\), where \(y\) represents the value of the car at a given time \(t\), \(m\) is the slope of the line, and \(c\) is the y-intercept (original value of the car), then we can say:At \(t = 0\), \(y = c = \$25,000\)At \(t = 3\), \(y = \$20,000\)Since \(m = (y2 - y1) / (x2 - x1)\),So, \(m = (\$20,000 - \$25,000) / (3 - 0) = -\$1,666.67\)So, the linear function can be modelled as \(y = -1666.67t + 25000\).

Step 2: Calculation of Exponential Depreciation

For the exponential function model, we can write the value as \(y = Ae^{kt}\), where \(A = \$25,000\), the initial price of the car, \(y\) is the value of the car after \(t\) years, and \(k\) is the constant we desire to find.We know the value after three years is \$20,000. We can represent this as \(20000 = 25000 * e^{3k}\).Dividing both sides by 25,000 gives us \(0.8 = e^{3k}\).Taking the natural log of both sides will solve for \(k\). So, \(ln(0.8) = 3k \Rightarrow k = ln(0.8) / 3 = -0.074\).Therefore, the exponential function will be \(y = 25000 * e^{-0.074t}\).

Step 3: Calculation of the Value After 5 years

Using the linear depreciation function, the value of the car 5 years after purchase becomes: \[y = -1666.67 * 5 + 25000 = \$15,833.33\].Using the exponential depreciation function, the value of the car 5 years after purchase becomes:\[y = 25000 * e^{-0.074 * 5} = \$16,213.15\].

Step 4: Graph & Comparison

For part (d) the graphs of the two functions can be plotted using a graphing utility. On comparing two models, the exponential depreciation model is more realistic because in reality, a car depreciates rapidly in the initial years and then the rate of depreciation slows down, which is correctly shown by the exponential function.

Key Concepts

Linear Depreciation

Linear depreciation is a method used to calculate the reduction in value of an asset over time at a constant rate. Imagine dividing a car's value into equal slices that you remove each year until it's valueless. In our exercise, we start with a car worth \(25,000. Over 3 years, the car's value decreases linearly to 80% of its initial value, settling at \)20,000. This straight-line depreciation is modeled by the function

\( y = mt + c \),

where \( y \) is the car's value at time \( t \), \( m \) is the rate of depreciation, and \( c \) is the initial value. After plugging in the known values, we find the rate \( m = -\$1666.67 \) per year, leading to the depreciation function

\( y = -1666.67t + 25000 \).

This linear model suggests that, regardless of how old the car is, it will lose the same value each year.

Exponential Decay

Exponential decay, unlike its linear counterpart, describes a situation where the value of an asset decreases by a set percentage over equal time periods. It's akin to a cake that loses half its size every day—always shrinking by proportion, not a fixed slice. A car's value dropping swiftly in early years then more slowly over time is well-captured by this model.

Representing this with the equation \( y = Ae^{kt} \), where \( A \) is the original value and \( k \) is the constant determining the rate of decay, we're tasked to find \( k \). By recognizing that the car is worth $20,000 after 3 years—a clue to the puzzle—we can reverse-engineer, finding \( k = -0.074 \) and the decay function

\( y = 25000e^{-0.074t} \).

This model reflects depreciation as it typically unfolds: quicker at the start, easing off with time.

Modeling Real-world Situations with Functions

In precalculus, functions are more than abstract math—they're tools we use to describe real-life scenarios, like how a car's worth shrinks over time. Linear and exponential functions help us paint a mathematical picture of this phenomenon. Envisioning the car's value as a mathematical journey over time, we can select the route—linear or exponential—based on how the vehicle weathers the years.

You can think of functions as 'stories' we tell about what happens as circumstances change—like a narrator detailing the protagonist's fortune dwindling at a consistent rhythm (linear) or accelerating and then slowing down (exponential). These functions not only aid in predictions but also allow for strategic decision-making, such as determining the best time to sell the car.

Graphical Representation of Functions

Graphing functions is a powerful way to visualize equations and compare their behavior. When we chart the depreciation functions for our car, the distinct paths they carve through the graph unveil their nature.

The linear function draws a straight, steadily descending line, symbolizing uniform year-on-year value loss. Meanwhile, the exponential function curves downward, a steeper slope at first softening as time progresses, mirroring how a vehicle's value plummets quicker early on, then more gently. Observing these graphs side by side, we not only grasp their differences but also gauge which aligns better with reality, aiding in choosing the most accurate model for our car's depreciation.