Problem 2 - Intermediate Algebra

Question

Suppose it is estimated that the value of a car depreciates \(30 \%\) per year for the first 5 years. The equation \(A=P_{0}(0.7)^{t}\) yields the value \((A)\) of a car after \(t\) years if the original price is \(P_{0}\). Find the value (to the nearest dollar) of each of the following cars after the indicated time. (a) $$\$ 16,500$$ car after 4 years (b) $$\$ 22,000$$ car after 2 years (c) $$\$ 27,000$$ car after 5 years (d) $$\$ 40,000$$ car after 3 years

Step-by-Step Solution

Answer

(a) $3962; (b) $10780; (c) $4538; (d) $13720.

Step 1: Understand the Formula

We are given the formula \( A = P_0 (0.7)^t \) where \( A \) is the value of the car after \( t \) years, and \( P_0 \) is the original price of the car. This formula represents a depreciation of 30% per year, leaving the car worth 70% of its value each year.

Step 2: Calculate for $16,500 Car After 4 Years

Use the formula with \( P_0 = 16500 \) and \( t = 4 \): \\[ A = 16500 \times (0.7)^4 \]. \First, calculate \( (0.7)^4 = 0.2401 \), \then multiply: \( A = 16500 \times 0.2401 = 3961.65 \). \Round to nearest dollar: \( A \approx 3962 \) dollars.

Step 3: Calculate for $22,000 Car After 2 Years

Use the formula with \( P_0 = 22000 \) and \( t = 2 \): \\[ A = 22000 \times (0.7)^2 \]. \Calculate \( (0.7)^2 = 0.49 \), \then multiply: \( A = 22000 \times 0.49 = 10780 \). \No rounding needed as value is already a whole number: \( 10780 \) dollars.

Step 4: Calculate for $27,000 Car After 5 Years

Use the formula with \( P_0 = 27000 \) and \( t = 5 \): \\[ A = 27000 \times (0.7)^5 \]. \First, calculate \( (0.7)^5 = 0.16807 \), \then multiply: \( A = 27000 \times 0.16807 = 4537.89 \). \Round to nearest dollar: \( A \approx 4538 \) dollars.

Step 5: Calculate for $40,000 Car After 3 Years

Use the formula with \( P_0 = 40000 \) and \( t = 3 \): \\[ A = 40000 \times (0.7)^3 \]. \Calculate \( (0.7)^3 = 0.343 \), \then multiply: \( A = 40000 \times 0.343 = 13720 \). \No rounding needed as value is already a whole number: \( 13720 \) dollars.

Key Concepts

Exponential Decay

Exponential decay is a mathematical concept that describes the process of a quantity decreasing at a constant rate over time. It is represented by the formula \( A = P_0 b^t \), where:

• \( A \) is the remaining quantity after time \( t \).
• \( P_0 \) is the initial quantity.
• \( b \) is the decay factor.
• \( t \) represents time passed.

In our car depreciation example, the decay factor \( b \) is \( 0.7 \), because the cars retain 70% of their value each passing year, meaning they lose 30%. Exponential decay processes occur in many real-life situations such as radioactive decay, population decline, and cooling of substances. This formula allows us to predict how much of a quantity will remain after a certain amount of time, which is why it's so useful in various fields including finance and physics.
Understanding and applying exponential decay helps us make informed decisions, anticipate future outcomes, and effectively manage resources. By using this model, we can clearly see how each year contributes to the overall rate of depreciation and plan accordingly for the lifespan and value of an object.

Car Depreciation

Car depreciation refers to the reduction in a car's value over time. This happens because of factors like wear and tear, new models being released, and the general decline in desirability as the car ages. In the context of our exercise, car depreciation is modeled using exponential decay.

• The formula \( A = P_0 (0.7)^t \) helps calculate the car's value after \( t \) years.
• Each year, the car retains just 70% of its previous year's value, reflecting a 30% depreciation rate.

This kind of depreciation model is standard for many durable goods that lose value predictably over time. For instance, when you buy a car, it's important to consider how its value will change in the future. This is crucial for decisions like selling or trading later on, obtaining car insurance, or understanding interest rates if financed. By grasping car depreciation, consumers can make better financial decisions, plan for future purchases, and understand how different factors like make, model, and condition impact a car's worth over time.

Mathematical Modeling

Mathematical modeling is the process of using mathematical equations and concepts to represent real-world phenomena. It allows us to predict behavior, understand relationships, and solve problems in various contexts. In our exercise, mathematical modeling is used to represent how a car's value declines over time due to depreciation.

• The model chosen, \( A = P_0 (0.7)^t \), effectively simplifies the real-world issue of car value depreciation into a manageable equation.
• This model helps sketch a clear picture of devaluation over time, making complex phenomena more comprehensible.

Mathematical models can often be simplified versions of reality, but they provide valuable insights and allow for precise computations when actual outcomes can't be easily observed. In areas such as finance, environmental science, and engineering, these models are indispensable for planning and predicting future trends.
Incorporating mathematical modeling into problems enhances our analytical skills and understanding, enabling us to manipulate situations confidently and effectively. Through these models, we gain the power to predict, strategize, and make better decisions based on anticipated future behaviors.