Problem 107 - Calculus Volume 1

Question

A car was purchased for $$\$ 26,000$$. The value of the car depreciates by $$\$ 1500$$ per year. a. Find a linear function that models the value \(V\) of the car after \(t\) years. b. Find and interpret \(V(4)\).

Step-by-Step Solution

Answer

a. \(V(t) = 26000 - 1500t\) b. \(V(4) = 20000\); after 4 years, the car is worth \(\$ 20,000\).

Step 1: Understand the Problem

The car's value decreases at a constant rate, which indicates a linear relationship between time and value. We need to create a function that represents this relationship.

Step 2: Identify Initial Value and Rate of Change

The initial value of the car when it was purchased is \(\\( 26,000\). The car depreciates by \(\\) 1500\) each year, so the rate of change is \(-1500\).

Step 3: Formulate the Linear Function

The value of the car after \(t\) years can be expressed with the linear function \(V(t) = 26000 - 1500t\). This equation reflects the initial value and the annual depreciation.

Step 4: Substitute to Find \(V(4)\)

Replace \(t\) with \(4\) in the function to find \(V(4)\). The calculation is: \(V(4) = 26000 - 1500 \times 4\).

Step 5: Compute \(V(4)\)

Perform the arithmetic: \(V(4) = 26000 - 6000 = 20000\). Thus, the value of the car after 4 years is \(\$ 20,000\).

Step 6: Interpret \(V(4)\)

The result, \(V(4) = 20000\), indicates that after 4 years, the car is worth \(\$ 20,000\).

Key Concepts

Depreciation

Depreciation refers to the reduction in the value of an asset over time. This can occur due to wear and tear, usage, or age. In the context of the given car example, depreciation is calculated as a consistent decline in value each year.
This is why we describe it as a linear function; each year the car's worth reduces by a fixed amount. Understanding depreciation is crucial, especially for assets like vehicles that lose value over time.

• The car begins with a value of \(26,000\).
• It loses \(1,500\) per year due to depreciation.
• This results in a steady decrease in the car's value annually.

By recognizing the depreciation, one can predict future values, such as what the car would be worth in subsequent years. This concept is applied not just to cars, but to many other assets as well.

Rate of Change

The rate of change in a linear function is a crucial concept as it tells us how much the dependent variable (in this case, the car's value) changes with respect to the independent variable (time). In the example provided, the rate of change is \-1500\, meaning for each year that passes, the car's value is reduced by \(1,500\).
This rate of change is identified as the slope in the linear function equation.

• It is a negative value, indicating a decrease.
• This constant figure simplifies predicting future values.
• Understanding this can help with financial planning or deciding future sales.

The rate of change is influential in determining how quickly or slowly an asset depreciates. For cars and other assets, this offers a clear picture of their future worth trajectory over time.

Initial Value

The initial value in a linear function represents the starting point of the dependent variable before any changes occur. It is the "y-intercept" of the graph of the function. In our scenario, the initial value is quite significant.

• For the car, this initial value is set at \(26,000\).
• This figure is crucial as our baseline, from which all subsequent depreciation is subtracted.

This is the value of the car when it was first purchased, and it forms the constant term in our linear equation \(V(t) = 26000 - 1500t\). Knowing the initial value lets us establish the starting point for our calculations and understand how much value is lost over time.

Function Evaluation

Function evaluation involves substituting a specific value for the independent variable into the function formula. This helps us find the corresponding dependent variable value.
For the car example, evaluating the function allows us to find out the car's value after a certain number of years. By replacing \(t\) with \(4\) in the equation \(V(t) = 26000 - 1500t\), we perform what is known as function evaluation:

• Substitute to find \(V(4): V(4) = 26000 - 1500 \times 4\).
• Calculate: \(V(4) = 26000 - 6000 = 20000\).

This process enables us to determine that after 4 years, the car's value is \(\$20,000\). Function evaluation is a powerful tool for calculating specific outcomes and making data-driven decisions.