Problem 24 - Applied Calculus

Question

A new bus worth $$ 100,000\( in 2010 depreciates linearly to $$ 20,000 in 2030 (a) Find a formula for the value of the bus, \)V\(, as a function of time, \)t,\( in years since 2010 (b) What is the value of the bus in \)2015 ?$ (c) Find and interpret the vertical and horizontal intercepts of the graph of the function. (d) What is the domain of the function?

Step-by-Step Solution

Answer

(a) \( V(t) = -4,000t + 100,000 \). (b) $80,000. (c) Vertical: $100,000 (2010), Horizontal: 2035. (d) Domain: \( t \in [0, 25] \).

Step 1: Understand the Problem

We need to find a linear depreciation formula for the bus's value, given two points in time: it was worth $100,000 in 2010 and depreciates to $20,000 by 2030. This requires creating a linear equation.

Step 2: Identify Required Formula

The value of the bus, \( V(t) \), can be expressed as a linear function \( V(t) = mt + b \), where \( m \) is the rate of depreciation per year and \( b \) is the initial value of the bus in 2010.

Step 3: Calculate the Slope (m)

The slope \( m \) is calculated as the change in value over the change in time, \( m = \frac{20,000 - 100,000}{2030 - 2010} = \frac{-80,000}{20} = -4,000 \). This means the bus depreciates $4,000 per year.

Step 4: Write the Linear Equation

Insert the initial value and slope into the equation: \( V(t) = -4,000t + 100,000 \). This represents the value of the bus \( t \) years after 2010.

Step 5: Calculate Value in 2015

To find \( V(2015) \), substitute \( t = 5 \) (since 2015 is 5 years after 2010): \( V(5) = -4,000(5) + 100,000 = -20,000 + 100,000 = 80,000 \). The value in 2015 is $80,000.

Step 6: Identify Vertical and Horizontal Intercepts

Vertical intercept occurs where \( t = 0 \). Here, \( V(0) = 100,000 \), representing the value in 2010. Horizontal intercept occurs where \( V(t) = 0 \). Solve \( -4,000t + 100,000 = 0 \) to find \( t = 25 \), meaning the bus will be worthless in 2035.

Step 7: Determine Domain of the Function

Since the bus is new in 2010 and naturally cannot be negative in value, the m domain is from 0 to 25 years (2010 to 2035) after which it is worthless. The domain is \( t \in [0,25] \).

Key Concepts

Understanding Linear Functions in Depreciation

A linear function is a mathematical expression used to describe a constant rate of change. In the context of depreciation, it's a valuable tool to model how the value of an asset decreases over time. A linear function can be represented as \( V(t) = mt + b \), where \( V(t) \) is the value at time \( t \), \( m \) is the slope, representing the annual depreciation rate, and \( b \) is the initial value. In terms of depreciation, this means that every year, the asset loses a fixed amount of its value. This makes it easy to predict future values without needing to recalculate past values repeatedly. Linear functions are particularly useful because they are straightforward to graph, allowing for visual interpretation.

Calculating the Slope of a Depreciation Function

The slope of a linear function is crucial as it tells us how steep the line is, which directly translates to the rate of change. In a depreciation scenario, the slope \( m \) indicates how much value the asset loses each year. Calculating the slope involves finding the change in value over a given time period. For example, consider an asset that decreases from \(100,000 to \)20,000 over 20 years. The slope \( m \) is calculated with the formula:

• \( m = \frac{20,000 - 100,000}{2030 - 2010} \)
• \( m = \frac{-80,000}{20} \)
• \( m = -4,000 \)

This slope means the value of the bus decreases by $4,000 each year. Calculating the slope can help predict how long an asset will last and assist in making informed financial decisions.

Interpreting Values in a Depreciation Context

Value interpretation involves understanding the practical meaning of different points on the graph of a depreciation function. For instance, the y-intercept \( V(0) \) shows the initial value of the asset, here $100,000 in 2010. Meanwhile, the horizontal intercept point \( t = 25 \) (solving \( -4,000t + 100,000 = 0 \)) represents when the asset will have no monetary value left, indicating that in 2035, the bus will be worthless. This interpretation helps stakeholders by providing clear indicators of both the current and future state of the asset, which can be crucial for strategic planning, budgeting, and decision-making in any financial setting. Understanding these intercepts can offer insights into the longevity and valuation timelines of the asset.

Establishing the Domain of a Depreciation Function

The domain of a function is a set of all possible input values (in this case, \( t \), time in years) for which the function is defined. For the depreciation function \( V(t) = -4,000t + 100,000 \), the domain is from 0 to 25 years. This range is determined by the points of when the bus's initial value is calculated until it theoretically becomes worthless. The domain can be expressed as \( t \in [0, 25] \). The lower limit of 0 pertains to the year 2010 when the bus is newly purchased at its original value. The upper limit of 25 implies that by the year 2035, the asset will have depreciated entirely to $0 value. Establishing this domain ensures that our depreciation function logically corresponds to real-world scenarios. It guides how these calculations can be practically applied to represent real asset behavior over time.