Problem 19 - Applied Calculus

Question

A $$\$ 15,000$$ robot depreciates linearly to zero in 10 years. (a) Find a formula for its value as a function of time. (b) How much is the robot worth three years after it is purchased?

Step-by-Step Solution

Answer

(a) The formula is \( V(t) = -1500t + 15000 \). (b) The robot is worth \( \$10,500 \) after 3 years.

Step 1: Identify the Starting Conditions

The robot is initially worth \( \\(15,000 \) and depreciates to \( \\)0 \) in 10 years. This provides two key points: at \( t = 0 \), the value is \( 15,000 \), and at \( t = 10 \), the value is \( 0 \). These points can be used to determine the depreciation formula.

Step 2: Determine the Slope of Depreciation

Linear depreciation means the value decreases at a constant rate. The slope \( m \) of this linear depreciation is given by \( m = \frac{\text{change in value}}{\text{time duration}} = \frac{0 - 15000}{10 - 0} = -1500 \).

Step 3: Write the Linear Depreciation Formula

With the slope \( m = -1500 \) and the initial value \( b = 15000 \), the depreciation formula, using the slope-intercept form, is \( V(t) = -1500t + 15000 \), where \( V(t) \) is the value at time \( t \).

Step 4: Calculate the Value After 3 Years

Substitute \( t = 3 \) into the formula: \( V(3) = -1500 \times 3 + 15000 \). Calculate this to find \( V(3) = -4500 + 15000 = 10500 \). So, the robot is worth \( \$10,500 \) after 3 years.

Key Concepts

Depreciation Formula

Depreciation represents how the value of an asset decreases over time. When something depreciates linearly, it means that the value decreases at a constant rate each year.
In our robot's case, which is worth \(15,000 initially and goes to \)0 over 10 years, the depreciation can be modeled using a straight line. The formula for linear depreciation follows a slope-intercept style, similar to a regular line equation in algebra. This formula is:

• \( V(t) = mt + b \)

where:

• \( V(t) \) is the value of the asset at time \( t \)
• \( m \) represents the slope (or rate) of depreciation
• \( b \) indicates the initial value or y-intercept

The robot's depreciation formula is thus \( V(t) = -1500t + 15000 \). This equation shows that for every year \( t \), the robot’s value decreases by \(1,500, starting from \)15,000.

Slope-Intercept Form

The slope-intercept form is a way to describe a straight line on a graph, using the formula: \( y = mx + b \). Here, \( m \) is the slope, indicating how steep the line is, and \( b \) is the y-intercept, where the line crosses the y-axis.
In the context of depreciation, this model is used to show how the value (\( V(t) \)) of an asset changes over time (\( t \)).
Essentially:

• The **slope (\( m \))** tells you the rate of change. In our robot example, \( m = -1500 \) means the value decreases by \(1,500 per year.
• The **y-intercept (\( b \))** is the starting point. For the robot, \( b = 15000 \) means the initial value is \)15,000.

The usefulness of the slope-intercept form lies in its simplicity and clarity when showing linear relationships, like constant depreciation rates.

Calculation of Value After Time

To find out how much an asset is worth after a certain period, the depreciation formula \( V(t) = mt + b \) can be easily applied. Let's calculate how much the robot is worth after three years.
We simply plug \( t = 3 \) into the formula:

• \( V(3) = -1500 \times 3 + 15000 \)
• This results in \( -4500 + 15000 = 10500 \).

This calculation shows that after three years, the robot's value has depreciated to $10,500.
The procedure is straightforward and highly effective for determining an asset's worth after any specified interval, so long as the depreciation rate is consistent.