Problem 21 - Applied Calculus

Question

A $$\$ 50,000$$ tractor has a resale value of $$\$ 10,000$$ twenty years after it was purchased. Assume that the value of the tractor depreciates linearly from the time of purchase. (a) Find a formula for the value of the tractor as a function of the time since it was purchased. (b) Graph the value of the tractor against time. (c) Find the horizontal and vertical intercepts, give units, and interpret them.

Step-by-Step Solution

Answer

(a) \( V(t) = -2000t + 50000 \). (b) Refer to a graph plot based on the function. (c) Vertical: \( (0, 50000) \), Horizontal: \( (25, 0) \).

Step 1: Understand Linear Depreciation

The problem assumes that the tractor's value depreciates linearly, meaning the loss in value is constant each year. We need to find the rate of depreciation and express the value as a function of time.

Step 2: Calculate Yearly Depreciation

The tractor depreciates from \( \\(50,000 \) to \( \\)10,000 \) over 20 years. The total depreciation is \( \\(50,000 - \\)10,000 = \\(40,000 \). Therefore, the rate of depreciation per year is \( \frac{\\)40,000}{20} = \$2,000 \) per year.

Step 3: Determine the Linear Function

Since the depreciation is linear, we use a linear equation of the form \( V(t) = mt + b \), where \( V(t) \) is the value after \( t \) years, \( m \) is the slope, and \( b \) is the initial value. Here, \( m = -2000 \) and \( b = 50000 \). Thus, the formula for the value is \( V(t) = -2000t + 50000 \).

Step 4: Graph the Equation

To graph \( V(t) = -2000t + 50000 \), plot points by choosing values for \( t \) (such as 0, 10, and 20) and calculating \( V(t) \). Connect these points with a straight line. The graph will be a downward-sloping line, starting at \( \$50,000 \) when \( t = 0 \).

Step 5: Find the Horizontal and Vertical Intercepts

The vertical intercept occurs when \( t = 0 \). By inserting \( t = 0 \) into \( V(t) = -2000t + 50000 \), we get \( V(0) = 50000 \); this represents the original purchase price.To find the horizontal intercept, set \( V(t) = 0 \):\[-2000t + 50000 = 0\]Solving for \( t \), we find \[t = \frac{50000}{2000} = 25\]. This means after 25 years, the tractor will have no value.

Step 6: Interpret the Intercepts

The vertical intercept (0, 50000) shows the initial value of the tractor at the time of purchase. The horizontal intercept (25, 0) indicates the time when the tractor's value will be zero, 25 years after purchase.

Key Concepts

Graphing Linear Functions

Graphing linear functions allows us to visualize how one quantity changes with respect to another in a consistent manner. In the case of linear depreciation, we are looking at how the value of an asset decreases over time. The simplest visual representation of this is a straight line on a graph.

To graph the function for the tractor, start by understanding the formula derived in the solution: \( V(t) = -2000t + 50000 \), where \( V(t) \) represents the value of the tractor, and \( t \) is time in years since purchase. By plotting this on a graph, with time on the x-axis and value on the y-axis, you can clearly see the depreciation trend.

• Start by plotting the y-intercept at \( t = 0 \), which gives you \( V(0) = 50000 \).
• Calculate additional points, like \( V(10) = 30000 \), which will show the tractor's value after 10 years.
• The final value of \( V(20) = 10000 \) after 20 years is also plotted to confirm the line's slope.

Connecting these points with a straight line illustrates the consistent, linear depreciation of the tractor's value over time.

Slope-Intercept Form

The slope-intercept form is a way of writing linear equations using the format \( y = mx + b \), where \( m \) represents the slope and \( b \) is the y-intercept. This form is particularly useful in understanding and graphing linear functions, like our depreciation problem.

In our tractor's valuation problem, we use this form to represent the linear decrease in value over time. The equation \( V(t) = -2000t + 50000 \) matches this pattern:

• The slope \( m = -2000 \) reflects the depreciation rate, indicating the tractor loses $2,000 in value each year.
• The y-intercept \( b = 50000 \) represents the starting value of the tractor when \( t = 0 \).

This straightforward form makes it easy to graph and analyze as it directly shows how the dependent variable, \( V(t) \), changes with variations in \( t \, \).

Interpreting Intercepts

Understanding intercepts gives us valuable insights into the origins and future of our linear function. In a graph, intercepts are the points where the line crosses the x-axis and y-axis. These can tell us about starting values and endpoint predictions.

For the vertical or y-intercept in our tractor example, it occurs when \( t = 0 \). At this point, \( V(t) = 50000 \), signifying the tractor's initial value. This is crucial as it sets the baseline for the linear depreciation.

• The calculation of the y-intercept was straightforward, as it is the number added or subtracted in the slope-intercept form \( y = mx + b \).

The horizontal or x-intercept provides another piece of the puzzle, occurring where \( V(t) = 0 \). Calculating this involves setting the linear function to zero and solving for \( t \):
\[-2000t + 50000 = 0\]
Solves to
\[t = 25\]
This indicates the tractor's value reaches zero after 25 years, offering both a practical and theoretical endpoint for its life.