Problem 12 - Calculus Concepts

Question

Distance Clinton County, Michigan, is mostly flat farmland partitioned by straight roads (often gravel) that run either north/south or east/west. A tractor driven north on Lowell Road from the Schafers farm's mailbox is $$ f(t)=0.28 t+0.6 \text { miles } $$ north of Howe Road \(t\) minutes after leaving the farm's mailbox. a. How far is the Schafers' mailbox from Howe Road? b. Use the limit definition of the derivative to show that the tractor is moving at a constant speed. c. How quickly (in miles per hour) is the tractor moving?

Step-by-Step Solution

Answer

a. 0.6 miles; b. The derivative is constant: 0.28; c. 16.8 miles per hour.

Step 1: Determine the Initial Distance

The initial distance from the Schafers' mailbox to Howe Road is given by evaluating the function \( f(t) \) at \( t = 0 \). This is the constant term in the equation for \( f(t) \). Thus, \( f(0) = 0.28 \times 0 + 0.6 = 0.6 \) miles. Therefore, the mailbox is 0.6 miles north of Howe Road.

Step 2: Understand the Function Form

The function \( f(t) = 0.28t + 0.6 \) is of the form \( f(t) = mt + b \), where \( m \) represents the rate of change. This linear function suggests that the rate of change or the speed of the tractor could be constant, as \( m \) does not depend on \( t \).

Step 3: Use the Limit Definition of Derivative

The derivative \( f'(t) \) represents the speed of the tractor. To find this, use the limit definition of the derivative: \[f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h}\]Apply it to our function: \[f'(t) = \lim_{h \to 0} \frac{(0.28(t+h) + 0.6) - (0.28t + 0.6)}{h} = \lim_{h \to 0} \frac{0.28h}{h} = 0.28\]Therefore, the derivative is 0.28, indicating the tractor is moving at a constant speed.

Step 4: Convert Speed to Miles per Hour

The derivative tells us the speed in miles per minute. To convert this to miles per hour, multiply by 60, since there are 60 minutes in an hour. \[0.28 \text{ miles per minute} \times 60 = 16.8 \text{ miles per hour}\]Hence, the tractor is moving at 16.8 miles per hour.

Key Concepts

Limit Definition

The limit definition of a derivative is a cornerstone concept in calculus. It helps us understand how functions change at a specific point. Suppose you have a function like \( f(t) = mt + b \), which describes a linear relationship. To find the rate at which the function changes, or its derivative, you use the limit definition. This involves considering a tiny change, \( h \), and seeing how the function \( f(t+h) \) behaves as \( h \) approaches zero. In mathematical terms, this is:

• \[ f'(t) = \lim_{h \to 0} \frac{f(t+h) - f(t)}{h} \]

This equation calculates the instantaneous rate of change of \( f(t) \) at any point \( t \). It essentially tells us how fast \( f(t) \) is changing with respect to changes in \( t \). By using this definition for our linear function, we can confirm that certain changes, like constant speeds, can be accurately described.

Linear Functions

Linear functions are simple yet powerful. They're expressed in the form \( f(t) = mt + b \), where \( m \) and \( b \) are constants. Here, \( m \) is the slope, or rate of change, and \( b \) is the intercept. These functions graph as straight lines.
In practical contexts, linear functions help define relationships with constant rates of change. For example, in the tractor problem, the linear function \( f(t) = 0.28t + 0.6 \) models how distance changes over time. The slope \( m = 0.28 \) tells us how many miles the tractor covers every minute. Since \( m \) doesn’t change over \( t \), we conclude that the tractor speeds at a constant rate. Therefore, linear functions simplify predicting these properties without complex calculations.
Linear functions are foundational for understanding calculus and other mathematical concepts. They provide a simple view into more complex relationships.

Rate of Change

The rate of change in a function refers to how fast or slow something varies with respect to time. In mathematics, this is often represented as the derivative of a function. For linear functions, the rate of change is constant and equal to the slope \( m \). If we consider the speed of a tractor, the rate of change represents how quickly it covers distance over time.
For the given function \( f(t) = 0.28t + 0.6 \), the rate of change is \( 0.28 \), reflecting its speed. The derivative \( f'(t) \) helps verify this, as calculating it reveals the same rate. This confirms that every minute, the tractor moves north by 0.28 miles. converting this to hours by multiplying by 60 gives us a broader perspective: 16.8 miles per hour.
Thus, understanding rate of change is crucial for real-world applications, such as determining speeds, predicting movements, and analyzing temporal changes. It serves as a pivotal mathematical concept for various scientific disciplines.

Distance and Speed Calculations

Finding the distance and speed of a moving object is vital in many scenarios. The tractor exercise reflects this need through the equation \( f(t) = 0.28t + 0.6 \). The given equation helps dictate where the tractor is at any time \( t \). Initially, at \( t = 0 \), the tractor is 0.6 miles from the starting point.
Every subsequent minute, the distance increases at a rate dictated by the linear function's slope, 0.28 miles per minute. From this, you can calculate the speed by finding the derivative, which provides the rate of change. Converting the constant speed from miles per minute to miles per hour involves multiplying by 60. This results in a speed of 16.8 miles per hour.
Numerous real-world applications depend on these calculations. They're fundamental in transportation, navigation, and even sports. Understanding how to interpret these equations allows us to track movements precisely, estimate arrival times, and optimize routes efficiently.