Problem 8 - Active Calculus

Question

The value, \(V,\) of a particular automobile (in dollars) depends on the number of miles, \(m\), the car has been driven, according to the function \(V=h(m)\). a. Suppose that \(h(40000)=15500\) and \(h(55000)=13200\). What is the average rate of change of \(h\) on the interval \([40000,55000],\) and what are the units on this value? b. In addition to the information given in (a), say that \(h(70000)=11100\). Determine the best possible estimate of \(h^{\prime}(55000)\) and write one sentence to explain the meaning of your result, including units on your answer. c. Which value do you expect to be greater: \(h^{\prime}(30000)\) or \(h^{\prime}(80000)\) ? Why? d. Write a sentence to describe the long-term behavior of the function \(V=h(m)\), plus another sentence to describe the long-term behavior of \(h^{\prime}(m) .\) Provide your discussion in practical terms regarding the value of the car and the rate at which that value is changing.

Step-by-Step Solution

Answer

Average rate of change: -0.1533 dollars per mile. Estimate of \( h'(55000) \): -0.1467 dollars per mile. \( h'(30000) < h'(80000) \). Long-term: \( h(m) \rightarrow 0 \) and \( h'(m) \rightarrow 0 \).

Step 1 – Calculate Average Rate of Change (Part a)

The average rate of change of a function between two points (m_1, m_2) is given by the formula \[ \text{Average Rate of Change} = \frac{h(m_2) - h(m_1)}{m_2 - m_1} \] Plugging in the given values: \[ \text{Average Rate of Change} = \frac{h(55000) - h(40000)}{55000 - 40000} = \frac{13200 - 15500}{55000 - 40000} = \frac{-2300}{15000} = -0.1533 \] The units are dollars per mile.

Step 2 – Estimate Derivative (Part b)

To estimate the derivative at a point, we use the average rate of change around that point. For the best possible estimate of \(h^{\prime}(55000)\), let's consider the interval between 40000 and 70000 miles: \[ \text{Estimate of } h^{\prime}(55000) = \frac{h(70000) - h(40000)}{70000 - 40000} = \frac{11100 - 15500}{70000 - 40000} = \frac{-4400}{30000} = -0.1467 \] This means at 55000 miles, the car's value is decreasing at a rate of approximately $0.1467 per mile.

Step 3 – Compare Derivatives (Part c)

If miles driven are fewer (closer to 0), the car loses value more rapidly, because it's newer and has a higher value. As more miles are driven, the rate of value loss decreases. Therefore, it is expected \[ h^{\prime}(30000) < h^{\prime}(80000) \] because the car value decreases more slowly at higher mileages.

Step 4 – Long-term Behavior of the Function (Part d)

As the number of miles driven increases, the car's value continues to decrease, approaching but never reaching zero. In long-term behavior, \( h(m) \rightarrow 0 \) as \( m \rightarrow \infty \). Meanwhile, the rate at which the car's value decreases (the derivative \( h'(m) \)) also approaches zero as miles increase since the car value diminishes at a slower pace over time. This means, in practical terms, that the depreciation of the car becomes less severe as the car ages.

Key Concepts

Average Rate of Change

The average rate of change gives us a snapshot of how a function's output changes between two input values. We calculate it using the formula:










The units of this rate depend on the input and output units of the function. In this example, the output is dollars and the input is miles. Therefore, the rate tells how many dollars of value the car loses per mile driven.

Instantaneous Rate of Change

The instantaneous rate of change at a particular point of a function is essentially the derivative at that point. It gives a more precise understanding of how the car's value is changing at exactly that number of miles. We estimate it by averaging rates of change over small intervals around the point of interest:

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In practical terms, this rate tells us how fast the car is losing value per additional mile driven at the exact number of miles referenced. For example, at 55000 miles, the value is decreasing at approximately $0.1467 per mile.

Long-term Behavior of Functions

When examining long-term behavior, we're interested in what happens to the function as the input grows very large. For the car's value function, as the miles driven increases, the value of the car decreases, approaching zero but never quite getting there. This is because as cars age and accumulate miles, their values drop significantly. Function: As the number of miles driven grows very large (approaches infinity), the value of the car ( V) gets closer and closer to zero. Derivative: As the number of miles driven increases, the rate of value loss ( h'(m)) also approaches zero, but remains negative, representing a slowing rate of depreciation. This means that over time, the car depreciates slower and slower.

First Derivative

The first derivative of a function gives us important information about the function's rate of change at any point. In this context, it tells us how fast the car's value is decreasing per mile driven. To find the first derivative, we often look at the slope of the tangent line at any given point: The first derivative helps us understand how the rate of value loss is evolving as the mileage increases. Initially, the car's value decreases rapidly when it's newer. Over time, the loss rate slows down.