Problem 86 - Applied Calculus

Question

An automobile dealer estimates that the newest model car will sell at the rate of \(30 / t\) cars per month, where \(t\) is measured in months and \(t=1\) corresponds to the beginning of January. Find the number of cars that will be sold from the beginning of January to the beginning of May.

Step-by-Step Solution

Answer

Approximately 48 cars will be sold from January to May.

Step 1: Identify the time period

We are asked to find the number of cars sold from the beginning of January to the beginning of May. Since January is month 1 and May is month 5, the time period of interest is from \( t = 1 \) to \( t = 5 \).

Step 2: Write down the rate function

The rate of car sales per month is given by \( \frac{30}{t} \). This is the rate at which cars are sold as a function of time \( t \).

Step 3: Set up the integral

To find the total number of cars sold, we need to integrate the rate function \( \frac{30}{t} \) over the interval \([1, 5]\). This is represented as: \[ \int_{1}^{5} \frac{30}{t} \, dt \].

Step 4: Calculate the integral

The integral of \( \frac{30}{t} \) with respect to \( t \) is \( 30 \ln |t| + C \). We apply the bounds from 1 to 5: \[ 30 \ln |5| - 30 \ln |1| \].

Step 5: Simplify the expression

We know \( \ln |1| = 0 \) and \( \ln |5| \approx 1.60944 \). Thus, the expression simplifies to \( 30 \times 1.60944 \approx 48.2832 \).

Step 6: Interpret the result

The result of the integral, approximately 48.28, represents the total number of cars sold between the beginning of January and the beginning of May.

Key Concepts

Rate of Change

In calculus, the rate of change is a crucial concept. It measures how a quantity changes over time. In real-world terms, imagine you're tracking sales of a product month by month. The rate of change tells you how quickly sales are increasing or decreasing each month.
Here, the rate of change is given by the function \( \frac{30}{t} \). This tells us the number of cars the dealer expects to sell each month. As \( t \) (time in months) changes, the rate changes. A larger \( t \) results in a smaller fraction, meaning fewer sales per month as the months go by.
To understand better, here's what the formula implies:

• When \( t \) is small (like 1 or 2), the rate \( \frac{30}{t} \) is high.
• When \( t \) increases, the rate decreases. For instance, at \( t = 3 \), the rate is 10; at \( t = 5 \), it's 6.

This tells us that car sales are expected to be brisker early on and taper off over time.

Definite Integral

A definite integral is like a magic tool in calculus. It helps us add up a whole bunch of tiny slices to get a total quantity, such as total sales or total growth. In simple terms, it lets us find the area under a curve, between two points. This "area" can represent various things; in our car sales example, it represents the total number of cars sold over a given period.
For the car dealership, we are integrating the rate function \( \frac{30}{t} \) over the interval \([1, 5]\).
In mathematical terms, this is:

• \( \int_{1}^{5} \frac{30}{t} \, dt \)

The calculated result, about 48.28, shows the total cars sold over four months, from January to May. This integral transformation from a rate (cars per month) to a total (cars sold) is a prime example of how definite integrals operate.

Logarithmic Function

Logarithmic functions are a special family of functions representing processes that grow slowly over time. When you're solving problems involving rates that decrease over time, like our car sales, you often use logarithms.
The integral of \( \frac{30}{t} \) ends up being expressed with logarithms: \( 30 \ln |t| + C \). The logarithm \( \ln |t| \) reflects the slow change as time \( t \) grows. It helps adjust our calculations for this decreasing rate of sales.
Here's why it matters:

• \( \ln |5| \), which approximately equals 1.60944, calculates the effect over five months.
• The term \( \ln |1| = 0 \) perfectly fits our initial month since there's no time yet for decay in rate.

Understanding logarithms gives us insight into real-life scenarios where things don't grow or shrink linearly, like car sales trailing off over time.