Problem 67 - Applied Calculus: For Business, Economics, and the Social and Life Sciences

Question

TICKET SALES The promoters of a county fair estimate that \(t\) hours after the gates open at \(9: 00\) A.M. visitors will be entering the fair at the rate of \(-4(t+2)^{3}+54(t+2)^{2}\) people per hour. How many people will enter the fair between 10:00 A.M. and noon?

Step-by-Step Solution

Answer

1220 people

Step 1: Define Variables and Limits

Define the time interval for which the number of people entering the fair has to be calculated. From 10:00 A.M. to noon is a 2-hour interval. If the gates open at 9:00 A.M., then at 10:00 A.M., \( t = 1 \) hour, and at noon, \( t = 3 \) hours. So, we will integrate from \( t = 1 \) to \( t = 3 \).

Step 2: Set Up the Integral

The rate at which people are entering the fair is given by the function \( -4(t+2)^{3}+54(t+2)^{2} \). To find the total number of people who enter between \( t = 1 \) and \( t = 3 \), we need to integrate this function over that interval: \[ \text{Total People} = \int_{1}^{3} -4(t+2)^{3} + 54(t+2)^{2} \, dt \]

Step 3: Simplify the Expression Inside the Integral

To help integrate, simplify the expression inside the integral first: Let \(u = t + 2\), hence \(du = dt\). The limits of integration change accordingly: when \(t = 1\), \(u = 3\), and when \(t = 3\), \(u = 5\). Rewrite the integral using \(u\): \[ \int_{1}^{3} -4(t+2)^{3} + 54(t+2)^{2} \, dt = \int_{3}^{5} -4u^{3} + 54u^{2} \, du \]

Step 4: Integrate the Function

Integrate the function \( -4u^{3} + 54u^{2} \) with respect to \( u \): The integral of \( -4u^{3} \) is \( -4 \frac{u^{4}}{4} = -u^{4} \), and the integral of \( 54u^{2} \) is \( 54 \frac{u^{3}}{3} = 18u^{3} \). So, \[ \int -4u^{3} + 54u^{2} \, du = -u^{4} + 18u^{3} + C \]

Step 5: Calculate the Definite Integral

Evaluate the definite integral from \( u = 3 \) to \( u =5 \): \[(-5^{4} + 18\times5^{3}) - (-3^{4} + 18\times3^{3}) = (-625 + 2250) - (-81 + 486) \] Simplify the results: -625 + 2250 = 1625 -81 + 486 = 405 So, 1625 - 405 = 1220 Therefore, the total number of people entering the fair between 10:00 A.M. and noon is 1220 people.

Key Concepts

Definite Integrals

When we talk about definite integrals in calculus, we are referring to the integral of a function over a specific interval. In this case, the definite integral helps us calculate the total number of people entering the fair between two time points. The given function indicates how fast people enter the fair, and integrating this function tells us the total number over a period. The definite integral \(\text{Total People} = \int_{1}^{3} -4(t+2)^{3} + 54(t+2)^{2} \text{d}t\) represents this calculation. This approach allows us to sum up the continuous rate of change to find out the overall amount.

Rate of Change

The concept of the rate of change is fundamental in calculus and real-world applications. In this exercise, the rate of people entering the fair is given functions as \(-4(t+2)^{3} + 54(t+2)^{2}\), which describes how quickly people are added to the fair each hour. By understanding this rate, we can make various predictive analyses. Rates of change are expressed in different fields like physics as speed, in finance as growth rates, and here, for ticket sales at a fair. The rate gives us instantaneous values, but we often need the total effect over time, addressed by integrating the rate function.

U-Substitution

U-substitution is a critical technique in integral calculus used to simplify the integration process. Here, we used it to change variables for easier computation. The substitution \(u = t + 2\) transforms a complicated integral into a simpler form. After making this substitution, our variable of integration changes, and accordingly, limits are adjusted from \(t = 1 \text{to} 3\) to \(u = 3 \text{to} 5\). Consequently, our integral becomes \(\text{Total People} = \int_{3}^{5} -4u^{3} + 54u^{2} \text{d}u\). This process makes it easier to handle and solve complex integrals that would otherwise be cumbersome to work with directly in their original form.

Time Intervals

Working with time intervals is crucial in problems dealing with rate functions. Time intervals define the specific periods over which we measure or calculate changes. In this problem, we consider the interval from 10:00 A.M. to noon. Since the gates opened at 9:00 A.M., 10:00 A.M. corresponds to \(t = 1\) hour, and noon corresponds to \(t = 3\) hours. This 2-hour window serves as our limits for the definite integral \(\text{from } t = 1 \text{to } 3\). Clearly defining the time intervals helps us set up the integral correctly and calculate the total number of people entering the fair within the given period.