Problem 80 - Applied Calculus

Question

After \(t\) hours of work, a bank clerk can process checks at the rate of \(r(t)\) checks per hour for the function \(r(t)\) given below. How many checks will the clerk process during the first three hours (time 0 to time 3 )? $$ r(t)=-t^{2}+60 t+9 $$

Step-by-Step Solution

Answer

The clerk processes 288 checks in the first three hours.

Step 1: Define the Problem

The rate of processing checks is given by the function \( r(t) = -t^2 + 60t + 9 \). We need to find the total number of checks processed from \( t = 0 \) to \( t = 3 \). This involves integrating the rate function \( r(t) \) over this interval.

Step 2: Set Up the Integral

To find the total number of checks processed, we will integrate the function \( r(t) = -t^2 + 60t + 9 \) from \( t = 0 \) to \( t = 3 \). This means computing the definite integral: \[ \int_{0}^{3} (-t^2 + 60t + 9) \, dt \]

Step 3: Integrate the Function

Integrate \( -t^2 + 60t + 9 \):- The integral of \( -t^2 \) is \( -\frac{t^3}{3} \).- The integral of \( 60t \) is \( 30t^2 \).- The integral of \( 9 \) is \( 9t \).Thus, the indefinite integral is: \[ -\frac{t^3}{3} + 30t^2 + 9t + C \] (where \( C \) is the constant of integration).

Step 4: Evaluate the Definite Integral

Evaluate from \( t = 0 \) to \( t = 3 \):\[ \left[ -\frac{t^3}{3} + 30t^2 + 9t \right]_{0}^{3} = \left(-\frac{27}{3} + 30 \times 9 + 9 \times 3\right) - \left(0\right) \]This simplifies to: \[ -9 + 270 + 27 = 288 \]

Step 5: Interpret the Result

The result of the integral evaluation is 288. This means the bank clerk processes 288 checks during the first three hours.

Key Concepts

Calculus Integration

Calculus integration is a fundamental concept used to find the total accumulation of a quantity, often represented as the area under a curve. In our exercise, we apply definite integration to determine the total number of checks processed by a bank clerk over a specific interval. Here, the rate function, \( r(t) = -t^2 + 60t + 9 \), shows how many checks are processed per hour. By integrating this function with respect to time from 0 to 3 hours, we can find the cumulative number.To set up the integral, we use

• the integral sign \( \int \),
• the limits of integration from 0 to 3 hours, and
• the function \( (-t^2 + 60t + 9) \) as the integrand.

The process of solving this involves finding an antiderivative of the function, followed by evaluating it at the bounds 0 and 3 and subtracting the two results.

Rate Functions

A rate function describes how quickly something happens over time. In calculus, this is typically represented by a function like \( r(t) \), where \( t \) denotes time. For our example, \( r(t) = -t^2 + 60t + 9 \) gives the rate at which checks are processed per hour by the clerk. It's important because it provides insight into how the process changes over time, especially if the rate isn't constant.To use a rate function effectively:

• Identify the variable involved, which is \( t \) for time.
• Recognize that the function value changes as \( t \) changes; for example, fewer checks might be processed if the rate slows down.
• Integrate the function over the desired time interval to find the total number of events, or in this case, checks processed.

Understanding rate functions help in planning and optimizing activities based on how they evolve over time.

Area Under the Curve

The concept of the area under the curve is pivotal in calculus integration. Finding the area under the rate function curve gives us the total number of checks processed. In mathematical terms, this corresponds to evaluating the definite integral of the rate function over the given time interval.In our exercise:

• The curve represented by \( r(t) \) changes with \( t \), capturing the rate dynamics.
• The area under this curve, from time \( t = 0 \) to \( t = 3 \), represents the cumulative checks processed.

Graphically, you could imagine this as slicing a piece of paper along the x-axis (time) and y-axis (rate), where the enclosed area is our target: the total checks processed. Integrating helps to precisely calculate this area.

Applied Calculus

Applied calculus involves using mathematical concepts in real-world scenarios to solve practical problems. Our exercise with the bank clerk is a classic example where calculus integration is used to compute something tangible: the total number of checks processed over time.In applied calculus:

• We take theoretical concepts like integration and apply them to real-world functions, such as \( r(t) = -t^2 + 60t + 9 \).
• This application allows for tangible insights into processes that have financial, engineering, or scientific implications.
• It supports making informed decisions based on mathematical analysis of rates and cumulative totals.

Through exercises like these, you can see how calculus provides essential tools for solving everyday problems, emphasizing not only the theory but its tangible impact.