Problem 52 - Calculus of a Single Variable

Question

A company introduces a new product for which the number of units sold \(S\) is \(S(t)=200\left(5-\frac{9}{2+t}\right)\) where \(t\) is the time in months. (a) Find the average rate of change of \(S(t)\) during the first year. (b) During what month of the first year does \(S^{\prime}(t)\) equal the average rate of change?

Step-by-Step Solution

Answer

The average rate of change during the first year is approximately -15 units per month. The instantaneous rate of change equals this average rate in approximately the 8th month.

Step 1: Calculate the average rate of change

Average rate of change is calculated using the difference quotient formula which is \(\frac{f(b) - f(a)}{b - a}\). In this case from \(t=0\) to \(t=12\) (as the question is asking for the average rate of change during the first year): \(\frac{S(12) - S(0)}{12 - 0}\)

Step 2: Evaluate \(S(t)\) at \(t=0\) and \(t=12\)

Substitute 0 and 12 into \(S(t)=200\left(5-\frac{9}{2+t}\right)\) to find \(S(0)\) and \(S(12)\), then substitute these results into the average rate of change formula.

Step 3: Calculate the derivative of \(S(t)\)

To calculate \(S'(t)\), the derivative of \(S(t)\), use the rules of differentiation. Remember to apply the chain rule correctly when differentiating terms like \(\frac{9}{2+t}\).

Step 4: Equate \(S'(t)\) to the average rate

We now solve for \(t\) by setting the derivative \(S'(t)\) equal to the average rate of change calculated in step 1. This will provide us with the month where the instantaneous rate of change equals the average rate of change.

Step 5: Solve for \(t\)

After equating the derivative to the average rate, simplify and solve the equation for \(t\).

Key Concepts

Rate of Change

Understanding the rate of change is crucial for analyzing the behavior of functions in calculus. It refers to how a function's output varies as its input changes. For instance, in real-life terms, it could describe how quickly a car's speed increases over time or how the sales of a product grow month by month. In the given exercise, the rate of change sheds light on how the number of units sold by a company changes over time.

By looking at the average rate of change during the first year, you can understand the overall performance of the product within that timeframe. This concept is foundational before delving into more complex aspects of calculus, as it provides a general sense of the function's behavior over a specified interval.

Difference Quotient Formula

To calculate the average rate of change, mathematicians use the difference quotient formula, which is given by \(\frac{f(b) - f(a)}{b - a}\). It essentially measures the slope of the secant line that connects two points on the graph of a function. This formula serves as the basis for understanding the average rate of change between two points.

Within the context of our exercise, we apply the difference quotient to the function \(S(t)\) to find the average rate of change over the first year, which is from \(t=0\) to \(t=12\). This calculation gives us a broad view of the product's performance over this interval.

Instantaneous Rate of Change

The instantaneous rate of change is the rate at which a function is changing at any given point, and is represented by the derivative of the function. Think of it as a snapshot of the rate of change at a specific instant - similar to how a speedometer shows your current speed at a precise moment. It is the limit of the average rate of change as the interval gets infinitely small.

In our exercise, we're interested in pinpointing the exact month during the first year when the instantaneous rate of change of sales, denoted as \(S'(t)\), equates to the average rate of change over the entire year. Solving this gives us deeper insight into the sales dynamics at a particular point in time.

Derivative of a Function

The derivative of a function at a point is the core concept in calculus, representing the instantaneous rate of change. It's like zooming in on the graph of the function until it looks straight, and then measuring the slope of that line. In mathematical terms, the derivative at a point is the limit of the difference quotient as the interval approaches zero.

In practice, we find the derivative using the rules of differentiation. For our exercise, to get \(S'(t)\), we apply these rules to the given function \(S(t)\). By understanding how to differentiate functions correctly, students can then calculate how fast the sales are changing at any given moment.

Chain Rule in Differentiation

Differentiation can be straightforward until you meet functions within functions. That's where the super handy chain rule comes into play. It's a method in calculus for finding the derivative of composite functions. To put it simply, if a function \(y\) is composed of another function \(u\), so \(y = f(u)\) and \(u = g(x)\), then the derivative of \(y\) with respect to \(x\) is the derivative of \(f\) with respect to \(u\) multiplied by the derivative of \(u\) with respect to \(x\), or \(\frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx}\).

When we differentiate the sales function in our exercise, we use the chain rule to handle the term \(\frac{9}{2+t}\). Differentiating complex functions like this without the chain rule would be tough, if not impossible. By breaking it down, the chain rule makes it manageable and is a testament to the power and elegance of calculus.