Problem 16 - Calculus Concepts

Question

Flu Shots The percentage of adults who said they got a flu shot before the winter of year \(t\) is given by $$ S(t)=-0.18 t^{2}+5.24 t+9 \text { percent } $$ where \(t\) is the number of years since 2000 , data from \(2004 \leq t \leq 2009 .\) (Source: Based on data in USA Today, p. \(1 \mathrm{~A}, 5 / 18 / 2009)\) a. Find the derivative formula using the algebraic method. b. Evaluate the derivative of \(s\) in \(2007 .\) Interpret the result.

Step-by-Step Solution

Answer

The derivative formula is \( S'(t) = -0.36t + 5.24 \). In 2007, the rate of increase in flu shots was 2.72% per year.

Step 1: Find the Derivative Formula

To find the derivative, we take the derivative of the given function \( S(t) = -0.18t^2 + 5.24t + 9 \) with respect to \( t \). Use the power rule, which states that if \( f(t) = at^n \), then \( f'(t) = nat^{n-1} \). Applying the power rule: 1. The derivative of \( -0.18t^2 \) is \( -0.36t \). 2. The derivative of \( 5.24t \) is \( 5.24 \). 3. The derivative of the constant \( 9 \) is \( 0 \).Therefore, the derivative \( S'(t) = -0.36t + 5.24 \).

Step 2: Evaluate the Derivative in 2007

To evaluate the derivative at the year 2007, first determine the value of \( t \) for the year 2007. Since \( t \) is the number of years since 2000, for 2007, \( t = 2007 - 2000 = 7 \). Substitute \( t = 7 \) into the derivative:\[ S'(7) = -0.36(7) + 5.24 \]Calculate:\[ S'(7) = -2.52 + 5.24 = 2.72 \]

Step 3: Interpret the Result

The derivative evaluated in 2007 gave us \( S'(7) = 2.72 \). This means that in 2007, the percentage of adults who got a flu shot was increasing at a rate of 2.72% per year. This rate of change indicates how quickly the percentage of adults receiving flu shots was rising at that time.

Key Concepts

Derivative Formula

In calculus, a derivative represents the rate at which a quantity changes. For the function given, which models the percentage of adults getting flu shots, finding the derivative helps us understand how this percentage is changing over time. To find a derivative algebraically, follow a systematic approach. You take the derivative of each term in the function separately.

In our example, the function is given as \( S(t) = -0.18t^2 + 5.24t + 9 \). Derivatives are found using rules like the power rule, which we'll explain in the next section. Once you apply these rules, you will get the derivative, which in this case is \( S'(t) = -0.36t + 5.24 \). This formula allows you to find the rate of change at any point \( t \).

Rate of Change

The derivative can be interpreted as a rate of change. For example, when we found \( S'(t) = -0.36t + 5.24 \), we could use it to determine how quickly the percentage of adults receiving flu shots changed over the years.

At any given point \( t \), the output of the derivative formula tells us the change per unit time. In other words, it shows how much the percentage is increasing or decreasing year by year.

• For \( t = 7 \) (the year 2007), we calculated \( S'(7) = 2.72 \).
• This means the percentage increased by 2.72% that year.

This understanding of derivatives as rates of change is crucial in fields like economics, physics, and biology, as it provides insights into trends over time.

Power Rule

The power rule is one of the simplest but most important rules for finding derivatives. It states: if you have a function \( f(t) = at^n \), then its derivative is \( f'(t) = nat^{n-1} \). This rule makes it straightforward to handle polynomials.

In our example, applying the power rule:

• For the term \( -0.18t^2 \), we multiply \(-0.18\) by the power \(2\) and subtract \(1\) from the power, resulting in \(-0.36t\).
• For \( 5.24t \), the derivative is simply the coefficient \( 5.24 \) because the power of \( t \) is \( 1 \).
• Constants like \( 9 \) vanish in derivatives because their rate of change is zero.

Using the power rule is fundamental not only in calculus but also in understanding how quantities change with respect to time.

Algebraic Method

The algebraic method involves using algebraic rules to manipulate an expression into its derivative form. This method requires understanding each component’s contribution to the overall change in the function.

Think of the algebraic method as breaking down a function into parts and then applying rules to each part sequentially. For the function \( S(t) = -0.18t^2 + 5.24t + 9 \).

• Begin with each component separately, first addressing \( -0.18t^2 \), then \( 5.24t \), and lastly the constant \( 9 \).
• Utilize the power rule for terms involving \( t \), and remember constants simply become \( 0 \).

After you finish, gather the results to form the complete derivative: \( S'(t) = -0.36t + 5.24 \). The algebraic method is especially useful in straightforward calculus problems where direct computation leads to clear insights.