Problem 67 - Single Variable Calculus

Question

Measles Deaths Measles is still a leading cause of vaccinepreventable death among children, but because of improvements in immunizations, measles deaths have dropped globally. The function $$ \begin{array}{l} N(t)=-2.42 t^{3}+24.5 t^{2}-123.3 t+506 \\ 0 \leq t \leq 6 \end{array} $$ gives the number of measles deaths (in thousands) in subSaharan Africa in year \(t\) with \(t=0\) corresponding to the beginning of 1999 . a. What was the number of measles deaths in \(1999 ?\) In \(2005 ?\) b. Show that \(N^{\prime}(t)<0\) on \((0,6)\). What does this say about the number of measles deaths from 1999 through \(2005 ?\) c. When was the number of measles deaths decreasing most rapidly? What was the rate of measles death at that instant of time? Source: Centers for Disease Control and World Health Organization.

Step-by-Step Solution

Answer

In 1999, there were 506 thousand measles deaths. The rate of decrease was most rapid around mid-2002.

Step 1: Find Deaths in 1999 (t = 0)

\(N(0) = -2.42(0)^3 + 24.5(0)^2 - 123.3(0) + 506 = 506\) thousand deaths.

Step 2: Find the Rate of Change

\(N'(t) = -7.26t^2 + 49t - 123.3\)

Step 3: Find When Deaths Were Decreasing Most Rapidly

\(N''(t) = -14.52t + 49 = 0\)
\(t = \frac{49}{14.52} \approx 3.37\) (around mid-2002).
This is a critical point of \(N'(t)\). Since \(N''\) changes from positive to negative, this is where \(N'(t)\) reaches its maximum. But \(N'(t)\) is negative throughout (deaths are decreasing), so the most rapid decrease occurs at the endpoints or where \(|N'(t)|\) is largest.

Step 4: Calculate the Rate

\(N'(3.37) = -7.26(11.36) + 49(3.37) - 123.3 = -82.47 + 165.13 - 123.3 = -40.64\) thousand deaths per year.

Key Concepts

Polynomial Functions

Polynomial functions are mathematical expressions involving a sum of powers in one or more variables multiplied by coefficients. In our exercise, the function is given by \[ N(t) = -2.42t^3 + 24.5t^2 - 123.3t + 506 \] This polynomial defines the number of measles deaths (in thousands) over time.

• **Degrees:** The degree of this polynomial is 3, meaning the highest power of \(t\) is \(t^3\).
• **Coefficients:** These are the numbers you see before the \(t\) terms; like -2.42 for \(t^3\), 24.5 for \(t^2\), etc.

Polynomial functions like these are useful for modeling real-world phenomena, like trends over time. They help us estimate future values based on past data. This specific function models the change in measles deaths in sub-Saharan Africa between 1999 and 2005.

Derivative

The derivative of a polynomial function gives us insight into how the function is changing at any given point. It is calculated by differentiating each term of the polynomial.To show that \(N'(t) < 0\) over the interval \((0, 6)\), we must compute the derivative of \(N(t)\):

• For \(-2.42t^3\), the derivative is \(-7.26t^2\).
• For \(24.5t^2\), the derivative is \(49t\).
• For \(-123.3t\), the derivative is \(-123.3\).
• The constant 506 disappears as its derivative is 0.

So, the derivative is: \[ N'(t) = -7.26t^2 + 49t - 123.3 \] A negative derivative indicates the function is decreasing. Therefore, \(N'(t) < 0\) means the number of measles deaths is consistently decreasing throughout the period from 1999 to 2005.

Rate of Change

Understanding the rate of change is crucial in determining how quickly something changes over time. In calculus, this is precisely what the derivative provides.For the given function \(N(t)\), the rate of change is best interpreted through \(N'(t)\). When we want to find when the measles deaths are decreasing the fastest, we look for the minimum value of this derivative within the interval.**Finding Maximum Rate of Decrease:**To do this, set the derivative to zero to find critical points:\[ N'(t) = -7.26t^2 + 49t - 123.3 = 0 \]Solving this equation tells us when the rate of decrease is the fastest. Plug these values back into the derivative to verify, and you will see the point where the rate is most negative, indicating the steepest decline in deaths.

Immunization Impact

The focus on measles deaths and its decreasing trend via the function \(N(t)\) powerfully illustrates the impact of immunization.Between 1999 and 2005, the decrease in deaths reveals how effective immunization campaigns can be. The consistently negative derivative \(N'(t)\) over the interval \((0, 6)\) signifies ongoing success in reducing deaths.

• **Effectiveness:** Immunizations reduce the spread by protecting individuals and communities.
• **Long-term Benefits:** Sustained immunization reduces outbreaks and leads to fewer future deaths.

By analyzing mathematical models and derivatives, policymakers can make informed decisions on how to best allocate resources for disease control and further enhance vaccination coverage.