Problem 19 - Calculus: Early Transcendentals

Question

In someone affected with measles, the virus level \( N \) (measured in number of infected cells per mL of blood plasma) reaches a peak density at about \( t = 12 \) days (when a rash appears) and then decreases fairly rapidly as a result of immune response. The area under the graph of \( N(t) \) from \( t = 0 \) to \( t = 12 \) (as shown in the figure) is equal to the total amount of infection needed to develop symptoms (measured in density of infected cells x time). The function \( N \) has been modeled by the function $$ f(t) = -t(t - 21)(t + 1) $$ Use this model with six subintervals and their midpoints to estimate the total amount of infection needed to develop symptoms of measles.

Step-by-Step Solution

Answer

The total infection required is approximately 4992.

Step 1: Understanding the Problem

We are given a function \( f(t) = -t(t - 21)(t + 1) \) that models the virus level \( N(t) \) in an individual with measles. Our task is to estimate the area under the curve from \( t = 0 \) to \( t = 12 \) using six subintervals.

Step 2: Choosing Subintervals and Midpoints

To approximate the area, divide the interval \([0, 12]\) into six subintervals. Each subinterval has a width of \( \Delta t = \frac{12 - 0}{6} = 2 \). The subinterval midpoints are \( t = 1, 3, 5, 7, 9, \text{ and } 11 \).

Step 3: Calculating Function Values at Midpoints

Compute \( f(t) \) at each midpoint.- For \( t = 1 \), \( f(1) = -(1)(1 - 21)(1 + 1) = 40 \)- For \( t = 3 \), \( f(3) = -(3)(3 - 21)(3 + 1) = 216 \)- For \( t = 5 \), \( f(5) = -(5)(5 - 21)(5 + 1) = 480 \)- For \( t = 7 \), \( f(7) = -(7)(7 - 21)(7 + 1) = 672 \)- For \( t = 9 \), \( f(9) = -(9)(9 - 21)(9 + 1) = 648 \)- For \( t = 11 \), \( f(11) = -(11)(11 - 21)(11 + 1) = 440 \)

Step 4: Estimating the Total Infection

Apply the midpoint rule for the approximation of the total infection required to develop symptoms. The approximate area is given by the sum of the areas of the rectangles:\[ \text{Area} \approx \Delta t \times [f(1) + f(3) + f(5) + f(7) + f(9) + f(11)] \]\[ \text{Area} \approx 2 \times [40 + 216 + 480 + 672 + 648 + 440] \]\[ \text{Area} \approx 2 \times 2496 = 4992 \]

Key Concepts

Understanding Definite Integrals

Definite integrals are a fundamental part of calculus and play a key role in solving problems related to accumulation and total change. They allow us to find the total area under a curve within a defined interval, thereby measuring quantities that aren't easily accessible through simple computations.
The idea is to break a complex problem into smaller, manageable parts. This helps in understanding the behavior of the function over a particular interval.

• A definite integral is represented as \( \int_{a}^{b} f(t) \, dt \), where \( a \) and \( b \) are the limits of integration.
• The definite integral gives us the net area under the curve between the limits \( a \) and \( b \).
• For our problem, the function \( f(t) = -t(t - 21)(t + 1) \) is evaluated between \( t = 0 \) and \( t = 12 \).

By understanding this concept, it becomes easier to estimate how much infestation a virus causes before it peaks by calculating the area under the curve.

Approach to Numerical Integration

Numerical integration is used when a problem does not allow for the exact computation of an integral. It is a technique to approximate the area under a curve. There are several methods to achieve this, with the midpoint rule being one of the simpler and effective ones.

• The goal is to approximate the integral of a function over a specified interval using discrete sample points.
• The midpoint rule involves dividing the interval into smaller intervals or subintervals. The sample points are taken from the midpoint of these subintervals.
• For each midpoint, we evaluate the function and multiply by the width of the subintervals to get an approximate area which is akin to calculating the area of rectangles under the curve.
• In our example, we used six subintervals of width 2 and midpoints at \( t = 1, 3, 5, 7, 9, \text{ and } 11 \).

This makes the midpoint rule a practical choice in many real-world scenarios where exact calculations are cumbersome or impossible.

Applications in Calculus Problems

Calculus has extensive applications across different domains, such as physics, biology, economics, and engineering due to its ability to deal with change and motion. Calculus applications involve finding solutions to problems related to rate of change and accumulation.
For instance, in the context of disease spread, as in the original exercise about measles, calculus can describe how the infection develops and what levels it reaches over time.

• Definite integrals can help model the total accumulation of an infection over a specific timeframe.
• Using numerical methods like the midpoint rule can aid in estimating this total when functions aren't straightforward to integrate by hand.
• Such estimations can be very valuable in healthcare for predicting trends and making informed decisions regarding disease control and resource allocation.

By understanding and applying these calculus principles, one can gain insights into various dynamic systems encountered in everyday life and scientific research.