Problem 32 - Precalculus Mathematics for Calculus

Question

A cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after \(t\) minutes. When the data in the table are graphed, the slope of the tangent line represents the heart rate in beats per minute. $$\begin{array}{|c|c|}\hline t \text { (min) } & \text { Heartbeats } \\\\\hline 36 & 2530 \\\38 & 2661 \\\40 & 2806 \\\42 & 2948 \\\44 & 3080 \\\\\hline\end{array}$$ (a) Find the average heart rates (slopes of the secant lines over the time intervals \([40,42]\) and \([42,44]\) (b) Estimate the patient's heart rate after 42 minutes by averaging the slopes of these two secant lines.

Step-by-Step Solution

Answer

The estimated heart rate at 42 minutes is 68.5 beats per minute.

Step 1: Understand the Problem

We are given a table of heartbeats at different times. We need to find the average heart rates over certain intervals. These rates are found by calculating the slopes of secant lines on the graph of the data.

Step 2: Calculate the Slope for Interval [40,42]

The slope of a secant line over the interval \( [40, 42] \) is given by the formula \( m = \frac{\Delta y}{\Delta x} = \frac{y_2 - y_1}{x_2 - x_1} \). Here, \( (x_1, y_1) = (40, 2806) \) and \( (x_2, y_2) = (42, 2948) \). Calculate the slope: \[ m = \frac{2948 - 2806}{42 - 40} = \frac{142}{2} = 71 \] heartbeats per minute.

Step 3: Calculate the Slope for Interval [42,44]

Similarly, find the slope for the interval \( [42, 44] \) using \( (x_1, y_1) = (42, 2948) \) and \( (x_2, y_2) = (44, 3080) \). The formula gives: \[ m = \frac{3080 - 2948}{44 - 42} = \frac{132}{2} = 66 \] heartbeats per minute.

Step 4: Average the Slopes

To estimate the heart rate at 42 minutes, average the slopes from Steps 2 and 3. This is calculated as: \[ \text{Average heart rate} = \frac{71 + 66}{2} = \frac{137}{2} = 68.5 \] heartbeats per minute.

Key Concepts

Secant Line

The concept of a secant line is fundamental in calculus. A secant line is a straight line that intersects a curve at two points. In the context of our exercise, the heartbeats recorded over time can be plotted on a graph forming a curve. The secant line represents the straight path connecting two points on this curve.
This line gives an average rate of change — in this case, the average heart rate over specific intervals. Using two points from the data table, such as

• [40, 2806] and [42, 2948]
• [42, 2948] and [44, 3080]

you can draw secant lines. The slope of these lines will help us understand how the heart rate changes over each time interval. The secant line simplifies complex curves into easier-to-understand linear segments.

Slope Calculation

Calculating the slope of a secant line involves determining the rate at which a dependent variable changes concerning a change in the independent variable. The slope formula is given by \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] where

• \( y_2 \) and \( y_1 \) are the values of the dependent variable (heartbeats here),
• \( x_2 \) and \( x_1 \) are the values of the independent variable (time in minutes).

For instance, to find the slope over the interval [40, 42], you use the given heartbeat numbers 2806 and 2948: \[ m = \frac{2948 - 2806}{42 - 40} = \frac{142}{2} = 71 \text{ heartbeats per minute} \] This process gives us the average heart rate over that interval. By doing the same for all intervals, you get a better understanding of how the heart rate varies over time.

Average Heart Rate

The average heart rate between intervals provides a more comprehensive picture of the body's responses. To estimate the heart rate at a specific time, such as 42 minutes, we average the calculated slopes of the secant lines from intervals around that time. In this exercise, we calculated the slopes for the intervals [40, 42] and [42, 44] as

• 71 heartbeats per minute,
• 66 heartbeats per minute, respectively.

To find the heart rate exactly at 42 minutes, compute the mean of these two values: \[ \text{Average heart rate} = \frac{71 + 66}{2} = 68.5 \text{ heartbeats per minute} \] This average provides a more stable estimation than any single slope, and helps to smoothen out the variations that might occur at a precise moment. Understanding this average helps in monitoring health and vitally important metrics efficiently.