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Chapter 13: Problem 42

Heart Rate \(A\) cardiac monitor is used to measure the heart rate of a patient after surgery. It compiles the number of heartbeats after \(t\) min. 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 min by averaging the slopes of these two secant lines.

Short Answer

Expert verified
The estimated heart rate after 42 minutes is 68.5 beats per minute.

Step by step solution

01

Identify the Intervals

We are given two intervals: \([40, 42]\) and \([42, 44]\). For each interval, we will calculate the average heart rate using the heartbeats data provided.
02

Calculate the Slope for [40, 42]

The slope of the secant line over the interval \([40, 42]\) is calculated by \( \frac{\text{change in heartbeats}}{\text{change in time}} = \frac{2948 - 2806}{42 - 40} = \frac{142}{2} = 71 \text{ beats per minute} \).
03

Calculate the Slope for [42, 44]

The slope of the secant line over the interval \([42, 44]\) is calculated by \( \frac{3080 - 2948}{44 - 42} = \frac{132}{2} = 66 \text{ beats per minute} \).
04

Find the Average of the Two Slopes

To estimate the patient's heart rate after 42 min, average the two slopes: \( \frac{71 + 66}{2} = 68.5 \text{ beats per minute} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Heart Rate Calculation
Understanding how to calculate heart rate is crucial, especially in a medical setting where monitoring a patient's recovery progress is essential. After surgery, doctors often keep an eye on the patient's heart rate to ensure that the heart is functioning properly. Here, the cardiac monitor collects data over a given period, capturing the total heartbeats as time progresses.
The heart rate, in this context, is calculated by finding the change in the number of heartbeats over a set interval of time. This change helps us assess how the heart is responding and recovering from the stress of surgery. By breaking this down into simple calculations, we can estimate the average beats per minute during a given time frame.
For example, if a patient has 2806 heartbeats at 40 minutes and 2948 heartbeats at 42 minutes, the heart rate between these times can be calculated as the difference in beats divided by the difference in time. This gives a clearer understanding of the average heart rate during that specific interval.
Secant Line
The secant line plays an important role when we want to find the average rate of change in a situation. In the context of heart rate calculations, the secant line helps us visualize how the heartbeats develop over a set period. By examining the slope of this line, we can determine the average heart rate between two points in time.
A secant line is essentially a straight line that connects two points on a curve. By finding the slope of this line, we determine the "average" behaviour of the data between those two points. For instance, connecting the heartbeats at 40 minutes and 42 minutes creates a secant line, helping us find the slope which serves as an indicator of average heart rate for that timeframe.
This approach is useful because it simplifies complex data into a straightforward calculation, helping healthcare professionals make informed decisions based on the patient's current heart function.
Slope Estimation
Slope estimation is a fundamental concept used in numerous fields, including medicine, to interpret data effectively. In this heart rate calculation exercise, the slope indicates the rate at which the heartbeats change over the given time intervals. The slope itself is calculated by taking the difference in the heartbeat count over a specified time span.
To put it simply, the slope in this scenario tells us how many beats per minute the heart is averaging over the interval. For the interval from 40 to 42 minutes, the slope is determined as follows: take the difference in heartbeats (142) and divide it by the difference in time (2 minutes) to get a slope of 71 beats per minute.
This slope estimation gives a numerical summary of how quickly or slowly the heartbeat is changing, providing an essential insight into the patient's heart condition. By averaging multiple slopes, as done with the estimation at 42 minutes, healthcare providers gain a reliable heartbeat estimate to monitor patient recovery accurately.

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Most popular questions from this chapter

Estimating Limits Numerically Complete the table of values (to five decimal places), and use the table to estimate the value of the limit. $$\lim _{x \rightarrow 1} \frac{x-1}{x^{3}-1}$$ $$\begin{array}{|c|c|c|c||c|c|c|} \hline x & 0.9 & 0.99 & 0.999 & 1.001 & 1.01 & 1.1 \\ \hline f(x) & & & & & & \\ \hline \end{array}$$

Does the Limit Exist? Let $$f(x)=\left\\{\begin{array}{ll} x-1 & \text { if } x<2 \\ x^{2}-4 x+6 & \text { if } x \geq 2 \end{array}\right.$$ (a) Find \(\lim _{x \rightarrow 2^{-}} f(x)\) and \(\lim _{x \rightarrow 2^{+}} f(x)\) (b) Does \(\lim _{x \rightarrow 2} f(x)\) exist? (c) Sketch the graph of \(f\)

DISCUSS: The Lorentz Contraction In the theory of relativity the Lorentz contraction formula $$L=L_{0} \sqrt{1-v^{2} / c^{2}}$$ expresses the length \(L\) of an object as a function of its velocity \(v\) with respect to an observer, where \(L_{0}\) is the length of the object at rest and \(c\) is the speed of light. Find \(\lim _{v \rightarrow c^{-}} L,\) and interpret the result. Why is a left-hand limit necessary?

Water Flow \(A\) tank holds 1000 gal of water, which drains from the bottom of the tank in half an hour. The values in the table show the volume \(V\) of water remaining in the tank (in gal) after \(t\) minutes. $$\begin{array}{c|c}\hline t \text { (min) } & V \text { (gal) } \\\\\hline 5 & 694 \\\10 & 444 \\ 15 & 250 \\\20 & 111 \\\25 & 28 \\\30 & 0 \\\\\hline\end{array}$$ (a) Find the average rates at which water flows from the tank (slopes of secant lines) for the time intervals \([10,15]\) and \([15,20]\) (b) The slope of the tangent line at the point \((15,250)\) represents the rate at which water is flowing from the tank after 15 min. Estimate this rate by averaging the slopes of the secant lines in part (a).

Find the limit, if it exists. If the limit does not exist, explain why. $$\lim _{x \rightarrow-4}|x+4|$$
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