A cardiac monitor is used to measure the heart rate of a patient after surgery.
ID: 2858240 • Letter: A
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. t (min) 36 38 40 42 44 Heartbeats 2530 2661 2806 2918 3080 The monitor estimates this value by calculating the slope of a secant line. Use the data to estimate the patient?s heart rate after 42 minutes using the secant line between the points with the given values of t. Round your answers to the nearest hundredth. (a) t = 36 and t = 42 (b) t = 38 and t = 42 (c) t =40 and t =42 (d) t = 42 and t = 44 (What are your conclusions?) (a) (b) (c) (d)Explanation / Answer
A line is tangent to a curve if they both pass through a point. The line doesn't intersect the curve, it just "touches" it. The tangent line is the best linear approximation of the curved line, at that point.
For your problem:
Pt -t(min) - heartbeats
1 -- 36 --- 2530
2 -- 38 --- 2661
3 -- 40 --- 2806
4 -- 42 --- 2948
5 -- 44 --- 3080
Slope = y/x
Let x = time, y = heartbeats
Slope(1~4) = (2948-2530) / (42-36)
Slope(1~4) = 418/6 = 69.666
Slope(2~4) = (2948-2661) / (42-38)
Slope(2~4) = 287/4 = 71.75
Slope(3~4) = (2948-2806) / (42-40)
Slope(3~4) = 142/2 = 71
Slope(4~5) = (3080 - 2948) / (44-42)
Slope(4~5) = 132/2 = 66
The best approximation of the heartrate at 42 minutes would be determined by using the secant line passing through the closest points to either side of 42 minutes; i.e. 44 and 40 minutes. This line should parallel the line tangent at 42 minutes.
Slope(4~5) = (3080 - 2806) / (44-40)
Slope(4~5) = 274/4 = 68.5
Note the slope of this secant line is the same as the average of the secant lines to either side of t = 42 minutes : (71 + 66) / 2 = 68.5 .
As to the conclusions, the only thing apparent to me is that the slope of the lines before 42 minutes are about 70-71 heartbeats/minute, and the slope after 42 minutes is 66 heartbeats/minute. That appears to be a fairly substantial difference, but nothing in the data indicates a cause. Mathematically, there is no reason the slopes should be different - the data appear to be continuous and the time interval remains constant.
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