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    

2543

2681

2819

2949

3086

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 one decimal place.)

(a)    *t* = 36 and *t* = 42


(b)    *t* = 38 and *t* = 42


(c)    *t* = 40 and *t* = 42


(d)    *t* = 42 and *t* = 44


To estimate the patient's heart rate after 42 minutes using the secant line between different points, we calculate the slope of the secant line for each pair of points given. The slope of the secant line represents the average rate of change (heart rate) between two points.

The formula for the slope of a secant line between two points \((t_1, H_1)\) and \((t_2, H_2)\) is:

\[
\text{slope} = \frac{H_2 - H_1}{t_2 - t_1}
\]

where:
- \( t_1 \) and \( t_2 \) are the times in minutes,
- \( H_1 \) and \( H_2 \) are the number of heartbeats at those times.

Let's calculate the slope for each pair of times given:

### (a) \( t = 36 \) and \( t = 42 \)

\[
\text{slope} = \frac{H(42) - H(36)}{42 - 36} = \frac{2949 - 2543}{42 - 36} = \frac{406}{6} \approx 67.7 \, \text{beats per minute}
\]

### (b) \( t = 38 \) and \( t = 42 \)

\[
\text{slope} = \frac{H(42) - H(38)}{42 - 38} = \frac{2949 - 2681}{42 - 38} = \frac{268}{4} = 67.0 \, \text{beats per minute}
\]

### (c) \( t = 40 \) and \( t = 42 \)

\[
\text{slope} = \frac{H(42) - H(40)}{42 - 40} = \frac{2949 - 2819}{42 - 40} = \frac{130}{2} = 65.0 \, \text{beats per minute}
\]

### (d) \( t = 42 \) and \( t = 44 \)

\[
\text{slope} = \frac{H(44) - H(42)}{44 - 42} = \frac{3086 - 2949}{44 - 42} = \frac{137}{2} = 68.5 \, \text{beats per minute}
\]

### Summary of Results:

- **(a) \( t = 36 \) and \( t = 42 \):** \( 67.7 \, \text{beats per minute} \)
- **(b) \( t = 38 \) and \( t = 42 \):** \( 67.0 \, \text{beats per minute} \)
- **(c) \( t = 40 \) and \( t = 42 \):** \( 65.0 \, \text{beats per minute} \)
- **(d) \( t = 42 \) and \( t = 44 \):** \( 68.5 \, \text{beats per minute} \)

These values provide an estimate of the patient's heart rate after 42 minutes based on the secant line slopes between the given points.

Would you like more details on any of these calculations, or do you have any questions?

### Related Questions:

1. How would you find the instantaneous heart rate at exactly 42 minutes using calculus?
2. What would the average heart rate be over the entire 8-minute period from 36 to 44 minutes?
3. How does the heart rate change as time progresses, and is there a pattern?
4. How could you estimate the heart rate if you only had the value at 42 minutes?
5. What is the significance of using secant lines to estimate instantaneous rates of change?

### Tip:
Secant lines are useful for approximating instantaneous rates when you don't have a continuous function but instead have discrete data points.

Learn how to estimate a patient's heart rate after 42 minutes using secant lines with this detailed math problem solution. Calculate the slopes between different time intervals (36-42, 38-42, 40-42, 42-44) to determine the heart rate in beats per minute. Suitable for high school students learning about calculus and rate of change.

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