Problem 61 - Understanding Elementary Algebra with Geometry

Question

Round off to the nearest hundredth when necessary. In physiology a jogger's heart rate \(N\), in beats per minute, is related linearly to the jogger's speed \(s .\) A certain jogger's heart rate is 80 beats per minute at a speed of \(15 \mathrm{ft} / \mathrm{sec}\) and 82 beats per minute at a speed of \(18 \mathrm{ft} / \mathrm{sec} .\) (a) Write an equation relating the jogger's speed and heart rate. (b) Predict this jogger's heart rate if she jogs at a speed of \(20 \mathrm{ft} / \mathrm{sec}\). (c) According to the equation obtained in part (a), what is the jogger's heart rate at rest? [Hint: At rest the jogger's speed is 0.]

Step-by-Step Solution

Answer

a) Equation: \( N = \frac{2}{3}s + 70 \); b) 83.33 beats per minute; c) 70 beats per minute.

Step 1: Identify the variables

Let the heart rate be denoted by \(N\) and the speed be denoted by \(s\). We are given two points: (15, 80) and (18, 82).

Step 2: Find the slope of the line

The slope (m) of the line can be found using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] where \((x_1, y_1) = (15, 80)\) and \((x_2, y_2) = (18, 82)\). Plugging in these values: \[ m = \frac{82 - 80}{18 - 15} = \frac{2}{3} \]

Step 3: Write the equation of the line

Using the slope-intercept form of the equation \( y = mx + b \), where \(m = \frac{2}{3}\), and using one of the points (15, 80) to find \(b\), we get: \[ 80 = \frac{2}{3}(15) + b \] Solving for \(b\): \[ 80 = 10 + b \] \[ b = 70 \] So, the equation relating heart rate and speed is: \[ N = \frac{2}{3}s + 70 \]

Step 4: Predict the heart rate at 20 ft/sec

Using the equation from Step 3: \[ N = \frac{2}{3}s + 70 \] Plug in \(s = 20\): \[ N = \frac{2}{3}(20) + 70 = \frac{40}{3} + 70 \] \[ N = 13.33 + 70 = 83.33 \] Therefore, the jogger's heart rate at 20 ft/sec is approximately 83.33 beats per minute.

Step 5: Find the heart rate at rest

At rest, the speed \(s = 0\). Using the equation: \[ N = \frac{2}{3}(0) + 70 = 70 \] Thus, the jogger's heart rate at rest is 70 beats per minute.

Key Concepts

slope-intercept form

In algebra, the slope-intercept form is a way to describe a straight line on a graph following the equation: The 'slope' (\text{'\(m\)})') indicates the steepness of the line, and 'intercept' (\text{'\(b\)'}) is where the line crosses the y-axis. To use this form, you need two pieces of information: a pair of coordinates (\text{'\((x_1, y_1)\)'}) and the slope (\text{'\(m\)'}) of the line. Here is how you find the slope: For example, if a jogger’s heart rate increases linearly with speed, knowing two points on the graph representing heart rate at different speeds allows you to find the slope (\text{'\(m\)'}), The equation transforms these two data points (\text{'\(x_1, y_1) = (15, 80)\)'} and (\text{'\(x_2, y_2) = (18, 82)\)'}), into the line equation: After finding the slope (\text{'\(m = \frac{2}{3}\)'})). Using one of the points, (e.g., (15, 80)) helps to find the intercept 'b' by solving: Thus, you get the full equation: \text{'\(N = \frac{2}{3}s + 70\)'}}, indicating the joggers' heart rate related to speed.

solving for variables

Solving for variables is an essential algebraic skill, where you isolate a variable on one side of an equation. Given \text{'\(N = \frac{2}{3}s + 70\)'})', for instance, we might want the heart rate \text{'\(N\)'} for a given speed (\text{'\(s\)'}). Let’s use a practical scenario: Predict the jogger’s heart rate at \text{'\(s = 20\)'} feet per second. Plugging \text{'\(s = 20\)'} into the equation: Using basic arithmetic, solve for \text{'\(N\)'}) Thus, the result: \text{'\(N = 83.33\)'} beats per minute when the jogger’s speed \text{'\(s\)'} is \text{'\(20\)'} feet per second. To predict another value, input a different \text{'\(s\)'}) and follow similar steps!

predicting values

Predictive analysis means using a formula to guess or estimate data points within or outside of the given range. With a linear equation like \text{'\(N = \frac{2}{3}s + 70\)'})', For instance, predicting the jogger’s heart rate at a speed not given: Either lower or higher than known values. Let's predict the heart rate at \text{'\(20 ft/sec\)'}: So, \text{'\(20\)'} plugged in \text{'\(s\)'} results in \text{83.33 beats\footnotesize integration method approach. This method showcases how predictability from a line equation based on two known results and establishing new or unfamiliar results' confidence.