Problem 54 - A Graphical Approach to College Algebra

Question

(Modeling) In Exercises \(53-56\), assume that a linear relationship exists between the two quantities. Cricket Chirping At \(68^{\circ} \mathrm{F},\) a certain species of cricket chirps 112 times per minute. At \(46^{\circ} \mathrm{F},\) the same cricket chirps 24 times per minute. (a) Express the number of chirps, \(y,\) as a linear function of the Fahrenheit temperature. (b) If the temperature is \(60^{\circ} \mathrm{F},\) how many times will the cricket chirp per minute? (c) If you count the number of cricket chirps in one-half minute and hear 40 chirps, what is the temperature?

Step-by-Step Solution

Answer

(a) \(y = 4x - 160\), (b) 80 chirps, (c) \(60^{\circ}F\).

Step 1: Identify Given Points

First, identify the given points that represent the temperature and the corresponding number of cricket chirps. These are (68, 112) and (46, 24), where the first number in each pair is the temperature in Fahrenheit and the second number is the chirps per minute.

Step 2: Find the Slope

Use the formula for the slope of a line, \(m = \frac{y_2 - y_1}{x_2 - x_1}\), to find the slope of the line representing the relationship between chirps and temperature. Substituting the given points: \(m = \frac{112 - 24}{68 - 46} = \frac{88}{22} = 4\).

Step 3: Write the Equation of the Line

Using the point-slope form of a line, \(y - y_1 = m(x - x_1)\), substitute one of the given points, such as (68, 112), and the slope: \(y - 112 = 4(x - 68)\). Simplifying this gives the linear equation, \(y = 4x - 160\).

Step 4: Calculate Chirps at 60°F

Substitute \(x = 60\) into the linear equation to find the number of chirps: \(y = 4(60) - 160 = 240 - 160 = 80\). So, at \(60^{\circ}F\), the cricket will chirp 80 times per minute.

Step 5: Find Temperature for 40 Chirps in Half a Minute

If there are 40 chirps in half a minute, this means there are 80 chirps per minute. Set \(y = 80\) and solve for \(x\) using the linear equation: \(80 = 4x - 160\). Solving gives \(4x = 240\), hence \(x = 60\). Therefore, the temperature is \(60^{\circ}F\).

Key Concepts

Slope Calculation

Calculating the slope of a line is a fundamental aspect of understanding linear functions. The "slope" essentially reflects how steep the line is, or how much one variable changes in relationship to another. In this context, we are examining the relationship between temperature and cricket chirping rates. You can calculate the slope using the formula: \[ m = \frac{y_2 - y_1}{x_2 - x_1} \] Here,

• $m$ is the slope, representing the rate of change of chirps per temperature degree
• $(x\_1,\; y\_1)$ and $(x\_2,\; y\_2)$ are two pairs of data points: with $x$ as temperature and $y$ as chirps per minute

Substituting the data points (68, 112) and (46, 24), the slope calculation becomes: \[ m = \frac{112 - 24}{68 - 46} = \frac{88}{22} = 4 \] Therefore, for every degree Fahrenheit, the chirping rate affects by four chirps per minute.

Temperature Conversion

Understanding temperature conversion is essential when dealing with problems connected to chirping rates because it helps us bridge the gap between different unit systems, should the need arise. However, in this exercise, we deal solely in Fahrenheit. Sometimes, though, you might need to convert between Fahrenheit and Celsius. The conversion formula between Fahrenheit

• To convert Celsius to Fahrenheit: \[ F = \frac{9}{5}C + 32 \]
• To convert Fahrenheit to Celsius: \[ C = \frac{5}{9}(F - 32) \]

This exercise doesn't require these conversions directly but understanding these can help relate findings across different contexts or further exercises.

Point-Slope Form

The point-slope form of a linear function is a convenient way to express the equation of a line when you know one point on the line and the slope. It’s especially useful in this exercise to establish the relationship between cricket chirps and temperature. The point-slope form is expressed as: \[ y - y_1 = m(x - x_1) \] Here,

• $m$ is the slope calculated previously
• $(x\_1,\; y\_1)$ is a point on the line, such as (68, 112)

Thus, using the point (68, 112) and the slope of 4, the equation is: \[ y - 112 = 4(x - 68) \]Simplifying this equation gives you the linear equation: \[ y = 4x - 160 \] With this equation, we can predict chirping rates at different temperatures or determine the temperature based on a chirping rate.