Problem 67 - Algebra and Trigonometry

Question

Biologists have observed that the chirping rate of crickets of a certain species is re- lated to temperature, and the relationship appears to be very nearly linear. A cricket produces 120 chirps per minute at \(70^{\circ} \mathrm{F}\) and 168 chirps per minute at \(80^{\circ} \mathrm{F} .\) (a) Find the linear equation that relates the temperature \(t\) and the number of chirps per minute \(n\) . (b) If the crickets are chirping at 150 chirps per minute, estimate the temperature.

Step-by-Step Solution

Answer

Equation: \( n = 4.8t - 216 \); Temperature: \( 76.25^{\circ} \)F.

Step 1: Understand Points and Slope Formula

We have two points related to chirps per minute and temperature: (70, 120) and (80, 168). We will use the slope formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \) where \( (x_1, y_1) = (70, 120) \) and \( (x_2, y_2) = (80, 168) \).

Step 2: Calculate the Slope

Substitute the values into the slope formula: \( m = \frac{168 - 120}{80 - 70} = \frac{48}{10} = 4.8 \). The slope of the line is 4.8 chirps per degree Fahrenheit.

Step 3: Use Point-Slope Form to Find the Equation

Using point-slope form \( y - y_1 = m(x - x_1) \), where \( m = 4.8 \), \( x_1 = 70 \), and \( y_1 = 120 \), substitute these values into the equation: \( n - 120 = 4.8(t - 70) \).

Step 4: Simplify to Find the Linear Equation

Distribute and simplify the equation: \( n - 120 = 4.8t - 336 \). Then add 120 to both sides to solve for \( n \): \( n = 4.8t - 216 \). This is the linear equation relating temperature and chirps.

Step 5: Substitute to Estimate Temperature

Given \( n = 150 \) chirps per minute, use the equation \( n = 4.8t - 216 \). Substitute \( 150 \) for \( n \): \( 150 = 4.8t - 216 \).

Step 6: Solve for Temperature

Add 216 to both sides: \( 150 + 216 = 4.8t \). Calculate the left side: \( 366 = 4.8t \). Divide both sides by 4.8: \( t = \frac{366}{4.8} = 76.25 \).

Step 7: State the Final Results

The linear equation is \( n = 4.8t - 216 \). If crickets are chirping at 150 chirps per minute, the temperature is approximately \( 76.25^{\circ} \)F.

Key Concepts

slope

The slope is an essential concept in understanding linear equations. It tells us how steep a line is, or in other words, the rate of change between two variables. When we look at our cricket example, the slope indicates how many more chirps occur per minute with each increase of one degree Fahrenheit.
To find the slope, we use the formula \( m = \frac{y_2 - y_1}{x_2 - x_1} \), where each \( (x, y) \) is a point on the line. In this exercise, \( (70, 120) \) and \( (80, 168) \) are such points, representing the temperature in degrees and the number of chirps. By substituting these into the formula, we find that the slope \( m \) is 4.8. This means that for every 1-degree Fahrenheit increase in temperature, crickets chirp 4.8 more times per minute.

• Slope is the rate of change between two variables.
• In this example, the slope tells us the increase in chirps per change in temperature.
• The calculated slope is 4.8 chirps per degree Fahrenheit.

point-slope form

Once you have the slope, the point-slope form becomes useful for forming the linear equation. This form is represented as \( y - y_1 = m(x - x_1) \). In our example, \( m \) is the slope of 4.8, while \( (x_1, y_1) \) can be any of our points—in this case, \( (70, 120) \).
Plug these values into the formula, resulting in the equation \( n - 120 = 4.8(t - 70) \). This equation illustrates how the number of chirps, \( n \), changes in relation to the temperature, \( t \). By simplifying this equation and solving for \( n \), we get the linear equation \( n = 4.8t - 216 \).

• Point-slope form helps you create the linear equation once the slope is known.
• Substituting the slope and a point gives us the relationship between chirps and temperature.
• Simplifying leads to our final equation, \( n = 4.8t - 216 \).

temperature estimation

After finding the linear equation, we can use it to estimate unknown values. In this case, we want to estimate the temperature when crickets chirp 150 times per minute. We start by using the equation \( n = 4.8t - 216 \) and substitute \( n = 150 \).
This gives us \( 150 = 4.8t - 216 \). By solving for \( t \), we add 216 to both sides, giving \( 366 = 4.8t \). Then, divide both sides by 4.8 to isolate \( t \), resulting in \( t = 76.25 \). This shows that when crickets chirp 150 times per minute, the temperature is approximately 76.25°F.

• Substituting known values into the linear equation helps estimate unknowns.
• For 150 chirps, we estimate a temperature of about 76.25°F from our equation.
• This highlights the practical use of linear equations in real-world scenarios.