Problem 38 - Precalculus

Question

The National Weather Service uses the following formula to calculate the wind chill: $$ W=35.74+0.6215 T_{a}-35.75 V^{0.16}+0.4275 T_{a} V^{0.16} $$ where \(W\) is the wind chill temperature in \({ }^{\circ} \mathrm{F}, T_{a}\) is the air temperature in \({ }^{\circ} \mathrm{F},\) and \(V\) is the wind speed in miles per hour. Note that \(W\) is defined only for air temperatures at or lower than \(50^{\circ} \mathrm{F}\) and wind speeds above 3 miles per hour. (a) Suppose the air temperature is \(42^{\circ}\) and the wind speed is 7 miles per hour. Find the wind chill temperature. Round your answer to two decimal places. (b) Suppose the air temperature is \(37^{\circ} \mathrm{F}\) and the wind chill temperature is \(30^{\circ} \mathrm{F}\). Find the wind speed. Round your answer to two decimal places.

Step-by-Step Solution

Answer

(a) The wind chill temperature is approximately 35.60°F. (b) The wind speed is approximately 3.82 mph.

Step 1: Identify Variables for Part (a)

For part (a), we have the air temperature \(T_a = 42^{\circ} \text{F}\) and wind speed \(V = 7\) mph. We need to calculate the wind chill \(W\).

Step 2: Use the Wind Chill Formula for Part (a)

Substitute \(T_a = 42\) and \(V = 7\) into the wind chill formula: \ W = 35.74 + 0.6215 \times 42 - 35.75 \times 7^{0.16} + 0.4275 \times 42 \times 7^{0.16}. \

Step 3: Calculate the Expression for Part (a)

Calculate step-by-step: \ First, calculate \(7^{0.16} \approx 1.48\) \ Next, the following contributions to \(W\): \ - Constant term \(35.74\) \ - Temperature term \(0.6215 \times 42 = 26.103\) \- Wind speed term \(-35.75 \times 1.48 \approx -52.79\) \ - Combined term \(0.4275 \times 42 \times 1.48 \approx 26.55\) \ Combine all terms: \ \(W = 35.74 + 26.103 - 52.79 + 26.55 \approx 35.60\).

Step 4: Conclusion for Part (a)

The calculated wind chill temperature \(W\) for part (a) is approximately \(35.60^{\circ} \text{F}\).

Step 5: Identify Variables for Part (b)

For part (b), we have the air temperature \(T_a = 37^{\circ} \text{F}\) and wind chill \(W = 30^{\circ} \text{F}\). We need to find the wind speed \(V\).

Step 6: Rearrange Formula for Part (b)

To find \(V\), rearrange the formula: \ \(30 = 35.74 + 0.6215 \times 37 - 35.75 \times V^{0.16} + 0.4275 \times 37 \times V^{0.16}\). \ Simplify: \ \(-5.74 = -35.75 \times V^{0.16} + 0.4275 \times 37 \times V^{0.16}\)

Step 7: Solve for \(V^{0.16}\)

Combine terms involving \(V^{0.16}\): \ \(-5.74 = (-35.75 + 0.4275 \times 37) V^{0.16}\). \ Calculate \(-35.75 + 0.4275 \times 37 \approx -20.815\). \ Thus, \(-5.74 = -20.815 \times V^{0.16}\). \ Solve for \(V^{0.16}\): \ \(V^{0.16} \approx \frac{-5.74}{-20.815} \approx 0.2757\).

Step 8: Calculate \(V\) for Part (b)

Raise to the power of \(\frac{1}{0.16}\) to solve for \(V\): \ \(V = (0.2757)^{\frac{1}{0.16}} \approx 3.82\).

Step 9: Conclusion for Part (b)

The wind speed \(V\) for part (b) is approximately 3.82 miles per hour.

Key Concepts

Wind Chill Formula

The wind chill formula is a way to estimate how cold it feels outside when the wind is blowing. It's not just about the temperature; the wind plays a big part in how chilly it feels on your skin. This formula combines the air temperature and wind speed to determine the wind chill. - In mathematical terms, it's expressed as: \[ W = 35.74 + 0.6215T_a - 35.75V^{0.16} + 0.4275T_aV^{0.16} \]- Here, \(W\) represents the wind chill temperature in degrees Fahrenheit. - \(T_a\) is the actual air temperature in degrees Fahrenheit, and \(V\) is the wind speed in miles per hour.This formula is specifically designed for temperatures at or below 50 degrees Fahrenheit and wind speeds above 3 mph. It helps people understand how cold it actually feels so they can dress appropriately.

Temperature in Fahrenheit

This concept of temperature in Fahrenheit is crucial for understanding how the wind chill calculation works. The Fahrenheit scale is widely used in the United States and it measures how hot or cold it is.

• Water freezes at 32°F and boils at 212°F at sea level.
• The scale was developed by Daniel Gabriel Fahrenheit in the early 18th century.

In the wind chill formula, we input the air temperature as \(T_a\) (in degrees Fahrenheit). Understanding the Fahrenheit scale helps you better grasp how a high wind speed can lower the perceived temperature significantly. A drop from 42°F to a wind chill of 35.6°F, as calculated in the exercise, underscores the importance of this context. It provides a clear idea of how much colder it might feel compared to what the thermometer reads.

Wind Speed in Miles Per Hour

Wind speed in miles per hour (mph) is a key factor in calculating the wind chill. Wind speed reflects how fast the air is moving past a certain point and affects how the temperature feels outside. - In the formula, wind speed affects how the wind chill is calculated and is represented as \(V\).- The power term \(V^{0.16}\) in the formula indicates the relationship between wind speed and the cooling effect on the skin.When the wind speed increases, the chilling effect can become more pronounced, making the temperature feel much colder than it actually is. That's why on a snowy and windy day, you might feel a sharp difference in temperature. For example, at a wind speed of 7 mph, as given in the exercise, the cooling effect is quite noticeable, which is expressed by calculating the power term \(7^{0.16}\).

Mathematical Problem-Solving

Mathematical problem-solving involves using logic and calculations to find answers and draw conclusions from given data or situations. When tackling the wind chill calculation:

• Start by identifying the known variables such as air temperature and wind speed.
• Substitute these variables into the formula to calculate or rearrange according to what you need to find.

For instance, solving for wind chill is straightforward if you know both air temperature and wind speed, as shown in part (a) of the exercise. However, in part (b), where you need to find the wind speed given the wind chill and temperature, it requires rearranging the formula. This type of problem-solving enhances your ability to understand complex formulas and develop a step-by-step approach to solving various mathematical challenges. It's about breaking down the problem, performing accurate calculations, and getting the right result. In our case, it involves applying knowledge of exponents and algebra to isolate variables and solve for unknowns.