Problem 52 - Applied Calculus

Question

If you are outdoors, the wind may make it feel a lot colder than the thermometer reads. You feel the windchill temperature, which, if the air temperature is \(20^{\circ} \mathrm{F}\), is given in \({ }^{\circ} \mathrm{F}\) by \(W(v)=48.17-27.2 v^{0.16}\), where \(v\) is the wind velocity in mph for \(5 \leq v \leq 60^{2}\) (a) If the air temperature is \(20^{\circ} \mathrm{F}\), and the wind is blowing at \(40 \mathrm{mph}\), what is the windchill temperature, to the nearest degree? (b) Find \(W^{\prime}(40)\), and explain what this means in terms of windchill.

Step-by-Step Solution

Answer

(a) The windchill temperature is \(-13^{\circ} F\). (b) \(W'(40) \approx -0.108\) indicates that every additional mph increases the chill by about 0.108 degrees.

Step 1: Understanding the Formula

The windchill formula is given by \( W(v) = 48.17 - 27.2 v^{0.16} \). This formula calculates the perceived temperature based on actual temperature and wind velocity. Here, \( W(v) \) is the windchill temperature in \( ^\circ F \), and \( v \) is the wind velocity in mph.

Step 2: Substituting into the Formula

For part (a), substitute \( v = 40 \) into the formula to find the windchill temperature: \[ W(40) = 48.17 - 27.2 \times (40)^{0.16} \].

Step 3: Calculating the Windchill Temperature

Calculate \( (40)^{0.16} \) and then \( 27.2 \times (40)^{0.16} \). Subtract this value from 48.17 to find \( W(40) \). Upon calculation, you find \( (40)^{0.16} \approx 2.241 \), and \( 27.2 \times 2.241 \approx 60.97 \). So, \( W(40) = 48.17 - 60.97 = -12.8 \). Round this to the nearest degree to get \(-13^{\circ} \mathrm{F}\).

Step 4: Differentiate the Windchill Formula

For part (b), find the derivative \( W'(v) \) to understand how windchill changes with changes in wind velocity. Differentiate \( W(v) = 48.17 - 27.2 v^{0.16} \) with respect to \( v \). Apply the power rule: \[ W'(v) = -27.2 \times 0.16 \times v^{0.16 - 1} \], which simplifies to \[ W'(v) = -4.352 \times v^{-0.84} \].

Step 5: Evaluate the Derivative at 40 mph

Substitute \( v = 40 \) into \( W'(v) \) to find \( W'(40) \). Calculate \( W'(40) = -4.352 \times (40)^{-0.84} \). Evaluating this gives \( W'(40) \approx -0.108 \).

Step 6: Interpret the Derivative

This means that at 40 mph, each 1 mph increase in wind speed decreases the windchill temperature by approximately 0.108 degrees Fahrenheit.

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with continuous change. It plays a crucial role in understanding dynamic systems, such as the way we perceive temperature in a windy environment. Calculus helps us model and calculate changes in conditions by using formulas and functions like the windchill formula in this exercise. This allows us to analyze how different factors, such as wind velocity, affect the way we experience temperature. By applying calculus, specifically through differentiation, we can more accurately predict and interpret variations in perceived temperature as conditions change.

Derivatives

A derivative represents the rate at which a quantity changes. In this exercise, the derivative of the windchill formula, noted as \( W'(v) \), shows how the windchill temperature changes as the wind velocity increases or decreases.

• To compute the derivative, we apply the power rule. This involves taking the exponent of the variable \( v \), multiplying it by the coefficient of \( v \), and subtracting one from the exponent.
• For example, to find the derivative of \( W(v) = 48.17 - 27.2 v^{0.16} \), we apply the power rule to \( -27.2 v^{0.16} \) which results in \( W'(v) = -4.352 imes v^{-0.84} \).

By evaluating this derivative at a specific wind speed, such as 40 mph, we gain insight into how sensitive the windchill temperature is to changes in wind velocity. This helps explain how even small variations in wind speed can significantly affect our perception of coldness.

Temperature Conversion

Temperature conversion is a critical concept when calculating windchill because it helps us compare temperatures using different scales, like Fahrenheit and Celsius. This exercise focuses on calculations in Fahrenheit, which is often used in the U.S. to report weather conditions. When performing calculations, knowing how to convert temperatures between different units is essential for clarity, especially if subsequent comparisons or applications require an alternative scale. Although this particular exercise does not require conversion between Fahrenheit and Celsius, understanding the concept ensures that you can extend these kinds of calculations to different scenarios. Here are some quick conversion formulas to keep in mind:

• From Celsius to Fahrenheit: \( F = \frac{9}{5}C + 32 \)
• From Fahrenheit to Celsius: \( C = \frac{5}{9}(F - 32) \)

These conversions maintain consistency in scientific and everyday discussions, ensuring a clear understanding of temperature comparisons across global regions.

Wind Velocity

Wind velocity, measured in miles per hour (mph) in this exercise, is a key factor in determining windchill temperature. The formula \( W(v) = 48.17 - 27.2 v^{0.16} \) shows that as wind speed increases, the windchill temperature decreases. Wind velocity impacts how quickly heat is removed from the body, which is why windchill is often dramatically lower than the actual temperature during windy conditions. Understanding the role of wind velocity in windchill calculations is crucial to predict how cold it will feel outdoors. When solving the exercise, we evaluate specific wind speeds, such as 40 mph, to determine the corresponding windchill temperature. The formula relies on calculating the power of wind velocity, showing the nonlinear relationship between the increase in wind speed and the substantial effect on perceived coldness. Thus, wind velocity is a pivotal parameter in understanding how harsh winter conditions can become.