Problem 223 - Calculus Volume 1

Question

[T] The temperature (in degrees Celsius) of a city in the northern United States can be modeled by the function $$ T(x)=5+18 \sin \left[\frac{\pi}{6}(x-4.6)\right] $$ where \(x\) is time in months and \(x=1.00\) corresponds to January \(1 .\) Determine the month and day when the temperature is \(21^{\circ} \mathrm{C} .\)

Step-by-Step Solution

Answer

The temperature is 21°C on June 22.

Step 1: Set the function equal to the given temperature

We are given that the temperature, \( T(x) = 5 + 18 \sin \left[\frac{\pi}{6}(x-4.6)\right] \), should equal \( 21 \degree \text{C} \). Set the function equal to 21:\[21 = 5 + 18 \sin \left[\frac{\pi}{6}(x-4.6)\right]\]

Step 2: Subtract 5 from both sides

To isolate the sine function, subtract 5 from both sides of the equation:\[21 - 5 = 18 \sin \left[\frac{\pi}{6}(x-4.6)\right]\]Simplifying gives:\[16 = 18 \sin \left[\frac{\pi}{6}(x-4.6)\right]\]

Step 3: Solve for the sine function

Divide both sides by 18 to find the sine value:\[\sin \left[\frac{\pi}{6}(x-4.6)\right] = \frac{16}{18}\]Simplify the fraction:\[\sin \left[\frac{\pi}{6}(x-4.6)\right] = \frac{8}{9}\]

Step 4: Solve for the angle inside the sine

Take the inverse sine to solve for the angle:\[\frac{\pi}{6}(x-4.6) = \arcsin\left(\frac{8}{9}\right)\]Calculate the arcsin value, which is approximately:\[\arcsin\left(\frac{8}{9}\right) \approx 1.1198 \]

Step 5: Solve for x

Multiply both sides by \( \frac{6}{\pi} \) to solve for \( x \):\[x - 4.6 = \frac{6}{\pi} \times 1.1198\]Calculate \( \frac{6}{\pi} \times 1.1198 \), which is approximately:\[x - 4.6 \approx 2.137\]

Step 6: Add 4.6 to solve for x

Add 4.6 to both sides to find the value of \( x \):\[x \approx 2.137 + 4.6\]This gives:\[x \approx 6.737\]

Step 7: Convert x to a date

The value \( x \approx 6.737 \) corresponds to the 6th month plus a fraction of the month. Convert this to a specific day in June.Since June has 30 days, calculate \(0.737 imes 30 \approx 22\).Thus, the temperature is \( 21^{\circ} \text{C} \) around June 22nd.

Key Concepts

Temperature Modeling

Temperature modeling involves using mathematical functions to describe how temperature varies over time. In this exercise, the temperature of a city is modeled using a trigonometric function of the form \[T(x) = 5 + 18 \sin \left[ \frac{\pi}{6}(x - 4.6) \right]\]This is a sinusoidal function, which is often used in modeling because it naturally reflects periodic changes, such as seasonal variations in temperature.

• **Amplitude**: The coefficient of the sine function, 18 in this case, is the amplitude, representing the maximum deviation from the average temperature.
• **Vertical Shift**: The constant 5 indicates a shift of the sine wave upward, setting the midpoint of the temperature at 5°C.
• **Period**: The factor \(\frac{\pi}{6}\) in the function affects the period of the sine wave. Since the argument of the sine is multiplied by \(\frac{\pi}{6}\), it specifies the function completes one cycle over 12 months (a year).
• **Horizontal Shift**: The \(x - 4.6\) inside the sine function indicates the wave is shifted 4.6 months to the right. This corresponds to seasonal temperature changes starting slightly after January.

Modeling temperature as a trigonometric function offers a predictable pattern, making it a practical tool for forecasters.

Inverse Trigonometric Functions

Inverse trigonometric functions enable us to determine angles or inputs based on the output value of a trigonometric function. For instance, if we know the sine of an angle, we can use the inverse sine or arcsin function to find the corresponding angle.In our solution, we reach a step where we have\[\sin \left[ \frac{\pi}{6}(x - 4.6) \right] = \frac{8}{9}\]To find the angle \( \frac{\pi}{6}(x - 4.6) \), we use the inverse sine function:\[\frac{\pi}{6}(x - 4.6) = \arcsin\left(\frac{8}{9}\right)\]The value of \(\arcsin\left(\frac{8}{9}\right)\) is approximately 1.1198 radians. The inverse trigonometric functions, including arcsin, arccos, and arctan, are crucial in problems like this as they allow us to

• Determine specific angle measures given the sine, cosine, or tangent ratio.
• Solve for unknown variables in equations involving trigonometric functions.

Understanding inverse trigonometric functions is key to addressing a range of problems, including those involving periodic patterns and waveforms.

Solving Trigonometric Equations

When solving trigonometric equations, the goal is to find the values of the variable that satisfy the equation. The process often involves several steps and transformations.First, isolate the trigonometric function on one side of the equation. In our example, the equation is:\[T(x) = 21 = 5 + 18 \sin \left[ \frac{\pi}{6}(x - 4.6) \right]\]We begin by simplifying it to:\[16 = 18 \sin \left[ \frac{\pi}{6}(x - 4.6) \right]\]Divide both sides by 18 to further isolate the sine;\[\sin \left[ \frac{\pi}{6}(x - 4.6) \right] = \frac{8}{9}\]Next, the inverse sine is used to find the angle.Once the angle is known, convert it to solve for the variable \(x\). This involves turning the angle equation back into the x-domain.

• Multiply both sides by the reciprocal of the coefficient of \(x\).
• Solve for the variable after adjusting for any shifts or additions noted in the equation.

In the end, recognize when the solution corresponds to a real-world context, such as time or distance. Ultimately, solving trigonometric equations involves understanding both algebraic manipulations and the geometric implications of these functions.