Problem 223 - Calculus Volume 1

Question

The temperature (in degrees Celsius) of a city in the northern United States can be modeled by the function \(T(x)=5+18 \sin \left[\frac{\pi}{6}(x-4.6)\right], \quad\) 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 reaches 21°C on July 11.

Step 1: Set the equation equal to 21

We need to find when the temperature is exactly 21°C. Set the given temperature function equal to 21:\[21 = 5 + 18 \sin\left[\frac{\pi}{6}(x-4.6)\right]\]

Step 2: Solve for the sine term

Subtract 5 from both sides to isolate the sine function:\[16 = 18 \sin\left[\frac{\pi}{6}(x-4.6)\right]\]Then divide both sides by 18:\[\sin\left[\frac{\pi}{6}(x-4.6)\right] = \frac{16}{18}\]Simplify the fraction:\[\sin\left[\frac{\pi}{6}(x-4.6)\right] = \frac{8}{9}\]

Step 3: Solve for the angle

Take the inverse sine of both sides to solve for the angle:\[\frac{\pi}{6}(x-4.6) = \sin^{-1}\left(\frac{8}{9}\right)\]Calculate \(\sin^{-1}\left(\frac{8}{9}\right)\), which is approximately \(0.9273\) radians. Thus,\[\frac{\pi}{6}(x-4.6) = 0.9273\]

Step 4: Solve for x

Multiply both sides by \(\frac{6}{\pi}\) to solve for \(x-4.6\):\[x-4.6 = \frac{6}{\pi} \times 0.9273\]Calculate:\[x-4.6 \approx 1.77\]Finally, solve for \(x\):\[x \approx 6.37\]

Step 5: Convert decimal month to date

Since \(x = 1\) corresponds to January 1, if \(x = 6.37\), we are in the 7th month (approximately mid-July). The part \(.37\) of a month translates to about 11 days (as \(0.37 \times 30\) is approximately 11). Therefore, \(x = 6.37\) corresponds to July 11.

Key Concepts

Temperature Modeling

Temperature modeling helps us understand how temperatures change over time. This is particularly useful in regions with distinct seasons, where temperature variations are cyclical.

• Temperature is often modeled with mathematical functions that represent its periodic nature.
• In the given model, the temperature of a northern U.S. city is expressed as a function of time in months.
• The equation used is: \(T(x) = 5 + 18 \sin \left[\frac{\pi}{6}(x-4.6)\right]\). This captures the cyclical nature of temperature changes throughout the year.

By using such a model, we can predict temperatures and understand patterns over several months or even years. The goal often is to identify specific time points, like when the temperature reaches a given degree.

Sinusoidal Functions

Sinusoidal functions are used to model periodic phenomena. These functions are key in representing repeating cycles, like seasons, tides, or sound waves. Let's see why they are useful:

• They have a characteristic "wave-like" shape, which repeats at regular intervals.
• The basic form of a sine function is \(y = A \sin(B(x - C)) + D\), where:
• \(A\) is the amplitude, indicating the peak deviation from the mean value.
• \(B\) affects the period, determining how quickly the cycles repeat.
• \(C\) shifts the graph horizontally, delaying or advancing the cycle.
• \(D\) shifts the graph vertically, moving the entire wave up or down.

In temperature modeling, the sinusoidal shape helps reflect the rise and fall of temperatures over months. The amplitude and period adjust according to the region's specific climate patterns.

Inverse Trigonometric Functions

Inverse trigonometric functions are essential tools for solving equations involving trigonometric functions. They allow us to find angles when we know the sine, cosine, or tangent values. This is particularly useful in problems involving periodic functions:

• The inverse sine function, \(\sin^{-1}\), is used to determine the angle whose sine is a given number.
• For example, in our temperature model, we isolated the sine term and solved \(\sin \left[\frac{\pi}{6}(x-4.6)\right] = \frac{8}{9}\).
• We used \(\sin^{-1}\left(\frac{8}{9}\right)\) to find the angle that corresponds to this sine value, which is approximately 0.9273 radians.

After finding this angle, we continued solving for \(x\) to determine the specific month and day. This approach is fundamental in scenarios where you need to work backward from trigonometric values to actual time or angles.