Problem 76 - Precalculus: Functions and Graphs

Question

In the study of frost penetration problems in highway engineering, the temperature \(T\) at time \(t\) hours and depth \(x\) feet is given by $$ T=T_{0} e^{-\lambda r} \sin (\omega t-\lambda x) $$ where \(T_{0}, \omega,\) and \(\lambda\) are constants and the period of \(T\) is 24 hours. (a) Find a formula for the temperature at the surface. (b) At what times is the surface temperature a minimum? (b) If \(\lambda=2.5,\) find the times when the temperature is a minimum at a depth of I foot.

Step-by-Step Solution

Answer

(a) Surface temperature formula: \( T = T_{0} \sin \left( \frac{\pi}{12} t \right) \). (b) Minimum surface temperature every 24 hours starting at \( t = 18 \) hours. (c) Minimum temperature at 1 foot depth at approximately \( t = 27.55 + 24k \) hours.

Step 1: Surface Temperature Formula

To find the formula for the surface temperature, we set the depth \( x = 0 \). Substitute \( x = 0 \) into the given formula:\[ T = T_{0} e^{-rac{c}{k}r} \sin(\omega t - \lambda \times 0) = T_{0} e^{0} \sin(\omega t) = T_{0} \sin(\omega t) \]This is the formula for the temperature at the surface.

Step 2: Determine Angular Frequency

We know the period \( T \) is 24 hours, which allows us to find the angular frequency \( \omega \):\[ \omega = \frac{2\pi}{T} = \frac{2\pi}{24} = \frac{\pi}{12} \]Substitute this value into our surface temperature formula to have:\[ T = T_{0} \sin\left(\frac{\pi}{12}t\right) \]

Step 3: Finding Minimum Surface Temperature

The surface temperature will be a minimum when \( \sin\left(\frac{\pi}{12}t\right) = -1 \). This occurs when:\[ \frac{\pi}{12}t = \frac{3\pi}{2} + 2k\pi \quad \text{for integer } k \]This simplifies to:\[ t = 18 + 24k \]where \( k \) is any integer. Therefore, the surface temperature is a minimum every 24 hours starting at \( t = 18 \) hours.

Step 4: Temperature at Depth of 1 Foot

Set \( x = 1 \) foot for the given equation:\[ T = T_{0} e^{-\lambda \times 1} \sin(\omega t - \lambda \times 1) \]\[ T = T_{0} e^{-\lambda} \sin(\omega t - \lambda) \]

Step 5: Frequency of Minimums at 1 Foot Depth

To find when this temperature is minimum, set \( \sin(\omega t - \lambda) = -1 \):\[ \omega t - \lambda = \frac{3\pi}{2} + 2k\pi \]Substituting \( \omega = \frac{\pi}{12} \) and \( \lambda = 2.5 \), solve:\[ \frac{\pi}{12}t - 2.5 = \frac{3\pi}{2} + 2k\pi \]\[ \frac{\pi}{12}t = \frac{3\pi}{2} + 2k\pi + 2.5 \]\[ t = 18 + 24k + \frac{30}{\pi} \approx 18 + 24k + 9.55 \]Thus, the times are approximately \( t = 27.55, 51.55, \ldots \).

Key Concepts

Understanding Periodic Functions

Periodic functions are mathematical functions that repeat their values in regular intervals or periods. They are integral in describing repetitive behavior in various physical phenomena, such as the swing of a pendulum or the oscillation of a wave. In the context of this exercise, the temperature as a function of time, given by T, is periodic because it repeats every 24 hours.

• To determine the periodic nature, look for the smallest interval where the function repeats its values.
• In this problem, the period is 24 hours, which means after every 24 hours the temperature profile at any depth remains the same.
• Many trigonometric functions, such as sine and cosine, naturally exhibit periodic behavior, which links them closely to periodic functions.

Recognizing periodicity is fundamental in solving real-world problems where conditions cycle predictably over time.

Equation Solving for Temperature

Equation solving involves finding values that satisfy given equations. It is a critical skill in understanding mathematical models used in engineering, such as the temperature model in this exercise.

• To solve for the surface temperature, we simplify the equation by letting the depth \( x = 0 \). This reduction helps isolate the varying component: the time \( t \).
• Solving the equation for minimum temperature at the surface involves finding when the sine function reaches its minimum value, \( -1 \).
• Similarly, to find when the temperature at 1-foot depth is at a minimum, we set the expression involving sine equal to \( -1 \) and solve for \( t \).

These solutions involve understanding and manipulating trigonometric identities and properties, especially shifts and stretches caused by multiplying the angle inside the sine function.

Angular Frequency and Its Implications

Angular frequency is a measure that indicates how fast something is oscillating in a cycle. In our temperature model, \( \omega \), the angular frequency, brings insight into how rapidly the temperature changes over time.

• The formula for angular frequency is \( \omega = \frac{2\pi}{T} \), where \( T \) is the period of the cycle in any oscillating system.
• In this situation, since the temperature's period is 24 hours, the angular frequency \( \omega \) is calculated as \( \frac{\pi}{12} \).
• Angular frequency plays a vital role in determining how compressed or stretched the cycles of our sinusoidal function are, directly affecting where maxima and minima occur.

Understanding angular frequency helps in predicting cyclic changes and fine-tuning models for better accuracy in predicting physical systems behaviors.