Problem 13
Question
Suppose that the temperature \(T\) (measured in degrees Fahrenheit) in a growing chamber varies over a 24 -hour period according to $$ T(t)=68+\sin \left(\frac{\pi}{12} t\right) $$ for \(0 \leq t \leq 24\). (a) Graph the temperature \(T\) as a function of time \(t\). (b) Find the average temperature and explain your answer graphically.
Step-by-Step Solution
Verified Answer
(a) Sine wave around 68°F; max 69°F at 6h, min 67°F at 18h. (b) Average 68°F over 24 hours.
1Step 1: Understand the Function
The given function is \( T(t)=68+\sin\left(\frac{\pi}{12}t\right) \). This represents the temperature in the chamber over time \( t \), where \( t \) ranges from 0 to 24 hours. The expression \( \sin\left(\frac{\pi}{12}t\right) \) means the sine wave completes a full cycle over 24 hours, as \( \frac{\pi}{12}\times 24 = 2\pi \). The temperature oscillates around 68°F, shifting by the sine's range: +1 to -1.
2Step 2: Graph the Function
To graph \( T(t) \), you can plot points at several values of \( t \) within the interval [0, 24]. For key points: at \( t=0, 12, 24 \), the function will take specific values: \( T(0)=T(24)=68 \) and \( T(12)=68 \). At \( t=6 \), \( \sin(\frac{\pi}{2})=1 \), so \( T(6)=69 \); at \( t=18 \), \( \sin(\frac{3\pi}{2})=-1 \), so \( T(18)=67 \). The sine function provides periodic peaks (highest at 69°F) and troughs (lowest at 67°F). The graph will be a sine wave with a period of 24 hours centered around 68°F.
3Step 3: Calculate the Average Temperature
The average temperature over a cycle is computed by integrating one full period of \( T(t) \) and dividing by the period. Compute: \[ \text{Average Temperature} = \frac{1}{24} \int_{0}^{24} T(t) \, dt = \frac{1}{24} \int_{0}^{24} \left(68 + \sin\left(\frac{\pi}{12}t\right)\right) \, dt. \] Separate the integral: \[ \frac{1}{24} \left( \int_{0}^{24} 68 \, dt + \int_{0}^{24} \sin\left(\frac{\pi}{12}t\right) \, dt \right). \] The first term integrates to 68(24), and the integral of \( \sin(\frac{\pi}{12}t) \) over its period is zero.
4Step 4: Solve the Integrals and Find the Average
Calculate the integrals separately: \( \int_{0}^{24} 68 \, dt = 68 \times 24 = 1632 \). For \( \int_{0}^{24} \sin\left(\frac{\pi}{12}t\right) \, dt \) use the substitution \( u = \frac{\pi}{12}t \), then \( du = \frac{\pi}{12} \, dt \), giving \( \int \sin(u) \, \frac{12}{\pi} \, du \). This will calculate to zero over the interval. Thus, \[ \text{Average Temperature} = \frac{1}{24} \times 1632 = 68. \] The average temperature is 68°F, demonstrating that the sine's variation balances over its cycle to affect no change in the mean.
Key Concepts
Sine FunctionTemperature ModelingIntegration
Sine Function
A sine function is a mathematical curve that describes a repetitive oscillation. It is typically expressed as \( \sin(x) \), where \( x \) is an angle measured in radians. The classic properties of the sine function—its periodicity, amplitude, and oscillation—make it well-suited for modeling phenomena that exhibit wave-like patterns.
A complete sine wave cycles between its maximum value of +1 and its minimum value of -1. This cycle is called a period. In our exercise, the sine function is given by \( \sin\left( \frac{\pi}{12} t \right) \), which completes a full cycle as \( t \) goes from 0 to 24 hours.
A complete sine wave cycles between its maximum value of +1 and its minimum value of -1. This cycle is called a period. In our exercise, the sine function is given by \( \sin\left( \frac{\pi}{12} t \right) \), which completes a full cycle as \( t \) goes from 0 to 24 hours.
- The coefficient \( \frac{\pi}{12} \) affects the frequency of the oscillation. More specifically, multiplying the input \( t \) by \( \frac{\pi}{12} \) ensures that the sine completes one full period in 24 hours.
- This sine wave is added to a constant, 68, which shifts the center of the oscillation up to 68°F. Thus, the temperature oscillates upward and downward around 68°F.
Temperature Modeling
Temperature modeling means creating a mathematical representation of temperature changes over time. In the exercise, the temperature \( T(t) \) is modeled with the function \( 68 + \sin\left( \frac{\pi}{12}t \right) \). This formula uses a sine function to capture how the temperature fluctuates
throughout a day.
Here’s how temperature modeling helps us understand this variation:
throughout a day.
Here’s how temperature modeling helps us understand this variation:
- By adding a constant (68) to the sine function, the model anchors the temperature around 68°F. This is referred to as the average temperature or baseline.
- The sine function itself, \( \sin\left( \frac{\pi}{12}t \right) \), introduces periodic fluctuations, accounting for temperature changes due to daily cycles.
- The temperature variation is capped at -1 and +1 from the baseline, representing a symmetric fluctuation around the average value. These fluctuations reflect the natural temperature peaks and troughs that occur throughout a typical 24-hour cycle.
Integration
Integration is a fundamental concept in calculus that represents the accumulation of quantities, often visualized as the area under a curve. In this exercise, we use integration to find the average temperature over a whole day.
To find the average temperature, we compute the integral of the given temperature function over one full period (from 0 to 24 hours) and divide by the duration of that period.
The integral is broken down into parts:
Integration, therefore, shows us both the total accumulation and the average effect across time, affirming the balance in cyclical models like temperature variations.
To find the average temperature, we compute the integral of the given temperature function over one full period (from 0 to 24 hours) and divide by the duration of that period.
The integral is broken down into parts:
- The constant term (68) contributes \( 68 \times 24 \) to the integral, representing the area under a constant curve.
- The sine component \( \sin\left( \frac{\pi}{12}t \right) \) integrates to zero over one complete cycle. This is due to the symmetric nature of the sine wave, where the positive area exactly cancels the negative area over one period.
Integration, therefore, shows us both the total accumulation and the average effect across time, affirming the balance in cyclical models like temperature variations.
Other exercises in this chapter
Problem 12
Find \(\frac{d y}{d x}\) \(y=\int_{-1}^{x} \frac{2}{t^{2}+t} d t\)
View solution Problem 12
Approximate $$ \int_{-2}^{2}\left(2+x^{2}\right) d x $$ using six equal subintervals.
View solution Problem 13
Find \(\frac{d y}{d x}\) $$ y=\int_{\pi / 2}^{x} \sin \left(t^{2}+1\right) d t $$
View solution Problem 13
Approximate $$ \int_{-1}^{2} e^{-x} d x $$ using three equal subintervals.
View solution