Problem 43 - Precalculus: Functions and Graphs

Question

Determining temperatures A meteorologist determines that the temperature \(T\) (in 'F) for a certain 24-hour period in winter was given by the formula \(T=\frac{1}{20} r(t-12)(t-24)\) for \(0 \leq t \leq 24,\) where \(t\) is time in hours and \(t=0\) corresponds to 6 A.M. (a) When was \(T>0,\) and when was \(T<0 ?\) (b) Sketch the graph of \(T\). Show that the temperature was \(32^{\circ} \mathrm{F}\) sometime between 12 noon and 1 P.M. (Hint: Use the intermediate value theorem.)

Step-by-Step Solution

Answer

(a) T>0 before noon; T<0 after noon. (b) T was 32°F between noon and 1 PM.

Step 1: Setting the Expression for T

The given temperature formula is \( T = \frac{1}{20} r(t-12)(t-24) \). To solve parts (a) and (b), note that you need the conditions when \( T > 0 \) and \( T < 0 \), and to eventually find when \( T = 32 \).

Step 2: Finding Critical Points

To understand when \( T = 0 \), solve \( \frac{1}{20} r(t-12)(t-24) = 0 \). The roots here are \( t = 12 \) and \( t = 24 \), indicating critical points at 12 PM and 6 AM.

Step 3: Analyzing the Intervals

Examine intervals between the critical points to decide when the temperature is positive or negative:1. For \( 0 \leq t < 12 \), \( (t-12) < 0 \) and \( (t-24) < 0 \), so \( T > 0 \).2. For \( 12 < t < 24 \), \( (t-12) > 0 \) and \( (t-24) < 0 \), so \( T < 0 \).

Step 4: Sketching the Graph

The graph of \( T \) is a downward opening parabola because the product \( (t-12)(t-24) \) is negative between roots. The temperature peaks at \( t = 18 \), halfway between the roots. Sketch this parabola with intercepts at \( t = 12 \) and \( t = 24 \).

Step 5: Verifying Temperature of 32°F

To show \( 32 = \frac{1}{20} r(t-12)(t-24) \), rearrange to: \( 640 = r(t-12)(t-24) \). As the temperature is increasing when \( t \) approaches noon from morning and then decreasing, use the Intermediate Value Theorem between \( t = 12 \) and \( t = 13 \). Solve \( 640 = r(t-12)(t-24) \) within these bounds to check the occurrence of \( T = 32 \).

Key Concepts

Functions

In precalculus, a function is a relationship between a set of inputs and a set of possible outputs where each input is related to exactly one output. It's like a machine that takes an input, processes it, and gives an output.
For the meteorologist's formula, we have the temperature function given by \( T = \frac{1}{20} r(t-12)(t-24) \). This function provides the temperature \( T \) at any time \( t \) during the day.

• The variable \( t \) is our input, representing the time in hours since 6 A.M.
• \( T \) is our output, which shows the temperature at that hour.
• The expression \( r(t-12)(t-24) \) affects how the temperature changes throughout the day.

Understanding functions involves recognizing how each part influences the output. Here, you need to see how changes in \( t \) impact \( T \). For example, \( (t-12)(t-24) \) determines where the function crosses the temperature axis, hitting zero at \( t = 12 \) and \( t = 24 \). Each part of the equation helps you predict the weather pattern for the day.

Graphing

Graphing a function helps visualize how it behaves over its domain, giving us a clearer picture of patterns or changes.
To graph the temperature function \( T = \frac{1}{20} r(t-12)(t-24) \), it resembles a parabola that opens downward. This is because of the quadratic terms that multiply together in the expression.

• The graph will cross the time axis at \( t = 12 \) and \( t = 24 \) (the critical points), where the temperature is zero.
• Between these points, the parabola peaks around \( t = 18 \), showing the warmest part of the day.
• To sketch this graph, plot the points at \( t = 12 \) and \( t = 24 \), and then draw a curve that peaks and falls between them.

This visual helps to determine visually when \( T > 0 \) and when \( T < 0 \). This makes it easier to understand the day's temperature dynamics.

Intermediate Value Theorem

The intermediate value theorem (IVT) is a fundamental concept in calculus, which says that if a function is continuous on a closed interval, then it takes on every value between its values at the ends of the interval.
The Intermediate Value Theorem is handy for determining specific values within a continuous function's range. In our exercise, it's used between 12 noon (\( t = 12 \)) and 1 PM (\( t = 13 \)) to prove the temperature reaches 32°F.

• The graph of \( T \) is continuous, meaning it doesn't jump or break anywhere within the time domain.
• We calculate the values of \( T \) at \( t=12 \) and \( t=13 \), ensuring that 32°F must occur between these times.
• Since the temperature is higher at noon and decreases by 1 PM, IVT guarantees there must be a moment during this hour when the temperature was exactly 32°F.

This theorem confirms changes without calculating explicit times, making it a powerful verification tool in precalculus.