Problem 22 - Applied Calculus

Question

Yesterday's temperature at \(t\) hours past midnight was \(f(t)^{\circ} \mathrm{C} .\) At noon the temperature was \(20^{\circ} \mathrm{C} .\) The first derivative, \(f^{\prime}(t),\) decreased all morning, reaching a low of \(2^{\circ} \mathrm{C} /\) hour at noon, then increased for the rest of the day. Which one of the following must be correct? (a) The temperature fell in the morning and rose in the afternoon. (b) At 1 pm the temperature was \(18^{\circ} \mathrm{C}\) (c) At 1 pm the temperature was \(22^{\circ} \mathrm{C}\) (d) The temperature was lower at noon than at any other time. (e) The temperature rose all day.

Step-by-Step Solution

Answer

(c) At 1 pm the temperature was \(22^{\circ} C\).

Step 1: Understanding the problem

The problem gives us the temperature function \( f(t) \) which represents temperature in degrees Celsius hours after midnight. We know that at noon (12 hours past midnight), the temperature is \(20^{\circ} C\) and the rate of change of temperature, given by the first derivative \( f'(t) \), decreases until noon and then starts increasing.

Step 2: Analyze derivative behavior

The derivative \( f'(t) \) measures the rate at which temperature changes. It's given that \( f'(t) \) decreased during the morning and reached a minimum value of \(2^{\circ} C/hour\) at noon. After noon, \( f'(t) \) starts to increase.

Step 3: Evaluate temperature change over time

Since \( f'(t) \) is decreasing in the morning and becomes positive at noon, the temperature could have been either increasing or decreasing in the morning depending on the initial value of \( f'(t) \). However, since \( f'(t) = 2^{\circ} C/hour\) at noon, the temperature increases after noon.

Step 4: Determine conditions at specific times

At 13:00 (1 pm), 1 hour past noon, \( f'(t) \) indicates an increase in the temperature because \( f'(t) \) is rising. Therefore, the temperature at 1 pm must be higher than it was at noon, which was \(20^{\circ} C\).

Step 5: Eliminate incorrect answers

Based on the provided information and the behavior of \( f'(t) \): (a) is incorrect because \( f(t) \) didn't necessarily fall in the morning if \( f'(t) \) might have started positive; (b) is incorrect because the temperature increases after noon; (d) is incorrect as \( f(t) \) increased after noon; (e) is incorrect because the derivative was decreasing in the morning, suggesting a possible temperature drop. Hence, (c) is plausible because \( f(t) \) rises due to \( f'(t) > 0 \) after noon, making the 1 pm temperature consistent with being \( 22^{\circ} C \).

Key Concepts

Temperature Function

The concept of a temperature function, denoted as \( f(t) \), is central to applied calculus, especially when analyzing how weather conditions change over time. In this case, \( f(t) \) represents the temperature in degrees Celsius at any given hour, \( t \), after midnight. This provides a continuous model for understanding temperature fluctuations throughout the day.

This function helps to visualize and predict the temperature by providing a mathematical model. It is particularly useful in predicting later temperatures based on known values or past trends, like those at noon in the exercise. By plugging in a value for \( t \), like 12 hours for noon, you can directly obtain the temperature for that specific time.

In real-world applications, temperature functions are often derived from empirical data collected over time, giving meteorologists and other scientists a reliable tool for forecasting weather patterns and temperature changes.

First Derivative Analysis

The first derivative of a function, denoted \( f'(t) \), is an essential tool in understanding how a function behaves over time. In the context of our temperature function, the first derivative indicates the rate of change of the temperature. This means it tells us how quickly the temperature is rising or falling at any given moment.

Understanding \( f'(t) \) can help us make informed guesses about temperature patterns. If \( f'(t) \) is positive, the temperature is increasing. Conversely, if \( f'(t) \) is negative, the temperature is decreasing. This can be crucial for making predictions, such as determining whether it will be colder or hotter over the next hour.

• When \( f'(t) \) decreases, the function's rate of change is slowing down. So, if it started off positive, it might still be increasing but at a slower pace.
• If \( f'(t) \) becomes negative, it would indicate that the temperature starts to decrease.
• For this problem, \( f'(t) \) reaches a low point at \(2^{\circ} \mathrm{C/hour}\) at noon before increasing.

This behavior helps us assess when the temperature changes direction, which is central to solving the exercise.

Rate of Change

The rate of change is a vital concept not only in calculus but in any situation involving dynamic systems. It speaks to how one quantity changes in relation to another. In this exercise, it describes how temperature responds over time, specifically measured by the first derivative \( f'(t) \).

The rate of change provides insight into how rapidly or slowly the temperature dynamics occur across the day. By analyzing the rate of change through the first derivative, we gain detailed information about the nature of temperature shifts:

• A positive rate indicates the temperature is increasing.
• A negative rate indicates a decreasing temperature.
• If the derivative is zero, it suggests no temperature change at that moment.

At noon, the temperature's rate of change was at its minimum, which was \(2^{\circ} \mathrm{C/hour}\), and it signified an increase post-noon. Hence, the temperature starts to rise after this point, making it crucial to forecasting afternoon temperatures accurately.