Problem 21 - 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 I 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

(a) The temperature fell in the morning and rose in the afternoon.

Step 1: Understand the given information

We are given that the temperature at noon is 20°C and at that time, the rate of change of temperature, \(f'(t)\), is 2°C/hour. The derivative decreased all morning and reached a low at noon, indicating the rate of temperature change is increasing after noon.

Step 2: Analyze the derivative behavior

Before noon, the rate \(f'(t)\) is decreasing and reaches its minimum at noon with \(f'(t) = 2\text{°C/hour}\). This decrease suggests temperatures were likely rising but possibly slowly or even falling. However, at noon it then starts to increase, indicating rising temperatures.

Step 3: Evaluate each option

Option (a) cannot be confirmed because the information does not specify if the temperature actually fell; \(f'(t)\) was still positive before noon. Option (b) and Option (c) are not addressed by the given \(f'(t)\) details meaningfully. Option (d) is incorrect since noon has \(f'(t) > 0\), suggesting temperature increase afterward. Option (e) is incorrect since the claim contradicts decreasing derivative value in the morning, possibly indicating a slowing rise or even a potential fall.

Step 4: Determine the correct statement

None of the options conclusively match the described behavior of \(f(t)\) given the derivative conditions alone, except that (a) seems the closest because while \(f'(t)\) behavior implies possible decrease or slower increase in the morning, it increases in the afternoon.

Key Concepts

Understanding the Derivative

In calculus, one of the most fundamental concepts is the derivative. It represents the rate at which a function is changing at any given point. Consider the function representing temperature over time, such as in our example. The derivative of this function, denoted as \( f'(t) \), tells us how quickly the temperature is changing as time progresses.

A positive derivative means that the function (temperature, in this case) is increasing. Conversely, a negative derivative implies a decrease in the function's value. When discussing temperatures over time, if the derivative \( f'(t) \) is positive, the temperature rises. This is important because even a decreasing derivative can still be positive, indicating that the temperature is rising, albeit at a slower rate.

• A decreasing derivative indicates the slope (or speed of temperature change) is becoming less steep.
• If the derivative equals zero, it suggests a turning point where the temperature may stop rising or begin to fall.
• An increasing derivative after a minimum implies that the function's value is likely to rise.

Exploring Temperature Change

Temperature change is a common real-life example where calculus, specifically derivatives, are applied. Understanding how derivatives affect temperature helps us interpret changes throughout the day.

In this exercise, the problem explored how the temperature (\( f(t) \)) evolved over time, specifically past midnight. The first derivative decreased during the morning, reaching its lowest at noon, and then increased, indicating possible trends in temperature change.

• Decreasing first derivative: Temperatures might rise slower or even fall depending on the starting point and curving nature.
• Increasing first derivative after midday: Likely indicates rising temperatures, as the rate of change becomes more significant (or steeper).
• To fully understand, we'd consider the context: a slow downtrend may mean cooling weather, but not if the baseline is risen.

Examining Rate of Change

Rate of change is another fundamental calculus concept helping us understand dynamic systems like temperature fluctuations in our exercise. The rate of change essentially tells us how a quantity changes concerning another parameter, typically time.

For temperatures throughout the day, the rate of change lets us predict how quickly it might get warmer or cooler.
- At any point, analyzing \( f'(t) \) gives insight into whether the environment is heating or cooling.
- In the exercise, the temperature's rate of change decreased up to noon but remained positive, signaling slower increasing temperatures or potential stabilization.
- After this point, as \( f'(t) \) increased, the temperature began rising more quickly, indicating a warmer afternoon.

Understanding how fast temperatures change helps us better prepare and predict daily weather shifts.