Problem 4 - Active Calculus

Question

The temperature, \(H\), in degrees Celsius, of a cup of coffee placed on the kitchen counter is given by \(H=f(t),\) where \(t\) is in minutes since the coffee was put on the counter. (a) Is \(f^{\prime}(t)\) positive or negative? [Choose: positive | negative] (b) What are the units of \(f^{\prime}(25) ?\) \(\square\) Suppose that \(\left|f^{\prime}(25)\right|=0.6\) and \(f(25)=65 .\) Fill in the blanks (including units where needed) and select the appropriate terms to complete the following statement about the temperature of the coffee in this case. At \(\square\) minutes after the coffee was put on the counter, its [Choose: derivative | temperature | change in temperature] is \(\square\) and will [Choose: increase | decrease] by about \(\square\) in the next 60 seconds.

Step-by-Step Solution

Answer

Negative, degrees Celsius per minute; at 25 minutes, its temperature is 65 °C and will decrease by 0.6°C in the next minute.

Step 1 - Determine the Sign of the Derivative

The derivative, denoted as \(f'(t)\), represents the rate of change of temperature with respect to time. Since the coffee is cooling down, the rate of change of temperature will be negative. Thus, \(f'(t)\) is negative.

Step 2 - Determine the Units of the Derivative

The units of \(f'(t)\) can be derived from the units of \(H\) and \(t\). Since \(H\) is in degrees Celsius and \(t\) is in minutes, the units of the derivative \(f'(t)\) will be 'degrees Celsius per minute' (\(\frac{\text{°C}}{\text{min}}\)).

Step 3 - Analyze Given Information

We are given that \(\left|f'(25)\right|=0.6\) and \(f(25)=65\). This means the absolute value of the rate of change in temperature at \(t = 25\) minutes is 0.6 °C/min. Since the derivative is negative, \(f'(25) = -0.6\).

Step 4 - Translate Information into Statement

At 25 minutes after the coffee was put on the counter, its temperature is 65 °C and will decrease by about 0.6 degrees Celsius in the next 60 seconds.

Key Concepts

temperature change

When discussing the temperature change of coffee as it cools, it's essential to understand how temperature varies over time. As soon as coffee is poured, it starts losing heat to its surroundings.
This cooling process is influenced by the temperature difference between the coffee and the ambient environment.
The greater the initial difference in temperature, the quicker the coffee will cool down initially. As time progresses, this rate decreases. This natural decline is captured in the function representing temperature change. Over time, each minute's temperature tends to drop, showing negative growth.

derivative units

In calculus, derivatives are fundamental to understanding how functions change. The derivative, denoted as \(f'(t)\), represents the rate of change of a function.
For the coffee temperature problem, \(f(t)\) is given in degrees Celsius (°C), while time \(t\) is in minutes.
Thus, the derivative \(f'(t)\) measures how fast the temperature changes over time. The units of \(f'(t)\) come from the function \(f(t)\) itself. Since temperature is measured in °C and time in minutes, \(f'(t)\) is expressed in degrees Celsius per minute (°C/min). This tells us how many degrees the temperature changes per minute.

rate of cooling

The rate of cooling refers to how quickly the coffee loses temperature. By analyzing the given data, we find that the magnitude of the rate of cooling at 25 minutes, \(|f'(25)|\), is 0.6 °C/min.
This value indicates the speed at which the coffee's temperature decreases. Since the derivative is negative, \(f'(25) = -0.6\).
This negative sign shows that the coffee is cooling, not warming. Knowing the rate of cooling, we can predict the temperature drop in a given period. For instance, within the next 60 seconds from 25 minutes, the temperature decreases by 0.6 °C.

calculus application

Calculus helps us understand and predict changes in various contexts. In the case of the cooling coffee, calculus is used to determine how quickly the temperature drops over time.
The derivative \(f'(t)\) shows the temperature change rate at any given moment \(t\).
By evaluating \(f'(25) = -0.6\), we gain insight into the cooling process exactly 25 minutes after the coffee was made.
Predicting temperature variations is crucial in many real-world scenarios, from industrial processes to everyday situations like understanding how quickly a hot drink might become undrinkable.