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Chapter 1: Problem 4

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.

Short Answer

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

Step by step solution

01

- 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.
02

- 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' (\(\frac{\text{掳C}}{\text{min}}\)).
03

- 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 掳C/min. Since the derivative is negative, \(f'(25) = -0.6\).
04

- Translate Information into Statement

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

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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 (掳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 掳C and time in minutes, \(f'(t)\) is expressed in degrees Celsius per minute (掳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 掳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 掳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.

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Most popular questions from this chapter

For each of the following prompts, give an example of a function that satisfies the stated criteria. A formula or a graph, with reasoning, is sufficient for each. If no such example is possible, explain why. a. A function \(f\) that is continuous at \(a=2\) but not differentiable at \(a=2\). b. A function \(g\) that is differentiable at \(a=3\) but does not have a limit at \(a=3\). c. A function \(h\) that has a limit at \(a=-2,\) is defined at \(a=-2,\) but is not continuous at \(a=-2\) d. A function \(p\) that satisfies all of the following: \- \(p(-1)=3\) and \(\lim _{x \rightarrow-1} p(x)=2\) $$\text { - } p(0)=1 \text { and } p^{\prime}(0)=0$$ \- \(\lim _{x \rightarrow 1} p(x)=p(1)\) and \(p^{\prime}(1)\) does not exist

The table below shows the number of calories used per minute as a function of an individual's body weight for three sports: $$\begin{array}{|l|l|l|l|l|l|l|} \hline \text { Activity } & 100 \mathrm{lb} & 120 \mathrm{lb} & 150 \mathrm{lb} & 170 \mathrm{lb} & 200 \mathrm{lb} & 220 \mathrm{lb} \\ \hline \text { Walking } & 2.7 & 3.2 & 4 & 4.6 & 5.4 & 5.9 \\ \hline \text { Bicycling } & 5.4 & 6.5 & 8.1 & 9.2 & 10.8 & 11.9 \\ \hline \text { Swimming } & 5.8 & 6.9 & 8.7 & 9.8 & 11.6 & 12.7 \\ \hline \end{array}$$ a) Determine the number of calories that a 200 lb person uses in one half-hour of walking. b) Who uses more calories, a \(120 \mathrm{lb}\) person swimming for one hour, or a \(220 \mathrm{lb}\) person bicycling for a half-hour? c) Does the number of calories of a person swimming increase or decrease as weight increases?

The temperature change \(T\) (in Fahrenheit degrees), in a patient, that is generated by a dose \(q\) (in milliliters), of a drug, is given by the function \(T=f(q)\). a. What does it mean to say \(f(50)=0.75 ?\) Write a complete sentence to explain, using correct units. b. A person's sensitivity, \(s,\) to the drug is defined by the function \(s(q)=f^{\prime}(q) .\) What are the units of sensitivity? c. Suppose that \(f^{\prime}(50)=-0.02 .\) Write a complete sentence to explain the meaning of this value. Include in your response the information given in (a).

Suppose that the population, \(P\), of China (in billions) can be approximated by the function \(P(t)=1.15(1.014)^{t}\) where \(t\) is the number of years since the start of \(1993 .\) a. According to the model, what was the total change in the population of China between January 1,1993 and January \(1,2000 ?\) What will be the average rate of change of the population over this time period? Is this average rate of change greater or less than the instantaneous rate of change of the population on January \(1,2000 ?\) Explain and justify, being sure to include proper units on all your answers. b. According to the model, what is the average rate of change of the population of China in the ten-year period starting on January 1, \(2012 ?\) c. Write an expression involving limits that, if evaluated, would give the exact instantaneous rate of change of the population on today's date. Then estimate the value of this limit (discuss how you chose to do so) and explain the meaning (including units) of the value you have found. d. Find an equation for the tangent line to the function \(y=P(t)\) at the point where the \(t\) -value is given by today's date.

An object moving along a straight line path has a differentiable position function \(y=s(t) ;\) \(s(t)\) measures the object's position relative to the origin at time \(t\). It is known that at time \(t=9\) seconds, the object's position is \(s(9)=4\) feet (i.e., 4 feet to the right of the origin). Furthermore, the object's instantaneous velocity at \(t=9\) is -1.2 feet per second, and its acceleration at the same instant is 0.08 feet per second per second. a. Use local linearity to estimate the position of the object at \(t=9.34\). b. Is your estimate likely too large or too small? Why? c. In everyday language, describe the behavior of the moving object at \(t=9 .\) Is it moving toward the origin or away from it? Is its velocity increasing or decreasing?
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