Skip to main content
Logo image

Active Calculus

Matthew Boelkins

Section 1.5 Interpreting, estimating, and using the derivative

It is a powerful feature of mathematics that it can be studied both as abstract discipline and as an applied one. For instance, calculus can be developed almost entirely as an abstract collection of ideas that focus on properties of functions. At the same time, if we consider functions that represent meaningful processes, calculus can describe our experience of physical reality. We have already seen that for the position function \(y = s(t)\) of a ball being tossed straight up in the air, the derivative of the position function, \(v(t) = s'(t)\text{,}\) gives the ball’s velocity at time \(t\text{.}\)
In this section, we investigate several functions with specific physical meaning, and consider how the units on the independent variable, dependent variable, and the derivative function add to our understanding. To start, we consider the familiar problem of a position function of a moving object.

Preview Activity 1.5.1.

One of the longest stretches of straight (and flat) road in North America can be found on the Great Plains in the state of North Dakota on state highway 46, which lies just south of the interstate highway I-94 and runs through the town of Gackle. A car leaves town (at time \(t = 0\)) and heads east on highway 46; its position in miles from Gackle at time \(t\) in minutes is given by the graph of the function in Figure 1.5.1. Three important points are labeled on the graph; where the curve looks linear, assume that it is indeed a straight line.
Figure 1.5.1. The graph of \(y = s(t)\text{,}\) the position of the car along highway 46, which tells its distance in miles from Gackle, ND, at time \(t\) in minutes.
  1. In everyday language, describe the behavior of the car over the provided time interval. In particular, discuss what is happening on the time intervals \([57,68]\) and \([68,104]\text{.}\)
  2. Find the slope of the line between the points \((57,63.8)\) and \((104,106.8)\text{.}\) What are the units on this slope? What does the slope represent?
  3. Find the average rate of change of the car’s position on the interval \([68,104]\text{.}\) Include units on your answer.
  4. Estimate the instantaneous rate of change of the car’s position at the moment \(t = 80\text{.}\) Write a sentence to explain your reasoning and the meaning of this value.

Subsection 1.5.1 Units of the derivative function

As we now know, the derivative of the function \(f\) at a fixed value \(x\) is given by
\begin{equation*} f'(x) = \lim_{h \to 0} \frac{f(x+h)-f(x)}{h}\text{,} \end{equation*}
and this value has several different interpretations. If we set \(x = a\text{,}\) one meaning of \(f'(a)\) is the slope of the tangent line at the point \((a,f(a))\text{.}\)
We also sometimes write \(\frac{df}{dx}\) or \(\frac{dy}{dx}\) instead of \(f'(x)\text{,}\) and these alternate notations help us see the units (and thus the meaning) of the derivative as the instantaneous rate of change of \(f\) with respect to \(x\). The units on the slope of the secant line, \(\frac{f(x+h)-f(x)}{h}\text{,}\) are “units of \(y\) per unit of \(x\text{,}\)” and when we take the limit as \(h\) goes to zero, the derivative \(f'(x)\) has the same units: units of \(y\) per unit of \(x\text{.}\) It is helpful to remember that the units on the derivative function are “units of output per unit of input,” for the variables of the original function.
For example, suppose that the function \(y = P(t)\) measures the population of a city (in thousands) at the start of year \(t\) (where \(t = 0\) corresponds to 2010 AD). We are told that \(P'(2) = 21.37\text{.}\) What is the meaning of this value? Well, since \(P\) is measured in thousands and \(t\) is measured in years, we can say that the instantaneous rate of change of the city’s population with respect to time at the start of 2012 is 21.37 thousand people per year. We therefore expect that in the coming year, about 21,370 people will be added to the city’s population.

Subsection 1.5.2 Toward more accurate derivative estimates

Recall that to estimate the value of \(f'(x)\) at a given \(x\text{,}\) we calculate a difference quotient \(\frac{f(x+h)-f(x)}{h}\) with a relatively small value of \(h\text{.}\) We should use both positive and negative values of \(h\) in order to account for the behavior of the function on both sides of the point of interest. To that end, we introduce the notion of a central difference and its role in estimating derivatives.

Example 1.5.2.

Suppose that \(y = f(x)\) is a function for which three values are known: \(f(1) = 2.5\text{,}\) \(f(2) = 3.25\text{,}\) and \(f(3) = 3.625\text{.}\) Estimate \(f'(2)\text{.}\)
Solution.
We know that \(f'(2) = \lim_{h \to 0} \frac{f(2+h) - f(2)}{h}\text{.}\) But since we don’t have a graph or a formula for the function, we can neither sketch a tangent line nor evaluate the limit algebraically. We can’t even use smaller and smaller values of \(h\) to estimate the limit. Instead, we have just two choices: using \(h = -1\) or \(h = 1\text{,}\) depending on which point we pair with \((2,3.25)\text{.}\)
So, one estimate is
\begin{equation*} f'(2) \approx \frac{f(1)-f(2)}{1-2} = \frac{2.5-3.25}{-1} = 0.75\text{.} \end{equation*}
The other is
\begin{equation*} f'(2) \approx \frac{f(3)-f(2)}{3-2} = \frac{3.625-3.25}{1} = 0.375\text{.} \end{equation*}
Because the first approximation looks backward from the point \((2,3.25)\) and the second approximation looks forward, it makes sense to average these two estimates in order to account for behavior on both sides of \(x=2\text{.}\) Doing so, we find that
\begin{equation*} f'(2) \approx \frac{0.75 + 0.375}{2} = 0.5625\text{.} \end{equation*}
The intuitive approach to average the two estimates found in Example 1.5.2 is in fact the best possible way estimate to a derivative when we have just two function values for \(f\) on opposite sides of the point of interest.
Figure 1.5.3. At left, the graph of \(y = f(x)\) along with the secant line through \((1,2.5)\) and \((2,3.25)\text{,}\) the secant line through \((2, 3.25)\) and \((3,3.625)\text{,}\) as well as the tangent line. At right, the same graph along with the secant line through \((1,2.5)\) and \((3,3.625)\text{,}\) plus the tangent line.
To see why, we think about the diagram in Figure 1.5.3. On the left, we see the two secant lines with slopes that come from computing the backward difference \(\frac{f(1)-f(2)}{1-2} = 0.75\) and from the forward difference \(\frac{f(3)-f(2)}{3-2} = 0.375\text{.}\) Note how the first slope over-estimates the slope of the tangent line at \((2,f(2))\text{,}\) while the second slope underestimates \(f'(2)\text{.}\) On the right, we see the secant line whose slope is given by the central difference
\begin{equation*} \frac{f(3)-f(1)}{3-1} = \frac{3.625-2.5}{2} = \frac{1.125}{2} = 0.5625\text{.} \end{equation*}
Note that this central difference has the same value as the average of the forward and backward differences (and it is straightforward to explain why this always holds). The central difference yields a very good approximation to the derivative’s value, because it yields a line closer to being parallel to the tangent line.
The central difference approximation to the value of the first derivative is given by
\begin{equation*} f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}\text{.} \end{equation*}
This quantity measures the slope of the secant line to \(y = f(x)\) through the points \((a-h, f(a-h))\) and \((a+h, f(a+h))\text{.}\)

Activity 1.5.2.

A potato is placed in an oven, and the potato’s temperature \(F\) (in degrees Fahrenheit) at various points in time is taken and recorded in the following table. Time \(t\) is measured in minutes.
Table 1.5.4. Temperature data in degrees Fahrenheit.
\(t\) \(0\) \(15\) \(30\) \(45\) \(60\) \(75\) \(90\)
\(F(t)\) \(70\) \(180.5\) \(251\) \(296\) \(324.5\) \(342.8\) \(354.5\)
  1. Use a central difference to estimate the instantaneous rate of change of the temperature of the potato at \(t= 30\text{.}\) Include units on your answer.
  2. Use a central difference to estimate the instantaneous rate of change of the temperature of the potato at \(t= 60\text{.}\) Include units on your answer.
  3. Without doing any calculation, which do you expect to be greater: \(F'(75)\) or \(F'(90)\text{?}\) Why?
  4. Suppose it is given that \(F(64) = 330.28\) and \(F'(64) = 1.341\text{.}\) What are the units on these two quantities? What do you expect the temperature of the potato to be when \(t = 65\text{?}\) when \(t = 66\text{?}\) Why?
  5. Write a couple of careful sentences that describe the behavior of the temperature of the potato on the time interval \([0,90]\text{,}\) as well as the behavior of the instantaneous rate of change of the temperature of the potato on the same time interval.

Activity 1.5.3.

A company manufactures rope, and the total cost of producing \(r\) feet of rope is \(C(r)\) dollars.
  1. What does it mean to say that \(C(2000) = 800\text{?}\)
  2. What are the units of \(C'(r)\text{?}\)
  3. Suppose that \(C(2000) = 800\) and \(C'(2000) = 0.35\text{.}\) Estimate \(C(2100)\text{,}\) and justify your estimate by writing at least one sentence that explains your thinking.
  4. Do you think \(C'(2000)\) is less than, equal to, or greater than \(C'(3000)\text{?}\) Why?
  5. Suppose someone claims that \(C'(5000) = -0.1\text{.}\) What would the practical meaning of this derivative value tell you about the approximate cost of the next foot of rope? Is this possible? Why or why not?

Activity 1.5.4.

Researchers at a major car company have found a function that relates gasoline consumption to speed for a particular model of car. In particular, they have determined that the consumption \(C\text{,}\) in liters per kilometer, at a given speed \(s\text{,}\) is given by a function \(C = f(s)\text{,}\) where \(s\) is the car’s speed in kilometers per hour.
  1. Data provided by the car company tells us that \(f(80) = 0.015\text{,}\) \(f(90) = 0.02\text{,}\) and \(f(100) = 0.027\text{.}\) Use this information to estimate the instantaneous rate of change of fuel consumption with respect to speed at \(s = 90\text{.}\) Be as accurate as possible, use proper notation, and include units on your answer.
  2. By writing a complete sentence, interpret the meaning (in the context of fuel consumption) of “\(f(80) = 0.015\text{.}\)”
  3. Write at least one complete sentence that interprets the meaning of the value of \(f'(90)\) that you estimated in (a).
In Section 1.4, we learned how use to the graph of a given function \(f\) to plot the graph of its derivative, \(f'\text{.}\) It is important to remember that when we do so, the scale and the units on the vertical axis often have to change to represent \(f'\text{.}\) For example, suppose that \(P(t) = 400-330e^{-0.03t}\) tells us the temperature in degrees Fahrenheit of a potato in an oven at time \(t\) in minutes. In Figure 1.5.5, we sketch the graph of \(P\) on the left and the graph of \(P'\) on the right.
Figure 1.5.5. Plot of \(P(t) = 400-330e^{-0.03t}\) at left, and its derivative \(P'(t)\) at right.
Notice that the vertical scales are different in size and different in units, as the units of \(P\) are °F, while those of \(P'\) are °F/min.

Subsection 1.5.3 Summary

  • The derivative of a given function \(y=f(x)\) measures the instantaneous rate of change of the output variable with respect to the input variable.
  • The units on the derivative function \(y = f'(x)\) are units of \(y\) per unit of \(x\text{.}\) Again, this measures how fast the output of the function \(f\) changes when the input of the function changes.
  • The central difference approximation to the value of the first derivative is given by
    \begin{equation*} f'(a) \approx \frac{f(a+h) - f(a-h)}{2h}\text{.} \end{equation*}
    This quantity measures the slope of the secant line to \(y = f(x)\) through the points \((a-h, f(a-h))\) and \((a+h, f(a+h))\text{.}\) The central difference generates a good approximation of the derivative’s value.

Exercises 1.5.4 Exercises

1. A cooling cup of coffee.

The temperature, \(H\text{,}\) in degrees Celsius, of a cup of coffee placed on the kitchen counter is given by \(H = f(t)\text{,}\) where \(t\) is in minutes since the coffee was put on the counter.
  1. Is \(f'(t)\) positive or negative?
    • positive
    • negative
    (Be sure that you are able to give a reason for your answer.)
  2. What are the units of \(f'(35)\text{?}\)
  3. Suppose that \(\lvert f'(35)\rvert = 1.5\) and \(f(35) = 68\text{.}\) 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 minutes after the coffee was put on the counter, its
    • derivative
    • temperature
    • change in temperature
    is and will
    • increase
    • decrease
    by about in the next 30 seconds.

2. A cost function.

The cost, \(C\) (in dollars) to produce \(g\) gallons of ice cream can be expressed as \(C = f(g)\text{.}\)
(a) In the expression \(f(100) = 250\text{,}\)
what are the units of 100?
  • dollars
  • gallons
  • dollars*gallons
  • dollars/gallon
  • gallons/dollar
what are the units of 250?
  • dollars
  • gallons
  • dollars*gallons
  • dollars/gallon
  • gallons/dollar
(b) In the expression \(f'(100) = 1.2\text{,}\)
what are the units of 100?
  • dollars
  • gallons
  • dollars*gallons
  • dollars/gallon
  • gallons/dollar
what are the units of 1.2?
  • dollars
  • gallons
  • dollars*gallons
  • dollars/gallon
  • gallons/dollar
(Be sure that you can carefully put into words the meanings of each of these statement in terms of ice cream and money.)

3. Weight as a function of calories.

A laboratory study investigating the relationship between diet and weight in adult humans found that the weight of a subject, \(W\text{,}\) in pounds, was a function, \(W=f(c)\text{,}\) of the average number of Calories, \(c\text{,}\) consumed by the subject in a day.
(a) In the statement \(f(1600) = 165\)
what are the units of 1600?
  • lb
  • cal
  • day
  • lb/cal
  • cal/lb
  • cal/day
  • lb/day
  • day/lb
  • day/cal
what are the units of 165?
  • lb
  • cal
  • day
  • lb/cal
  • cal/lb
  • cal/day
  • lb/day
  • day/lb
  • day/cal
(Think about what this statement means in terms of the weight of the subject and the number of calories that the subject consumes.)
(b) In the statement \(f'(2000)=0\text{,}\)
what are the units of 2000?
  • lb
  • cal
  • day
  • lb/cal
  • cal/lb
  • cal/day
  • lb/day
  • day/lb
  • day/cal
what are the units of 0?
  • lb
  • cal
  • day
  • lb/cal
  • cal/lb
  • cal/day
  • lb/day
  • day/lb
  • day/cal
(Think about what this statement means in terms of the weight of the subject and the number of calories that the subject consumes.)
(c) In the statement \(f^{-1}(173) = 2400\text{,}\)
what are the units of 173?
  • lb
  • cal
  • day
  • lb/cal
  • cal/lb
  • cal/day
  • lb/day
  • day/lb
  • day/cal
what are the units of 2400?
  • lb
  • cal
  • day
  • lb/cal
  • cal/lb
  • cal/day
  • lb/day
  • day/lb
  • day/cal
(Think about what this statement means in terms of the weight of the subject and the number of calories that the subject consumes.)
(d) What are the units of \(f'(c)=dW/dc\text{?}\)
  • lb
  • cal
  • day
  • lb/cal
  • cal/lb
  • cal/day
  • lb/day
  • day/lb
  • day/cal
(e) Suppose that Sam reads about \(f'\) in this study and draws the following conclusion: If Sam increases her average calorie intake from 2800 to 2840 calories per day, then her weight will increase by approximately 0.8 pounds.
Fill in the blanks below so that the equation supports her conclusion.
\(f'\Big(\) \(\Big)=\)

4. Displacement and velocity.

The displacement (in meters) of a particle moving in a straight line is given by
\begin{equation*} s = t^2 - 5 t + 16, \end{equation*}
where \(t\) is measured in seconds.
(A)
(i) Find the average velocity over the time interval [3,4].
Average Velocity = meters per second.
(ii) Find the average velocity over the time interval [3.5,4].
Average Velocity = meters per second.
(iii) Find the average velocity over the time interval [4,5].
Average Velocity = meters per second.
(iv) Find the average velocity over the time interval [4,4.5].
Average Velocity = meters per second.
(B) Find the instantaneous velocity when \(t=4\text{.}\)
Instantaneous velocity = meters per second.

5.

A cup of coffee has its temperature \(F\) (in degrees Fahrenheit) at time \(t\) given by the function \(F(t) = 75 + 110 e^{-0.05t}\text{,}\) where time is measured in minutes.
  1. Use a central difference with \(h = 0.01\) to estimate the value of \(F'(10)\text{.}\)
  2. What are the units on the value of \(F'(10)\) that you computed in (a)? What is the practical meaning of the value of \(F'(10)\text{?}\)
  3. Which do you expect to be greater: \(F'(10)\) or \(F'(20)\text{?}\) Why?
  4. Write a sentence that describes the behavior of the function \(y = F'(t)\) on the time interval \(0 \le t \le 30\text{.}\) How do you think its graph will look? Why?

6.

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 \(T = f(q)\text{.}\)
  1. What does it mean to say \(f(50) = 0.75\text{?}\) Write a complete sentence to explain, using correct units.
  2. A person’s sensitivity, \(s\text{,}\) to the drug is defined by the function \(s(q) = f'(q)\text{.}\) What are the units of sensitivity?
  3. Suppose that \(f'(50) = -0.02\text{.}\) Write a complete sentence to explain the meaning of this value. Include in your response the information given in (a).

7.

The velocity of a ball that has been tossed vertically in the air is given by \(v(t) = 16 - 32t\text{,}\) where \(v\) is measured in feet per second, and \(t\) is measured in seconds. The ball is in the air from \(t = 0\) until \(t = 2\text{.}\)
  1. When is the ball’s velocity greatest?
  2. Determine the value of \(v'(1)\text{.}\) Justify your thinking.
  3. What are the units on the value of \(v'(1)\text{?}\) What does this value and the corresponding units tell you about the behavior of the ball at time \(t = 1\text{?}\)
  4. What is the physical meaning of the function \(v'(t)\text{?}\)

8.

The value, \(V\text{,}\) of a particular automobile (in dollars) depends on the number of miles, \(m\text{,}\) the car has been driven, according to the function \(V = h(m)\text{.}\)
  1. Suppose that \(h(40000) = 15500\) and \(h(55000) = 13200\text{.}\) What is the average rate of change of \(h\) on the interval \([40000,55000]\text{,}\) and what are the units on this value?
  2. In addition to the information given in (a), say that \(h(70000) = 11100\text{.}\) Determine the best possible estimate of \(h'(55000)\) and write one sentence to explain the meaning of your result, including units on your answer.
  3. Which value do you expect to be greater: \(h'(30000)\) or \(h'(80000)\text{?}\) Why?
  4. Write a sentence to describe the long-term behavior of the function \(V = h(m)\text{,}\) plus another sentence to describe the long-term behavior of \(h'(m)\text{.}\) Provide your discussion in practical terms regarding the value of the car and the rate at which that value is changing.
You have attempted of activities on this page.
PrevTopNext
Feedback