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## Section1.6Interpreting, Estimating, and Using the Derivative

###### Motivating Questions
• In contexts other than the position of a moving object, what does the derivative of a function measure?

• What are the units of the derivative function $f'\text{,}$ and how are they related to the units of the original function $f\text{?}$

• Given the value of the derivative of a function at a point, what can we infer about how the value of the function changes nearby?

A powerful feature of mathematics is that it can be studied both as an 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.

###### Example1.69

One of the longest stretches of straight (and flat) road in North America can be found on the Great Plains as part of North Dakota state Highway 46, which lies just south of the interstate highway I-94 and runs through the town of Gackle. Suppose 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 Figure1.70 below. Three important points are labeled on the graph; where the curve looks linear, you may assume that it is indeed a straight line.

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 in 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.

Hint
1. When is the car moving? When is the car stopped? How do you know? (What features of the graph tell you this information?)

2. Units of slope should be units of $s$ per unit of $t$. Think about what the slope of a secant line represents.

3. Use the formula for average rate of change.

4. Close to the time $t=80$ minutes, the function is linear. What does this mean for the instantaneous rate of change of the car's position at this instant?

1. The car drives away from Gackle at a constant speed for $57$ minutes, stops for $11$ minutes (on the interval $[57,68]$), and then resumes driving away from Gackle at a constant speed for another $36$ minutes (on the interval $[68,104]$).

2. The slope between these points is $m=\frac{43}{57}\approx0.75$ miles per minute.

3. $AV_{[68,104]}=\frac{43}{36}\approx1.194$ miles per minute.

4. $s'(80)=\frac{43}{36}$ miles per minute, the speed of the car after $80$ minutes of driving.

Solution
1. For the first $57$ minutes, the car travels away from Gackle at a constant speed. Since it travels a total of $63.8$ miles during this time, its velocity is $\frac{63.8}{57}$ miles per minute, or approximately $67.16$ miles per hour. The car is stopped between minutes $57$ and $68\text{,}$ and then resumes driving away from Gackle, again at a constant speed. Between $68$ and $104$ minutes, the car travels $106.8-63.8=43$ miles, so its (average) velocity during this time is $\frac{43}{104-68}=\frac{43}{36}$ miles per minute, or about $71.67$ miles per hour.

2. The slope of the line between the points $(57,63.8)$ and $(104,106.8)$ is $m=\frac{106.8-63.8}{104-57}=\frac{43}{47}\approx0.915$ miles per minute. This value represents the average speed of the car between $57$ and $104$ minutes of driving. Because the car is stopped for the first $11$ minutes of this time interval, the average speed on this interval is down to approximately $45$ miles per hour.

3. The average rate of change of the car's position on the interval $[68,104]$ is

\begin{equation*} AV_{[68,104]}=\frac{106.8-63.8}{104-68}=\frac{43}{36}\approx1.194 \end{equation*}

miles per minute. On this interval, the car travels at an average speed of $72$ miles per hour.

4. At $t=80$ minutes, we are solidly in the middle of the time interval $[68,104]\text{,}$ during which the car appears to be traveling at a constant velocity of about $72$ miles per hour. A constant velocity (and in general, a linear position function) on an interval means the instantaneous velocity of the car at a point on that interval is the same as the average velocity of the car over that interval. Therefore, the instantaneous rate of change of the car's position at the moment $t=80$ minutes is about $72$ miles per hour, or $1.2$ miles per minute.

### SubsectionUnits 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.

###### Example1.71

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?

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.

### SubsectionTangent Lines

Recall that a line can be written as $y=m(x-x_0)+y_0$ where $m$ is the slope of the line and $(x_0,y_0)$ is a point on the line. Using this information we are in a position to quickly find the equation for the tangent line to a curve at a point. Specifically, we have the following definition.

###### The Tangent Line

Given a differentiable function $f$ and a point $(x_0,y_0)$ the equation for the tangent line to the function $f$ at $(x_0,y_0)$ is given by

\begin{equation*} y=f'(x_0)(x-x_0)+y_0. \end{equation*}

###### Example1.72

We can find the equation of the line tangent to the curve given by graphing $y=f(x)$ where $f(x)=x^2+2$ at $(2,3)$ by first noting that

\begin{equation*} f'(x)=\lim_{h \rightarrow 0} \frac{(x+h)^2+2-x^2-2}{h}=2x\text{.} \end{equation*}

Therefore the equation of the tangent line to $f$ at $(2,3)$ is

\begin{equation*} y=f'(2)(x-2)+3=4(x-2)-3=4x-5\text{.} \end{equation*}

We will look more at tangent lines in future sections but the basic ideas appear here. Specifically, if we know data about a function at a specific point such as it's value at that point and rate of change at that point then we can estimate its value at a future point.

###### Example1.73

A potato is placed in an oven, and the potato's temperature $F$ (in degrees Fahrenheit) at various points in time (measured in minutes after the potato goes in the oven) is taken and recorded.

1. 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?

2. 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.

Hint
1. Remember that the units on $F'$ will be degrees Fahrenheit per minute.

2. Be careful to distinguish between the temperature, $F\text{,}$ and the rate of change of temperature, $F'\text{,}$ in your commentary.

1. After $64$ minutes in the oven, the temperature of the potato is $330.28$ degrees Fahrenheit and is increasing at a rate of $1.341$ degrees per minute. We expect $F(65)\approx 331.625$ degrees, and $F(66)\approx 332.962$ degrees.

2. The temperature of the potato should be increasing throughout the $90$ minute period, but the rate at which it increases should decrease as time passes.

Solution
1. The value $F(64) = 330.28$ is the temperature of the potato in degrees Fahrenheit after $64$ minutes in the oven, while $F'(64) = 1.341$ measures the instantaneous rate of change of the potato's temperature with respect to time at the instant $t = 64$ minutes, and its units are degrees per minute. Because at time $t = 64$ the potato's temperature is increasing at a rate of $1.341$ degrees per minute, we expect that at $t = 65\text{,}$ the temperature will be about $1.341$ degrees greater than at $t = 64\text{,}$ or in other words $F(65) \approx 330.28 + 1.341 = 331.621$ degrees. Similarly, at $t = 66\text{,}$ two minutes have elapsed from $t = 64\text{,}$ so we expect an increase of $2 \cdot 1.341$ degrees: $F(66) \approx 330.28 + 2 \cdot 1.341 = 332.962$ degrees.

2. Throughout the time interval $[0,90]\text{,}$ the temperature $F$ of the potato is increasing. But as time goes on, the rate at which the temperature is rising should decrease. That is, while the values of $F$ continue to get larger as time progresses, the values of $F'$ are getting smaller (while still remaining positive). We thus might say that the temperature of the potato is increasing, but at a decreasing rate.

###### Example1.74

Suppose 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?

Hint
1. The cost of producing 2000 feet of rope is $\ldots$

2. Remember that the units on any derivative are units of output per unit of input.

3. How much do you expect $C$ to increase for each additional foot away from 2000? By how many feet will the input increase to get to 2100?

4. In manufacturing processes, there is often a decrease in cost per unit as the number of units increases.

5. Is it possible for the total cost function to be decreasing at some point?

1. It costs $800 to make 2000 feet of rope. 2. dollars per foot. 3. $C(2100) \approx 835\text{.}$ 4. Either $C'(2000) = C'(3000)$ or $C'(2000) > C'(3000)\text{.}$ 5. Impossible. The total cost function $C(r)$ can never decrease. Solution 1. $C(2000) = 800$ means that it costs$800 to make 2000 feet of rope.

2. The units of $C'(r)$ are dollars per foot.

3. If $C(2000) = 800$ and $C'(2000) = 0.35\text{,}$ then we know that once 2000 feet of rope are produced, the total cost function is increasing at $0.35 per additional foot of rope. So, if we manufacture an additional 100 feet of rope, the additional total cost will be approximately \begin{equation*} 100 \ \text{feet} \cdot 0.35 \ \frac{\text{dollars} }{\text{foot} } = 35 \ \text{dollars}\text{.} \end{equation*} Therefore, we find that $C(2100) \approx C(2000) + 35 = 835\text{,}$ or that the cost to make 2100 feet of rope is about$835.

4. Either $C'(2000) = C'(3000)$ or $C'(2000) > C'(3000)\text{,}$ since we expect the cost per foot of additional rope to either stay constant or to get smaller as the production volume increases. Said differently, the instantaneous rate of change of the total cost function should either be constant or decrease due to economy of scale.

5. It is impossible to have $C'(5000) = -0.1$ and indeed to have any negative derivative value for the total cost function. The total cost function $C(r)$ can never decrease, because it doesn't make sense for the total cost of producing 5001 feet of rope to be less than the total cost of producing 5000 feet of rope.

###### Example1.75

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 in 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).

Hint
1. How can you use a secant line to make an estimate?

2. What is happening when the car is traveling at 80 km/hr?

3. Remember that units on the derivative are units of output per unit of input.

1. $f'(90) \approx 0.0006$ liters per kilometer per kilometer per hour.

2. At 80 kilometers per hour, the car is using fuel at a rate of 0.015 liters per kilometer.

3. When the car is traveling at 90 kilometers per hour, its rate of fuel consumption per kilometer is increasing at a rate of 0.0006 liters per kilometer per kilometer per hour.

Solution
1. It is reasonable to estimate $f'(90)$ using $f(100)$ and $f(80)\text{.}$

\begin{equation*} f'(90) \approx \frac{f(100) - f(80)}{100-80} = \frac{0.027 - 0.015}{20} = \frac{0.012}{20} = 0.0006 \end{equation*}

which tells us that $f'(90) \approx 0.0006$ liters per kilometer per kilometer per hour.

2. When the car is traveling at 80 kilometers per hour, it is using fuel at a rate of 0.015 liters per kilometer. That is, at the given speed, for each additional kilometer the car travels, it uses an additional 0.015 liters of fuel.

3. To say that $f'(90) = 0.0006$ liters per kilometer per kilometer per hour means that when the car is traveling at 90 kilometers per hour, its rate of fuel consumption per kilometer is increasing at a rate of 0.0006 liters per kilometer per kilometer per hour. If we increase our speed from 90 to 91 km/hr, we would expect our rate of fuel consumption to rise by 0.0006 liters for each additional kilometer driven.

### SubsectionGraphing the Derivative

In Section1.5, 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 Figure1.76 below, we sketch the graph of $P$ on the left and the graph of $P'$ on the right.

Notice that both the size and the units of the vertical scales differ, as the units of $P$ are $^{\circ}$F, while those of $P'$ are $^{\circ}$F/min.

### SubsectionSummary

• 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 $f(x)$ per unit of $x\text{.}$ Again, this measures how fast the output of the function $f$ changes when the input of the function changes.

### SubsectionExercises

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 secant line on the interval $[9.99,10.01]$ 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?

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).

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 $v'(t)\text{?}$

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.