Chapter 11
Mechanics 1: Linear Kinematics and Calculus
Never mistake motion for action.
— Ernest Hemingway (1899–1961
“Ladies and gentlemen, may I direct your attention to the center ring. Witness before you two
ordinary textbooks, one labeled College Physics and the other Calculus. Their
combined 2,500+ pages weigh over 25 lbs. Yet in this chapter and the next, your brave
stuntauthors will attempt a most deathdefying and impossible spectacle of mysticism and
subterfuge: to reduce these two massive books into a mere 150 pages!”
Just like any good circus act, this one is prefaced with a lot of build up to set your
expectations. The difference here is that the purpose of our preface is to lower your
expectations.
11.1Overview and Other ExpectationReducing
Remarks
OK, there's no way we can really cover all of physics and calculus in two chapters. As any
politician knows, the secret to effectively communicate complicated subject matter in a short
amount of time is to use lies, both the omission and commission kind. Let's talk about each of
these kinds of lies in turn, so you will know what's really in store.
11.1.1What is Left Out?
Just about everything—let's talk about what we are leaving out of physics first. To put the
word “physics” on this chapter would be even more of an insult to people who do real physics
than this chapter already is. We are concerned only with
mechanics, and very simple mechanics of rigid bodies at that. Some topics traditionally
found in a firstyear physics textbook that are not discussed in this book include:

energy and work

temperature, heat transfer, thermodynamics, entropy

electricity, magnetism, light

gases, fluids, pressure

oscillation and waves.
A note about energy and work is in order, because even in the limited context of mechanics, the
fundamental concept of energy plays a central role in traditional presentations. Many problems are easier
to solve by using conservation of energy than by considering the forces and applying Newton's
laws. (In fact, an alternative to the Newtonian dynamics that we study in this book exists. It
is known as
Lagrangian dynamics and focuses on energy rather than forces. When used properly, both
systems produce the same results, but Lagrangian dynamics can solve certain problems more
elegantly and is especially adept at handling friction, compared to
Newtonian dynamics.) However, at the time of this writing, basic general purpose digital
simulations are based on Newtonian dynamics, and energy does not play a direct role. That isn't
to say an understanding of energy is useless; indeed disobedience of the conservation of energy
law is at the heart of many simulation problems! Thus, energy often arises more as a way to
understand the (mis)behavior of a digital simulation, even if it doesn't appear in the simulation
code directly.
Now let's talk about the ways in which this book will irritate calculus professors. We think
that a basic understanding of calculus is really important to fully grasp many of the concepts
from physics. Conversely, physics provides some of the best examples for explaining calculus.
Calculus and physics are often taught separately, usually with calculus coming first. It is our
opinion that this makes calculus harder to learn, since it robs the student of the most intuitive
examples—the physics problems for which calculus was invented to solve! We hope interleaving
calculus with physics will make it easier for you to learn calculus.
Our calculus needs are extremely modest in this book, and we have left out even more from
calculus than we did from physics. After reading this chapter, you should know:

The basic idea of what a derivative measures and what it is used for.

The basic idea of what an integral measures and what it is used for.

Derivatives and integrals of trivial expressions containing polynomials and trig
functions.
Of course, we are aware that a large number of readers may already have this knowledge. Take a
moment to put yourself into one of the following categories:

I know absolutely nothing about derivatives or integrals.

I know the basic idea of derivatives and integrals, but probably couldn't solve any
freshman calculus problems with a pencil and paper.

I have studied some calculus.
Level 2 knowledge of calculus is sufficient for this book, and our goal is to move everybody who
is currently in category 1 into category 2. If you're in category 3, our calculus discussions
will be a (hopefully entertaining) review.We have no delusions that we can move
anyone who is not already there into category 3.
11.1.2Some Helpful Lies about Our Universe
The universe is commonly thought to be discrete in both space and time. Not only is matter broken
up into discrete chunks called atoms, but there is evidence that the very fabric of space and
time is broken up into discrete pieces also. Now, there is a difference of opinion as to whether
it's really that way or just appears that way because the only way we can interact with space is
to throw particles at it, but it's our opinion that if it looks like a
duck, walks like a duck, quacks like a duck, has webbed feet and a beak, then it's a good working
hypothesis that it tastes good when put into eggrolls with a little dark sauce.
For a long time, the mere thought that the universe might not be continuous had not even
considered the slightest possibility of crossing anybody's mind, until the ancient Greeks got a
harebrained and totally unjustified idea that things might be made up of atoms. The fact that
this later turned out to be true is regarded by many as being good luck rather then good
judgment. Honestly, who would have thought it? After all, everyday objects, such as the desk on
which one of the authors is currently resting his wrists as he types this sentence, give every
appearance of having smooth, continuous surfaces. But who cares? Thinking of the desk as having a
smooth, continuous surface is a harmless but useful delusion that lets the author rest his wrists
comfortably without worrying about atomic bond energy and quantum uncertainty theory at all.
Not only is this trick of thinking of the world as continuous a handy psychological
rationalization, it's also good mathematics. It turns out that the math of continuous things is a
lot less unwieldy than the math of discrete things. That's why the people who were thinking about
how the world works in the 15th century were happy to invent a mathematics for a continuous
universe; experimentally, it was a good approximation to reality, and theoretically the math
worked out nicely.
Sir Isaac Newton was thus able to discover a lot of fundamental results about continuous
mathematics, which we call “calculus,” and its application to the exploration of a continuous
universe, which we call “physics.”
Now, we're mostly doing this so that we can model a game world inside a computer, which is
inherently discrete. There's a certain amount of cognitive dissonance involved with programming a
discrete simulation of a continuous model of a discrete universe, but we'll try not to let it
bother us. Suffice it to say that we are in complete control of the discrete universe inside our
game, and that means that we can choose the kind of physics that applies inside that universe.
All we really need is for the physical laws to be sufficiently like the ones we're used to for
the player to experience willing suspension of disbelief, and hopefully say, “Wow! Cool!” and
want to spend more money. For almost all games that means a cozy Newtonian universe without the
nasty details of quantum mechanics or relativity. Unfortunately, that means also that there are a
pair of nasty trolls lurking under the bridge, going by the names of chaos and instability, but
we will do our best to appease them.
For the moment, we are concerned about the motion of a small object called a “particle.” At any
given moment, we know its position and velocity. The particle has mass. We do not concern ourselves with the
orientation of the particle (for now), and thus we don't think of the particle as spinning. The
particle does not have any size, either. We will defer adding those elements until later, when we
shift from particles to rigid bodies.
We are studying classical mechanics, also known as Newtonian mechanics, which has several
simplifying assumptions that are incorrect in general but true in everyday life in most ways that
really matter to us. So we can darn well make sure they are true inside our computer world, if
we please. These assumptions are:

Time is absolute.

Space is Euclidian.

Precise measurements are possible.

The universe exhibits causality and complete predictability.
The first two are shattered by relatively, and the second two by quantum mechanics.
Thankfully, these two subjects are not necessary for video games, because your authors do not
have more than a pedestrian understanding of them.
We will begin our foray into the field of mechanics by learning about kinematics, which is
the study of the equations that describe the motion of a particle in various simple but
commonplace situations. When studying kinematics, we are not concerned with the causes of
motion—that is the subject of
dynamics, which will be covered in Chapter 12. For now, “ours is not to
question why,” ours is just to do the math to get equations that predict the position, velocity,
and acceleration of the particle at any given time
$t$
, or die. Well, forget about the last part
anyway.
Because we are treating our objects as particles and tracking their position only, we will not
consider their orientation or rotational effects until Chapter 12. When rotation is
ignored, all of the ideas of linear kinematics extend into 3D in a straightforward way, and so
for now we will be limiting ourselves to 2D (and 1D). This is convenient, since the authors
do not know how to design those little origamilike things that lay flat and then pop up when you
open the book, and the publisher wouldn't let us even if we were compulsive enough to learn how
to do it. Later we'll see why treating objects as particles is perfectly justifiable.
11.2Basic Quantities and Units
Mechanics is concerned with the relationship among three fundamental quantities in nature:
length, time, and mass. Length is a quantity you are no doubt familiar
with; we measure length using units such as centimeters, inches, meters, feet, kilometers, miles,
and
astronomical units. Time is another quantity we are very comfortable with
measuring, in fact most of us probably learned how to read a clock before we learned how to
measure distances. The units used to measure time are the familiar
second, minute, day, week,
fortnight, and so on. The month and the year are often not good units to use for
time because different months and years have different durations.
The quantity mass is not quite as intuitive as length and time. The measurement of an
object's mass is often thought of as measuring the “amount of stuff” in the object. This is
not a bad (or at least, not completely terrible) definition, but its not quite right, either
[1]. A more precise definition might be that mass is a measurement of
inertia, that is, how much resistance an object has to being accelerated. The more
massive an object is, the more force is required to start it in motion, stop its motion, or
change its motion.
Mass is often confused with weight, especially since the units used to measure mass are
also used to measure weight: the gram, pound, kilogram, ton, and so forth. The mass of an object
is an intrinsic property of an object, whereas the weight is a local phenomenon that depends on
the strength of the gravitational pull exerted by a nearby massive object. Your mass will be the
same whether you are in Chicago, on the moon, near Jupiter, or lightyears away from the nearest
heavenly body, but in each case your weight will be very different. In this book and in most
video games, our concerns are confined to a relatively small patch on a flat Earth, and
we approximate
gravity by a constant downward pull. It won't be too harmful to confuse mass and weight because
gravity for us will be a constant. (But we couldn't resist a few cool exercises about the
International Space Station.)
In many situations, we can discuss the relationship between the fundamental quantities without
concern for the units of measurement we are using. In such situations, we'll find it useful to
denote length, time, and mass by
$L$
,
$T$
, and
$M$
, respectively. One important such case is in
defining
derived quantities. We've said that length, time, and mass are the fundamental
quantities—but what about other quantities, such as area, volume, density, speed, frequency,
force, pressure, energy, power, or any of the numerous quantities that can be measured in
physics? We don't give any of these their own capital letter, since each of these can be defined
in terms of the fundamental quantities.
For example, we might express a measurement of area as a number of “square feet.” We have
created a unit that is in terms of another unit. In physics, we say that a measurement of area
has the unit “length squared,” or
${L}^{2}$
. How about speed? We measure speed using the units such
as miles per hour or meters per second. Thus speed is the ratio of a distance per unit time, or
$L/T$
.
One last example is frequency. You probably know that frequency measures how many times
something happens in a given time interval (how “frequently” it happens). For example, a
healthy adult has an average heart rate of around 70 beats per minute (BPM). The motor in a car
might be rotating at a rate of 5,000 revolutions per minute (RPM). The
NTSC television standard is defined as 29.97 frames per second (FPS). Note that in each of
these, we are counting how many times something happens within a given duration of time. So we
can write frequency in generic units as
$1/T$
or
${T}^{1}$
, which you can read as “per unit
time.” One of the most important measurements of frequency is the
Hertz, abbreviated Hz, which means “per second.” When you express a frequency in Hz, you
are describing the number of events, oscillations, heartbeats, video frames, or whatever
per second. By definition,
$1\text{}\mathrm{H}\mathrm{z}=1\text{}{\mathrm{s}}^{1}$
.
Table 11.1 summarizes several quantities that are measured in physics, their relation
to the fundamental quantities, and some common units used to measure them.
Quantity 
Notation 
SI unit 
Other units 
Length 
$L$

$\mathrm{m}$

$\mathrm{c}\mathrm{m}$
,
$\mathrm{k}\mathrm{m}$
,
$\mathrm{i}\mathrm{n}$
,
$\mathrm{f}\mathrm{t}$
,
$\mathrm{m}\mathrm{i}$
, light year, furlong

Time 
$T$

$\mathrm{s}$

$\mathrm{m}\mathrm{i}\mathrm{n}$
,
$\mathrm{h}\mathrm{r}$
,
$\mathrm{m}\mathrm{s}$

Mass 
$M$

$\mathrm{k}\mathrm{g}$

$\mathrm{g}$
, slug,
$\mathrm{l}\mathrm{b}$
(poundmass) 
Velocity 
$L/T$

$\mathrm{m}/\mathrm{s}$

$\mathrm{f}\mathrm{t}/\mathrm{s}$
,
$\mathrm{m}/\mathrm{h}\mathrm{r}$
,
$\mathrm{k}\mathrm{m}/\mathrm{h}\mathrm{r}$

Acceleration 
$L/{T}^{2}$

${\mathrm{m}/\mathrm{s}}^{2}$

${\mathrm{f}\mathrm{t}/\mathrm{s}}^{2}$
,
$(\mathrm{m}/\mathrm{h}\mathrm{r})/\mathrm{s}$
,
$(\mathrm{k}\mathrm{m}/\mathrm{h}\mathrm{r})/\mathrm{s}$

Force 
$ML/{T}^{2}$

$\mathrm{N}$
(Newton) =
$\mathrm{k}\mathrm{g}\cdot \mathrm{m}/{\mathrm{s}}^{2}$

$\mathrm{l}\mathrm{b}$
(poundforce), poundal 
Area 
${L}^{2}$

${\mathrm{m}}^{2}$

$\mathrm{m}{\mathrm{m}}^{2}$
,
$\mathrm{c}{\mathrm{m}}^{2}$
,
$\mathrm{k}{\mathrm{m}}^{2}$
,
$\mathrm{i}{\mathrm{n}}^{2}$
,
$\mathrm{f}{\mathrm{t}}^{2}$
,
$\mathrm{m}{\mathrm{i}}^{2}$
,acre, hectare

Volume 
${L}^{3}$

${\mathrm{m}}^{3}$

$\mathrm{m}{\mathrm{m}}^{3}$
,
$\mathrm{c}{\mathrm{m}}^{3}$
, L (liter),
$\mathrm{i}{\mathrm{n}}^{3}$
,
$\mathrm{f}{\mathrm{t}}^{3}$
,
teaspoon,fl oz (fluid ounce), cup, pint, quart, gallon

Pressure 
Force/Area
=
$(ML/{T}^{2})/{L}^{2}$
=
$M/(L{T}^{2})$

$\mathrm{P}\mathrm{a}$
(Pascal)
=
$\mathrm{N}/{\mathrm{m}}^{2}$
=
$\mathrm{k}\mathrm{g}/(\mathrm{m}\cdot {\mathrm{s}}^{2})$

$\mathrm{p}\mathrm{s}\mathrm{i}$
(
$\mathrm{l}\mathrm{b}\mathrm{s}/\mathrm{i}{\mathrm{n}}^{2}$
), millibar, inch of mercury,
$\mathrm{a}\mathrm{t}\mathrm{m}$
(atmosphere)

Energy 
Force
$\times $
Length
=
$(ML/{T}^{2})\cdot L$
=
$M{L}^{2}/{T}^{2}$

$\mathrm{J}$
(Joule)
=
$\mathrm{N}\cdot {\mathrm{m}}^{2}$
=
$\frac{\mathrm{k}\mathrm{g}\cdot \mathrm{m}}{{\mathrm{s}}^{2}}\cdot \mathrm{m}$
=
$\frac{\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^{2}}{{\mathrm{s}}^{2}}$

$\mathrm{k}\mathrm{W}\cdot \mathrm{h}\mathrm{r}$
(kilowatthour), footpound, erg,
calorie,
$\mathrm{B}\mathrm{T}\mathrm{U}$
(British thermal unit),
ton of TNT

Power 
Energy / Time
=
$(M{L}^{2}/{T}^{2})/T$
=
$M{L}^{2}/{T}^{3}$

$\mathrm{W}$
(Watt)
=
$\mathrm{J}/\mathrm{s}$
=
$\frac{\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^{2}}{{\mathrm{s}}^{2}}\cdot {\mathrm{s}}^{1}$
=
$\frac{\mathrm{k}\mathrm{g}\cdot {\mathrm{m}}^{2}}{{\mathrm{s}}^{3}}$

$\mathrm{h}\mathrm{p}$
(horsepower) 
Frequency 
$1/T={T}^{1}$

$\mathrm{H}\mathrm{z}=1/\mathrm{s}={\mathrm{s}}^{1}$
= “per second” 
$\mathrm{K}\mathrm{H}\mathrm{z}=1,000\text{}\mathrm{H}\mathrm{z}$
,
$\mathrm{M}\mathrm{H}\mathrm{z}=1,000,000\text{}\mathrm{H}\mathrm{z}$
, “per minute”, “per annum” 
Table 11.1Selected physical quantities and common units of measurements
Of course, any real measurement doesn't make sense without attaching specific units to it. One
way to make sure that your calculations always make sense is to carry around the units at all
times and treat them like algebraic variables. For example, if you are computing a pressure and
your answer comes out with the units m/s, you know you have done something wrong; pressure has
units of force per unit area, or
$ML/({T}^{2}{L}^{2})$
. On the other hand, if you are solving a problem
and you end up with an answer in pounds per square inch (psi), but you are looking for a value in
Pascals, your answer is probably correct, but just needs to be converted to the desired units.
This sort of reasoning is known as
dimensional analysis. Carrying around the units and treating them as algebraic variables
quite often highlights mistakes caused by different units of measurement, and also helps make
unit conversion a snap.
Because unit conversion is an important skill, let's briefly review it here. The basic concept is
that to convert a measurement from one set of units to another, we multiply that measurement by a
wellchosen fraction that has a value of 1. Let's take a simple example: how many feet is
14.57 meters? Looking up the conversion factor, we see that
$1\text{}\mathrm{m}\approx 3.28083\text{}\mathrm{f}\mathrm{t}$
. This means
that
$1\text{}\mathrm{m}/3.28083\text{}\mathrm{f}\mathrm{t}\approx 1$
. So let's take our measurement and multiply
it by a special value of “1:”
$$\begin{array}{}\text{(11.1)}& 14.57\text{}\mathrm{m}=14.57\text{}\mathrm{m}\times 1\approx 14.57\text{}\mathrm{m}\times {\displaystyle \frac{3.28083\text{}\mathrm{f}\mathrm{t}}{1\text{}\mathrm{m}}}\approx 47.80\text{}\mathrm{f}\mathrm{t}.\end{array}$$
Our conversion factor tells us that the numerator and denominator of the fraction in
Equation (11.1) are equal: 3.28083 feet is equal to 1 meter. Because
the numerator and denominator are equal, the “value” of this fraction is 1. (In a physical
sense, though, certainly numerically the fraction doesn't equal 1.) And we know that multiplying
anything by 1 does not change its value. Because we are treating the units as algebraic
variables, the m on the left cancels with the m in the bottom of the fraction.
Of course, applying one simple conversion factor isn't too difficult, but consider a more
complicated example. Let's convert 188 km/hr to ft/s. This time we need to multiply by “1”
several times:
$$188{\displaystyle \frac{\mathrm{k}\mathrm{m}}{\mathrm{h}\mathrm{r}}}\times {\displaystyle \frac{1\text{}\mathrm{h}\mathrm{r}}{3600\text{}\mathrm{s}}}\times {\displaystyle \frac{1000\text{}\mathrm{m}}{1\text{}\mathrm{k}\mathrm{m}}}\times {\displaystyle \frac{3.28083\text{}\mathrm{f}\mathrm{t}}{1\text{}\mathrm{m}}}\approx 171{\displaystyle \frac{\mathrm{f}\mathrm{t}}{\mathrm{s}}}.$$
11.3Average Velocity
We begin our study of kinematics by taking a closer look at the simple concept of speed. How do
we measure speed? The most common method is to measure how much time it takes to travel a fixed
distance. For example, in a race, we say that the fastest runner is the one who finishes the
race in the shortest amount of time.
Consider the fable of the tortoise and the hare. In the story, they decide to have a race, and
the hare, after jumping to an early lead, becomes overconfident and distracted. He stops during
the race to take a nap, smell the flowers, or some other form of lollygagging. Meanwhile, the
tortoise plods along, eventually passing the hare and crossing the finish line first. Now this
is a math book and not a selfhelp book, so please ignore the moral lessons about focus and
perseverance that the story contains, and instead consider what it has to teach us about average
velocity. Examine Figure 11.1, which shows a plot of the position of each animal
over time.
A playbyplay of the race is as follows. The gun goes off at time
${t}_{0}$
, and the hare sprints
ahead to time
${t}_{1}$
. At this point his hubris causes him to slow his pace, until time
${t}_{2}$
when
a cute female passes by in the opposite direction. (Her position over time is not depicted in
the diagram.) At this point a different
tragic male trait causes the hare to turn around and walk with her, and he proceeds to chat her
up. At
${t}_{3}$
, he realizes that his advances are getting him nowhere, and he begins to pace back
and forth along the track dejectedly until time
${t}_{4}$
. At that point, he decides to take a nap.
Meanwhile, the tortoise has been making slow and steady progress, and at time
${t}_{5}$
, he catches
up with the sleeping hare. The tortoise plods along and crosses the tape at
${t}_{6}$
. Quickly
thereafter, the hare, perhaps awakened by the sound of the crowd celebrating the tortoise's
victory, wakes up at time
${t}_{7}$
and hurries in a frenzy to the finish. At
${t}_{8}$
, the hare crosses
the finish line, where he is humiliated by all his peers, and the cute girl bunny, too.
To measure the average velocity of either animal during any time interval, we divide the
animal's displacement by the duration of the interval. We'll be focusing on the hare, and we'll
denote the position of the hare as
$x$
, or more explicitly as
$x(t)$
, to emphasize the fact that
the hare's position varies as a function of time. It is a common convention to use the capital
Greek letter delta (“
$\mathrm{\Delta}$
”) as a prefix to mean “amount of change in.” For example,
$\mathrm{\Delta}x$
would mean “the change in the hare's position,” which is a displacement of the hare.
Likewise
$\mathrm{\Delta}t$
means “the change in the current time,” or simply, “elapsed time between
two points.” Using this notation, the average velocity of the hare from
${t}_{a}$
to
${t}_{b}$
is given
by the equation
Definition of average velocity
$$\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}\text{}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}={\displaystyle \frac{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{p}\mathrm{s}\mathrm{e}\mathrm{d}\text{}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}}}={\displaystyle \frac{\mathrm{\Delta}x}{\mathrm{\Delta}t}}={\displaystyle \frac{x({t}_{b})x({t}_{a})}{{t}_{b}{t}_{a}}}.$$
This is the definition of average velocity. No matter what specific units we use, velocity
always describes the ratio of a length divided by a time, or to use the notation discussed in
Section 11.2, velocity is a quantity with units
$L/T$
.
If we draw a straight line through any two points on the graph of the hare's position, then the
slope of that line measures the average velocity of the hare over the interval between the two
points. For example, consider the average velocity of the hare as he decelerates from time
${t}_{1}$
to
${t}_{2}$
, as shown in Figure 11.2. The slope of the line is
the ratio
$\mathrm{\Delta}x/\mathrm{\Delta}t$
. This slope is also equal to the tangent of the angle marked
$\alpha $
, although for now the values
$\mathrm{\Delta}x$
and
$\mathrm{\Delta}t$
are the ones we will have at our
fingertips, so we won't need to do any trig.
Returning to Figure 11.1, notice that the hare's average velocity from
${t}_{2}$
to
${t}_{3}$
is negative. This is because velocity is defined as the ratio of
net displacement over time. Compare this to speed, which is the total distance
divided by time and cannot be negative. The sign of displacement and velocity are sensitive to
the direction of travel, whereas distance and speed are intrinsically nonnegative. We've already
spoken about these distinctions way back in Section 2.2. Of
course it's obvious that the average velocity is negative between
${t}_{2}$
and
${t}_{3}$
, since the hare
was going backwards during the entire interval. But average velocity can also be negative on an
interval even in situations where forward progress is being made for a portion of the interval,
such as the larger interval between
${t}_{2}$
and
${t}_{4}$
. It's a case of “one step forward, two steps
back.”
Average velocity can also be zero, as illustrated during the hare's nap from
${t}_{4}$
to
${t}_{7}$
. In fact, the average velocity will be zero any time an
object starts and ends at the same location, even if it was it motion during
the entire interval! (“Two steps forward, two steps back.”) Two such
intervals are illustrated in
Figure 11.3.
And, of course, the final lesson of the fable is that the average velocity of the tortoise is
greater than the average velocity of the hare, at least from
${t}_{0}$
to
${t}_{7}$
, when the
tortoise crosses the finish line. This is true despite the fact that the hare's average
speed was higher, since he certainly traveled a larger distance with all the female
distractions and pacing back and forth.
One last thing to point out. If we assume the hare learned his lesson and congratulated the
tortoise (after all, let's not attribute to the poor animal all the negative personality
traits!), then at
$t={t}_{8}$
they were standing at the same place. This means their net
displacements from
${t}_{0}$
to
${t}_{8}$
are the same, and thus they have
the same average velocity during this interval.
11.4Instantaneous Velocity and the Derivative
We've seen how physics defines and measures the average velocity of an object over an interval,
that is, between two time values that differ by some finite amount
$\mathrm{\Delta}t$
. Often, however,
it's useful to be able to speak of an object's instantaneous velocity, which means the
velocity of the object for one value of
$t$
, a single moment in time. You can see that this is
not a trivial question because the familiar methods for measuring velocity, such as
$$\mathrm{a}\mathrm{v}\mathrm{e}\mathrm{r}\mathrm{a}\mathrm{g}\mathrm{e}\text{}\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}={\displaystyle \frac{\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}}{\mathrm{e}\mathrm{l}\mathrm{a}\mathrm{p}\mathrm{s}\mathrm{e}\mathrm{d}\text{}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}}}={\displaystyle \frac{\mathrm{\Delta}x}{\mathrm{\Delta}t}}={\displaystyle \frac{x({t}_{b})x({t}_{a})}{{t}_{b}{t}_{a}}},$$
don't work when we are considering only a single instant in time. What are
${t}_{a}$
and
${t}_{b}$
, when
we are looking at only one time value? In a single instant, displacement and elapsed time are
both zero; so what is the meaning of the ratio
$\mathrm{\Delta}x/\mathrm{\Delta}t$
? This section introduces a
fundamental tool of calculus known as the
derivative. The derivative was invented by Newton to investigate precisely the kinematics
questions we are asking in this chapter. However, its applicability extends to virtually every
problem where one quantity varies as a function of some other quantity. (In the case of velocity,
we are interested in how position varies as a function of time.)
Because of the vast array of problems to which the derivative can be applied, Newton was not the
only one to investigate it. Primitive applications of integral calculus to compute volumes and
such date back to ancient Egypt. As early as the 5th century, the Greeks were exploring the
building blocks of calculus such as infinitesimals and the method of exhaustion. Newton usually
shares credit with the German mathematician
Gottfried Leibniz (1646–1716) for inventing calculus in the 17th century,
although Persian and Indian writings contain examples of calculus concepts being used. Many
other thinkers made significant contributions, includingFermat, Pascal, and Descartes. It's somewhat
interesting that many of the earlier applications of calculus were integrals, even though most calculus
courses cover the “easier” derivative before the “harder” integral.
We first follow in the steps of Newton and start with the physical example of velocity, which we
feel is the best example for obtaining intuition about how the derivative works. Afterwards, we
consider several other examples where the derivative can be used, moving from the physical to the
more abstract.
11.4.1Limit Arguments and the Definition of the Derivative
Back to the question at hand: how do we measure instantaneous velocity? First, let's observe one
particular situation for which it's easy: if an object moves with constant velocity over an
interval, then the velocity is the same at every instant in the interval. That's the very
definition of constant velocity. In this case, the average velocity over the interval must be
the same as the instantaneous velocity for any point within that interval. In a graph such as
Figure 11.1, it's easy to tell when the object is moving at constant velocity
because the graph is a straight line. In fact, almost all of Figure 11.1 is made up
of straight line segments, so determining instantaneous velocity is as easy as
picking any two points on a straightline interval (the endpoints of the interval seem like a
good choice, but any two points will do) and determining the average velocity between those
endpoints.
But consider the interval from
${t}_{1}$
to
${t}_{2}$
, during which the hare's
overconfidence causes him to gradually decelerate. On this interval, the
graph of the hare's position is a curve, which means the slope of the line,
and thus the velocity of the hare, is changing continuously. In this
situation, measuring instantaneous velocity requires a bit more finesse.
For concreteness in this example, let's assign some particular numbers. To keep those numbers
round (and also to stick with the racing theme), please allow the whimsical choice to measure
time in minutes and distance in
furlongs. We will assign
${t}_{1}=1\text{}\mathrm{m}\mathrm{i}\mathrm{n}$
and
${t}_{2}=3\text{}\mathrm{m}\mathrm{i}\mathrm{n}$
, so the total duration is 2 minutes. Let's say that
during this interval, the hare travels
from
$x(1)=4\text{}\mathrm{f}\mathrm{u}\mathrm{r}$
to
$x(3)=8\text{}\mathrm{f}\mathrm{u}\mathrm{r}$
. For purposes of illustration, we will set our sights on the answer
to the question: what is the hare's instantaneous velocity at
$t=2.5\text{}\mathrm{m}\mathrm{i}\mathrm{n}$
? This is all
depicted in Figure 11.4.
It's not immediately apparent how we might measure or calculate the velocity at the exact
moment
$t=2.5$
, but observe that we can get a good approximation by computing the average
velocity of a very small interval near
$t=2.5$
. For a small enough interval, the graph is nearly
the same as a straight line segment, and the velocity is nearly constant, and so the
instantaneous velocity at any given instant within the interval will not be too far off from the
average velocity over the whole interval.
In Figure 11.5, we fix the left endpoint of a line segment at
$t=2.5$
and move the right endpoint closer and closer. As you can see, the shorter the interval, the
more the graph looks like a straight line, and the better our approximation becomes. Thinking
graphically, as the second endpoint moves closer and closer to
$t=2.5$
, the slope of the line
between the endpoints will converge to the slope of the line that is tangent to the curve
at this point. A tangent line is the graphical equivalent of instantaneous velocity, since it
measures the slope of the curve just at that one point.
Let's carry out this experiment with some real numbers and see if we cannot
approximate the instantaneous velocity of the hare. In order to do this,
we'll need to be able to know the position of the hare at any given time, so
now would be a good time to tell you that the position of the hare is given
by the function
$$x(t)={t}^{2}+6t1.$$
Table 11.2 shows tabulated calculations for average
velocity over intervals with a right hand endpoint
$t+\mathrm{\Delta}t$
that moves closer and closer to
$t=2.5$
.
The rightmost column, which is the average velocity, appears to be converging to a velocity of
1 furlong/minute. But how certain are we that this is the correct value? Although we do not
have any calculation that will produce a resulting velocity of exactly 1 furlong/minute, for all
practical purposes, we may achieve any degree of accuracy desired by using this approximation
technique and choosing
$\mathrm{\Delta}t$
sufficiently small. (We are ignoring issues related to the
precision of floating point representation of numbers in a computer.)
$t$

$\mathrm{\Delta}t$

$t+\mathrm{\Delta}t$

$x(t)$

$x(t+\mathrm{\Delta}t)$

$x(t+\mathrm{\Delta}t)x(t)$

$\frac{x(t+\mathrm{\Delta}t)x(t)}{\mathrm{\Delta}t}$

[6pt]
2.500

0.500 
3.000 
7.750 
8.0000 
0.2500 
0.5000 
2.500 
0.100 
2.600 
7.750 
7.8400 
0.0900 
0.9000 
2.500 
0.050 
2.550 
7.750 
7.7975 
0.0475 
0.9500 
2.500 
0.010 
2.510 
7.750 
7.7599 
0.0099 
0.9900 
2.500 
0.005 
2.505 
7.750 
7.7549 
0.0049 
0.9950 
2.500 
0.001 
2.501 
7.750 
7.7509 
0.0009 
0.9990 
Table 11.2Calculating average velocity for intervals of varying durations
This is a powerful argument. We have essentially assigned a value to an expression that we
cannot evaluate directly. Although it is mathematically illegal to substitute
$\mathrm{\Delta}t=0$
into the expression, we can argue that for smaller and smaller values of
$\mathrm{\Delta}t$
, we converge
to a particular value. In the parlance of calculus, this value of 1 furlong/minute is a
limiting value, meaning that as we take smaller and smaller positive values for
$\mathrm{\Delta}t$
, the result of our computation approaches 1, but does not cross it (or reach it exactly).
Convergence arguments such as this are defined with rigor in calculus by using a formalized tool
known as a limit. The mathematical notation for this is
$$\begin{array}{}\text{(11.2)}& v(t)=\underset{\mathrm{\Delta}t\to 0}{lim}{\displaystyle \frac{x(t+\mathrm{\Delta}t)x(t)}{\mathrm{\Delta}t}}.\end{array}$$
The notation `
$\to $
' is usually read as “approaches” or “goes to.” So
the right side of Equation (11.2) might be read
as
$$\u2018\u2018\mathrm{T}\mathrm{h}\mathrm{e}\text{}\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}\text{}\mathrm{o}\mathrm{f}\text{}\frac{x(t+\mathrm{\Delta}t)x(t)}{\mathrm{\Delta}t}\text{}\mathrm{a}\mathrm{s}\text{}\mathrm{\Delta}t\text{}\mathrm{a}\mathrm{p}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{a}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}\text{}\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o},{\textstyle \text{''}}$$
or
$$\u2018\u2018\mathrm{T}\mathrm{h}\mathrm{e}\text{}\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{i}\mathrm{t}\text{}\mathrm{a}\mathrm{s}\text{}\mathrm{\Delta}t\text{}\mathrm{a}\mathrm{p}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{a}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}\text{}\mathrm{z}\mathrm{e}\mathrm{r}\mathrm{o}\text{}\mathrm{o}\mathrm{f}\text{}\frac{x(t+\mathrm{\Delta}t)x(t)}{\mathrm{\Delta}t}.{\textstyle \text{''}}$$
In general, an expression of the form
$\underset{a\to k}{lim}[\mathrm{b}\mathrm{l}\mathrm{a}\mathrm{h}]$
is
interpreted to mean “The value that [blah] converges to, as
$a$
gets closer
and closer to
$k$
.”
This is an important idea, as it defines what we mean by instantaneous
velocity.
Instantaneous velocity at a given time
$t$
may be interpreted as the
average velocity of an interval that contains
$t$
, in the limit as the
duration of the interval approaches zero.
We won't have much need to explore the full power of limits or get bogged down in the finer
points; that is the mathematical field of
analysis, and would take us a bit astray from our current, rather limited, objectives.
We are glossing over some important details
so that we can
focus on one particular case, and that is the use of limits to define the
derivative.
The derivative measures the rate of change of a function. Remember that
“function” is just a fancy word for any formula, computation, or procedure that takes an input
and produces an output. The derivative quantifies the rate at which the output of the function
will change in response to a change to the input. If
$x$
denotes the value of a function at a
specific time
$t$
, the derivative of that function at
$t$
is the ratio
$dx/dt$
. The symbol
$dx$
represents the change in the output produced by a very small change in the input, represented by
$dt$
. We'll speak more about these “small changes” in more detail in just a moment.
For now, we are in an imaginary racetrack where rabbits and turtles race and moral lessons are
taught through metaphor. We have a function with an input of
$t$
, the number of minutes elapsed
since the start of the race, and an output of
$x$
, the distance of the hare along the racetrack.
The rule we use to evaluate our function is the expression
$x(t)={t}^{2}+6t1$
. The derivative
of this function tells us the rate of change of the hare's position with respect to time and is
the definition of instantaneous velocity. Just previously, we defined instantaneous
velocity as the average velocity taken over smaller and smaller intervals, but this is
essentially the same as the definition of the derivative. We just phrased it the first time
using terminology specific to position and velocity.
When we calculate a derivative, we won't end up with a single number. Expecting the answer to
“What is the velocity of the hare?” to be a single number makes sense only if the velocity is
the same everywhere. In such a trivial case we don't need derivatives, we can just use average
velocity. The interesting situation occurs when the velocity varies over time. When we calculate
the derivative of a position function in such cases, we get a velocity function, which
allows us to calculate the instantaneous velocity at any point in time.
The previous three paragraphs express the most important concepts in this section, so please
allow us to repeat them.
A derivative measures a rate of change. Since velocity is the rate of
change of position with respect to time, the derivative of the position
function is the velocity function.
The next few sections discuss the mathematics of derivatives in a bit more detail, and we return
to kinematics in Section 11.5. This material is aimed at those who have not
had firstyear calculus. If you already have a calculus background, you can safely
skip ahead to Section 11.5 unless you feel in need of a refresher.
Section 11.4.2 lists several examples of derivatives to give you a better
understanding of what it means to measure a rate of change, and also to back up our claim that
the derivative has very broad applicability. Section 11.4.3
gives the formal mathematical definition of the derivative and shows how to use this definition to solve problems. We also
finally figure out how fast that hare was moving at
$t=2.5$
. Section 11.4.4
lists various commonly used alternate notations for derivatives, and finally,
Section 11.4.5 lists just enough rules about derivatives to satisfy the very
modest differential calculus demands of this book.
11.4.2Examples of Derivatives
Velocity may be the easiest introduction to the derivative, but it is by no means the only
example. Let's look at some more examples
to give you an idea of the wide array of problems to which the derivative is applied.
The simplest types of examples are to consider other quantities that vary with time. For
example, if
$R(t)$
is the reading of a rain meter at a given time
$t$
, then the derivative,
denoted
${R}^{\prime}(t)$
, describes how hard it was raining at time
$t$
. Perhaps
$P(t)$
is the reading of
a pressure valve on a tank containing some type of gas. Assuming the pressure reading is
proportional to the mass of the gas inside the chamber, the rate of change
${P}^{\prime}(t)$
indicates how fast gas is flowing into or out of the chamber attime
$t$
.
There are also physical examples for which the independent variable is not time. The prototypical
case is a function
$y(x)$
that gives the height of some surface above a reference point at the
horizontal position
$x$
. For example, perhaps
$x$
is the distance along our metaphorical
racetrack and
$y$
measures the height at that point above or below the altitude at the starting
point. The derivative
${y}^{\prime}(x)$
of this function is the slope of the surface at
$x$
, where positive
slopes mean the runners are running uphill, and negative values indicate a downhill portion of
the race. This example is not really a new example, because we've looked at graphs of functions
and considered how the derivative is a measure of the slope of the graph in 2D.
Now let's become a bit more abstract, but still keep a physical dimension as the independent
variable. Let's say that for a popular rockclimbing wall, we know a function
$S(y)$
that
describes, for a given height
$y$
, what percentage of rock climbers are able to reach that height
or higher. If we assume the climbers start at
$y=0$
, then
$S(0)=100\mathrm{\%}$
. Clearly
$S(y)$
is a
nonincreasing function that eventually goes all the way down to 0%at some maximum height
${y}_{\mathrm{m}\mathrm{a}\mathrm{x}}$
that nobody has ever reached.
Now consider the interpretation of derivative
${S}^{\prime}(y)$
. Of course,
${S}^{\prime}(y)\le 0$
, since
$S(y)$
is
nonincreasing. A large negative value of
${S}^{\prime}(y)$
is an indication that the height
$y$
is an area
where climbers are likely to drop out. Perhaps the wall at that height is
a challenging area.
${S}^{\prime}(y)$
closer to zero is an indication that fewer climbers drop out at
height
$y$
. Perhaps there is a plateau that climbers can reach, and there they rest. We might
expect
${S}^{\prime}(y)$
to decrease just after this plateau, since the climbers are more rested. In fact,
${S}^{\prime}(y)$
might also become closer to zero just before the plateau, because as climbers
begin to get close to this milestone, they push a bit harder and are more reluctant to give
up.
One last example. Figure 11.6 shows happiness as a function of salary. In this case, the
derivative is essentially the same thing as what economists would call “marginal utility.” It's
the ratio of additional units of happiness per additional unit of income. According this figure,
the marginal utility of income decreases, which of course is the famous law of diminishing
returns. According to our research, it even becomes negative after a certain point, where the troubles
associated with high income begin to outweigh the psychological benefits. The economistspeak
phrase “negative marginal utility” is translated into everyday language as “stop doing that.”
11.4.3Calculating Derivatives from the Definition
Now we're ready for the official definition of the derivative found in most math textbooks, and
to see how we can compute derivatives using the definition. A derivative can be understood as
the limiting value of
$\mathrm{\Delta}x/\mathrm{\Delta}t$
, the ratio of the change in output divided by the
change in input, taken as we make
$\mathrm{\Delta}t$
infinitesimally small. Let's repeat this
description using mathematical notation. It's an equation we gave earlier in the chapter, only
this time we put a big box around it, because that's what math books do to equations that are
definitions.
The Definition of a Derivative
$$\begin{array}{}\text{(11.3)}& {\displaystyle \frac{dx}{dt}}=\underset{\mathrm{\Delta}t\to 0}{lim}{\displaystyle \frac{\mathrm{\Delta}x}{\mathrm{\Delta}t}}=\underset{\mathrm{\Delta}t\to 0}{lim}{\displaystyle \frac{x(t+\mathrm{\Delta}t)x(t)}{\mathrm{\Delta}t}}.\end{array}$$
Here the notation for the derivative
$dx/dt$
is known as
Leibniz's notation. The symbols
$dx$
and
$dt$
are known as
infinitesimals. Unlike
$\mathrm{\Delta}x$
and
$\mathrm{\Delta}t$
, which are variables representing finite
changes in value,
$dx$
and
$dt$
are symbols representing “an infinitesimally small change.” Why
is it so important that we use a very small change? Why can't we just take the ratio
$\mathrm{\Delta}x/\mathrm{\Delta}t$
directly? Because the rate of change is varying continuously. Even within a very
small interval of
$\mathrm{\Delta}t=.0001$
, it is not constant. This is why a limit argument is used,
to make the interval as small as we can possibly make it—infinitesimally small.
In certain circumstances, infinitesimals may be manipulated like algebraic variables (and you can
also attach units of measurement to them and carry out dimensional analysis to check your work).
The fact that such manipulations are often correct is what gives Leibniz notation its intuitive
appeal. However, because they are infinitely small values, they require special handling, similar
to the symbol
$\mathrm{\infty}$
, and so should not be tossed around willynilly. For the most part, we
interpret the notation
$\frac{dx}{dt}$
not as a ratio of two variables, but as a single symbol
that means “the derivative of
$x$
with respect to
$t$
.” This is the safest procedure and
avoids any chance of the aforementioned willynilliness. We have more to say later on Leibniz
and other notations, but first, let's finally calculate a derivative and answer the burning
question: how fast was the hare traveling at
$t=2.5$
?
Differentiating a simple function by using the definition Equation (11.3) is an
important rite of passage, and we are proud to help you cross this threshold. The typical
procedure is this:

Substitute
$x(t)$
and
$x(t+\mathrm{\Delta}t)$
into the definition.
(In our case,
$x(t)={t}^{2}+6t1$
).

Perform algebraic manipulations until it is legal to
substitute
$\mathrm{\Delta}t=0$
. (Often this boils down to getting
$\mathrm{\Delta}t$
out of the denominator.)

Substitute
$\mathrm{\Delta}t=0$
, which evaluates the expression
“at the limit,” removing the limit notation.

Simplify the result.
Applying this procedure to our case yields
$$\begin{array}{rl}v(t)={\displaystyle \frac{dx}{dt}}& =\underset{\mathrm{\Delta}t\to 0}{lim}{\displaystyle \frac{x(t+\mathrm{\Delta}t)x(t)}{\mathrm{\Delta}t}}\\ & =\underset{\mathrm{\Delta}t\to 0}{lim}{\displaystyle \frac{[(t+\mathrm{\Delta}t{)}^{2}+6(t+\mathrm{\Delta}t)1]({t}^{2}+6t1)}{\mathrm{\Delta}t}}\\ & =\underset{\mathrm{\Delta}t\to 0}{lim}{\displaystyle \frac{({t}^{2}2t(\mathrm{\Delta}t)(\mathrm{\Delta}t{)}^{2}+6t+6(\mathrm{\Delta}t)1)+({t}^{2}6t+1)}{\mathrm{\Delta}t}}\\ & =\underset{\mathrm{\Delta}t\to 0}{lim}{\displaystyle \frac{2t(\mathrm{\Delta}t)(\mathrm{\Delta}t{)}^{2}+6(\mathrm{\Delta}t)}{\mathrm{\Delta}t}}\\ & =\underset{\mathrm{\Delta}t\to 0}{lim}{\displaystyle \frac{\mathrm{\Delta}t(2t\mathrm{\Delta}t+6)}{\mathrm{\Delta}t}}\\ \text{(11.4)}& & =\underset{\mathrm{\Delta}t\to 0}{lim}2t\mathrm{\Delta}t+6.\end{array}$$
Now we are at step 3. Taking the limit in
Equation (11.4) is now easy; we simply substitute
$\mathrm{\Delta}t=0$
. This substitution was not legal earlier because there was a
$\mathrm{\Delta}t$
in the
denominator:
$$\begin{array}{rl}v(t)={\displaystyle \frac{dx}{dt}}& =\underset{\mathrm{\Delta}t\to 0}{lim}2t\mathrm{\Delta}t+6\\ & =2t(0)+6\\ \text{(11.5)}& & =2t+6.\end{array}$$
Finally! Equation (11.5) is the velocity
function we've been looking for. It allows us to plug in any value of
$t$
and
compute the instantaneous velocity of the hare at that time. Putting in
$t=2.5$
, we arrive at the answer to our question:
$$\begin{array}{rl}v(t)& =2t+6,\\ v(2.5)& =2(2.5)+6=1.\end{array}$$
So the instantaneous velocity of the hare at
$t=2.5$
was precisely 1 furlong per minute, just as
our earlier arguments predicted. But now we can say it with confidence.
Figure 11.7 shows this point and several others along the interval we've
been studying. For each point, we have calculated the instantaneous velocity at that point
according to Equation (11.5) and have drawn the tangent line
with the same slope.
It's very instructive to compare the graphs of position and velocity side by side.
Figure 11.8 compares the position and velocity of our fabled
racers.
There are several interesting observations to be made about
Figure 11.8.

When the position graph is a horizontal line, there is zero
velocity, and the velocity graph traces the
$v=0$
horizontal axis
(for example, during the hare's nap).

When the position is increasing, the velocity is positive,
and when the position is decreasing (the hare is moving the wrong
way) the velocity is negative.

When the position graph is a straight line, this constant velocity is
indicated by a horizontal line in the velocity graph.

When the position graph is curved, the velocity is changing continuously, and the
velocity graph will not be a horizontal line. In this case, the velocity graph
happens to be a straight line, but later we'll examine situations where the velocity
graph is curved.

When the position function changes slope at a “corner,” the
velocity graph exhibits a discontinuity. In fact, the derivative at
such points does not exist, and there is no way to define the instantaneous
velocity at those points of discontinuity. Fortunately, such situations
are nonphysical—in the real world, it is impossible for an object to
change its velocity instantaneously. Changes to velocity always occur
via an acceleration over a (potentially brief, but finite) amount of
time.
Later we show that such rapid accelerations over short durations
are often approximated by using
impulses.

There are sections on the velocity graph that look identical to each other even
though the corresponding intervals on the position graph are different from one
another. This is because the derivative measures only the rate of change of a
variable. The absolute value of the function does not matter. If we add a constant
to a function, which produces a vertical shift in the graph of that function, the
derivative will not be affected. We have more to say on this when we talk about
the relationship between the derivative and integral.
At this point, we should acknowledge a few ways in which our explanation of the derivative
differs from most calculus textbooks. Our approach has been to focus on one specific example,
that of instantaneous velocity. This has led to some cosmetic differences, such as notation. But
there were also many finer points that we are glossing over.
For example, we have not bothered defining continuous functions, or given rigorous definitions
for when the derivative is defined and when it is not defined. We have discussed the idea
behind what a limit is, but have not provided a formal definition or considered limits when
approached from the left and right, and the criteria for the existence of a welldefined limit.
We feel that leading off with the best intuitive example is always the optimum way to teach
something, even if it means
“lying” to the reader for a short while. If we were writing a calculus textbook, at this point
we would back up and correct some of our lies, reviewing the finer points and giving more precise
definitions.
However, since this is not a calculus textbook, we will only warn you that what we said
above is the big picture, but isn't sufficient to handle many edge cases when functions do weird
things like go off into infinity or exhibit “jumps” or “gaps.” Fortunately, such edge cases
just don't happen too often for functions that model physical phenomena, and so these details
won't become an issue for us in the context of physics.
We do have room, however, to mention alternate notations for the derivative that you are
likely to encounter.
11.4.4Notations for the Derivative
Several different notations for derivatives are in common use. Let's point out some ways that
other texts might look different from what we've said here. First of all, there is a trivial
issue of naming. Most calculus textbooks define the derivative in very general terms, where the
output variable is named
$y$
, the symbol
$x$
refers to the input variable rather than the output
variable, and the function is simply named
$f$
. In other words, the function being differentiated
is
$y=f(x)$
. Furthermore, many will assign the shrinking “step amount” to the variable
$h$
rather than using the
$\mathrm{\Delta}$
notation, which has advantages when solving the equations that
result when you work out derivatives from the definition. With
these variables, they would define the derivative as
Definition of a derivative using variables in most calculus
textbooks
$$\begin{array}{}\text{(11.6)}& {\displaystyle \frac{dy}{dx}}=\underset{h\to 0}{lim}{\displaystyle \frac{y(x+h)y(x)}{h}}.\end{array}$$
The differences between Equations (11.3) and
(11.6) are clearly cosmetic.
A variation on the Leibniz notation we prefer in this book is to prefix an
expression with
$d/dt$
to mean “the derivative with respect to
$t$
of this
thing on the right.” For example
$$\frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}({t}^{2}+5t)$$
can be read as “the derivative with respect to
$t$
of
${t}^{2}+5t$
.” This is
a very descriptive and intuitive notation. If we call the expression on the
right
$x$
, and interpret the juxtaposition of symbols as multiplication, we
can pull the
$x$
back on top of the fraction to get our original notation, as
in
$$\frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}({t}^{2}+5t)={\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}x={\displaystyle \frac{dx}{dt}}.$$
It's important to interpret these manipulations as notational manipulations
rather than having any real mathematical meaning. The notation is attractive
because such algebraic manipulations with the infinitesimals often work out.
But we reiterate our warning to avoid attaching much mathematical meaning to
such operations.
Another common notation is to refer to the derivative of a function
$f(x)$
with a prime:
${f}^{\prime}(x)$
.
This is known as
prime notation or Lagrange's
notation. It's used when the independent variable that
we are differentiating with respect to is implied or understood by context. Using this notation
we would define velocity as the derivative of the position function by
$v(t)={x}^{\prime}(t)$
.
One last notation, which was invented by Newton and is used mostly when the independent variable
is time (such as in the physics equations Newton invented), is dot notation. A derivative
is indicated by putting a dot over the variable; for example,
$v(t)=\dot{x}(t)$
.
Here is a summary of the different notations for the derivative you will see, using velocity and
position as the example:
$$v(t)=\frac{dx}{dt}=\frac{d}{dt}x(t)={x}^{\prime}(t)=\dot{x}(t).$$
11.4.5A Few Differentiation Rules and Shortcuts
Now let's return to calculating derivatives. In practice, it's seldom necessary to go back to
the definition of the derivative in order to differentiate an expression. Instead, there are
simplifying rules that allow you to break down complicated functions into smaller pieces that can
then be differentiated. There are also special functions, such as
$\mathrm{ln}x$
and
$\mathrm{tan}x$
, for
which the hard work of applying the definition has already been done and written down in those
tables that line the insides of the front and back covers of calculus books. To differentiate
expressions containing such functions, one simply refers to the table (although we're going to do
just a bit of this “hard work” ourselves for sine and cosine).
In this book, our concerns are limited to the derivatives of a very small set of functions, which
luckily can be differentiated with just a few simple rules. Unfortunately, we don't have the
space here to develop the mathematical derivations behind these rules, so we are simply going to
accompany each rule with a brief explanation as to how it is used, and a (mathematically
nonrigorous) intuitive argument to help you convince yourself that it works.
Our first rule, known as the constant rule, states that the derivative of a constant
function is zero. A constant function is a function that always produces the same value. For
example,
$x(t)=3$
is a constant function. You can plug in any value of
$t$
, and this function
outputs the value 3. Since, the derivative measures how fast the output of a function changes in
response to changes in the input
$t$
, in the case of a constant function, the output never
changes, and so the derivative is
${x}^{\prime}(t)=0$
.
The Constant Rule
$$\frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}k=0,{\textstyle \phantom{\rule{1em}{0ex}}}k\text{}\text{is any constant.$$
The next rule, sometimes known as the sum rule, says that differentiation is a
linear operator. The meaning of “linear” is essentially identical to our definition
given in Chapter 5, but let's review it in the context of the derivative. To
say that the derivative is a linear operator means two things. First, to take the derivative of a
sum, we can just take the derivative of each piece individually, and add the results together.
This is intuitive—the rate of change of a sum is the total rate of change of all the parts
added together. For example, consider a man who moves about on a train. His position in world
space can be described as the sum of the train's position, plus the man's position in the body
space of the train. Likewise, his velocity relative
to the ground is the sum of the train's velocity relative to the ground, plus his velocity
relative to the train.
Derivative of a Sum
$$\begin{array}{}\text{(11.7)}& {\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}[f(t)+g(t)]={\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}f(t)+{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}g(t).\end{array}$$
The second property of linearity is that if we multiply a function by some constant, the
derivative of that function gets scaled by that same constant. One easy way to see that this
must be true is to consider unit conversions. Let's return to our favorite function that yields
a hare's displacement as a function of time, measured in furlongs. Taking the derivative of this
function with respect to time yields a velocity, in furlongs per minute. If somebody comes along
who doesn't like furlongs, we can switch from furlongs to meters, by scaling the original
position function by a factor of 201.168. This must scale the derivative by the same
factor, or else the hare would suddenly change speed just because we switched to another unit.
Derivative of a Function Times a Constant
$$\begin{array}{}\text{(11.8)}& {\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}[kf(t)]=k[{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}f(t)],{\textstyle \phantom{\rule{1em}{0ex}}}k\text{}\text{is any constant}.\end{array}$$
If we combine Equations (11.7) and
(11.8), we can state the linearity rule in a more
general way.
The Sum Rule
$$\begin{array}{}\text{(11.4.5)}& {\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}[af(t)+bg(t)]=a[{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}f(t)]+b[{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}g(t)].\end{array}$$
The linear property of the derivative is very important since it allows us to break down many
common functions into smaller, easier pieces.
One of the most important and common functions that needs to be differentiated also happens to be
the easiest: the polynomial. Using the linear property of the derivative, we can break down, for
example, a fourthdegree polynomial with ease:
$$\begin{array}{rl}x(t)& ={c}_{4}{t}^{4}+{c}_{3}{t}^{3}+{c}_{2}{t}^{2}+{c}_{1}t+{c}_{0},\\ {\displaystyle \frac{dx}{dt}}& ={\displaystyle \frac{d}{dt}}[{c}_{4}{t}^{4}+{c}_{3}{t}^{3}+{c}_{2}{t}^{2}+{c}_{1}t+{c}_{0}]\\ \text{(11.9)}& & ={c}_{4}\left[{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{t}^{4}\right]+{c}_{3}\left[{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{t}^{3}\right]+{c}_{2}\left[{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{t}^{2}\right]+{c}_{1}\left[{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}t\right]+\left[{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{c}_{0}\right].\end{array}$$
The last derivative
$\frac{d}{dt}{c}_{0}$
is zero by the constant rule, since
${c}_{0}$
does not vary.
This leaves us with four simple derivatives, each of which can be plugged into the definition of
a derivative, Equation (11.3), without too much trouble. Solving each of these
four individually is considerably easier than plugging the original polynomial into
Equation (11.3). If you do go through this exercise (like every firstyear
calculus student does), you notice two things. First of all, the algebraic tedium increases as
the power of
$t$
gets higher. Second, a quite obvious pattern is revealed, known as the power
rule.
The Power Rule
$$\frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{t}^{n}=n{t}^{n1},{\textstyle \phantom{\rule{1em}{0ex}}}n\text{}\text{is an integer}.$$
This rule gives us the answers to the four derivatives needed above:
$$\begin{array}{rlrl}{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{t}^{4}& =4{t}^{3},& {\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{t}^{3}& =3{t}^{2},\\ {\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{t}^{2}& =2{t}^{1}=2t,& {\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}t& =1{t}^{0}=1.\end{array}$$
Notice in the last equation we used the identity
${t}^{0}=1$
. However, even without that
identity, it should be very clear
that
$\frac{d}{dt}t$
must be unity. Remember that the derivative answers the question,
“What is the rate of change of the output, relative to the rate of change of the input?” In the
case of
$\frac{d}{dt}t$
, the “output” and the “input” are both the variable
$t$
, and so their
rates of change are equal. Thus the ratio that defines the derivative is equal to one.
One last comment before we plug these results into
Equation (11.9) to differentiate our
polynomial. Using the identity
${t}^{0}=1$
, the power rule is brought into
harmony with the constant rule:
Derivative of a constant, using the power rule
$$\begin{array}{rlrl}{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}k& ={\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}(k{t}^{0})& & \mathrm{U}\mathrm{s}\mathrm{i}\mathrm{n}\mathrm{g}\text{}{t}^{0}=1,\\ & =k\left[{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{t}^{0}\right]& & \mathrm{L}\mathrm{i}\mathrm{n}\mathrm{e}\mathrm{a}\mathrm{r}\text{}\mathrm{p}\mathrm{r}\mathrm{o}\mathrm{p}\mathrm{e}\mathrm{r}\mathrm{t}\mathrm{y}\text{}\mathrm{o}\mathrm{f}\text{}\mathrm{d}\mathrm{e}\mathrm{r}\mathrm{i}\mathrm{v}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{v}\mathrm{e},\\ & =k[0({t}^{1})]& & \mathrm{P}\mathrm{o}\mathrm{w}\mathrm{e}\mathrm{r}\text{}\mathrm{r}\mathrm{u}\mathrm{l}\mathrm{e}\text{}\mathrm{f}\mathrm{o}\mathrm{r}\text{}n=0,\\ & =0.\end{array}$$
Let's get back to our fourthdegree polynomial. With the sum and power rule at our disposal, we
can make quick work of it:
$$\begin{array}{rl}x(t)& ={c}_{4}{t}^{4}+{c}_{3}{t}^{3}+{c}_{2}{t}^{2}+{c}_{1}t+{c}_{0},\\ {\displaystyle \frac{dx}{dt}}& =4{c}_{4}{t}^{3}+3{c}_{3}{t}^{2}+2{c}_{2}t+{c}_{1}.\end{array}$$
Below are several more examples of how the power rule can be used. Notice that the power rule
works for negative exponents as well:
$$\begin{array}{rl}{\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}(3{t}^{5}4t)& =15{t}^{4}4,\\ {\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}(\frac{{t}^{100}}{100}+\sqrt{\pi})& ={t}^{99},\\ {\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}(\frac{1}{t}+\frac{4}{{t}^{3}})={\displaystyle \frac{d}{dt}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}({t}^{1}+4{t}^{3})& ={t}^{2}12{t}^{4}=\frac{1}{{t}^{2}}\frac{12}{{t}^{4}}.\end{array}$$
11.4.6Derivatives of Some Special Functions
with Taylor Series
This section looks at some very special examples of differentiating polynomials. Given any
arbitrary function
$f(x)$
, the Taylor series of
$f$
is a way to express
$f$
as a
polynomial. Each successive term in the polynomial is determined by taking a higher order
derivative of the function, which is perhaps the main point of Taylor series that you should
learn when you take a real calculus class, but right now we're not interested in where Taylor
series come from, just that they exist. The Taylor series is a very useful tool in video games
because it provides polynomial approximations, which are “easy” to evaluate in a computer, for
functions that are otherwise “hard” to evaluate. We don't have the space to discuss much of
anything about Taylor series in general, but we would like to look at a few important examples of
Taylor series. The Taylor series for the sine and cosine functions are
Taylor series for
$\mathbf{sin}\mathbf{}\mathbf{(}\mathit{x}\mathbf{)}$
and
$\mathbf{cos}\mathbf{}\mathbf{(}\mathit{x}\mathbf{)}$
$$\begin{array}{rl}\mathrm{sin}x& =x{\displaystyle \frac{{x}^{3}}{3!}}+{\displaystyle \frac{{x}^{5}}{5!}}{\displaystyle \frac{{x}^{7}}{7!}}+{\displaystyle \frac{{x}^{9}}{9!}}+\cdots ,\\ \text{(11.10)}& \mathrm{cos}x& =1{\displaystyle \frac{{x}^{2}}{2!}}+{\displaystyle \frac{{x}^{4}}{4!}}{\displaystyle \frac{{x}^{6}}{6!}}+{\displaystyle \frac{{x}^{8}}{8!}}+\cdots .\end{array}$$
This pattern continues forever; in other words, to compute the exact value of
$\mathrm{sin}x$
would
require us to evaluate an infinite number of terms. However, notice that the denominators of the
terms are growing very rapidly, which means we can approximate
$\mathrm{sin}x$
simply by stopping after
a certain number of terms, and ignore the rest.
This is exactly the process by which trigonometric functions are computed inside a computer.
First, trig identities are used to get the argument into a restricted range (since the functions
are periodic). This is done because when the Taylor series is truncated, its accuracy is highest
near a particular value of
$x$
, and in the case of the trig functions, this point is usually
chosen to be
$x=0$
. Then the Taylor series polynomial with, say, four terms is evaluated.
This approximation is highly accurate. Stopping at the
${x}^{7}$
term is sufficient to calculate
$\mathrm{sin}x$
to about five and a half decimal digits for
$1<x<+1$
.
All this trivia concerning approximations is interesting, but our real reason for bringing up
Taylor series is to use them as nontrivial examples of differentiating polynomials with the power
rule, and also to learn some interesting facts about the sine, cosine, and exponential functions.
Let's use the power rule to differentiate the Taylor series expansion of
$\mathrm{sin}(x)$
. It's not that
complicated—we just have to differentiate each term by itself. We're not even intimidated by
the fact that there are an infinite number of terms:
Differentiating Taylor series for
$\mathbf{sin}\mathbf{}\mathbf{(}\mathit{x}\mathbf{)}$
$$\begin{array}{rl}{\displaystyle \frac{d}{dx}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}\mathrm{sin}x& ={\displaystyle \frac{d}{dx}}(x{\displaystyle \frac{{x}^{3}}{3!}}+{\displaystyle \frac{{x}^{5}}{5!}}{\displaystyle \frac{{x}^{7}}{7!}}+{\displaystyle \frac{{x}^{9}}{9!}}+\cdots )\\ & ={\displaystyle \frac{d}{dx}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}x{\displaystyle \frac{d}{dx}}{\displaystyle \frac{{x}^{3}}{3!}}+{\displaystyle \frac{d}{dx}}{\displaystyle \frac{{x}^{5}}{5!}}{\displaystyle \frac{d}{dx}}{\displaystyle \frac{{x}^{7}}{7!}}+{\displaystyle \frac{d}{dx}}{\displaystyle \frac{{x}^{9}}{9!}}+\cdots & & \text{(Sum rule)}\\ & =1{\displaystyle \frac{3{x}^{2}}{3!}}+{\displaystyle \frac{5{x}^{4}}{4!}}{\displaystyle \frac{7{x}^{6}}{7!}}+{\displaystyle \frac{9{x}^{8}}{9!}}+\cdots & & \text{(Power rule)}\\ \text{(11.11)}& & =1{\displaystyle \frac{{x}^{2}}{2!}}+{\displaystyle \frac{{x}^{4}}{4!}}{\displaystyle \frac{{x}^{6}}{6!}}+{\displaystyle \frac{{x}^{8}}{8!}}+\cdots \end{array}$$
In the above derivation, we first used the sum rule, which says that to differentiate the whole
Taylor polynomial, we can differentiate each term individually. Then we applied the power rule
to each term, in each case multiplying by the exponent and decrementing it by one. (And also
remembering that
$\frac{d}{dx}\text{}x=1$
for the first term.) To understand the last step, remember
the definition of the factorial operator:
$n!=1\times 2\times 3\times \cdots \times n$
. Thus
the constant in the numerator of each term cancels out the highest factor in the factorial in the
denominator.
Does Equation (11.11) the last look familiar? It should, because it's
the same as Equation (11.10), the Taylor series for
$\mathrm{cos}x$
. In other words, we
now know the derivative of
$\mathrm{sin}x$
, and by a similar process we can also obtain the derivative
of
$\mathrm{cos}x$
. Let's state these facts formally.
Derivatives of Sine and Cosine
$$\begin{array}{rlr}{\displaystyle \frac{d}{dx}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}\mathrm{sin}x=\mathrm{cos}x,& & {\displaystyle \frac{d}{dx}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}\mathrm{cos}x=\mathrm{sin}x.\end{array}$$
The derivatives of the sine and cosine functions will become useful in later sections.
Now let's look at one more important special function that will play an important role later in
this book, which will be convenient to be able to differentiate, and which also happens to have a
nice, tidy Taylor series. The function we're referring to is the exponential function,
denoted
${e}^{x}$
. The mathematical constant
$e\approx 2.718282$
has many well known and
interesting properties, and pops up in all sorts of problems from finance to signal processing.
Much of
$e$
's special status is related to the unique nature of the function
${e}^{x}$
. One
manifestation of this unique nature is that
${e}^{x}$
has such a beautiful Taylor series:
Taylor series of
${\mathit{e}}^{\mathit{x}}$
$$\begin{array}{}\text{(11.12)}& {e}^{x}=1+x+{\displaystyle \frac{{x}^{2}}{2!}}+{\displaystyle \frac{{x}^{3}}{3!}}+{\displaystyle \frac{{x}^{4}}{4!}}+{\displaystyle \frac{{x}^{5}}{5!}}+\cdots \end{array}$$
Taking the derivative gives us
$$\begin{array}{rl}{\displaystyle \frac{d}{dx}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{e}^{x}& ={\displaystyle \frac{d}{dx}}\text{}(1+x+{\displaystyle \frac{{x}^{2}}{2!}}+{\displaystyle \frac{{x}^{3}}{3!}}+{\displaystyle \frac{{x}^{4}}{4!}}+{\displaystyle \frac{{x}^{5}}{5!}}+\cdots )\\ & =0+1+{\displaystyle \frac{x}{1!}}+{\displaystyle \frac{{x}^{2}}{2!}}+{\displaystyle \frac{{x}^{3}}{3!}}+{\displaystyle \frac{{x}^{4}}{4!}}+\cdots \\ & =1+x+{\displaystyle \frac{{x}^{2}}{2!}}+{\displaystyle \frac{{x}^{3}}{3!}}+{\displaystyle \frac{{x}^{4}}{4!}}+\cdots \end{array}$$
But this result is equivalent to the definition of
${e}^{x}$
in
Equation (11.12); the only difference between them is the cosmetic issue of
when to stop listing terms explicitly and end with the “
$\cdots $
”. In other words, the
exponential function is its own derivative:
$d/dx\text{}{e}^{x}={e}^{x}$
. The exponential function is the only function that can boast this unique property. (To be more precise, any multiple of the exponential function, including zero, has this quality.)
The Exponential Function Is Its Own Derivative
$$\frac{d}{dx}}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{e}^{x}={e}^{x}.$$
It is this special property about the exponential function that makes it unique and causes it to
come up so frequently in applications. Anytime the rate of change of some value is proportionate
to the value itself, the exponential function will almost certainly arise somewhere in the math
that describes the dynamics of the system.
The example most of us are familiar with is compound interest. Let
$P(t)$
be the amount of money
in your bank account at time
$t$
; assume the amount is accruing interest. The rate of change per
time interval—the amount of interest earned—is proportional to the amount of money in your
account. The more money you have, the more interest you are earning, and the faster it grows.
Thus, the exponential function works its way into finance with the equation
$P(t)={P}_{0}{e}^{rt}$
,
which describes the amount of money at any given time
$t$
, assuming an initial amount
${P}_{0}$
grows
at an interest rate of
$r$
, where the interest is compounded continually.
You might have noticed that the Taylor series of
${e}^{x}$
is strikingly similar to the series
representation of
$\mathrm{sin}x$
and
$\mathrm{cos}x$
. This similarity hints at a deep and surprising
relationship between the exponential functions and the trig functions, which we explore in
Exercise 11.
We hope this brief encounter with Taylor series, although a bit outside of our main thrust, has
sparked your interest in a mathematical tool that is highly practical, in particular for its
fundamental importance to all sorts of approximation and numerical calculations in a computer. We
also hope it was an interesting nontrivial example of differentiation of a polynomial. It also
has given us a chance to discuss the derivatives of the sine, cosine, and exponential functions;
these derivatives come up again in latersections.
11.4.7The Chain Rule
The chain rule is the last rule of differentiation we discuss here. The chain rule tells us how
to determine the rate of change of a function when the argument to that function is itself some
other function we know how to differentiate.
In the race between the tortoise and hare, we never really thought much about exactly what our
function
$x(t)$
measured, we just said it was the “position” of the hare. Let's say that the
course was actually a winding track with hills and bridges and even a vertical loop, and that the
function that we graphed and previously named
$x(t)$
actually measures the linear distance
along this winding path, rather than, say, a horizontal position. To avoid the horizontal
connotations associated with the symbol
$x$
, let's introduce the variable
$s$
, which gives the
distance along the track (in furlongs, of course).
Let's say that we have a function
$y(s)$
that describes the altitude of the track at a given
distance. The derivative
$dy/ds$
tells us very basic things about the track at that location. A
value of zero means the course is flat at that location, a positive value means the runners are
running uphill, and a large positive or negative value indicates a location where the track is
very steep.
Now consider the composite function
$y(s(t))$
. You should be able to convince yourself that this
tells us the hare's altitude for any given time
$t$
. The derivative
$dy/dt$
tells us how fast
the hare was moving vertically, at a given time
$t$
. This is very different from
$dy/ds$
. How
might we calculate
$dy/dt$
? You might be tempted to say that to make this determination, we
simply find out where the hare was on the track at time
$t$
, and then the answer is the slope of
the track at this location. In math symbols, you are saying that the vertical velocity is
${y}^{\prime}(s(t))$
. But that isn't right. For example, while the hare was taking a nap (
$ds/dt=0$
), it
doesn't matter what the slope of the track was; since he wasn't moving along it, his vertical
velocity is zero! In fact, at a certain point in the race he turned around and ran on the track
in the wrong direction (
$ds/dt<0$
), so his vertical velocity
$dy/dt$
would be opposite of
the track slope
$dy/ds$
. And obviously if he sprints quickly over a place in the track, his
vertical velocity will be higher than if he strolled slowly over that same spot. But likewise, where the
track is flat, it doesn't matter how fast he runs across it, his vertical velocity will be zero.
So we see that the hare's vertical velocity is the product of his speed (measured
parametrically along the track) and the slope of the track at that point.
This rule is known as the chain rule. It is particularly intuitive when written in
Leibniz notation, because the
$ds$
infinitesimals appear to “cancel.”
The Chain Rule of Differentiation
$$\begin{array}{rl}\frac{dy}{dt}& =\frac{dy}{ds}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}\frac{ds}{dt}.\end{array}$$
Here are a few examples, using functions we now know how to differentiate:
Examples of the
chain rule
$$\begin{array}{rl}\frac{d}{dt}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}\mathrm{sin}3x& =3\mathrm{cos}3x,\\ \frac{d}{dt}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}\mathrm{sin}({x}^{2})& =2x\mathrm{cos}({x}^{2}),\\ \frac{d}{dt}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{e}^{\mathrm{cos}x+3x}& =(\mathrm{sin}x+3){e}^{\mathrm{cos}x+3x},\\ \frac{d}{dt}{\textstyle \phantom{\rule{thinmathspace}{0ex}}}{e}^{\mathrm{sin}3x+\mathrm{sin}({x}^{2})}& =(3\mathrm{cos}3x+2x\mathrm{cos}({x}^{2})){e}^{\mathrm{sin}3x+\mathrm{sin}({x}^{2})}.\end{array}$$
We're going to put calculus from a purely mathematical perspective on the shelf for a while and
return our focus to kinematics. (After all, our purpose in discussing calculus was, like Ike
Newton, to improve our understanding of mechanics.) However, it won't be long before we will
return to calculus with the discussion of the integral and the fundamental theorem of calculus.
11.5Acceleration
We've made quite a fuss about the distinction between instantaneous velocity and average
velocity, and this distinction is important (and the fuss is justified) when the velocity is
changing continuously. In such situations, we might be interested to know the rate at
which the velocity is changing. Luckily we have just learned about the derivative, whose
raison d'être is to investigate rates of change.
When we take the derivative of a velocity function
$v(t)$
we get a new function describing how
quickly the velocity is increasing or decreasing at that instant. This instantaneous rate of
change is an important quantity in physics, and it goes by a familiar name: acceleration.
In ordinary conversation, the verb “accelerate” typically means “speed up.” However, in
physics, the word “acceleration” carries a more general meaning and may refer to any
change in velocity, not just an increase in speed.
In fact, a body can undergo an acceleration even when its speed is constant! How can this be?
Velocity is a vector value, which means it has both magnitude and direction. If the direction of
the velocity changes, but the magnitude (its speed) remains the same, we say that the body is
experiencing an acceleration. Such terminology is not mere nitpicking with words, the
acceleration in this case is a very real sensation that would be felt by, say, two people riding
in the back seat of a swerving car who find themselves pressed together to one side. We have
more to say about this particular situation in Section 11.8.
We can learn a lot about acceleration just by asking ourselves what sort of units we should use
to measure it. For velocity, we used the generic units of
$L/T$
, unit length per unit time.
Velocity is a rate of change of position (
$L$
) per unit time (
$T$
), and so this makes sense.
Acceleration is the rate of change of velocity per unit time, and so it must be expressed in
terms of “unit velocity per unit time.” In fact, the units used to measure velocity are
$L/{T}^{2}$
. If you are disturbed by the idea of “time squared,” think of it instead as
$(L/T)/T$
,
which makes more explicit the fact that it is a unit of velocity
$(L/T)$
per unit time.
For example, an object in free fall near Earth's surface accelerates at a rate of about
$32\text{}\mathrm{f}\mathrm{t}/{\mathrm{s}}^{2}$
, or
$9.8\text{}\mathrm{m}/{\mathrm{s}}^{2}$
. Let's say that you are dangling
a metal bearing off the side of
Willis Tower. You drop the
bearing, and it begins accelerating, adding
$9.8\text{}\mathrm{m}/\mathrm{s}$
to its downward velocity
each second. (We are ignoring wind resistance.) After, say, 2.4 seconds, its velocity will be
$$2.4\text{}\mathrm{s}\times 32\text{}{\displaystyle \frac{\mathrm{f}\mathrm{t}}{{\mathrm{s}}^{2}}}=76.8\text{}{\displaystyle \frac{\mathrm{f}\mathrm{t}}{\mathrm{s}}}.$$
More generally, the velocity at an arbitrary time
$t$
of an object under
constant acceleration is given by the simple linear formula
$$\begin{array}{}\text{(11.13)}& v(t)={v}_{0}+at,\end{array}$$
where
${v}_{0}$
is the initial velocity at time
$t=0$
, and
$a$
is the constant acceleration. We
study the motion of objects in free fall in more detail in Section 11.6,
but first, let's look at a graphical representation of acceleration.
Figure 11.9 shows plots of a position function and the
corresponding velocity and acceleration functions.
You should study Figure 11.9 until it makes
sense to you. In particular, here are some noteworthy observations:

Where the acceleration is zero, the velocity is constant and the
position is a straight (but possibly sloped) line.

Where the acceleration is positive, the position graph is curved like
$\bigcup $
,
and where it is negative, the position graph is curved like
$\bigcap $
. The most
interesting example occurs on the right side of the graphs. Notice that at the time
when the acceleration graph crosses
$a=0$
, the velocity curve reaches its apex, and
the position curve switches from
$\bigcup $
to
$\bigcap $
.

A discontinuity in the velocity function causes a “kink” in the
position graph. Furthermore, it causes the acceleration
to become infinite (actually, undefined), which is why, as we said
previously, such discontinuities don't happen in the real world.
This is why the lines in the velocity graph are connected at those
discontinuities, because the graph is of a physical situation
being approximated by a mathematical model.

A discontinuity in the acceleration graph causes a kink in the
velocity graph, but notice that the position graph is still smooth.
In fact, acceleration can change instantaneously, and for this
reason we have chosen not to bridge the discontinuities in the
acceleration graph.
The accelerations experienced by an object can vary as a function of time, and indeed we can
continue this process of differentiation, resulting in yet another function of time, which some
people call the
“jerk” function. We stick with the position function and its first two derivatives in this
book. Furthermore, it's very instructive to consider situations in which the acceleration is
constant (or at least has constant magnitude). This is precisely what we're going to do in the
next few sections.
Section 11.6 considers objects under constant acceleration, such as objects
in free fall and projectiles. This will provide an excellent backdrop to introduce the integral,
the complement to the derivative, in Section 11.7.
Then Section 11.8 examines objects traveling in a circular path, which
experience an acceleration that has a constant magnitude but a direction that changes continually
and always points towards the center of the circle.
11.6Motion under Constant Acceleration
Let's look now at the trajectory an object takes when it accelerates at a constant rate over
time. This is a simple case, but a common one, and an important one to fully understand. In fact,
the equations of motion we present in this section are some of the most important mechanics
equations to know by heart, especially for video game programming.
Before we begin, let's consider an even simpler type of motion—motion with constant velocity.
Motion with constant velocity is a special case of motion with constant acceleration—the case
where the acceleration is constantly zero. The motion of a particle with constant velocity is an
intuitive linear equation, essentially the same as Equation (9.1),
the equation of a ray. In one dimension, the position of a particle as a function of time is
$$\begin{array}{}\text{(11.14)}& x(t)={x}_{0}+vt,\end{array}$$
where
${x}_{0}$
is the position of the particle at time
$t=0$
, and
$v$
is the constant velocity.
Now let's consider objects moving with constant acceleration. We've already mentioned at least
one important example: when they are in free fall, accelerating due to gravity. (We'll ignore
wind resistance and all other forces.) Motion in free fall is often called
projectile motion. We start out in one dimension here to keep things simple. Our goal is a
formula
$x(t)$
for the position of a particle at a given time.
Take our example of illegal ballbearingbombing off of Willis Tower. Let's set a reference
frame where
$x$
increases in the downward direction, and
${x}_{0}=0$
. In other words,
$x(t)$
measures
the distance the object has fallen from its drop height at time
$t$
. We also assume for now that
initial velocity is
${v}_{0}=0\text{}\mathrm{f}\mathrm{t}/\mathrm{s}$
, meaning you merely release the ball bearing and don't
throw it.
At this point, we don't even know what form
$x(t)$
should take, so we're a bit stuck. The
“front door” to this solution seems to be locked for us at the moment, so instead we try to
sneak around and enter through the back, using an approach similar to the one we used earlier to
define instantaneous velocity. We'll consider ways that we might approximate the answer and
then watch what happens as the approximations get better and better.
$$\begin{array}{cc}\begin{array}{c}6\text{}\mathrm{S}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{e}\mathrm{s},\mathrm{\Delta}t=0.40\\ \\ \begin{array}{rrr}{t}_{0}& {v}_{0}& \mathrm{\Delta}x\\ 0.00& 0.00& 0.00\\ 0.40& 12.80& 5.12\\ 0.80& 25.60& 10.24\\ 1.20& 38.40& 15.36\\ 1.60& 51.20& 20.48\\ 2.00& 64.00& 25.60\\ & \mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}& 76.80\end{array}\\ \\ \\ 12\text{}\mathrm{S}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{e}\mathrm{s},\mathrm{\Delta}t=0.20\\ \\ \begin{array}{rrr}{t}_{0}& {v}_{0}& \mathrm{\Delta}x\\ 0.00& 0.00& 0.00\\ 0.20& 6.40& 1.28\\ 0.40& 12.80& 2.56\\ 0.60& 19.20& 3.84\\ 0.80& 25.60& 5.12\\ 1.00& 32.00& 6.40\\ 1.20& 38.40& 7.68\\ 1.40& 44.80& 8.96\\ 1.60& 51.20& 10.24\\ 1.80& 57.60& 11.52\\ 2.00& 64.00& 12.80\\ 2.20& 70.40& 14.08\\ & \mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}& 84.48\end{array}\end{array}& \begin{array}{c}24\text{}\mathrm{S}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{e}\mathrm{s},\mathrm{\Delta}t=0.10\\ \\ \begin{array}{rrr}{t}_{0}& {v}_{0}& \mathrm{\Delta}x\\ 0.00& 0.00& 0.00\\ 0.10& 3.20& 0.32\\ 0.20& 6.40& 0.64\\ 0.30& 9.60& 0.96\\ 0.40& 12.80& 1.28\\ 0.50& 16.00& 1.60\\ 0.60& 19.20& 1.92\\ 0.70& 22.40& 2.24\\ 0.80& 25.60& 2.56\\ 0.90& 28.80& 2.88\\ 1.00& 32.00& 3.20\\ 1.10& 35.20& 3.52\\ 1.20& 38.40& 3.84\\ 1.30& 41.60& 4.16\\ 1.40& 44.80& 4.48\\ 1.50& 48.00& 4.80\\ 1.60& 51.20& 5.12\\ 1.70& 54.40& 5.44\\ 1.80& 57.60& 5.76\\ 1.90& 60.80& 6.08\\ 2.00& 64.00& 6.40\\ 2.10& 67.20& 6.72\\ 2.20& 70.40& 7.04\\ 2.30& 73.60& 7.36\\ & \mathrm{T}\mathrm{o}\mathrm{t}\mathrm{a}\mathrm{l}& 88.32\end{array}\end{array}\end{array}$$
Table 11.3Values for different numbers of slices
Let's make our example a bit more specific. Earlier, we computed that after being in free fall
for 2.4 seconds, the ball bearing would have a velocity of
$v(2.4)=76.8\text{}\mathrm{f}\mathrm{t}/\mathrm{s}$
.
However, we didn't say anything about how far it had traveled during that time. Let's try to
compute this distance, which is
$x(2.4)$
. To do this, we chop up the total 2.4 second interval
into a number of smaller “slices” of time, and approximate how far the ball bearing travels
during each slice. We can approximate the total distance traveled as the sum of the distances
traveled during each slice. To approximate how far the ball bearing travels during one single
slice, we first compute the velocity of the ball bearing at the start of the slice by using
Equation (11.13). Then we approximate the distance traveled during
the slice by plugging this velocity as the constant velocity for the slice into
Equation (11.14).
Table 11.3 shows tabulated values for 6, 12, and 24
slices. For each slice,
${t}_{0}$
refers to the starting time of the slice,
${v}_{0}$
is the velocity at
the start of the slice (computed according to Equation (11.13) as
${v}_{0}={t}_{0}\times 32\text{}\mathrm{f}\mathrm{t}/{\mathrm{s}}^{2}$
),
$\mathrm{\Delta}t$
is the duration of the slice, and
$\mathrm{\Delta}x$
is our approximation for the displacement during the slice (computed according to
Equation (11.14) as
$\mathrm{\Delta}x={v}_{0}\mathrm{\Delta}t$
).
Since each slice has a different initial velocity, we are accounting for the fact that the
velocity changes over the entire interval. (In fact, the computation of the starting velocity
for the slice is not an approximation—it is exact.) However, since we ignore the change in
velocity within a slice, our answer is only an approximation. Taking more and more slices,
we get better and better approximations, although it's difficult to tell to what value these
approximations are converging. Let's look at the problem graphically to see if we can gain some
insight.
In Figure 11.10, each rectangle represents one time interval in
our approximation. Notice that the distance traveled during an interval is the same as the area
of the corresponding rectangle:
$$\begin{array}{rl}(\mathrm{a}\mathrm{r}\mathrm{e}\mathrm{a}\text{}\mathrm{o}\mathrm{f}\text{}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e})& =(\mathrm{w}\mathrm{i}\mathrm{d}\mathrm{t}\mathrm{h}\text{}\mathrm{o}\mathrm{f}\text{}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e})\times (\mathrm{h}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}\text{}\mathrm{o}\mathrm{f}\text{}\mathrm{r}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{a}\mathrm{n}\mathrm{g}\mathrm{l}\mathrm{e})\\ & =(\mathrm{d}\mathrm{u}\mathrm{r}\mathrm{a}\mathrm{t}\mathrm{i}\mathrm{o}\mathrm{n}\text{}\mathrm{o}\mathrm{f}\text{}\mathrm{s}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{e})\times (\mathrm{v}\mathrm{e}\mathrm{l}\mathrm{o}\mathrm{c}\mathrm{i}\mathrm{t}\mathrm{y}\text{}\mathrm{u}\mathrm{s}\mathrm{e}\mathrm{d}\text{}\mathrm{f}\mathrm{o}\mathrm{r}\text{}\mathrm{s}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{e})\\ & =(\mathrm{d}\mathrm{i}\mathrm{s}\mathrm{p}\mathrm{l}\mathrm{a}\mathrm{c}\mathrm{e}\mathrm{m}\mathrm{e}\mathrm{n}\mathrm{t}\text{}\mathrm{d}\mathrm{u}\mathrm{r}\mathrm{i}\mathrm{n}\mathrm{g}\text{}\mathrm{s}\mathrm{l}\mathrm{i}\mathrm{c}\mathrm{e}).\end{array}$$
Now we come to the key observation. As we increase the number of slices, the total area of the
rectangles becomes closer and closer to the area of the triangle under the velocity curve. In the
limit, if we take an infinite number of rectangles, the two areas will be equal. Now, since
total displacement of the falling ball bearing is equal to the total area of the rectangles,
which is equal to the area under the curve, we are led to an important discovery.
The distance traveled is equal to the area under the velocity curve.
We have come to this conclusion by using a limit argument very similar to the one we made to
define instantaneous velocity—we consider how a series of approximations converges in the limit
as the approximation error goes to zero.
Notice that we have made no assumptions in this argument about
$v(t)$
. In the example at hand,
it is a simple linear function, and the graph is a straight line; however, you should be able to
convince yourself that this
procedure will work for any arbitrary velocity
function. This limit argument is a formalized tool in
calculus known as the
Riemann integral, which we will consider in Section 11.7. That will also be the
appropriate time to consider the general case of any
$v(t)$
. However, since there is so much we
can learn from this specific example, let's keep it simple as long as possible.
Remember the question we're trying to answer: how far does an object travel after being dropped
at an initial zero velocity and then accelerated due to gravity for 2.4 seconds at a constant
rate of
$32\text{}\mathrm{f}\mathrm{t}/{\mathrm{s}}^{2}$
? How does this new realization about the equivalence of
distance traveled and the area under the graph of
$v(t)$
help us? In this special case,
$v(t)$
is a simple linear function, and the area under the curve from
$t=0$
to
$t=2.4$
is a triangle.
That's an easy shape for us to compute an area. The base of this triangle has length
$2.4\text{}\mathrm{s}$
, and the height is
$v(2.4)=76.8\text{}\mathrm{f}\mathrm{t}/\mathrm{s}$
, so the area is
$$\frac{\mathrm{b}\mathrm{a}\mathrm{s}\mathrm{e}\times \mathrm{h}\mathrm{e}\mathrm{i}\mathrm{g}\mathrm{h}\mathrm{t}}{2}}={\displaystyle \frac{2.4\text{}\mathrm{s}\times 76.8\text{}\mathrm{f}\mathrm{t}/\mathrm{s}}{2}}=92.16\text{}\mathrm{f}\mathrm{t}.$$
Thus after a mere 2.4 seconds, the ball bearing had already dropped more than 92 feet!
That solves the specific problem at hand, but let's be more general. Remember that the larger
goal was a kinematic equation
$x(t)$
that predicts an object's position given any initial
position and any initial velocity. First, let's replace the constant 2.4 with an arbitrary time
$t$
. Next, let's remove the assumption that the object initially has zero velocity, and instead
allow an arbitrary initial velocity
${v}_{0}$
. This means the area under the curve
$v(t)$
is no
longer a triangle—it is a triangle on top of a rectangle, as shown in
Figure 11.11.
The rectangle has base
$t$
and height
${v}_{0}$
, and its area represents the distance that would be
traveled if there were no acceleration. The triangle on top of the rectangle also has base
$t$
,
and the height is
$at$
, the difference in
$v(t)$
compared to the initial velocity as a result of
the acceleration at the rate
$a$
over the duration of
$t$
seconds. Summing these two parts
together yields the total displacement, which we denote as
$\mathrm{\Delta}x$
:
$$\begin{array}{rl}\mathrm{\Delta}x& =(\text{Area~of~rectangle})+(\text{Area~of~triangle})\\ & =\left(\begin{array}{c}\text{Rectangle}\\ \text{base}\end{array}\right)\left(\begin{array}{c}\text{Rectangle}\\ \text{height}\end{array}\right)+{\displaystyle \frac{1}{2}}\left(\begin{array}{c}\text{Triangle}\\ \text{base}\end{array}\right)\left(\begin{array}{c}\text{Triangle}\\ \text{height}\end{array}\right)\\ & =(t)({v}_{0})+(1/2)(t)(at)\\ & ={v}_{0}t+(1/2)a{t}^{2}.\end{array}$$
We have just derived a very useful equation, so let's highlight it so that people who are
skimming will notice it.
Formula for Displacement Given Initial Velocity and Constant Acceleration
$$\begin{array}{}\text{(11.15)}& \mathrm{\Delta}x={v}_{0}t+(1/2)a{t}^{2}.\end{array}$$
Equation (11.15) is one of only a handful of equations in
this book that are worth memorizing. It is very useful for solving practical problems that arise
in physics simulations.
It's common that we only need the displacement
$\mathrm{\Delta}x$
, and the absolute position
$x(t)$
doesn't matter. However, since the function
$x(t)$
was our stated goal, we can easily express
$x(t)$
in terms of Equation (11.15) by adding the
displacement to our initial position, which we denote as
${x}_{0}$
:
$$x(t)={x}_{0}+\mathrm{\Delta}x={x}_{0}+{v}_{0}t+(1/2)a{t}^{2}.$$
Let's work through some examples to show the types of problems that can be solved by using
Equation (11.15) and its variants. One tempting scenario
is to let our ball bearing hit the ground. The observation deck on the 103rd floor of
Willis Tower is 1,353 ft above the sidewalk. If it is dropped from that height, how long will it
take to fall to the bottom?
Solving Equation (11.15) for
$t$
, we have
Solving for time
$$\begin{array}{rl}\mathrm{\Delta}x& ={v}_{0}t+(1/2)a{t}^{2}\\ 0& =(a/2){t}^{2}+{v}_{0}t\mathrm{\Delta}x\\ t& ={\displaystyle \frac{{v}_{0}\pm \sqrt{{v}_{0}^{2}4(a/2)(\mathrm{\Delta}x)}}{2(a/2)}}& & \text{(quadratic formula)}\\ \text{(11.16)}& t& ={\displaystyle \frac{{v}_{0}\pm \sqrt{{v}_{0}^{2}+2a\mathrm{\Delta}x}}{a}}.\end{array}$$
Equation (11.16) is a very useful general equation.
Plugging in the numbers specific to this problem, we have
$$\begin{array}{rl}t& ={\displaystyle \frac{{v}_{0}\pm \sqrt{{v}_{0}^{2}+2a\mathrm{\Delta}x}}{a}}\\ & ={\displaystyle \frac{(0)\pm \sqrt{(0{)}^{2}+2(32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2})(1{\textstyle \phantom{\rule{thinmathspace}{0ex}}}353\text{}\mathrm{f}\mathrm{t})}}{32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2}}}\end{array}$$
$$\begin{array}{rl}& =\pm {\displaystyle \frac{\sqrt{86{\textstyle \phantom{\rule{thinmathspace}{0ex}}}592\text{}(\mathrm{f}\mathrm{t}/\mathrm{s}{)}^{2}}}{32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2}}}\\ & \approx \pm {\displaystyle \frac{294.3\text{}\mathrm{f}\mathrm{t}/\mathrm{s}}{32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2}}}\\ & \approx \pm 9.197\text{}\mathrm{s}.\end{array}$$
The square root in Equation (11.16) introduces the
possibility for two solutions. We always use the root that results in a positive value for
$t$
.
Naturally, a person in the business of dropping ball bearings from great heights is interested in
how much damage he can do, so the next logical question is, “How fast is the ball bearing
traveling when it hits the sidewalk?” To answer this question, we plug the total travel time
into Equation (11.13):
$$v(t)={v}_{0}+at=0\text{}\mathrm{f}\mathrm{t}/\mathrm{s}+(32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2})(9.197\text{}\mathrm{s})=294.3\text{}\mathrm{f}\mathrm{t}/\mathrm{s}.$$
If we ignore wind resistance, at the moment of impact, the ball bearing is traveling at a speed
that covers a distance of roughly a football field in one second! You can see why the things we
are doing in our imagination are illegal in real life. Let's keep doing them.
Now let's assume that instead of just dropping the ball bearing, we give it an initial velocity
(we toss it up or down). It was our free choice to decide whether up or down is positive in
these examples, and we have chosen
$+x$
to be the downward direction, so that means the initial
velocity will be negative. What must the initial velocity be in order for the ball bearing to
stay in the air only a few seconds longer, say a total of 12 seconds? Once again, we'll first
manipulate Equation (11.15) to get a general solution; this
time we'll be solving for
${v}_{0}$
:
Solving for initial velocity
$$\begin{array}{rl}\mathrm{\Delta}x& ={v}_{0}t+(1/2)a{t}^{2},\\ {v}_{0}t& =\mathrm{\Delta}x+(1/2)a{t}^{2},\\ {v}_{0}& =\mathrm{\Delta}x/t(1/2)at.\end{array}$$
And now plugging in the numbers for our specific problem, we have
$$\begin{array}{rl}{v}_{0}& =\mathrm{\Delta}x/t(1/2)at\\ & =(1{\textstyle \phantom{\rule{thinmathspace}{0ex}}}353\text{}\mathrm{f}\mathrm{t})/(12.0\text{}\mathrm{s})(1/2)(32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2})(12.0\text{}\mathrm{s})\\ & =112.8\text{}\mathrm{f}\mathrm{t}/\mathrm{s}192\text{}\mathrm{f}\mathrm{t}/\mathrm{s}\\ & =79.2\text{}\mathrm{f}\mathrm{t}/\mathrm{s}.\end{array}$$
Notice that the result is negative, indicating an upwards velocity. If we give the ball bearing
this initial velocity, we might wonder how long it takes for the bearing to come back down to its
initial position. Using Equation (11.16) and
letting
$\mathrm{\Delta}x=0$
, we have
$$\begin{array}{rl}t& ={\displaystyle \frac{{v}_{0}\pm \sqrt{{v}_{0}^{2}+2a\mathrm{\Delta}x}}{a}}\\ & ={\displaystyle \frac{(79.2\text{}\mathrm{f}\mathrm{t}/\mathrm{s})\pm \sqrt{(79.2\text{}\mathrm{f}\mathrm{t}/\mathrm{s}{)}^{2}+2(32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2})(0\text{}\mathrm{f}\mathrm{t})}}{32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2}}}\\ & ={\displaystyle \frac{79.2\text{}\mathrm{f}\mathrm{t}/\mathrm{s}\pm \sqrt{(79.2\text{}\mathrm{f}\mathrm{t}/\mathrm{s}{)}^{2}}}{32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2}}}\\ & ={\displaystyle \frac{79.2\text{}\mathrm{f}\mathrm{t}/\mathrm{s}\pm 79.2\text{}\mathrm{f}\mathrm{t}/\mathrm{s}}{32\text{}{\mathrm{f}\mathrm{t}/\mathrm{s}}^{2}}}\\ & =0\text{}\text{or}\text{}4.95\text{}\mathrm{s}.\end{array}$$
It's no surprise that
$t=0$
is a solution; we were solving for the time values when the ball
bearing was at its initial position.
Examine the graph in Figure 11.12, which plots the position and
velocity of an object moving under constant velocity
$a$
with an initial velocity
${v}_{0}$
, where
${v}_{0}$
and
$a$
have opposite signs. Let's make three key observations. Although we use terms
such as “height,” which are specific to projectile motion, similar statements are true anytime
the signs of
${v}_{0}$
and
$a$
are opposite.
The first observation is that the projectile reaches its maximum height, denoted
${x}_{\mathrm{m}\mathrm{a}\mathrm{x}}$
, when the acceleration has consumed all of the velocity and
$v(t)=0$
. It's
easy to solve for the time when this will occur by using
Equation (11.13),
$v(t)={v}_{0}+at$
:
Time to reach apex
$$\begin{array}{rl}v(t)& =0,\\ {v}_{0}+at& =0,\\ t& ={v}_{0}/a.\end{array}$$
Right now we are in one dimension and considering only the height. But if we are in more than
one dimension, only the velocity parallel to the acceleration must vanish. There could be
horizontal velocity, for example. We discuss projectile motion in more than one dimension in just
a moment.
The second observation is that the time it takes for the object to travel from its maximum
altitude to its initial altitude, denoted
${t}_{e}$
in Figure 11.12, is
the same as the time taken to reach the maximum. In other words, the projectile reaches its apex
at
${t}_{e}/2$
.
The third and final observation is that the velocity at
$t={t}_{e}$
, which we have denoted
${v}_{e}$
, has
the same magnitude as the initial velocity
${v}_{0}$
, but the opposite sign.
Before we look at projectile motion in more than one dimension, let's summarize the formulas we
have derived in this section. The first two are the only ones worth memorizing; the others can
be derived from them.
Summary of Kinematics Equations Dealing with Constant Acceleration
$$\begin{array}{rl}v(t)& ={v}_{0}+at,\\ \mathrm{\Delta}x& ={v}_{0}t+(1/2)a{t}^{2},\\ x(t)& ={x}_{0}+\mathrm{\Delta}x={x}_{0}+{v}_{0}t+(1/2)a{t}^{2},\\ {v}_{0}& =\mathrm{\Delta}x/t(1/2)at,\\ \text{(11.17)}& t& ={\displaystyle \frac{{v}_{0}\pm \sqrt{{v}_{0}^{2}+2a\mathrm{\Delta}x}}{a}},a& =2{\displaystyle \frac{\mathrm{\Delta}x{v}_{0}t}{{t}^{2}}}.\end{array}$$
Extending the ideas from the previous section into 2D or 3D is mostly just a matter of switching
to vector notation;
$x$
,
$v$
, and
$a$
become
$\mathbf{p}$
,
$\mathbf{v}$
, and
$\mathbf{a}$
,
respectively.
Of course, the time
$t$
remains a scalar:
Equations for motion under constant acceleration, in
vector form
$$\begin{array}{rl}\mathbf{v}(t)& ={\mathbf{v}}_{0}+t\mathbf{a},\\ \text{(11.18)}& \mathrm{\Delta}\mathbf{p}& ={\mathbf{v}}_{0}t+({t}^{2}/2)\mathbf{a},\text{(11.19)}& \mathbf{p}(t)& ={\mathbf{p}}_{0}+\mathrm{\Delta}\mathbf{p}={\mathbf{p}}_{0}+t{\mathbf{v}}_{0}+({t}^{2}/2)\mathbf{a},{\mathbf{v}}_{0}& =\mathrm{\Delta}\mathbf{p}/t(1/2)at,\\ \mathbf{a}& =2{\displaystyle \frac{\mathrm{\Delta}\mathbf{p}t{\mathbf{v}}_{0}}{{t}^{2}}}.\end{array}$$
Note that we didn't make a vector version of
Equation (11.17); we'll get to that in a
moment.
This seemingly trivial change in notation is actually hiding two rather deep facts. First, in the
algebraic sense, the vector notation is really just shorthand for sets of parallel scalar
equations for
$x$
,
$y$
, and
$z$
. The important point is that the three (Cartesian) coordinates
are completely independent of one another. For example, we can make calculations
regarding
$y$
and completely ignore the other dimensions, provided that the hypothesis of
constant acceleration is met for the object's motion. If it were not for the independence of the
coordinates, we could not make this change in notation. The second fact hidden in this notation
is that, when we view the vectors in the equations above as geometric rather than algebraic
entities, the particular coordinate system used to describe those vectors is irrelevant. We
don't even need to specify one. Of course, this is a basic principle of physics: Mother Nature
doesn't know what coordinate system you are using.
We were able to leap from 1D to 3D mostly just by bolding a few letters due to the independence
of the coordinates. However, there is a bit more to say about projectile motion in multiple
dimensions because there are situations where we need to consider the effects of all the
coordinates at the same time. One situation has already been alluded to by the lack of a vector
equation corresponding to Equation (11.17).
In other words, how could we solve for time
$t$
given a displacement
$\mathrm{\Delta}\mathbf{p}$
,
acceleration
$\mathbf{a}$
, and initial velocity
${\mathbf{v}}_{0}$
? In one dimension, the projectile
is “confined” and basically cannot help but hitting the target implied by
$\mathrm{\Delta}x$
. But in
two or more dimensions, the situation is more complicated. The increase in complexity that
attends the increase in dimensions is analogous to computing the intersection of two rays (see
Section A.8). In 2D, any two rays must intersect unless they
are parallel, whereas in 3D, the possibility exists for skew rays, which are not parallel
but do not intersect.
For example, earlier we computed how long it would take for a ball bearing dropped from a great
height to hit the sidewalk below, which is a onedimensional problem. The corresponding
threedimensional problem would be to try to drop the ball bearing into a bucket which is free to
move around on the sidewalk. Let's say that the bucket is off to our left. Our initial velocity
had better have some leftward component then, or else the ball bearing won't land in the bucket.
Another indication that the multidimensional case is more complicated than 1D is that a direct
translation of Equation (11.17) into vector
form results in nonsensical operations of taking the square root of a vector and dividing one
vector by another.
The key to solving this problem is to realize that any horizontal changes (either to the bucket's
position or the initial velocity of the ball bearing) do not affect how long it takes the
ball bearing to reach the sidewalk. This is because the coordinates are independent from one
another. The horizontal velocity and acceleration do not interact with the vertical velocity and
acceleration. To be specific, let's switch to our standard 3D coordinate system, which has
$+y$
pointing up and
$x$
and
$z$
in the horizontal plane. The time it takes the ball bearing to reach
the altitude of the bucket depends only on the equations having to do with
$y$
; the
$x$

and
$z$
coordinates can be ignored for this purpose. In other words,
calculating the time when a projectile will reach a target is still a onedimensional
calculation—we just need to chose which direction to use. We can apply
Equation (11.17) to solve for a time of
impact
$t$
. But this solution is just a proposal. We know that if the projectile were to
hit the target, it would do so at this time. To make sure we really did hit the target, we must
plug this time of alleged impact into
Equation (11.19) to see where the projectile will be at
that location, and verify that the position of the projectile is within appropriate tolerances.
Let's talk a bit more about exactly what it means to “chose which direction to use,” as was
stated in the previous paragraph. In cases of simple projectile motion, such as the ballbearing
example, where gravity is the constant acceleration, the direction to choose is obvious: use the
direction of gravity. Furthermore, because coordinate systems are chosen such that “up” is one
of the cardinal axes, the process of solving a onedimensional problem in that direction is a
trivial matter of plucking out the appropriate Cartesian coordinate and discarding the others. In
general, however, the situation can be more complicated. But before we discuss the details of
the general case, there are a few more things we can say about this very important and common
special situation.
To study projectile motion where acceleration is solely due to gravity, which is a constant and
acts along a cardinal axis, let's establish a 2D coordinate space where
$+y$
is up and
$x$
is the
horizontal axis. Without loss of generality we can rotate our plane such that it contains the
initial velocity, and thus the entire trajectory of the particle. We choose
$+x$
in the
horizontal direction of the initial velocity. We also simplify things by setting the origin at
the object's initial position. This notation (along with a few other items that we'll need in a
moment) are illustrated in Figure 11.13.
Reviewing the notation in Figure 11.13, we see that we can express the
position of the particle as a function of time either as
$\mathbf{p}(t)$
, or we can refer to an
individual coordinate with
$x(t)$
and
$y(t)$
. Instantaneous velocity (not shown on the diagram),
can be denoted in vector form either as
$\mathbf{v}(t)$
or using derivative notation as
$\dot{\mathbf{p}}(t)$
. The scalar velocity components will be denoted using derivative notation
as
$\dot{x}(t)$
and
$\dot{y}(t)$
. The initial position and velocity will be denoted by adding a
subscript 0 (
${\dot{y}}_{0}$
is the initial vertical velocity). We denote the acceleration due to
gravity as either
$g$
or
$\mathbf{g}$
.
Let's state the equations for velocity and position using the notation just described:
$$\begin{array}{}\text{(11.20)}& \dot{\mathbf{p}}(t)& ={\mathbf{v}}_{0}+t\mathbf{g},& \dot{x}(t)& ={\dot{x}}_{0},& \dot{y}(t)& ={\dot{y}}_{0}+gt,\text{(11.21)}& \mathbf{p}(t)& =t{\mathbf{v}}_{0}+({t}^{2}/2)\mathbf{g},& x(t)& =t{\dot{x}}_{0},& y(t)& =t{\dot{y}}_{0}+(1/2)g{t}^{2}.\end{array}$$
The distances labeled
$h$
and
$d$
in Figure 11.13 are often of interest;
they are the apex height and horizontal travel distance, respectively. As discussed earlier in a
onedimensional context, the maximum height is reached when all of the initial velocity in the
upwards direction has been consumed by gravity, in other words when
$\dot{y}(t)=0$
. This occurs
at time
Time to reach apex
$${t}_{a}={\dot{y}}_{0}/g,$$
and at this time, the height is equal to
Altitude at apex
$$\begin{array}{rl}h& =y({t}_{a})={t}_{a}{\dot{y}}_{0}+(1/2)g{t}_{a}^{2}\\ & =({\dot{y}}_{0}/g){\dot{y}}_{0}+(1/2)g({\dot{y}}_{0}/g{)}^{2}\\ & =({\dot{y}}_{0}^{2}/g)+(1/2)({\dot{y}}_{0}^{2}/g)\\ & ={\dot{y}}_{0}^{2}/2g.\end{array}$$
We stated earlier that the time for the object to come back down to its initial height (which we
denoted
${t}_{e}$
) was twice the time needed to reach its apex; however, at that time we merely
appealed to a diagram. This time, let's verify it algebraically:
Time to return to
initial altitude.
$$\begin{array}{rl}y(t)& =t{\dot{y}}_{0}+(1/2)g{t}^{2},\\ 0& ={t}_{e}{\dot{y}}_{0}+(1/2)g{t}_{e}^{2},& & \text{(initial position is at the origin)}\\ (1/2)g{t}_{e}^{2}& ={t}_{e}{\dot{y}}_{0},\\ {t}_{e}& =2{\dot{y}}_{0}/g.& & \text{(divide by}(1/2)g{t}_{e}\text{)}\end{array}$$
As expected, the flight time
${t}_{e}$
is twice the time needed to reach the apex. Now, let's
compute
$d$
, the horizontal distance traveled:
Horizontal travel distance
$$d=x({t}_{e})={t}_{e}{\dot{x}}_{0}=2{\dot{y}}_{0}{\dot{x}}_{0}/g.$$
Of course,
${t}_{e}$
and
$d$
are based on the assumption that we want to know when the projectile
returns to its initial altitude. This is important when launching a projectile from a flat
ground plane. If the projectile isn't launched from the ground, or if the ground isn't flat,
then we'll need to consider where the parabola intersects the ground plane.
We often wish to specify the initial velocity in terms of an angle and speed,
rather than velocities along each axes. In other words, we wish to use polar coordinates rather
than Cartesian. As shown in Figure 11.13, we denote the initial launch
speed as
${s}_{0}$
(which is equal to the magnitude of
${\mathbf{v}}_{0}$
) and the launch angle as
$\theta $
. Converting the initial velocity from Cartesian to polar coordinates (see
Section 7.1.3 if you don't remember how), we get
$${\dot{x}}_{0}={s}_{0}\mathrm{cos}\theta ,\phantom{\rule{.5in}{0ex}}{\dot{y}}_{0}={s}_{0}\mathrm{sin}\theta .$$
Plugging this into our kinematics
Equations (11.20) and
(11.21), we get the equations of motion for a
projectile in terms of its launch angle and speed:
$$\begin{array}{rlrl}\dot{x}(t)& ={s}_{0}\mathrm{cos}\theta ,& \dot{y}(t)& ={s}_{0}\mathrm{sin}\theta +gt,\\ x(t)& =t{s}_{0}\mathrm{cos}\theta ,& y(t)& =t{s}_{0}\mathrm{sin}\theta +(1/2)g{t}^{2}.\end{array}$$
We can also express
${t}_{e}$
,
$h$
, and
$d$
in terms of
${s}_{0}$
and
$\theta $
:
Important quantities in projectile motion, expressed in terms of launch angle and speed
$$\begin{array}{rlrl}{t}_{a}& ={\dot{y}}_{0}/g=({s}_{0}\mathrm{sin}\theta )/g& & ={s}_{0}(\mathrm{sin}\theta )/g,\\ {t}_{e}& =2{\dot{y}}_{0}/g=2({s}_{0}\mathrm{sin}\theta )/g& & =2{s}_{0}(\mathrm{sin}\theta )/g,\\ d& =2{\dot{y}}_{0}{\dot{x}}_{0}/g=2({s}_{0}\mathrm{sin}\theta )({s}_{0}\mathrm{cos}\theta )/g& & =2{s}_{0}^{2}(\mathrm{sin}\theta )(\mathrm{cos}\theta )/g,\\ h& =(1/2){\dot{y}}_{0}^{2}/g=(1/2)({s}_{0}\mathrm{sin}\theta {)}^{2}/g& & ={s}_{0}^{2}({\mathrm{sin}}^{2}\theta )/{\textstyle \phantom{\rule{thinmathspace}{0ex}}}2g.\end{array}$$
These equations are highly practical because they directly capture the relationship between the
“userfriendly” quantities of launch speed, launch angle, flight time, and flight distance.
At this point, let's pause to make an interesting observation about the relationship between the
initial speed
${s}_{0}$
and the horizontal distance traveled
$d$
. It's a quadratic relationship,
meaning when we increase
${s}_{0}$
by a factor of
$k$
, we increase
$d$
by a factor of
${k}^{2}$
. It
might seem more natural for the relationship to be linear, meaning that
$d$
would increase by the
same factor
$k$
. We can understand the quadratic relationship by breaking the initial velocity
into its horizontal and vertical components, denoted earlier as
${\dot{x}}_{0}$
and
${\dot{y}}_{0}$
,
respectively. It's not difficult to see that increasing
${\dot{x}}_{0}$
will increase
$d$
by the
same factor. Less obvious is that the same is true for
${\dot{y}}_{0}$
. This is true because the
duration that the object is airborne is proportional to
${\dot{y}}_{0}$
. So if we increase the
vertical velocity, we give the object more time to travel. Thus any scale factor we apply to
$s$
will affect the distance twice, once as a result of the increased ground velocity due to
${\dot{x}}_{0}$
, and again as a result of the increased travel time due to
${\dot{y}}_{0}$
. This
produces a quadratic relationship between
$s$
and
$d$
.
Now let's return to a question we put on hold from earlier: how might we determine the point of
impact for any arbitrary vectors
$\mathrm{\Delta}\mathbf{p}$
,
$\mathbf{a}$
, and
${\mathbf{v}}_{0}$
? We said
before that the key was to “choose a direction” and solve a onedimensional problem in that
direction. If a cardinal direction is chosen, we just throw out the other coordinates. For an
arbitrary direction, we project the problem onto a line in that direction. Any component of
displacement, velocity, or acceleration perpendicular to that line is discarded during the
projection. We learned how to project onto a line
and measure displacement in a particular direction by using the dot product in
Section 2.11. All that is left is to select a direction.
Assuming the projectile hits the target, we will get the same value for
$t$
no matter what
direction we choose. But that doesn't mean the choice is irrelevant. For example, in the
ballbearing example, it would be a disaster to chose the
$+x$
or
$+z$
directions, since there is
no acceleration in either of those directions and application of
Equation (11.17) would result in a division
by zero. This suggests the strategy of simply using
$\mathbf{a}$
itself as the direction of
projection. To do this, we dot each vector quantity with
$\mathbf{a}$
, making the substitutions
$\mathrm{\Delta}x=\mathrm{\Delta}\mathbf{p}\cdot \mathbf{a}$
,
$v=\mathbf{v}\cdot \mathbf{a}$
, and
$a=\mathbf{a}\cdot \mathbf{a}$
. Then these scalar quantities can be plugged into
Equation (11.17).
Exercise 10 explores this in more detail.
11.7The Integral
We have just showed that the total displacement of an object in a time interval is equal to the
area under the plot of the object's velocity. We used the example of constant acceleration, which
has a simple graph, and the area was easy to solve geometrically. We did not pursue in further
generality the limit argument that led us to the surprising equivalence, because this special
case has such compelling applications. Now we are ready to discuss more general cases. The need
to compute a “continuous summation,” where the rate of growth is a known function, is a common
concept in engineering and science. The calculus tool used to compute these sums is the
integral.
If you have already studied integral calculus and have a good intuition about what the integral
is used for, then you can safely skip ahead to Section 11.8, when our
focus returns to the subject of mechanics. However, if you've never had integral calculus or if
your intuition about the integral is a bit shaky, keep reading.
There are two important ways of approaching the integral. The first way is essentially to make
the notion of “summing up many tiny elements” a bit more precise and introduce some
mathematical formalism. The other way is to compare the integral to the derivative. It's
important to understand both interpretations. The integral is a bit more difficult to grasp than
the derivative, but for reasons that become apparent later, it plays a much greater role in
physics simulations and many other areas of video game programming. Understanding what the
integral does is very important, even if the vast assortment of penandpaper techniques
to compute integrals analytically is not very useful in our case, being replaced instead by
techniques of numerical integration.
Let's turn our informal summation into mathematics notation, in which we compute the area under
the curve
$f(x)$
in the interval
$a\le x\le b$
. We partition this interval into
$n$
slices,
each having the width
$\mathrm{\Delta}x=(ba)/n$
. The
$i$
th rectangle will have a lefthand
coordinate
${x}_{i}$
, a height equal to
$f({x}_{i})$
, and an area of
$f({x}_{i}){\textstyle \phantom{\rule{thinmathspace}{0ex}}}\mathrm{\Delta}x$
.
Using summation notation, we add up all these rectangles:
$$\mathrm{A}\mathrm{r}\mathrm{e}\mathrm{a}\approx \sum _{i=1}^{n}f({x}_{i})\mathrm{\Delta}x.$$
The error in this approximation decreases as we increase the number of slices
$n$
, and by now,
unless you're the new kid in town, you know that we need to take it to the limit one more
time.
By taking the limit as