University Physics

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University Physics I

Author: Jeﬀrey R. Schmidt Aﬃliation: University of Wisconsin

c

⃝ 2001, revision August 27, 2012

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Forward

Many of us, faculty and students, are weary of being forced to migrate to a new edition of university physics textbooks every year, diﬀering only in price and weight. There is virtually no diﬀerence between edition X and edition X + 1 of any of the standard texts, they are all very much the same, and the same from edition to edition, in all respects except for price and weight (I recently got an evaluation copy of a physics 201/202 text that weighs 15 pounds, and runs over 200.00 retail price). The only way out of this charade is to provide students with a free alternative, that I can update every year at no cost to anyone, and so here it is, University Physics I, a very bare-bones, work in progress, but free physics text. A lot of institutions and individuals around the country are engaged in similar projects. I make no apologies for the appearance that this text is highly mathematical. If you take any standard text and cut out all of “filler” and pointless anecdotes and superfluous photos, the results will be highly mathematical in appearance, and will be about the length of this book. I have concentrated on problem-solving and expression of the basic concepts, since that is what the student really needs, and that is what testing and assessment focus on. I will remind the student that doing the homework is not optional, but is mandatory. You can’t master something by watching others do it.

Times are definitely changing. In 1961 the number of hours per week spent on academic study by full-time

college students was 40 hours per week outside of class. In 2004 that number had dropped to 27 hours per

week outside of class, in both cases the full-time load was 16hours/credits. Expectations have evolved similarly:

in 1961 the average grade earned in a college class was a C (earned by over 40% of the students in the course), and by 2007 the average over 40% were awarded A-grades, and only 7% were awarded C-grades, with private universities and colleges leading the way in grade inflation, awarding 50% A-grades on average. Polls of students conducted in 2007 reveal that 40% of students expect to get B-grades for simply attending the course, without regard to

performance. This is an average over all disciplines, in the pure sciences the numbers are much less ludicrous.

College really does diﬀer from high-school. I hear a lot of students complain that they can’t put in study time because of work, or because they “have lives”. Here is some shocking news for those students; everybody has a job, and everybody “has a life”, yet the ones that will succeed will be the ones that take their studies as seriously as they take their jobs and any other aspects of their lives. The current generation of students is not unique, they don’t face hardships that previous generations did not, and nothing more is expected of them than what was expected of prior generations. And nothing less is expected of them! In fact the current generation has a lot of time-saving technology at its disposal, and good use should be made of this advantage that generations before did not have. I have faith in you, you can do it. All it takes is discipline.

Speaking of this technological advantage, this year (2010) I integrated quite a bit of symbolic computation involving free software such as REDUCE, axiom, maxima, and ode into the course, partly as a way of providing math assis-tance to students, and to provide the tools for pushing the basic physics forward into other disciplines, in support of those disciplines and the students studying them. Problems and examples labeled with ∗ are very similar to problems appearing on the physics GRE. I have added a great deal of optional material specifically for architecture, engineering, biomedical, chemistry and geoscience/hydrology students. Most of this material will not be covered in lecture, we only have time for the basics, and much of this optional material requires some mathematical background beyond that required for the standard portion of the course. This optional material is in chapters 10, 11, 13

and 14 and is clearly marked as “not covered in 201”, but is there for you if you need it. I regard the

required mathematical background for the standard 201 course to be rather minimal (one semester of calculus, what was called freshman math when your parents were in college), others might say one semester is inadequate, but one semester of calculus is the nation-wide norm for this course. Math refreshers and reviews are integrated into

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the text. The fact is that you can’t do anything of any significance in the sciences without mathematics. You must deal with it. Learn the mathematics, or get used to making way for those who have learned it. Harsh words,

but sound advice; I am on your side, which should be evident from the fact that I wrote this book for you, and I have combined it with as many tools as I can to make your job easier. But easier is not easy, and you will have to spend the time outside of class, do the homework, and learn the required mathematics.

In college at the University of Wisconsin-Madison in the late 70′s-early 80′s I had a good friend and classmate, Donn Henriksen. You will always treasure the short time that you spend in college, four (or five!) very short but important years. For me they passed quickly, I graduated, and lost track of many good people as we began our careers and families and adult lives. After two decades, I was able to re-established contact with some, Donn included, thanks to the Internet (you see, it is good for something more than copyright-violation).

In the fall of 2006 Donn was diagnosed with late-stage stomach cancer, and by May of 2007, he had passed away. Donn managed to remain a true intellectual all of his life, his thirst for learning never diminished. He read through these notes, solved the problems, and uncovered quite a few typos in the winter of 2006/2007 as he refreshed his understanding of the physics that we learned as classmates way back in 1976, and endured his treatments. These notes, which will soon become a real book (but a free one) are dedicated to the memory of my friend Donn, whose desire to keep learning as a life-long pursuit should serve as a model to which all students should aspire.

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Kinematics

Kinematics is the description of motion without any consideration as to what causes the motion. It is simply the mathematical description of where an object is at time t, and how fast it is moving, and in what direction.

The information about where the moving object is can be presented in two simple ways, as a graph of its position versus time, or as a mathematical function, position as a function of time. We will start with the very simplest type of motion, that which takes place on a straight line, which we will call one-dimensional motion.

1.1 Motion on a line

The mathematical description of motion in one dimension is incredibly simple, we specify the position x(t) of an object on a coordinate line as a function of time. This function can be arbitrary, but it must at least be unicursal (what we think of intuitively as continuous), meaning that it can be drawn without lifting the pen from the paper

lim dt→0 ( x(t + dt)− x(t) ) = 0, ∀ t (1.1)

so that if you are at x = 1.0 m (for example) at t = 0 s, you can’t spontaneously jump to some other point in an infinitismal time (no teleportation, no transporter beams, and so forth). We do not necessarily require x(t) to be

smooth, which would to say

lim dt→0 (_{x(t + dt)}_{− x(t)} dt ) ∀ t (1.2)

exists and is bounded. For example the motion

x(t) =    1.0 + 2.0 t 0≤ t ≤ 1 3 + 4(t− 1) 1 ≤ t ≤ 2 7 + 3(t− 2) 2 ≤ t (1.3)

is just fine, it describes motion at constant velocity, then at t = 1 s and again at t = 2 you change velocity. Motion graphs of x(t) versus t show this.

This brings us to a set of important ideas. The

dis-placement ∆x of an object between times t1and t2is the change in its position

∆x = x(t2)− x(t1) (1.4) and the average velocity of the object over this

inter-val is the rate with which its displacement changed 00.0 0.5 1.0 1.5 2.0 2.5 3.0 2 4 6 8 10 t, sec x(t), m 1

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¯

v(t1, t2) =

x(t2)− x(t1) t2− t1

(1.5) This is positive if the object moves to the right (x(t2) > x(t1)) and is negative if it moves left, and so has a magnitude called the average speed

¯ s(t1, t2) = x(t2)− x(t1) t2− t1 (1.6)

and a direction indicated by its sign. The best way to think of average speed is as arc-length traveled per unit time, in one and two dimensions respectively this will be

¯ s(t1, t2) = ∫ t2 t1 √ v2 _dt, _¯_s(t 1, t2) = ∫ t2 t1 √ v2 x+ v2y dt (1.7)

and you might need to be a bit careful in your calculation of√v2_.

Instantaneous velocity is the limit of average velocity as t2= t1+ dt→ t1, or as dt→ 0 v(t1) = lim dt→0 x(t1+ dt)− x(t1) dt = dx dt(t1) (1.8)

in other words, the derivative of x(t1) evaluated at t1. The end of this chapter contains a comprehensive diﬀerential calculus review.

Acceleration measures how fast the velocity changes. The average acceleration is

¯ a(t1, t2) = v(t2)− v(t1) t2− t1 (1.9) and instantaneous is a(t1) = lim dt→0 v(t1+ dt)− v(t1) dt = dv dt(t1) (1.10)

A position function must be continuous and smooth if you want to find its instantaneous velocity at any time.

Example 1. For the motion described by Eq. 1.3 find the displacement and average speed between t = 0 and t = 3.

Your first step is to get v(t) = dx

dt(t) for each segment v(t) =    2.0 0≤ t ≤ 1 4 1≤ t ≤ 2 3 2≤ t (1.11)

and integrate from ti = 0 to tf = 3 , breaking up the integral (see the Chapter 1 appendix, Eq. 1.131) into parts over which v(t) is constant

∆ = ∫ 3 0 v(t) dt = ∫ 1 0 2 dt + ∫ 2 1 4 dt + ∫ 3 2 3 dt = 9 m, v =¯ ∆ T = 9 3 = 3 m s (1.12)

(can you see why ¯s = ¯v for this example?).

Example 2. Lets extend this example, suppose that

v(t) =    2.0 0≤ t ≤ 1 −4 1 ≤ t ≤ 2 3 2≤ t (1.13) Then ∆ = ∫ 3 0 v(t) dt = ∫ 1 0 2 dt− ∫ 2 1 4 dt + ∫ 3 2 3 dt = 1 m, v =¯ ∆ T = 1 3 m s ¯ s = 1 3 s ( ∫ 3 0 v(t) dt = ∫ 1 0 2 dt + ∫ 2 1 4 dt + ∫ 3 2 3 dt ) = 3m s (1.14)

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1.1. MOTION ON A LINE 3

Example 3. Consider the one-dimensional motion

x(t) =    3.0 t2 ₀_{≤ t ≤ 1} 3 + 6(t− 1) 1 ≤ t ≤ 4 15 3≤ t (1.15)

To get v(t), we diﬀerentiate this once;

v(t) =    6.0 t 0≤ t ≤ 1 6 1≤ t ≤ 4 0 3≤ t (1.16)

Lets use gnuplot as described in problem 1.35 to graph both functions

f1(x)=(x<=1) ? 3*x**2 : 0 f2(x)=(x>1) ? 3+6*(x-1) : f1(x) f3(x)=(x>3) ? 15 : f2(x) plot[0:4][-1:7] f3(x) 0 2 4 6 8 10 12 14 16 0 0.5 1 1.5 2 2.5 3 3.5 4 f3(x) f1(x)=(x<=1) ? 6*x : 0 f2(x)=(x>1) ? 6 : f1(x) f3(x)=(x>3) ? 0 : f2(x) plot[0:4] f3(x) -1 0 1 2 3 4 5 6 7 0 0.5 1 1.5 2 2.5 3 3.5 4 f3(x)

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1.1.1 Motion at constant velocity

Motion at constant velocity is very simple, the position function has the generic form

x(t) = x0+ v0t (1.17)

Compute the average velocity

¯

v(t1, t2) =

(x0+ v0t2)− (x0+ v0t1) t2− t1

= v0 (1.18)

The object moves at a constant rate (speed) in the same direction, forever. The instantaneous velocity is exactly the same v(t1) = lim dt→0 (x0+ v0(t1+ dt))− (x0+ v0t1) (t1+ dt)− t1 = v0 (1.19)

and the accelerations (average and instantaneous) are of course zero. There is nothing more to this type of motion. What is the meaning of x0? It is the position when the clock reads zero, x(0) = x0+ v0· 0 = x0.

1.1.2 Motion at constant acceleration

Motion at constant acceleration is far more interesting (it is the motion occurring when constant forces are applied to a body), the position function has the generic form

x(t) = x0+ v0t + 1 2at

2 _(1.20)

Compute the average velocity

¯ v(t1, t2) = (x0+ v0t2+1₂at22)− (x0+ v0t1+1₂at21) t2− t1 = v0+ a( t2+ t1 2 ) (1.21) and instantaneous v(t1) = lim dt→0 (x0+ v0(t1+ dt) +1₂a(t1+ dt)2)− (x0+ v0t1+1₂at21) (t1+ dt)− t1 = v0+ a t1 (1.22) The average and instantaneous accelerations are the same

¯ a(t1, t2) = (v0+ at2)− (v0+ at1) t2− t1 = a = d dt(v0+ at) t=t1 (1.23)

What is the meaning of x0? It is the position when the clock reads zero, x(0) = x0+ v0· 0 +1₂a· 02= x0. What is the meaning of v0? It is the velocity when the clock reads zero v(0) = v0+ a· 0 = v0.

One of the most important generic classes of problems in one-dimensional kinematics is the problem of when, where and if two moving objects with prescribed trajectories will meet or collide.

Example 4. A ball is thrown from ground level straight up into the air. A time T later another is thrown up at the

same speed. At what height do they collide.

The first object, launched at t = 0, will have height versus time of y1(t) = v0t−

1 2gt

2 _(1.24)

The second ball travels the same path, but reaches each height at a time T after the first ball reaches that same height, and so

y2(t) = v0(t− T ) − 1

2g(t− T )

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1.1. MOTION ON A LINE 5

0

2

4

6

8

0

10

20

30

40

50

60 t, (s)

y

1

(t), y

2

(t), (m)

Plots of these for v0 = 30m_s and T = 2s are illustrated.

From the graph we can immediately obtain a useful bit of data; the collision takes place when the first projectile is on its way back, having al-ready peaked in height. The algebraic solution is gotten by finding the time at which both projectiles reach the same height;

y1(t∗) = y2(t∗) or v0t_∗− 1 2gt 2 ∗ = v0(t_∗− T ) − 1 2g(t∗− T ) 2_(1.26)

solving, we discover that the collision occurs at t_∗=v0 g + T 2, at a height y∗= v0t∗− 1 2gt 2 ∗= v2 0T 2g − gT2 8 (1.27)

0

2

4

6

8

0

50

100

150

200 t, (s)

y

1

(t), y

2

(t), (m)

Example 5. A car stopped at a traﬃc light

begins at rest. At time t = 0 the light turns green and the car accelerates with acceleration a. At just that instant a car traveling at speed v0 speeds through the light.

How much time does it take the accelerating car to catch the speeder, and how far from the light does this occur?

A graph of both position versus time curves for the vehicles y1(t) = 1 2at 2 _(1.28) and y2(t) = v0t (1.29) are below for v0= 30m_s and a = 10_sm2.

We can see from the slope of the graph that the accelerating car is traveling much faster than the speeder when they finally meet, which takes place when they arrive at the same point at the same time;

y1(t∗) = y2(t∗), or v0t∗= 1 2at

2

∗ giving t∗= 2v_a0, and location y∗= v0t∗= 2v2

0

a (1.30)

at which time the accelerating car has velocity

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which we could see from the graph.

Example 6. Train A traveling at speed vA is a distance D from a switch in the track. At that instant train B enters the track traveling in the same direction but at a greater speed vB. What is the minimal acceleration that the engineer of train B must apply in order to avoid a collision?

0

2

4

6

8

0

50

100

150

200 t, (s)

x

A

(t), x

B

(t), (m)

The position versus time of the trains is xA(t) = D + vAt, and xB(t) = vbt−1

2at 2 (1.32) graphs of these for vB = 2vA = 30m_s and a = 2m_s2 ,D = 50 are illustrated.

Our new condition is that when xA(t_∗) = xB(t_∗)

that vA(t∗) = vB(t∗) (1.33) so that there is no relative velocity between the trains, hence no collision. This means

vA = vB− at_∗) or t_∗ = vB− vA a (1.34)

0

2

4

6

8

0

50

100

150

200 t, (s)

x

A

(t), x

B

(t), (m)

inserting this into

D + vAt∗= vbt∗− 1 2at 2 ∗ (1.35) results in a =(vB− vA) 2 2D (1.36)

The slopes of the two curves must also match up when the trains meet, they will be tangent to one another as in the figure.

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1.2. ARBITRARY MOTION 7

1.2 Arbitrary motion

1

We will study motion at constant acceleration (due to constant forces acting on a body) for the first half of the course, but you have to realize that acceleration may not be constant, it can be arbitrary. The problem of describing the position of a body given its acceleration (given the forces acting on it) is one of the fundamental problems of this course. Mathematically, it amounts to solving diﬀerential equations, in fact this very problem is what calculus was invented for. Let me illustrate two rather simple and universal approaches based on the idea that the motion will be smooth, meaning that all of the derivatives of x(t) will exist.

Suppose that the acceleration of a body depends on its position; a simple example is motion caused by a spring or restoring force

a(t) =−ω2x(t) (1.37)

in which ω is a constant with units of 1/s, subject to x(0) = 1 and v(0) = 0. We will use one of the most powerful tools of calculus, the series expansion (see the appendix to this chapter), assume that all of the derivatives of x exist, and write

x(t) = x0+ x1t + 1 2x2t 2₊ 1 3!x3t 3₊ 1 4!x4t 4₊ 1 5!x5t 5₊_{· · ·} _(1.38)

in which xi are all constants. Note first that

x(0) = x0= 1 Then v(t) = dx dt(t) = x1+ x2t + 1 2!x3t 2₊ 1 3!x4t 3₊_{· · ·} _(1.39)

and again examine things at t = 0

v(0) = x1= 0 and compute a(t)

a(t) =d 2_x dt2(t) = x2+ x3t + 1 2!x4t 2₊ 1 3!x5t 3₊_{· · ·} _(1.40)

and simply put these expansions into the equation of motion and solve in succession for the constants

by matching coeﬃcients of like powers of t

a(t) = −ω2x(t) x2+ x3t + 1 2!x4t 2₊ 1 3!x5t 3₊_{· · · = −ω}2(_{1 + 0 t +}1 2x2t 2₊ 1 3!x3t 3₊ 1 4!x4t 4₊ 1 5!x5t 5₊_{· · ·}) _(1.41) We obtain coefficient of t0; x2 = −ω2 coefficient of t1_; _x 3 = 0 coefficient of t2; 1 2!x4 = −ω 21 2x2= ω 4 coefficient of t3; 1 3!x5 = −ω 21 3!x3= 0 · · · (1.42)

and if we put things together, and actually paid attention in our calculus courses, we might even recognize that the series adds up to some well-known function

x(t) = 1−ω 2 2!t 2₊ω 4 4!t 4₊_{· · · = cos(ωt)} _(1.43)

This technique almost never fails, and is very simple and mechanical to carry out, its only drawback is that the

answer is a series, which you might not recognize how to sum up.

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The second approach (very powerful and universal, see Chapter 6) makes use of the product rule of calculus. Begin with the same equation,

a(t) = −ω2x(t) multiply by v(t) a(t) v(t) = −ω2x(t) v(t) both sides are t derivatives d

dt (₁ 2v 2_(t)) ₌ _−ω2d dt (₁ 2x 2_(t)) integrate ∫ t 0 d dt (₁ 2v 2_(t))_dt ₌ _−ω2 ∫ t 0 d dt (₁ 2x 2_(t))_dt 1 2 ( v2(t)− v2(0) ) = −ω 2 2 ( x2(t)− x2(0) ) (1.44) and solve for v(t)

v(t) = dx dt = √ v2₍₀₎− ω2 ( x2_(t)− x2₍₀₎ ) ∫ dt = ∫ dx √ v2₍₀₎_{− ω}2(_x2_(t)_{− x}2₍₀₎) (1.45)

and you have reduced the problem to doing an integral to get x(t). This will be our preferred technique (the

energy method) which we will expand upon when the concept of energy is introduced. Suppose that v(0) = 0 and

x(0) = 1; ∫ t 0 dt = ∫ x 1 dx √ ω2(₁− x2), t = 1 ωcos −1_x, _{x = cos(ωt)}

We can reduce both of these approaches to a simple formula provided the acceleration is an explicit function of t (as opposed to implicit such as a = a(x) = a(x(t))). All we do is integrate;

dv dt = a(t), ∫ v(t) v(0) dv = ∫ t 0 a(t′) dt′ v(t) = dx dt = v(0) + ∫ t 0 a(t′) dt′ ∫ x(t) x(0) dx = ∫ t 0 v0dt + ∫ t 0 ( ∫ t′ 0 a(t′′) dt′′ ) dt′ x(t) = x(0) + v(0) t + ∫ t 0 ( ∫ t′ 0 a(t′′) dt′′ ) dt′ (1.46)

and all you need to do is the integrals, for example let a(t) = α t in which α is a constant

x(t) = x(0) + v(0) t + ∫ t 0 ( ∫ t′ 0 α t′′dt′′ ) dt′= x(0) + v(0) t + ∫ t 0 α 2t ′2_dt′ _{= x(0) + v(0) t +}α 6t 3

or suppose a(t) = a0, a constant

x(t) = x(0) + v(0) t + ∫ t 0 ( ∫ t′ 0 a0dt′′ ) dt′ = x(0) + v(0) t + ∫ t 0 a0t′dt′= x(0) + v(0) t + a0 2 t 2

Formulas! Hurrah, your favorite things! You can’t make an omelet without breaking some eggs, and you can’t do science without a little bit of math.

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1.3. VECTORS 9

1.3 Vectors

Most of the quantities that you will work with in this course have direction as well as a magnitude associated with them, since we will want to describe motion in two or three dimensions. Such quantities are called vectors. A quantity with no direction associated with it, that is unaltered by any changes in coordinate referential, is called a

scalar. The most important scalars in classical physics are mass and temperature.

Typical vectors are forces, positions, velocities and momenta. The common representation of a vector is as a point in space such as in the figure.

The point is specified by giving its x and y coordinates, either as an ordered pair (triple) or as multiples of unit vectors. We could specify the vector in the figure by giving Cartesian coordinate of its tip

v = (vx, vy) = vxi + vyj v = (vx, vy, vz) = vxi + vyj + vzk i = (1, 0, 0) j = (0, 1, 0) k = (0, 0, 1) (1.47)

from the Pythagorean theorem we obtain its length (magnitude)

|v| =√v2

x+ v2y, or |v| = √

v2

x+ vy2+ vz2 (1.48)

and its direction relative to the x-axis

θ = tan−1 vy

vx (1.49)

(thats inverse tangent, not one over tangent). In terms of the magnitude and direction we can express the compo-nents as

vx=|v| cos θ and vy =|v| sin θ (1.50)

Vectors as algebraic objects add component by component

a = (ax, ay), b = (bx, by), then a + b = (ax+ bx, ay+ by) (1.51)

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a + b = (ax+ bx, ay+ by, az+ bz) (1.52) It will prove to be very useful to be able to compute the angle between two vec-tors, from the figure below we see that

θab= θa− θb= tan−1 ay ax − tan−1by bx (1.53) take the cosine of both sides of this equa-tion and use

cos(x− y) = cos(x) cos(y) + sin(x) sin(y) cos(θab) = cos(θa− θb)

= cos(θa) cos(θb) + sin(θa) sin(θb) (1.54)

cos(θab) = AxBx |A||B|+

AyBy

|A||B| (1.55)

Define the dot product of two vectors, which produces a scalar

a· b = axbx+ ayby

a· b = axbx+ ayby+ azbz (1.56) and we see that the cosine of the angle between two vectors is

cos(θab) = a· b

|a||b| (1.57) which is very useful since it involves only the components of the two vectors.

1.3.1 Basic vector operations

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1.3. VECTORS 11

Scalar multiplication. When you multiply a vector by a scalar, each component is amplified, changing the length

of the vector but not its direction.

s r = s (rx, ry, rz) = (srx, sry, srz) (1.58)

for

r a vector, s a scalar (1.59)

Vector addition. Vectors add component by corresponding component

a + b = (ax, ay, az) + (bx, by, bz) = (ax+ bx, ay+ by, az+ bz) (1.60)

The length of a vector is computed via the Pythagorean theorem;

|a| =√a2

x+ a2y+ a2z (1.61)

Find a vector ⊥ to another vector Given a vector in two dimensions such as

a = (ax, ay) = axi + ayj (1.62) any vector b perpendicular to it is a vector such that

cos θab= 0 =

a· b

|a||b| = 0, a· b = 0 (1.63)

Write this equation out in components

axbx+ ayby = 0 (1.64)

This is one equation in two unknowns bx, by, and so cannot be uniquely solved. As long as both components axand ay are non-zero, we simply pick a value for bx and proceed;

Let bx= 1, ax+ ayby = 0, by=−ax

ay (1.65)

and therefore

b = (1,−ax

ay) (1.66)

However if b⊥ a, then so is s b for s any scalar, and we could even pick s = ay, making

sb = (ay,−ax) ⊥ a = (ax, ay) (1.67)

In three dimensions there is a whole plane of vectors⊥ to

a = (ax, ay, az), ax, ay, az̸= 0 (1.68) and so we proceed in steps. Let

b = (bx, by, bz), a· b = axbx+ ayby+ azbz= 0, if b⊥ a (1.69) There will be many such vectors, so for example let bx= 0, by = 1;

0 + ay+ azbz= 0, bz=− ay az

(1.70) and then any vector

sb = (0, s,−say az

)⊥ a (1.71)

This vector lies in the yz−plane. We could also set by= 0, bx= 1; to get sb = (s, 0,−sax

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another vector perpendicular to a but lying in the xz−plane.

Linear independence. Suppose that you wish to solve the equation y− ax = 0 for a; it is easy enough, you simply

divide by x, since numbers form a field, and division by anything but zero is legitimate in a field. But what about solving v1+ a v2 = 0? You can’t divide vectors because they do not form a field. It might be possible that the only solution to

av1+ bv2= 0 (1.73)

is a = b = 0. If this is true, then we say that{v1, v2} are linearly independent. For example

a (1, 2) + b (3, 4) = 0 (1.74)

implies that a + 3b = 0 and 2a + 4b = 0, solving we obtain−2b + 3b = b = 0, which makes a = 0 too, the vectors are linearly independent. The most important sets of linearly independent vectors are the bases{i, j} in two dimensions, and{i, j, k} in three.

Write a vector as a sum of vectors. This is tantamount to using a set of vectors as a basis, and so requires that

the set of vectors is a linearly independent set. Let

a = (ax, ay), b = (bx, by) (1.75) be two non-collinear vectors. They are non-collinear if

a̸= s b (1.76)

Pick another vector

c = (cx, cy) (1.77)

and determine the two scalars α, β such that

c = α a + β b (1.78)

We call this resolving vector c into a and b. Write it out in components

(cx, cy) = α(ax, ay) + β(bx, by) (1.79)

use the scalar multiplication rule

(cx, cy) = (αax, αay) + (βbx, βby) (1.80) use the vector addition rule

(cx, cy) = (αax+ βbx, αay+ βby) (1.81) and look at the two resulting equations

cx= αax+ βbx, cy = αay+ βby (1.82) which is two equations in two unknowns, which we solve; in the first solve for β

β = cx− αax bx

(1.83) and put it into the second

cy= αay+ (_c x− αax bx ) by (1.84)

and get α by itself;

cybx− cxby= α(aybx− axby), α = cybx− cxby aybx− axby

(1.85) and go back and get β;

β = cyax− cxay aybx− axby

(1.86) You will make extensive use of your abilities to solve two equations in two unknowns in this course.

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1.3. VECTORS 13

Dot products. The dot product (or scalar product) can be neatly summarized by the rules i· i = j · j = k · k = 1, i· j = j · i = 0

i· k = k · i = 0, k· j = j · j = 0 (1.87)

and the fact that it is distributive;

a·

(

b + c

)

= a· b + a · c (1.88)

Demonstration of these facts from the definition

a· b = axbx+ ayby+ azbz (1.89)

I will leave to you.

Cross products. The cross product is another vector product invented as a tool for computing areas. We know

that the area of a rectangle is base× height, and the same is true for a parallelogram. What if we want to compute the area of a parallelogram whose sides are given by two vectors A = axi + ayj and B = bxi + byj? Of course we can do it the long way and resolve the vectors into components, find the angle between them and construct the base and height, but the cross product is designed to do all of this from just the components themselves.

We want a vector product that represents the area of the parallelogram made from A and B, we call it A× B. Since there will be no area at all if A = B, we can ensure that the product gives us zero when we apply it to two copies of the same vector if we define× such that

A× B = −B × A (1.90)

since then A× A = −A × A = 0. Then use the distributive property

A× B = (axi + ayj)× (bxi + byj)

= axbxi× i + axbyi× j + aybxj× i + aybyj× j = (axby− aybx)i× j (1.91) Is this the area in question? It is, simply resolve the expression into magnitudes and phases;

(axby− aybx) (i× j) = |A||B| (

cos θasin θb(i× j) − cos θbsin θa(i× j)

)

= |A||B| sin(θa− θb) (i× j)) (1.92) which from the figure we see is precisely base × height. We now finish the job by assigning the cross product the status of a vector, so that i× j is a vector, by defining

i× j = k (1.93)

In order to have a consistent definition and make Eq. 1.91 true, we can show that this requires

i× i = j × j = k × k = 1, i× j = −j × i = k

k× i = −i × k = j, j× k = −k × j = i (1.94)

The cross product of two vectors lets us not only compute the area of the parallelogram made from the vectors entirely in terms of its components, but also the sine of the angle between the vectors

sin(θa− θb) =|A × B|

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and provides us with a test of whether or not two vectors are parallel to one another;

A× B = 0 if A|| B (1.96)

Perpendicularity

Two vectors a and b are⊥ to each other if there is a θab= 90 degree angle between them, making

cos θab= 0, a· b = 0 (1.97)

Collinearity

Two vectors a and b are collinear if there is a scalar s such that

a = sb, (ax, ay, az) = (sbx, sby, sbz) (1.98) Solve these three equations

s =ax bx =ay by = az bz (1.99) To test for collinearity, we compute all three ratios, if they are not all the same, the vectors are not collinear. Remember that you can divide numbers, such as individual components of vectors, but you cannot divide by the vectors themselves.

1.3.2 Examples

Example 7. Let

v = (1, 1, 1) = i + j + k, u = (1,−2, 1) = i − 2j + k, w = (1, 0,−1) = i − k

Find the lengths of these vectors.

|v| =√v· v =√1 + 1 + 1 =√3, |u| =√1 + 4 + 1 =√6, |w| =√1 + 1 =√2

Example 8. Show that these vectors have 90o_{angles between any pair.}

u· v = 0 = (1)(1) + (1)(−2) + (1)(1), w· v = 0 = (1)(1) + (−1)(1), w· u = 0 = (1)(1) + (−1)(1) Example 9. Show that these vectors are linearly independent.

0 = av + bu + cw = (a + b + c, a− 2b, a + b − c) = (0, 0, 0)

The middle one says a = 2b, the las one says c = 3b, and the first says (1 + 2 + 3)b = 0, so b = 0 making a = b = c = 0 the only solution.

Example 10. Let

z = (1, 2, 3), Show that z = αu + βv + γw, and find α, β, γ

We know this can be done uniquely since{u, v, c} are linearly independent. Because we can use the dot-products above, note z· u = (1)(1) + (−2)(2) + (3)(1) = 0 = ( αu + βv + γw ) · u = α|u|2_{= 6α,} _{we find} _{α = 0} z· v = (1)(1) + (1)(2) + (1)(3) = 6 = ( αu + βv + γw ) · v = β|v|2_{= 3β,} _{we find} _{β = 2} z· w = (1)(1) + (−1)(3) = −2 = ( αu + βv + γw ) · w = γ|w|2_{= 2γ,} _{so that} _{γ =}₋₁

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1.4. APPENDIX. DIFFERENTIAL CALCULUS 15

1.4 Appendix. Diﬀerential calculus

It is very important that you be able to perform the basic calculus operations;

1. computations of limits and derivatives, including partial derivatives 2. expansions of functions, parameterization of curves

3. simple integrations.

In this review we will make sure that you are up to speed on all of these basic skills, and we will review trigonometry while we are at it.

There is no substitute for a full year of calculus instruction, and so math 221 is an absolute prerequisite for

physics 201, and math 222 is an absolute prerequisite for physics 202. There is no back door. A smooth function y = f (x) of a variable x is diﬀerentiable at x if its derivative

f′(x) =df (x)

dx = limdx→0

f (x + dx)− f(x) dx exists (is a number) at x.

The limit-taking process is very simple; we expand the numerator in ascending powers of dx, perform the division, and take the limit by setting dx = 0 afterwards (in a nutshell). The geometrical interpretation of the deriva-tive of f (x) at x0 is that f′(x0) is the slope of the line tangent to f (x) at x0;

Therefore a useful formula for translating cal-culus to geometry is tan θ = (_{df (x)} dx ) x=x0 (1.100) for the tangent to the curve at x0.

θ

x₀

y=f(x)

The limit taking process is most easily handled for polynomial functions such as

f (x) = N ∑ n=0

anxn

by use of the following simple rule; d dx ( f (x) + g(x) ) = df (x) dx + dg(x) dx (1.101)

which is easy to prove from the definition above. This says that the derivative of a (finite) sum is the sum of the derivatives of the summands.

Example 11. Let f (x) = x3, the steps taken in computing the derivative are;

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f (x + dx)− f(x) dx = (x + dx)3_{− x}3 dx = (x 3_{+ 3x}2_{dx + 3x (dx)}2_{+ (dx)}3₎_{− x}3 dx

Perform the division (x

3_{+ 3x}2_{dx + 3x (dx)}2_{+ (dx)}3₎_{− x}3

dx =

3x2_{dx + 3x (dx)}2_{+ (dx)}3 dx

= 3x2+ 3x dx + (dx)2 Perform the limit (set dx = 0); dx

3 dx = dxlim_→0 ( 3x2+ 3x dx + (dx)2 ) = 3x2 (1.102)

Example 12. Let f (x) = a + b x2_{, in which a and b are constants.} Step 1. Write out the fraction

f (x + dx)− f(x) dx = (a + b(x + dx)2)− (a + bx2) dx = (a + b(x 2_{+ 2x dx + (dx)}2₎₎_{− (a + bx}2₎ dx

Perform the division; (a + b(x

2_{+ 2x dx + (dx)}2₎₎_{− (a + bx}2₎

dx =

2bx dx + b(dx)2

dx = 2bx + b dx Perform the limit (set dx = 0); d

dx ( a + bx2 ) = lim dx_→0 ( 2bx + b dx ) = 2bx (1.103)

which we can see is the sum of the derivatives of the two terms a and bx2_{, the derivative of a constant being zero.}

1.4.1 Partial derivatives versus total

Technically nearly all derivatives computed in this course are partial derivatives, derivatives of a function of more than one variable with respect to one variable. If the function depends on one variable, these derivatives

are the same. The definition is very simple

(_{∂f (x, y)} ∂x ) y= limdx_→0 f (x + dx, y)− f(x, y) dx , (_{∂f (x, y)} ∂y ) x= limdy_→0 f (x, y + dy)− f(x, y) dx (1.104)

so that partial derivation with respect to x regards everything but x as a constant. For example f (x, y) = x + 2y, (_{∂f (x, y)} ∂x ) y = 1, (_{∂f (x, y)} ∂y ) x = 2 f (x, y) = 3 x y (_{∂f (x, y)} ∂x ) y= 3y, (_{∂f (x, y)} ∂y ) x= 3x f (x, y) = x y, (_{∂f (x, y)} ∂x ) y = 1 y, (_{∂f (x, y)} ∂y ) x =−x y2 (1.105)

in which the basic rules for diﬀerentiating powers, products and quotients (which we are about to list) apply to partial as well as total derivatives.

1.4.2 Slopes of curves without calculus

Using nothing more that basic algebra, one can easily compute the slopes of lines tangent to arbitrary planar curves, you simply need to make a few simple observations about the curve in question, or even better yet, be able to graph it.

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1.4. APPENDIX. DIFFERENTIAL CALCULUS 17 -60 -40 -20 0 20 40 60 -4 -3 -2 -1 0 1 2 3 4 o o

Consider for example the curve y = ax2_{+ bx + c. Suppose that you want to find the slope of the line tangent to the} curve at some point x0. You can draw three types of straight lines relative to this curve; ones that don’t intersect the curve, ones that intersect it at one point, and lines that intersect the curve at two points. “Generic” lines will intersect the curve at two points, but only vertical lines (infinite slope) and tangent lines intersect at one point. This is the key observation, since we know that the slope of the curve is not infinite for any finite value of x.

To find the slope of the curve (slope of tangent line) at point (x, y) = (u, v) on the curve, notice that v = au2_{+ bu + c,} and rewrite the curve equation as

y− (ax2+ by + c) = 0 The parametric equation of a line of slope m through (u, v) is

x = u + t, y = mt + v, y = m(x− u) + v

(notice that when the parameter t = 0, you are at the point (u, v) on the curve). Substitute this into our equation (mt + v)− (a(t + u)2+ b(tu) + c) = 0, (mt + v)− (at2+ 2atu + au2+ bt + bu + c) = 0

Use v− (au2_{+ bu + c) = 0;}

mt− (at2+ 2atu + bt) = 0, at2+ (2au + b− m)t = 0

If this line is tangent to the curve at (u, v) then t = 0 must be a double-root of this equation! In other

words

m = 2au + b which is exactly the derivative of our curve at the point (u, v)

d dx

(

ax2+ bx + c

x=u= 2au + b

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Once again you see three types of lines, those that intersect the circle at 0, 1, 2 points. The lines through a single point of the circle are the tangent (or the vertical, which we don’t care about) lines. Play the same game;

y2+ x2− R2= 0, (mt + v)2+ (u + t)2− R2= 0, (m2t2+ 2mvt + t2+ 2ut) + (v2+ u2− R2) = 0 and look for the condition making the root a double root;

(m2+ 1)t2+ (2mv + 2u)t = 0, t = 0 is a double root iﬀ m =−u v Verify this with calculus;

y2= R2− x2, dy 2 dx = 2y dy dx = 0− 2x, dy dx x=u,y=v=− x y x=u,y=v=− u v

These methods work (the ideas are from the field of algebraic geometry) because the derivative has a geometrical meaning; it is the slope of the line tangent to the curve, the slope of the line through two points that are taken to be the same. The methods work just as well for rational functions as for polynomial.

Example. Find the slope of the line tangent to curve y = 1/x at the point (u, 1/u).

Once construct the line x = u + t, y = 1_u + mt of slope m that passes through both points (u, 1/u) and (u, 1/u) (that’s not a typo), the first point corresponding to t = 0, the second t = 0 (a double root);

xy− 1 = 0, (u + t)(1

u+ mt)− 1 = 0, mt

2_{+ (mu +} 1 u)t = 0 a quadratic equation with a double-rrot t t = 0 if m =−_u12, the correct derivative

dy dx =−

1

x2 at x = u.

1.4.3 The rules of diﬀerentiation

Derivatives of a product f (x) g(x) can be easily computed from the definition, d dx ( f (x)g(x) ) = lim dx_→0 f (x + dx)g(x + dx)− f(x)g(x) dx

by replacing f (x + dx) with f (x + dx)− f(x) + f(x) and rearranging

d dx ( f (x)g(x) ) = lim dx→0 ( f (x + dx)− f(x) + f(x) ) g(x + dx)− f(x)g(x) dx = lim dx→0 ( f (x + dx)− f(x) ) g(x + dx) + f (x) ( g(x + dx)− g(x) ) dx = lim dx→0 (_{f (x + dx)}_{− f(x)} dx g(x + dx) ) + f (x) lim dx→0 (_{g(x + dx)}_{− g(x)} dx ) = df (x) dx g(x) + f (x) dg(x) dx (1.106)

If f (x), f′(x), g(x) and g′(x) all exist.

We state this as being the product rule for derivatives. d dx ( f (x)g(x) ) =df (x) dx g(x) + f (x) dg(x) dx (1.107)

1.4.4 The binomial theorem

This theorem is of great antiquity, and is extremely useful for both algebraic and calculus applications. It says that

(a + b)N = N ∑ m=0 ( N m ) ambN−m (1.108)

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where the number ₍

N m

)

= N !

m!(N− m)! is a binomial coeﬃcient, and the factorial of an integer N is

N ! = N· (N − 1) · (N − 2)...2 · 1, 1! = 1, 0! = 1 (the last relation is a definition). For example

(a + b)2 = a2+ 2ab + b2

(a + b)3 = a3+ 3a2b + 3ab2+ b3

(a + b)4 = a4+ 4a3b + 6a2b2+ 4ab3+ b3 (1.109)

1.4.5 Series expansion

This last calculation was a little on the tricky side, but there exists a powerful tool for performing most of the operations of calculus in a simple way, the series expansion.

We suppose that the function f (x) exists at the point x0, and for that matter that it exists near x0. Let x− x0 be small, so that x is close to x0. The idea of the series expansion is that in the neighborhood of x0, we could replace f (x) with a polynomial Pf(x) = N ∑ n=0 an n!(x− x0) n _(1.110)

in which n! = n· (n − 1) · (n − 2)...2 · 1 is our factorial of the integer n.

The number of terms N that we need to calculate to get Pf depends on what we want to do with it, and is based on the following concept: the function f (x) and polynomial Pf(x) agree at x0, and have the same derivative at x0, and the same second derivative, and so on up to the Nthderivative. We would call Pf(x) an Nthorder series expansion of f (x) about the point x0.

Step 1. Both f (x) and Pf(x) agree at x0;

f (x0) = Pf(x0) = a0+ a1(x0− x0) + a2 2!(x0− x0) 2 + ... +aN N !(x0− x0) N = a0 requires that a0= f (x0).

Step 2. Both f′(x) and d

dxPf(x) agree at x0; f′(x0) = 0 + a1+ 2 a2 2!(x0− x0) + 3 a3 3!(x0− x0) 2 + ... + NaN N !(x0− x0) N₋₁ = a1 requires that a1= f′(x0) = _{dx x=x}df 0.

Step 3. Both f′′(x) and _dxd Pf(x) agree at x0; f′′(x0) = 0 + 0 + 2 a2 2! + 3· 2 a3 3!(x0− x0) + ... + N· (N − 1) aN N !(x0− x0) N₋₂ _{= a} 2 requires that a2= f′′(x0) =d 2_f dx2_x=x 0.

For ninety percent of all of the calculus applications in our physics text, this is enough; the polynomial Pf(x) that agrees with f (x) up to two derivatives in the neighborhood of x0 is

Pf(x) = f (x0) + f′(x0) (x− x0) + 1 2f

′′_(x

0) (x− x0)2+ ... (1.111) This is called the Euler-Maclaurin or Taylor series for f (x) near x0, and it may be substituted in place of f (x) in the neighborhood of x0.

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What do we use it for? For starters it can be used to get formulas for derivatives of products and quotients. In

most applications you only need to keep one or two terms in a Taylor series. For example, let x = t + dt

and x0= t, then

Pf(t + dt) = f (t) + f′(t) dt +1 2f

′′_{(t) (dt)}2_{+ ...} _(1.112)

and you can replace any occurrence of f (t+dt) in a formula that involves taking the limit dt→ 0 with this expression.

Example 13. d dt ( f (t)g(t) ) = lim dt_→0 ( f (t + dt)g(t + dt)− f(t)g(t) ) dt = lim dt_→0 ( f (t) + f′(t) dt + ... )( g(t) + g′(t) dt + ... ) − f(t)g(t) dt = lim dt→0 g(t) f′(t) dt + g′(t) f (t) dt + ... dt = g(t) f ′_{(t) + g}′_{(t) f (t)} _(1.113)

and we are done quickly and cleanly, all of the terms in (...) contain at least two factors of dt, and so in the limit become zero.

Example 14. l’Hospitals rule is a formula for computing the limit of the ratio of two functions that both vanish

at x0, lim x→x0 f (x) = 0 = lim x→x0 g(x) The limit of the ratio is then

lim x→x0 f (x) g(x) = limx→x0 f (x0) + f′(x0) (x− x0) +₂1f′′(x0)(x− x0)2+ ... g(x0) + g′(x0) (x− x0) +₂1g′′(x0)(x− x0)2+ ... but both f (x0) = 0 and g(x0) = 0, so

lim x→x0 f (x) g(x) = limx→x0 f′(x0) (x− x0) +1₂f′′(x0)(x− x0)2+ ... g′(x0) (x− x0) +1₂g′′(x0)(x− x0)2+ ... divide out the factor (x− x0) from numerator and denominator:

= lim x_→x0

f′(x0) +12f′′(x0)(x− x0) + ... g′(x0) +12g′′(x0)(x− x0) + ... and in the limit x→ x0, (x− x0)→ 0,

lim x_→x0 f′(x0) +1₂f′′(x0)(x− x0) + ... g′(x0) +1₂g′′(x0)(x− x0) + ... =f ′_(x 0) g′(x0) We restate this as l’Hospitals rule;

lim x→x0 f (x) g(x) = f′(x0) g′(x0) (1.114) provided g′(x0) is non-zero. If g′(x0) and f′(x0) are in fact both zero, we simply repeat the process noting that in

lim x_→x0

f′(x0) +12f′′(x0)(x− x0) + ... g′(x0) +12g′′(x0)(x− x0) + ...

the first term in both numerator and denominator are zero and we can divide out another factor of (x− x0);

lim x→x0 f (x) g(x) = limx→x0 0 +1 2f′′(x0) + ... 0 + 1 2g′′(x0) + ... = f ′′_(x 0) g′′(x0) , if f (x0) = g(x0) = f′(x0) = g′(x0) (1.115)

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1.4.6 Rational functions

Consider a function that is the ratio of two functions, both of which you can diﬀerentiate; h(x) = f (x)

g(x) To compute the derivative of h(x) we take these algebraic steps, first

h(x) g(x) = f (x) now apply the product rule

d dx ( h(x) g(x) ) = d dxf (x) = f ′_(x), _h′_{(x) g(x) + h(x) g}′_{(x) = f}′_(x) and rearrange h′(x) =f ′_(x)_{− h(x) g}′_(x) g(x) = f′(x)−f (x)_g(x)g′(x) g(x) = f′(x) g(x)− f(x) g′(x) g2_(x) (1.116)

which we will call the quotient rule.

1.4.7 Radicals

Consider the radical

f (x) = √n_x To compute its derivative, first raise both sides to the nth_power

fn(x) = f (x)fn−1(x) = x Now diﬀerentiate and apply the product rule repeatedly to the left side

d dxf n_{(x) =} d dxx = 1, n f n−1_(x)f′_{(x) = 1} solve for f′(x); f′(x) = 1 nfn−1_(x) = 1 n xn−1n = 1 nx 1 n−1 We have shown that

d dx n √ x = d dxx 1 n ₌ 1 nx 1 n−1 _(1.117)

and therefore for any power a, integral, rational or otherwise d

dxx

a_{= a x}a₋₁ which we call the power rule for diﬀerentiation.

Example 15. Find a series expansion for f (x) =√x valid near x0= 4. The first step is to compute a few derivatives, using the power rule with a = 1₂,

d dx √ x = 1 2 1 √ x, d2 dx2 √ x = d dx 1 2 1 √ x =− 1 22 1 √ x3 and so inserting this all into Eq. 5 we find that

√ 4 + dx =√4 + 1 2√4dx + 1 2 ( − 1 22 1 √ 43 ) (dx)2+ ... = 2 +dx 4 − (dx)2 64 + ...

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This should be written using x = 4 + dx, as √ x = 2 +(x− 4) 4 − (x− 4)2 64 + ...

and in any formula involving√x that will be used for x near 4, this is a valid replacement. In particular, this can be used to calculate square roots of numbers close to 4, such as 5 for which

√

5≈ 2 +1 4−

1

64+ ... = 2.234375...

which gives quite good accuracy (we are oﬀ in the third decimal place) with only these three terms in the series.

1.4.8 Basic integration

Integration is the anti-derivative; this is how we will define it. The process of integration must undo the process of diﬀerentiation, and so we define

∫ b a

df (x)

dx dx = f (b)− f(a) (1.118)

A Riemann sum will do the trick; consider ∫ b a f (x) dx = lim N_→∞ N ∑ n=0 f (a +(b− a) N n) (b− a) N (1.119)

which is the area under the curve f (x) from a to b, being made up of little strips of width (b−a)_N of height f (a + n(b−a)_N ).

How does this undo the derivative? Let ∆ = (b− a)

N (1.120)

and write the Riemann sum as ∫ b a f (x) dx = lim N→∞ N ∑ n=0 f (a + n∆) ∆ (1.121)

and insert into this a diﬀerentiated function f (x) = dg(x) dx = lim∆_→0 g(x + ∆)− g(x) ∆ a b ∫ b a f (x) dx = lim N→∞∆ ( f (a) + f (a + ∆) + ... + f (a + (N− 1)∆) + f(a + N∆) ) (1.122) = lim N→∞∆ (_{g(a + ∆)}_{− g(a)} ∆ + g(a + 2∆)− g(a + ∆) ∆ + ... + g(a + (N− 1)∆) − g(a + (N − 2)∆) ∆ + g(a + N ∆)− g(a + (N − 1)∆) ∆ ) (1.123)

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1.4. APPENDIX. DIFFERENTIAL CALCULUS 23 You can see that consecutive terms partially cancel, and so does the ∆ in the denominator, leaving

= lim N_→∞ ( g(a + N ∆)− g(a) ) = lim N_→∞ ( g(a + N(b− a) N )− g(a) ) = g(b)− g(a) (1.124) Integration is a harder problem than diﬀerentiation, since the only procedure for performing integrals is to either do the Riemann sum directly, or find a function whose derivative is the integrand, or to perform variable changes that put the integrand into a more readily recognized form.

Example 16. It is not hard to show that

N ∑ n=1

n = N (N + 1)

2 (1.125)

To do it lay out all numbers 1 through N in a row

1 + 2 + 3 + 4 + 5 + 6 + ... + (N− 2) + (N − 1) + N (1.126) and below it all numbers in reverse order

N + (N− 1) + (N − 2) + ... + 6 + 5 + 4 + 3 + 2 + 1 (1.127) and add the two rows

(N + 1) + (N + 1) + (N + 1) + ... + (N + 1) + (N + 1) + (N + 1) = N (N + 1) (1.128) but this is each number counted twice.

We can use this to integrate ∫ b a x dx = lim N→∞ N ∑ n=0 (a + n∆) ∆ = lim N→∞ ( (N + 1)a∆ + ∆2 N ∑ n=0 n ) (1.129)

with ∆ = b−a_N . Put this in; ∫ b a x dx = lim N→∞ ( (N + 1)a(b− a) N + 1 2N (N + 1) (_b_{− a} N )2) = a(b− a) +1 2(b− a) 2₌1 2b 2₋1 2a 2 _(1.130)

By considering more diﬃcult Riemann sums we can establish ∫ b a xndx = 1 n + 1 ( bn+1− an+1 ) , n̸= −1 (1.131)

This formula is valid for any n̸= −1, even irrational values. Like diﬀerentiation, integration is a linear operation

∫ b a (f (x) + g(x)) dx = ∫ b a f (x) dx + ∫ b a g(x) dx (1.132)

and another very useful property inherited from the Riemann sum definition is ∫ c a f (x) dx = ∫ b a f (x) dx + ∫ c b f (x) dx (1.133)

What is integration used for in 201? Suppose that you know the value of a function, such as the position of a body, at time t0, and the velocity at all times. Integration is used to recover the position at any time t from

v(t) =dx

dt, x(t) = x(t0) + ∫ t

t0

(32)

1.4.9 The logarithm and exponential

In Eq. 1.131 the case n =−1 is special. Consider the integral g(x) =

∫ x 1

dx′

x′ (1.135)

since for such a function _∫ ab 1 dx x = ∫ a 1 dx x + ∫ ab a dx x (1.136)

and the variable change

x = ay (1.137)

turns this into _∫ ab 1 dx x = ∫ a 1 dx x + ∫ b 1 dy y , or g(ab) = g(a) + g(b) (1.138)

This is the law of exponents, recall that xa_{· x}b _{= x}a+b_{, from which we conclude that g(a) is an anti or inverse} exponentiation, we need only to establish the base, which we will call “e”. In other words f (x) = ex _{and g(x) are} inverses of one another,

g(f (x)) = x = f (g(x)) (1.139)

From our integral representation

ln x = ∫ x 1 dy y (1.140) we see that d dxln x = 1 x (1.141)

Apply the chain rule to Eq. 1.139 d dx ( g(f (x) ) =dx dx = 1, 1 f (x) df dx = 1, df dx = f (x) (1.142)

which is the law of diﬀerentiation of the antilog or exponential function. We can use our power series methods to determine e; f (x) = ∞ ∑ n=0 1 n! (_dn_f dxn ) x=0x n = ∞ ∑ n=0 1 n! ( f (x) ) x=0 xn= ∞ ∑ n=0 1 n!x n e = f (1) = ∞ ∑ n=0 1 n!1 n_{= 1 +} 1 2!+ 1 3!+· · · = 2.71828 · · · (1.143) To summarize we call g(x) = ln(x), f (x) = exand we have

ln(x) = ∫ x 1 dx′ x′ , e ln(x)_{= ln(e}x_{) = x,} d dxe x_{= e}x_, d dxe ax _{= a e}ax _(1.144)

1.4.10 Quadratic forms

Solving an equation such as

ax2+ bx + c = 0

for x is relatively simple. We will follow a procedure that actually generalizes to cubic and even quartic equations. First divide by a and write the equation in the following way

x2+ b ax +

c

University Physics