# self-assessment ```1
PHY 3221
Spring 2010
Mathematics Self-Assessment
Steven Detweiler
The lack of mathematical sophistication is a leading cause of difficulty for students in
Classical Mechanics.
An official pre-requisite of phy3321 is phy2048, phy2049 and the math requirements
include MAC 2311, 2312 and 2313 (Vector Calculus). These math courses together cover
derivatives and integrals of trig and log functions, series and sequences, analytic geometry,
vectors and partial derivatives and multiple integrals. We will casually be using math from
all of these subjects. None of these should be completely unfamiliar to you.
The following discussions and questions are grouped by subject and in approximate order
of difficulty—easiest first. These are representative of the level of mathematics which I
expect. You should feel very comfortable with mathematics at this level, at least through
section C on calculus. Section D, on differential equations, is probably more difficult for
you but important. The answers to the questions are not always given. If you do not know
that your answer is correct then you are not comfortable with mathematics at this level.
Sections E, F, G, H and I are more advanced than is necessary as a prerequisite for
this course. But, these ideas will often be discussed in class. If you understand the Taylor
series in section G, then you are likely to find section H, on calculators, interesting and
amusing. Don’t be surprised if my discussions seem confusing at first: To understand math
and physics often requires multiple, multiple readings while working out algebraic details
with paper and pencil in hand.
Section I involves an ordinary differential equation that has an interesting application to
A.
Algebra
Solve for x:
f (x) = ax2 + bx + c = 0.
For what value of x is f (x) a maximum or a minimum?
Make a sketch of the function y(x) where y = mx + b and where m and b are constants.
What are the meanings of the constants m and b in terms of your sketch?
Factor:
(a2 + 4ab + 4b2 )
B.
Vector Algebra
~ = 1ı̂ + 2̂ + 3̂ and B
~ = 4ı̂ + 5̂ + 6̂.
Let A
~
What is |A|?
~ · B?
~
What is A
and
(a2 − 9b2 )
2
~ × B?
~
What is A
~ and B?
~
What is the cosine of the angle between A
~·C
~ = 10 and the angle between A
~ and C
~ is 30◦ , then what is the magnitude of C?
~
If A
C.
Calculus
If x0 , v0 and a are constants and
1
x(t) = x0 + v0 t + at2
2
then what is dx/dt? What is d2 x/dt2 ? If a < 0, does the function x(t) curve up or down?
If x is negative when t = 0 and x is positive when t is very large: then for precisely which
values of t is x positive?
Evaluate the derivative
d
A cos(ωt + φ)
dt
where A, ω and φ are constant.
Evaluate the following integrals
π
Z
sin θ dθ,
0
Z
where k is a constant, and
Z
1
D.
k
dx
x2
x
k
dx.
x
Differential equations
These next two problems might be difficult or possibly unfamiliar to you, but take a
careful look at them because these are very important in classical mechanics.
Find two different functions which satisfy the differential equation
d2 f (x)
− λ2 f (x) = 0.
dx2
Find two different functions which satisfy the differential equation
d2 f (x)
+ ω 2 f (x) = 0.
2
dx
3
E.
Trigonometry
Euler’s identity,
eiθ = cos θ + i sin θ,
is probably new to you. But it provides a convenient and easy way to derive some of the
basic trig identities such as
ei(α+β) = eiα eiβ
cos(α + β) + i sin(α + β) = (cos α + i sin α) × (cos β + i sin β)
or, after multiplying out the right hand side,
cos(α + β) + i sin(α + β) = cos α cos β − sin α sin β + i(cos α sin β + sin α cos β)
The real and the imaginary parts of this equation give the well known trig identities:
cos(α ± β) = cos α cos β ∓ sin α sin β
and
sin(α ± β) = sin α cos β ± cos α sin β.
Use the Euler identity to show that
sin2 θ + cos2 θ = 1.
Hint: start with eiα e−iα = 1 and then use the Euler Identity.
F.
Sums
Question: Evaluate the sum
S(x) =
∞
X
xn
for |x| < 1 .
n=0
Ans: Note that
S(x) =
∞
X
n=0
n
x = 1+
∞
X
xn
n=1
∞
X
= 1+x
xn
n=0
= 1 + xS(x)
So we have
and, finally
S = 1 + xS
(1 − x)S = 1
1
S(x) =
.
1−x
4
G.
Taylor expansions of a function
Any differentiable function f (x) may be approximated in the neighborhood of a point x0
by the Taylor expansion
2
3
df
1
1
2d f
3d f
f (x) = f (x0 ) + (x − x0 )
+ (x − x0 )
+ (x − x0 )
dx x=x0 2
dx2 x=x0 6
dx3 x=x0
n
1
d f
+ · · · + (x − x0 )n n
+ ···
n!
dx x=x0
For example, consider f (x) = 1/(1 − x), expanded about x0 = 0. Then
f (x) = 1/(1 − x)
df
= [1/(1 − x)2 ]x=0 = 1
dx x=0
d2 f
= 2[1/(1 − x)3 ]x=0 = 2
2
dx x=0
d3 f
= 6[1/(1 − x)4 ]x=0 = 6
dx3 x=0
dn f
= n![1/(1 − x)n+1 ]x=0 = n!
dxn x=0
The Taylor expansion for 1/(1 − x) with x0 = 0 is now
1
1
1
1
= 1 + x + x2 × 2 + x3 × 6 + · · · + xn × n! + · · ·
1−x
2
6
n!
And this is easily seen to be
∞
X
1
=
xn ,
1 − x n=0
the same as our example for doing sums above!
Taylor expansions of this sort are extremely useful in physics. For example in special
relativity when we are interested to see how close special relativity is to Newtonian physics
for small speeds v, we usually make the assumption that v/c 1. Then we make Taylor
expansions of the relevant formulae, and include only the terms proportional to v/c or maybe
also v 2 /c2 .
Common Taylor expansions give approximations such as
1
= 1 + + O(2 )
1−
(1 + )n = 1 + n + O(2 )
√
1
1 − = 1 − + O(2 )
2
1
1
√
= 1 + + O(2 )
2
1−
5
1
1
1
e = 1 + + 2 + 3 + 4 + O(5 )
2
6
24
1
1
1
ei = 1 + i − 2 − i 3 + 4 + O(5 ).
2
6
24
The O(n ) term here is standard mathematical notation to mean a function which is less
than some constant times n in the limit that → 0. In other words for small , O(n ) is no
bigger than something times n .
We can use the Euler identity ei = cos + i sin to easily pick off the purely real terms
from this last expansion which give the expansion of cos for a small angle , and the purely
imaginary terms, which give the expansion of sin for small :
and
H.
1
1
cos = 1 − 2 + 4 + O(6 )
2
24
1 3
sin = − + O(5 ).
6
Calculators
When solving a physics problem, think with your brain not with your calculator! Before
touching your calculator, check to see that your algebraic answer has the correct units and
that it has the expected behavior for various limits. It is nearly impossible to check the
correctness of an answer once you touch your calculator.
You might find it amusing that the number 10100 has been given the name googol, and
googol
10
is called googolplex—and these names were coined well before the internet was
invented. But, note the difference in spelling between googol and the name of the internet
search engine. As an aside: The internet was invented by physicists who wanted to exchange
easily experimental data between the United States and Europe.
Here are a couple examples which are relevant to one of the homework problems for this
course. Let f (n) = n2 , where n is an integer. First evaluate f (102 ) − f (102 − 1) on your
calculator. You should get 199.
Now try to evaluate f (10100 ) − f (10100 − 1). Your calculator will choke on this problem,
but your brain can easily find the answer to 100 significant digits. Note that
f (n) − f (n − 1) = n2 − (n − 1)2 = n2 − (n2 − 2n + 1) = 2n − 1.
With n = 10100 , it is easy to see that f (n) − f (n − 1) = 2 × 10100 − 1 ≈ 2 × 10100 with 100
significant digits.
Here is a second, more challenging, problem. Let f (n) = n−2 where n is an integer.
Evaluate f (10100 − 1) − f (10100 ). Your calculator will also choke on this problem, but again
you can easily find the answer to about 100 significant digits. Use the Taylor expansion
f (n + δn) = f (n) + δn
df
df
2
+ . . . ⇒ f (n + δn) − f (n) = δn
+ . . . = −δn 3 + . . .
dn
dn
n
With n = 10100 and δn = −1, we easily have f (10100 − 1) − f (10100 ) = 2 × 10−300 + . . .,
where the . . . represents terms which are comparable to 1/n4 = 10−400 or smaller. Thus, the
answer is correct for the first 100 digits.
6
For a final example which reveals the limitations of your calculator, evaluate
√
1 − 1 − 3 × 10−30
The answer is not zero. Analytically, find an approximation to the answer. In this context,
the word “analytically” means that you should use algebra and calculus to find the answer.
And you shouldn’t touch a calculator or computer.
Hint: use a Taylor expansion.
I.
Radioactivity and a simple differential equation
The radioactive nucleus 14 C spontaneously decays into 14 N a β − and a ν̄e . That is to
say, carbon–14 decays into nitrogen–14, a beta particle (also known as an electron), and an
anti-electron-neutrino, which is generically described as just a neutrino. If you start with
a glass full of 14 C today, then in 5730 years you will only have half a glass of 14 C. After a
total of 11460 years only a quarter of a glass will remain. And so forth. We say that the
half-life of 14 C is 5730 years.
In general for any radioactive particle, if we start at t = 0 with N0 particles, then after
a time t the number remaining is
N (t) = N0
1 t/t1/2
2
where t1/2 is the half-life.
Radioactive decay gives one example of a number N (t) whose rate-of-change in time
dN (t)/dt is proportional to the number N (t) itself. In other words,
dN (t)
∝ N (t).
dt
For definiteness assume that
dN (t)
= −λN (t),
dt
where λ is a constant. We solve this differential equation by rewriting it as
dN (t)
= −λ dt
N
and integrating both sides
Z
Z
dN (t)
= − λ dt
N
ln N = −λt + constant
or
N = N0 e−λt
is a general solution to the differential equation
where ln(N0 ) = constant is a constant of integration determined by the initial conditions.
The last line follows by taking the logarithm of both side of the previous equation.
7
With radioactivity, we often define the “e-folding time” τ ≡ 1/λ, which also happens to
be the “mean-lifetime” of the particle, so that
N (t) = N0 e−t/τ .
τ is called the e-folding time because the number of particles decreases by a factor of e after
a time τ .
It is easy to see the relationship between t1/2 and τ by starting with
−t/τ
N0 e
= N0
1 t/t1/2
2
.
Now divide out the N0 , and take the natural logarithm of both sides
1
t
t
. (We are using ln(AB ) = B ln(A) and ln e = 1)
ln
− =
τ
t1/2
2
Finally, cancel the t, invert each side of the equation, and use the fact that ln(1/2) = − ln 2.
The result is
t1/2 = τ ln 2.
Note that
τ (14 C) = 5370 yr/ ln 2 = 7750 yr
is the e-folding time of 14 C.
The mean-lifetime (e-folding time) of a muon is about 2 µs. So, the half-life of a muon is
about t1/2 (muon) = ln 2 × 2 µs ≈ 1.4 µs. .
1.
Carbon dating
Carbon–14 14 C and carbon–13 13 C interact chemically in nearly identical ways. 14 C is
created naturally by nuclear reactions from cosmic ray neutrons colliding with atmospheric
nitrogen 14 N; the nuclear event also gives off a free proton (hydrogen nucleus). Living things,
like you or like a tree, absorb chemically both 13 C and 14 C while they are growing. 13 C is
stable. But once the tree dies, 14 C is no longer absorbed by the tree and the amount of 14 C
decays with a half-life of 5730 yr. By measuring the ratio of 14 C to 13 C in the atmosphere,
and comparing it to the ratio of 14 C to 13 C in a wooden spear, archeologists can calculate
what fraction of the original 14 C in the spear has decayed, and from that fraction they can
determine how long ago the spear was a growing tree branch.
This same basic technique is used to date accurately many archeological relics over time
scales of hundreds of years to a few ten-thousands of years. For shorter times, not enough
of the 14 C has decayed to measure accurately, and for longer time not enough of the 14 C
remains to measure accurately.
Similar techniques, using different radioactive isotopes are used to determine the “lifetime” of rocks from the moon, rocks on earth and meteorites. In this context the “lifetime”
of a rock is the time since the rock solidified.
8
2.
A note on nuclear power
Today, we have the technology to manufacture nuclear power. By and large, this can
be done safely if it is done carefully. The typical daily radioactive waste emission from a
nuclear power plant, directly into the atmosphere, is actually less than daily radioactive
waste emission into the atmosphere from a coal-fired power plant. However, a significant
roadblock to “going nuclear” is the disposal of nuclear waste which is collected inside the
reactor and stored “off-site.”
Every nuclear reactor (or explosion of a nuclear weapon) creates nuclear waste with a
mixture of many different radioactive isotopes. Some of these have reasonably short halflifes such as iodine-126 with a half-life of 13 days—the iodine-126 from a nuclear test is
essentially gone after a year. (A year is about 24 half-lifes and (1/2)−24 ≈ 6 × 10−8 . Only
1 part in 10 million remains after a year, and only 1 part in 1014 after two years.) Some of
the radioactive isotopes have very long half-lifes, such as Uranium-238 with a half-life of 4.5
billion years. Essentially none of the 238 U decays over a time span comparable to the time
that Man has been walking on this planet.
The disposal of nuclear waste remains a significant technological problem.
```