Exam (elaborations) TEST BANK FOR Foundation Mathematics for the Physical Sciences By Riley K.F. and Hobson M.P.
Foundation Mathematics
for the
... [Show More] Physical Sciences
Student Solution Manual
K. F. RILEY
University of Cambridge
M. P. HOBSON
University of Cambridge
Contents
Preface page vii
1 Arithmetic and geometry 1
2 Preliminary algebra 14
3 Differential calculus 30
4 Integral calculus 43
5 Complex numbers and hyperbolic functions 54
6 Series and limits 67
7 Partial differentiation 82
8 Multiple integrals 99
9 Vector algebra 109
10 Matrices and vector spaces 122
11 Vector calculus 140
12 Line, surface and volume integrals 155
13 Laplace transforms 170
14 Ordinary differential equations 175
15 Elementary probability 198
A Physical constants 214
v
1 Arithmetic and geometry
Powers and logarithms
1.1 Evaluate the following to 3 s.f.:
(a) eπ, (b)πe, (c)log10(log2 32), (d) log2(log10 32).
Parts (a) and (b) do no more than test the understanding of notation, and are found directly
using a calculator. (a) eπ = 23.1, and (b) πe = 22.5. For the two other parts:
(c) log10(log2 32) = log10(5) = 0.699.
(d) log2(log10 32) = log2(1.505). We therefore need the value of x that satisfies 2x =
1.505. To find it, take logarithms and obtain
x ln 2 = ln 1.505 ⇒ x = ln 1.505
ln 2
= 0.4088
0.6931
= 0.590.
1.3 Find the number for which the cube of its square root is equal to twice the square of its cube root.
If a is the required number, then
a3/2 = 2a2/3 ⇒ 2 = a(3/2)−(2/3) = a5/6.
Now taking logarithms:
56
ln a = ln 2 ⇒ a = e(6 ln 2)/5 = e0.83177... = 2.297 . . .
1.5 By applying the rationalisation procedure twice, show that
131
3 −
√
5 +
√
7
= 9 − 11
√
5 + 7
√
7 + 6
√
35.
Initially treating
√
7 −
√
5 as one unit, we have
131
3 −
√
5 +
√
7
= 131[3 − (
√
7 −
√
5)]
9 − (
√
7 −
√
5)2
= 131[3 − (
√
7 −
√
5)]
9 − 7 − 5 + 2
√
35
.
1
2 Arithmetic and geometry
Since the denominator is now −3 + 2
√
35, as a second step we must multiply both
numerator and denominator by 3 + 2
√
35:
131
3 −
√
5 +
√
7
= 131[3 − (
√
7 −
√
5)] (3 + 2
√
35)
−9 + (4 × 35)
= 131(9 − 3
√
7 + 3
√
5 + 6
√
35 − 14
√
5 + 10
√
7)
131
= 9 − 11
√
5 + 7
√
7 + 6
√
35.
1.7 Solve the following for x:
(a) x = 1 + ln x, (b) ln x = 2 + 4 ln 3, (c) ln(ln x) = 1.
(a) By inspection of either the original equation or its exponentiated form, ex = e1eln x =
ex, we conclude that x = 1.
(b) By exponentiation, x = e2+4 ln 3 = e2e4 ln 3 = e234 = 81e2 = 598.5.
(c) ln(ln x) = 1 ⇒ ln x = e ⇒ x = ee = 15.15.
1.9 Express (2n + 1)(2n + 3)(2n + 5) . . . (4n − 3)(4n − 1) in terms of factorials.
Denoting the expression by f (n),
f (n) = (4n)!
(2n)!
1
(2n + 2)(2n + 4) . . . (4n − 2)(4n)
= (4n)!
(2n)! (n + 1)(n + 2) . . . (2n − 1)(2n) 2n
= (4n)! n!
(2n)! (2n)! 2n
.
1.11 Measured quantities x and y are known to be connected by the formula
y = ax
x2 + b
,
where a and b are constants. Pairs of values obtained experimentally are
x: 2.0 3.0 4.0 5.0 6.0
y: 0.32 0.29 0.25 0.21 0.18
Use these data to make best estimates of the values of y that would be obtained for (a) x = 7.0, and
(b) x = −3.5. As measured by fractional error, which estimate is likely to be the more accurate?
3 Arithmetic and geometry
In order to use this limited data to best advantage when estimating a and b graphically, the
equation needs to be arranged in the linear form v = mu + c, since a straight-line graph
is much the easiest form from which to extract parameters. The given equation can be
arranged as
x
y
= x2
a
+ b
a
,
which is represented by a line with slope a
−1 and intercept b/a when x2 is used as the
independent variable and x/y as the dependent one. The required tabulation is:
x 2.0 3.0 4.0 5.0 6.0
y 0.32 0.29 0.25 0.21 0.18
x2 4.0 9.0 16.0 25.0 36.0
x/y 6.25 10.34 16.00 23.81 33.33
Plotting these data as a graph for 0 ≤ x2 ≤ 40 produces a straight line (within normal
plotting accuracy). The line has a slope
1
a
= 28.1 − 2.7
30.0 − 0.0
= 0.847 ⇒ a = 1.18.
The intercept is at x/y = 2.7, and, as this is equal to b/a, it follows that b = 2.7 × 1.18 =
3.2. In fractional terms this is not likely to be very accurate, as b x2 for all but two of
the x-values used.
(a) For x = 7.0, the estimated value of y is
y = 1.18 × 7.0
49.0 + 3.2
= 0.158.
(b) For x = −3.5, the estimated value of y is
y = 1.18 × (−3.5)
12.25 + 3.2
= −0.267.
Although as a graphical extrapolation estimate (b) is further removed from the measured
values, it is likely to be the more accurate because, using the fact that y(−x) = −y(x),
it is effectively obtained by (visual) interpolation amongst measured data rather than by
extrapolation from it.
1.13 The variation with the absolute temperature T of the thermionic emission current i from a heated
surface (in the absence of space charge effects) is said to be given by
i = AT 2e
−BT ,
where A and B are both independent of T . How would you plot experimental measurements
of i as a function of T so as to check this relationship and then extract values for A
and B?
4 Arithmetic and geometry
The equation can be rearranged to read ln(i/T 2) = lnA − BT , and so y = ln(i/T 2)
should be plotted against T , obtaining a straight-line graph if the relationship is valid. If
so, the (negative) slope of the graph gives B and the intercept y0 on the y-axis gives A as
A = ey0 .
Dimensions
1.15 Three very different lengths that appear in quantum physics and cosmology are the Planck length
p, the Compton wavelength λm, and the Schwarzchild radius rs . Given that
p =
hG
2πcn
, λm = h
mc
, rs = 2GM
c2 ,
where m and M are masses, calculate the dimensions of the gravitational constant G and those of
the Planck constant h. Deduce the value of n in the formula for the Planck length.
Remembering that numerical constants do not contribute to ‘dimensional equations’, from
the Schwarzschild radius formula we have
[G] =
c2rs
M
= (LT
−1)2 L
M
= L3T
−2M
−1.
From the Compton wavelength formula it follows that, [h] = [λmc] = L2T
−1M. Finally,
from the Planck length formula
[cn] =
hG
l2p
= L2T
−1ML3T
−2M
−1
L2
= L3T
−3,
from which, since [c] = LT
−1, it follows that n = 3.
1.17 According to Bohr’s theory of the hydrogen atom, the ionisation energy of hydrogen is mee4/82
0h2.
Using Appendix A, show that this expression does have the dimensions of an energy and that its
value when expressed in electron-volts is 13.8 eV.
The dimensions of 0 seem difficult to determine, but one common equation containing it
is that for the force F between two charges separated by a distance r, i.e.
F = q1q2
4π0r2
⇒ [F] = [q]2
[0]L2 (∗).
The dimensions of the given expression for the ionization energy E contain
[q]4/[0]2, and by squaring (∗) we deduce that they are ([F]L2)2. Using this,
and the result from Problem 1.15 that [h] = ML2T
−1, the dimensions of [E] are
given by
[E] = M
(ML2T −1)2
× (MLT
−2 × L2)2 = M3L6T
−4
L4T −2M2
= ML2
T 2
= [energy].
5 Arithmetic and geometry
Substituting the numerical values given in Appendix A yields a value of 2.2 × 10−18 J,
which is 2.2 × 10−18 ÷ 1.60 × 10−19 = 13.8 when expressed in electron-volts.
1.19 The electrical conductivity σ of a metal is measured in siemens per metre (Sm−1), where 1 S is the
unit of conductance of an electrical component and is equivalent to 1AV−1. TheWiedemann–Franz
law states that at absolute temperature T , and under certain conditions, σ is related to the thermal
conductivity λ of the metal by the equation
λ
σT
= π2
3
k
e
2
.
Verify that this equation is dimensionally acceptable and, using Appendix A, estimate the thermal
conductivity of copper at room temperature, given that its electrical conductivity is 5.6 × 107 Sm−1.
Dealing first with the LHS of the equation, the thermal conductivity is measured in joules
per square metre per second for a unit temperature gradient (or watts per metre per kelvin).
Its dimensions are therefore given by
[λ] = ML2T
−2 L
−2 T
−1
θL−1
= MLT
−3θ
−1.
From the formula watts = volts × amps, we conclude that the dimensions of voltage
are ML2T
−2T
−1 × I
−1, and that those of a siemens are therefore I 2T 3M
−1L
−2. This
gives the dimensions of σ as [σ] = M
−1L
−3T 3I 2, from which it follows that [λ/σT ] =
M2L4T
−6I
−2θ
−2; recall that T in the given equation is a temperature, not a time.
On the RHS, since the Boltzmann constant k has units of joules per kelvin, and charge=
current × time, [k/e] = (ML2T
−2θ
−1)/(IT ) leading to [(k/e)2] = M2L4T
−6I
−2θ
−2.
Comparison with the result for the LHS shows that the stated formula is dimensionally
acceptable. Taking room temperature as 293 K, and substituting other values from
Appendix A, gives an estimate for λ of 402Wm−1 K−1.
1.21 The following is a student’s proposed formula for the energy flux S (the magnitude of the so-called
Poynting vector) associated with an electromagnetic wave in a vacuum, the electric field strength
of the wave being E and the associated magnetic flux density being B:
S = 1
2
0
μ0
1/2
E2 +
μ0
0
1/2
B2
.
The dimensions of 0, the permittivity of free space, areM
−1L
−3T 4I 2, and those of its permeability
μ0 are MLT
−2I
−2. Given, further, that the force acting on a rod of length that carries a current I
at right angles to a field of magnetic flux density B is BI, determine whether the student’s formula
could be correct and, if not, locate the error as closely as possible.
From the force on a current-carrying rod,
[B] = [force]
[current] × L
= MLT
−2
IL
= MT
−2I
−1,
6 Arithmetic and geometry
whilst energy flux, measured in joules per square metre per second, has dimensions
(ML2T
−2)L
−2T
−1 = MT
−3.
Since the electric field E has dimensions [V ]L
−1, and from the solution to Problem
1.19, [V ] = ML2T
−3I
−1, the dimensions of E are MLT
−3I
−1. Those of the ‘E2’ term
are therefore
M
−1L
−3T 4I 2
MLT −2I−2
1/2
(MLT
−3I
−1)2 = MT
−3.
The corresponding calculation for the ‘B2’ term is
MLT
−2I
−2
M−1L−3T 4I 2
1/2
(MT
−2I
−1)2 = L2M3T
−7I
−4.
Thus the electric term is compatible with an energy flux, but the magnetic one is not, and
the error is almost certainly in the latter.
Binomial expansion
1.23 Evaluate those of the following that are defined: (a) 5C3, (b) 3C5, (c) −5C3, (d) −3C5.
(a) 5C3 = 5!
3! 2!
= 10.
(b) 3C5. This is not defined as 5 > 3 > 0.
For (c) and (d) we will need to use the identity
−mCk = (−1)km(m + 1) · · · (m + k − 1)
k!
= (−1)k m+k−1Ck.
(c) −5C3 = (−1)3 5+3−1C3 = − 7!
3! 4!
= −35.
(d) −3C5 = (−1)5 5+3−1C5 = − 7!
5! 2!
= −21.
1.25 By applying the binomial expansion directly to the identity
(x + y)p(x + y)q ≡ (x + y)p+q ,
prove the result
r
t=0
pCr−t
qCt = p+qCr =
r
t=0
pCt
qCr−t
which gives a formula for combining terms from two sets of binomial coefficients in a particular
way (a kind of ‘convolution’, for readers who are already familiar with this term).
7 Arithmetic and geometry
First, we write each term in the form
(x + y)m =
m
s=0
mCsxsym−s ,
where m represents, in turn, p, q and p + q.
Next we consider all the terms in the product of sums on the LHS that lead to terms
containing xr . If the first sum contributes a term containing xr−t , with 0 ≤ t ≤ r, then the
second sum must contribute one containing xt . The power of y that is in the same product
term will be yp−(r−t ) × yq−t = yp+q−r . The full form of the term, including the relevant
binomial coefficients, is therefore pCr−txr−typ−r+t × qCtxtyq−t .
The sum of all these terms over t = 0, 1, . . . , r must also give the coefficient of
xr in the expansion of (x + y)p+q, i.e. p+qCr ; this establishes the left-hand equality.
The right-hand equality follows, either by symmetry or by interchanging the roles of p
and q.
Trigonometric identities
1.27 Prove that
cos
π
12
=
√
3 + 1
2
√
2
by considering
(a) the sum of the sines of π/3 and π/6,
(b) the sine of the sum of π/3 and π/4.
(a) Using
sinA + sinB = 2 sin
A + B
2
cos
A − B
2
,
we have
sin
π
3
+ sin
π
6
= 2 sin
π
4
cos
π
12
,
√
3
2
+ 1
2
= 2
1 √
2
cos
π
12
,
cos
π
12
=
√
3 + 1
2
√
2
.
(b) Using, successively, the identities
sin(A + B) = sinAcosB + cosAsin B,
sin(π − θ) = sin θ,
cos(12
π − θ) = sin θ,
8 Arithmetic and [Show Less]