TEST BANK FOR Foundation Mathematics for the Physical Sciences By Riley K.F. and Hobson M.P.

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