Calculation solutions
Chapter 1 solutions
Solution 1.1. Parts a. and c. can be expanded to confirm that they are identities. Parts b. and d. are
immediately seen not to be identities.
a. Identity
b. Equation
c. Identity
d. Equation
Solution 1.2. Each can be solved using the quadratic formula.
a. x =
1, ⇡
b. x = 1, 9.9
c. No real roots
d. x =
1
4
1±
p
21
Solution 1.3. Each can be expanded and manipulated into a simple polynomial expression that can
be solved directly.
a. x = 0
b. x =
3, 2
c. x = 0, 2
d. x = 0, ±1
Solution 1.4. Some manipulation confirms the following.
a.
y
(y 1)(y 2)
b.
z 2 +3
z 3 +2z
c.
(x 3)
(x+4)(x2 2)
⌘
d.
s2
s2 4
1
s+2
⌘
2
y 2
3
2z
⌘
1
y 1
z
2(z 2 +2)
⌘1
x 2
2(x2 2)
1
2(x+4)
1
s+2
+
Solution 1.5. In each case we manipulate the expression to the form f (x) = 0 and plot f (x), as shown
in Figure 1.1.
6
0
2
2x
5
10
2x2
5x
25
4
0
2
20
4
30
4
2
0
2
4
6
6
x
(a) 2x2
4
2
0
x
5x = 25
2
4
(b) 2x = 5
10
6
4
2
10
x2 + 2x
2x
5
2
0
ex
0
5
4
6
2
1
0
x
(c) 2x = x2
1
2
10
3
2
2.5, 5
b. x ⇡ 2.3
c. x ⇡
0.3
0
x
1
(d) ex = 10
2x
Figure 1.1: Sketches for Solution 1.5
a. x ⇡
1
2
3
d. x ⇡ 2.3
Solution 1.6. The given root is used to determine a factor of the form “(x
then be performed to determine the other factors.
a. x3 + 3x2 + 2x ⌘ x(x + 1)(x + 2)
b. y 3
2y 2
c. z 4
29z 2 + 100 ⌘ (z
y + 2 ⌘ (y
2)(y
5)(z
1)(y + 1)
2)(z + 2)(z + 5)
d. 2s3 + 6s2 + 12s + 8 ⌘ 2(s + 1)(s2 + 2s + 4)
Solution 1.7. Algebraic manipulations can be performed.
a. x 2 (1, 1] [ [1, 1)
b. x 2 (1, 5) [ (0, 5)
c. x 2 ( 1, 14 ( 1
p
17)) [ ( 14 ( 1 +
p
17), 1)
d. x 2 [ 10, 5] [ [ 2, 2] [ [5, 10]
Solution 1.8. Plots are shown in Figure 1.2.
c)” and long division can
20
40
15
10
20
5
0
0
4
2
0
x
2
5
4
6
4
2
0
2
4
x
(a) 5x2 + 2x <
10
(b) x2 + x
100
50
80
40
60
25
30
40
20
20
10
0
20
0
2
0
2
4
6
6
8
4
2
0
x
x
2
(d) x2 + 2x
(c) 2x + 5x > 10
4
6
2x
Figure 1.2: Sketches for Solution 1.8
a. No solution exists
b. An approximate interval is seen to be x 2 [ 3.2, 2.2]
c. An approximate interval is seen to be x 2 (1.5, 1)
d. An approximate interval is seen to be x 2 ( 1, 2.1] [ [0.6, 5.3]
Solution 1.9. We note that the curve has a root at x = 1 and so (x
demonstrates that it can be factorised as x3 + 4x2
x
4 = (x
1) is a factor. Some work
1)(x + 1)(x + 4). The curve therefore
crosses the x-axis at x = ±1 and 4. Furthermore, it tends to 1 as x ! 1, and
sketch is given in Figure 1.3.
1 as x !
1. A
30
20
f (x)
10
0
10
20
4
2
0
2
x
Figure 1.3: Plot of x3 + 4x2
x
4 required for Solution 1.9.
Solution 1.10. We note that the curve has a root at y = 1 and so (y
1) is a factor. Some work
demonstrates that it can be factorised as y 5 +4y 4 5y 3 20y 2 +4y+16 = (y 1)(y+1)(y+4)(y 2)(y+2).
The curve therefore crosses the y-axis at y = ±1, ±2 and 4. Furthermore, it tends to 1 as y ! 1, and
1 as y !
1. A sketch is given in Figure 1.4.
40
20
0
20
40
4
2
0
2
y
Figure 1.4: Plot of y 5 + 4y 4
5y 3
20y 2 + 4y + 16 required for Solution 1.10.
Chapter 2 solutions
Solution 2.1. The following should be clear from observation.
a. Many-to-one mapping and a function
b. Many-to-one mapping and a function
c. One-to-many mapping and not a function
d. Many-to-one mapping and a function
Solution 2.2. The following conclusions can be reached after some manipulation.
a. f (x) demonstrates positive symmetry about x = 2
1)3 ; it therefore demonstrates negative symmetry about
b. g(y) can be factorised as g(y) = (y
x=1
c. h(z) demonstrates positive symmetry about z = 0; it is therefore an even function
d. l(s) demonstrates negative symmetry about s = 0; it is therefore an odd function
Solution 2.3. We note that x =
2 is a root of both the numerator and denominator and so the
function can be written as,
f (x) =
3(x3
x2
+ 2x2
4
25x
50)
=
(x + 2)(x 2)
x 2
=
3(x + 2)(x 5)(x + 5)
3(x 5)(x + 5)
A plot of the function can be seen in Figure 2.1. From this we note the following properties.
0.4
f (x)
0.2
0
0.2
0.4
20
Figure 2.1: Plot of f (x) =
10
0
x
x2 4
3(x3 +2x2 25x 50)
10
20
required for Solution 2.3.
a. f (x) is undefined at x = ±5 and these are removed from R to form the domain; the range is given
by R
b. f (x) has a single root at x = 2
c. f (x) demonstrates no symmetries
d. f (x) ! 0 as x ! ±1 (horizontal asymptotes) and f (x) ! ±1 as x ! ±5 (vertical asymptotes)
Solution 2.4. We define the basic functions that contribute to the composite as follows.
a. f (x) = (m n) (x) where m(x) = ⇡x2 + 3x
4 and n(x) = e2x
b. g(x) = (m n) (x) where m(x) = sin x and n(x) = x2
2x + 2
c. h(x) = (m n p) (x) where m(x) = ex , n(x) = sin x and p(x) = x5
d. l(x) = (m n p q) (x) where m(x) = x2
2x + 5, n(x) = sin x, p(x) = ex and q(x) = x2
Solution 2.5. We note the following as the independent variable ! ±1.
a.
x2 2x
4x2 5x+3
!
b.
y 5 +⇡y 3 +1
y 3 +y+1
! ±1 as y ! ±1; i.e. there are no horizontal asymptotes
c.
50z 2
3z 2 z+2
d.
t
2(t2 +1)
!
1
4
50
3
as x ! ±1
as z ! ±1
! 0 as t ! ±1
Solution 2.6. We find the following after some manipulation.
p
a. f (x) = 0 when x = ± 1
⇡n for n 2 N
3/2
b. g(x) = 0 when x = ± e 2
c. h(x) 6= 0 for x 2 R
1
2
d. l(x) = 0 when x = ⇡ n
for n 2 Z
Solution 2.7. We find the following after some manipulation.
a. f (x) = 2x ) f
1
b. g(x) = log5 (x
1) ) g
(x) =
x
2
1
(x) = 1 + 5x
2x
1
2)(x + 4) ) h
c. h(x) = (x
1
(x) =
1+
p
x
7 for x
7
d. l(x) = x + x2 + x3 + x4 + x5 an explicit solution does not exist
Solution 2.8. We determine the required domains and ranges as follows.
a. f (x) = sin(x) has domain x 2 R and range {f 2 R :
1  f  1}
b. g(x) = ex has domain x 2 R and range {g 2 R : g > 0}
c. h(x) = ln x has domain {x 2 R : x > 0} and range h 2 R
d. l(x) =
x4 2x
x2 +x 2
has domain x 2 R and range {l 2 R : l
min(l)}
Solution 2.9. The composite functions are determined from the functions given.
⇣
2x+1
⇡x+1
a. (f
g) (x) = sin
b. (f
h f ) (x) = sin esin x
c. (g f
f ) (x) =
d. (g h f ) (x) =
3x2
⌘
2 sin sin x+1
3(sin sin x)2 ⇡ sin sin x+1
2esin x +1
3e2 sin x ⇡esin x +1
Solution 2.10. The domain and range of the composite functions are as follows.
a. f (x) =
e2x +2ex
e2x +ex 2
b. g(x) = sin x2
c. h(x) = ln
⇣
d. l(x) = cos
1
x 1
⇣
⌘
has domain {x 2 R : x 6= 0} and range {f 2 R : f 2 ( 1, 0) [ (1, 1)}
2 has domain x 2 R and range {g 2 R : g 2 [ 1, 1]}
has domain {x 2 R : x > 1} and range h 2 R
x2 +2x+1
x2 +3x+1
⌘
has domain {x 2 R : x 6=
1
2
3±
p
5 } and range {l 2 R : l 2 [ 1, 1]}
Chapter 3 solutions
Solution 3.1. Recall that we require the left and right limit to be equal to the value of the function
at all points in R. The properties of quadratic functions mean that the function is continuous below
x=
2 and above x =
2. However, at x =
lim f (x) = 14
x! 2
2, we note that f ( 2) = 14 and
6=
The function is therefore not continuous at this point.
lim f (x) = 13
x! 2+
Solution 3.2. We use the fundamental definition of a derivative in each case.
d
f (x + )
f (x) = lim
!0
dx
f (x)
Some manipulation leads to the following.
a. f 0 (x) = 3x2 + 2x
b. g 0 (y) = 2e2y
c. h0 (z) = (z + 1)ez
2⇡
p3
d. m0 (p) =
Solution 3.3. Each derivative can be found using the product rule.
a. f 0 (x) = x (2 sin(x) + x cos(x))
b. g 0 (x) = x4 + 4x3
c. h0 (x) =
d. m0 (x) =
x2
2x + 2 ex
ex (x ln x+1)
x
sec2 (x) tan(x)
ex
Solution 3.4. Each derivative can be found using the product or quotient rule.
a. f 0 (x) =
x2 +2x+6
x2 +4x+4
b. g 0 (x) =
3x2 (x8 2x6 +2x3 +10x2 4)
(x6 2)2
c. h0 (x) =
2x5 +8x4 +16x3 88x2 +82x+32
(x 5)2 (x 3)2 (x 1)2 (x+1)2
d. m0 (x) = 0
Solution 3.5. Each derivative can be found using the chain rule.
a. f 0 (x) = 3x2 cos x3
b. g 0 (x) =
tan(x)
c. h0 (x) =
ep
2 1 x
d. m0 (x) =
p
1
x
ex (e2x +1)
(e2x 1)2
Solution 3.6. Each derivative can be found using the chain rule.
2
a. f 0 (x) = 2xesin(x ) cos(x2 )
b. g 0 (x) = 3x2 sin(2
c. h0 (x) =
1
d. m0 (x) =
3
ln 4
x3 ) sec2 cos(2
x3 )
Solution 3.7. Each derivative can be found using the product and chain rules.
a. f 0 (x) = 2ex cos(x2 )
b. g 0 (x) =
c. h0 (x) =
d. m0 (x) =
x2 sin(x2 )
2ex + 2e2x tan(ex
2(x+1) cos(x2 5x+cos x)
x2 +2x 2
2x sin(2 x2 )
ex e x
x2 )
2ex + e2x
+ ln(x2 + 2x
2 (ex
2) (5
2x) sec2 ex
2x + sin x) sin x2
x2
5x + cos x
(ex +e x ) cos(2 x2 )
(ex e x )2
Solution 3.8. Simple application of the chain rule leads to
d
g 0 (x)
ln g(x) =
dx
g(x)
Solution 3.9. Each derivative can be found using the product and/or chain rules.
a. f 0 (x) = ex arccos(x)
b. g 0 (x) =
ex
e2x +1
c. h0 (x) = arcsin(cos(x))
d. m0 (x) =
x
pe
1 x2
p x sin(x)
2
1 cos (x)
2(2 2x) arctan(sin x2 )
p
1 (2x x2 )2
4x arcsin(2x x2 ) cos(x2 )
1+sin2 (x2 )
Solution 3.10. The derivative can be obtained from the product and chain rules.
o
d n g(x)
eg(x) sin(f (x))h0 (ln(x))
e
sin(f (x))h(ln x)) = eg(x) f 0 (x) cos(f (x))h(ln(x))+eg(x) sin(f (x))g 0 (x)h(ln(x))+
dx
x
Chapter 4 solutions
Solution 4.1. The continuity of all higher-order derivatives is required.
a. Smooth for all x 2 R
b. The first (and all other) derivatives are not equal either side of x = 0. The function is therefore
not smooth at this point.
c. The first (and all other) derivatives are not equal either side of x = 4. The function is therefore
not smooth at this point.
d. The first (and all other) derivatives are not equal either side of x = 0. The function is therefore
not smooth at this point.
Solution 4.2. The derivatives can be obtained from repeated use of the chain rule.
a. h0 (z) = sin(z) + z cos(z)
b. h00 (z) = 2 cos(z)
z sin(z)
c. h(4) (z) = z sin(z)
4 cos(z)
d. h(8) (z) = z sin(z)
8 cos(z)
Solution 4.3. The turning points coincide with roots of the first derivative function.
a. f (x) = x +
2
x
p
has turning points at x = ± 2
b. g(x) = x2 + 5x
c. h(x) = 2
2 has a turning point at x =
5
2
ln(x) has no turning points
d. m(x) = e2x
2ex + 2 has a turning point at x = 0
Solution 4.4. The 2nd derivative test can be applied to obtain the following.
a. f (x) = x +
2
x
has maximum at x =
b. g(x) = x2 + 5x
c. h(x) = 2
p
2 and a minimum at x =
2 has a minimum at x =
ln(x) has no turning points
5
2
p
2
d. m(x) = e2x
2ex + 2 has a minimum at x = 0
Solution 4.5. The turning points coincide with roots of the first derivative function.
1
2
a. f (x) = sin(x) has turning points at x = ⇡ n
b. m(x) = cos(x)
for n 2 Z
sin(x) has turning points at x = 2n
1
4
⇡ and x = 2n +
3
4
⇡
p
p
c. g(x) = cos(x2 ) has turning points at x = ± 2⇡n and x = ± ⇡(1 + 2n) for n 2 z
d. h(x) = sin2 (x)
2 sin(x) + 1 has turning points at x = 2n ±
1
2
⇡
Solution 4.6. The 2nd derivative test can be applied to obtain the following.
a. f (x) = sin(x) has maxima for x = ⇡ 2n +
b. m(x) = cos(x)
1
2
and minima for x = ⇡ 2n
sin(x) has maxima at x = 2n
1
4
1
2
for n 2 Z
⇡ and minima at x = 2n +
3
4
⇡
p
p
c. g(x) = cos(x2 ) has maxima x = ± 2⇡n and minima for x = ± ⇡(1 + 2n) for n 2 Z
d. h(x) = sin2 (x)
2 sin(x) + 1 has maxima at x = 2n
1
2
⇡ and minima at x = 2n +
1
2
⇡
Solution 4.7. The function of interest is
h(x) = (f
g g)(x) = (x4 )2 + (x4 ) + 1 = x8 + x4 + 1
which is such that
h0 (x) = 4(2x7 + x3 )
We therefore have a single turning point at x = 0. The 2nd derivative test is inconclusive and we
continue by sampling h0 (x) either side of x = 0 to conclude that the turning point is a minimum.
Solution 4.8. Neither of the functions demonstrate singularities and so we proceed by looking at the
asymptotic behaviour and local turning points.
a. Note that f (x) = xe1
x2
! 0 as x ! ±1, the range is therefore determined by turning points
p
at finite x. The function has a maximum at x = p12 such that f (x) = 2e , and a minumum at
⇥ pe pe⇤
pe
x = p12 such that f (x) =
.
2 . The range is therefore f 2 R|f 2
2,
2
b. Note that g(x) =
x2 2x+1
x2 +⇡
! 1 as x ! ±1, the range is therefore determined by turning points
at finite x. The function has a maximum at x =
⇡ such that g(x) = 1 +
⇥
⇤
x = 1 such that g(x) = 0. The range is therefore g 2 R|g 2 0, 1 + ⇡1 .
1
⇡
and a minimum at
c. Note that the function oscillates between fixed points as x ! ±1, the range is therefore determined by these fixed points that are given by the maximum and minimum values. The function
has maxima of 1 and 0 at various x and minima of
n
h
io
given by h 2 R|h 2 12 p12 , 1 .
1
2
p1
2
at other x. The range is therefore
d. Note that m(x) = x sin(x2 1) oscillates with increasing amplitude as x ! ±1 and is unbounded.
The range is therefore given by {m 2 R}.
Solution 4.9. The cylinder has unit volume given by
V = ⇡r2 h = 1
)
h=
1
⇡r2
and surface area given by
✓
◆
1
S(r) = 2⇡r r + 2
⇡r
)
S = 2⇡r (r + h)
We then seek the minimum value of S(r) which can be shown to be at r =
1
p
3
2⇡
p
where S(r) = 3 3 2⇡.
Solution 4.10. The investor is required to sell the asset for the maximum price within the time interval
q
5⇡
t 2 [0, 4]. P (t) has a maximum value of $1000 at t =
2 ⇡ 2.8 and so the maximum profit to be
made (ignoring the time value of money) is $500.
Chapter 5 solutions
Solution 5.1. Each can be written down from inspection.
a. 1 + 2 + 3 + 4 + . . . + 10 =
10
X
i
i=1
b. 1 + 3 + 9 + 27 + 81 =
5
X
3i
1
i=1
c. 1 + 2 + 5 + 12 + 27 + . . . =
1
X
2i
i
i=1
d. 1 + (x
1) + (x
1)2 + . . . + (x
1)10 =
11
X
(x
1)i
1
i=1
Solution 5.2. Each can be rewritten in terms of the summations given as follows.
a. 0.5 + 4 + 13.5 + 32 + . . . + 500 =
10
X
0.5k 3 = 0.5
k=1
b. ⇡ 3 + 4⇡ 3 + 9⇡ 3 + . . . + 81⇡ 3 = ⇡ 3
10
X
3025
2
k3 =
k=1
10
X
1)3 = 285⇡ 3
(n
k=1
c.
10
X
i3 + i2
10
X
2i + 1 =
i=1
1) =
i=1
1
X
d.
i3 + (i
n3 +
n= 10
10
X
=
12
X
4n2
10
X
i=1
24n + 36 =
n=3
10
X
( i)3 +
i=1
(i)3 + 4
i=1
10
X
(i
i3 +
2)2 =
10
X
1)2 = 3310
(i
i=1
10
X
4(i + 2)2
24(i + 2) + 36
i=1
1885
i=1
Solution 5.3. Recognise that
k
X
n=
n=1
and so f (n) =
k
(k + 1)
2
2n
7 .
Solution 5.4. We attempt to cast each expression in terms of the standard arithmetic or geometric
forms.
a.
10 ✓
X
4
n=1
b.
c.
50
X
◆
2) =
=
10 ✓
X
13
n=1
50
X
(0 + 2(k
k=1
100
X
X 1
1
1
=
⇥
i
1.07i
1.07
1.07
i=1
20
X
3(n
4
k=1
i=1
d.
(2k
3n
4
1)
4
◆
is a finite arithmetic progression
1)) is a finite arithmetic progression
100
2j + 0.5j+2 =
j=1
20
X
1
is a finite geometric progression
(2 + 2(j
1)) +
j=1
metic progression
20
X
j=1
0.53 ⇥ 0.5j
1
is the sum of a geometric and arith-
Solution 5.5. Using the above results, we apply the standard formulae for the sums of arithmetic and
geometric progressions.
◆ X
10 ✓
10 ✓
X
3n
13
a.
4
=
4
4
n=1
n=1
b.
50
X
(2k
2) =
k=1
c.
100
X
i=1
50
X
(0 + 2(k
3(n
1)
4
1)) =
k=1
1
=
1.07i
100
X
i=1
1
1
⇥
1.07 1.07i
1
=
◆
=
10
2
✓
2
13
4
9
3
4
◆
=
5
4
50
⇥ 98 = 2450
2
1
1 1.07 100
⇥
⇡ 14.269
1.07
1 1.07 1
d.
20
X
2j + 0.5j+2 =
j=1
20
X
(2 + 2(j
1)) +
j=1
20
X
j=1
0.53 ⇥ 0.5j
1
= 10 ⇥ 40 + 0.53 ⇥
1 0.520
= 400.25
1 0.5
Solution 5.6. Each summation is identified as a particular example of a standard form and conclusions
are then drawn as follows.
a.
1
1
X
X
1
1
1
=
n
n
1.1
1.1
1.1
n=1
n=1
1
. The summation is therefore an infinite geometric progression with
common ratio less than one. It has a finite value given by
b.
1
X
1 1 1.1
1.1 1 1.1
1
1
= 10.
(⇡ + 1)k is an infinite arithmetic progression and so does not have a finite value.
k=10
c.
1
X
1.07i =
i= 1
1
X
1.07
k
. The summation is therefore an infinite geometric progression with com-
k=1
1
1 1 1.07
100
mon ratio less than one. It has a finite value given by 1.07
1 1.07 1 = 7 .
✓ ◆j+2 !
1
X
1
d.
j+
is the sum of infinite geometric and arithmetic progressions. The arithmetic
4
j=100
component has infinite sum.
Solution 5.7. The standard approach is taken in each case.
a. f (x) = ex
1
4) + 12 e3 (x
= e3 + e3 (x
4)2 + 16 e3 (x
4)3 . . . =
1
X
e3
(x
n=0
b. g(y) = (4
5
3y) = 1024
c. h(z) = sin(z 2 ) =
d. m(p) =
1
2
1+( p
2)
=
2
4320y + 1620y
p
p
3840y + 5760y
p
2 ⇡(z
p
⇡)
8⇡(p ⇡)
(4+⇡ 2 )2
4
4+⇡ 2
(z
+
3
⇡)2 + 43 ⇡ 3/2 (z
4(3⇡ 2 4)(p ⇡)2
(4+⇡ 2 )3
4
243y 5
⇡)3 + . . .
16⇡(⇡ 2 4)(p ⇡)3
(4+⇡ 2 )4
+ ...
Solution 5.8. The ratio test is applied in each case to determine x such that
lim
n!1
a. We require |1
b. We require
1
x
an+1 (x)
<1
an (x)
x| < 1 and so x 2 (0, 2)
< 1 and so x 2 ( 1, 1) [ (1, 1)
c. We require lim
x
< 1 and so x 2 R
n+1
d. We require lim
2 x
23 1 +
n!1
n!1
1
n
< 1 and so x 2 ( 4, 4)
4)n
n!
Solution 5.9. Where appropriate, a Taylor expansion of the numerator is used in each of the following.
a. lim
3
x2
x2
x
x!0
= lim x
x!0
p
x =0
✓ ⇢
◆
1
x6
x10
2
x
+
+ ...
=0
x
6
120
✓
⇢
e2(x 1) 1
1
= lim
2(x 1) + 2(x 1)2 +
c. lim
x!1
x!1 x
x 1
1
✓
⇢
ln(x2 ) sin(x ⇡)
ln(x2 )
1
d. lim
= lim
(x ⇡)
(x
x!⇡
x!⇡
x ⇡
x ⇡
6
sin(x2 )
b. lim
= lim
x!0
x!0
x
4
(x
3
3
1) + . . .
⇡)3 + . . .
◆
◆
=2
= 2 ln ⇡
Solution 5.10. We take the first few terms of the Maclaurin expansion of f (x)
f (x) ' 1 + 4x2 +
22 4 241 6
x +
x
3
30
Further terms will improve the approximation for x where the series is convergent. Using the standard
form of the Maclaurin expansion for the exponential function, we have
f (x) =
1
X
an (x)
n=0
where
4n (x sin x)n
.
n!
The ratio test therefore requires
lim
n!1
4(x sin x)
<1
n+1
which is true for all x 2 R.
Chapter 6 solutions
Solution 6.1. The integrals can be performed from knowledge of the standard integrals.
a.
b.
c.
d.
R
R
R
1
2
sin(2x)dx =
e
5y
R⇣
1
5y
5e
dy =
p 1
dz
1 9z 2
1
2(p 2)
cos(2x) + c
=
1
3
+c
arcsin(3z) + c
1
2(p+2)
⌘
dp =
1
2
(ln(2
p)
ln(p + 2)) + c
Solution 6.2. The integrals can be performed with an appropriate substitution.
a.
b.
c.
d.
R
R
R
R
2ex
(ex +2)2 dx
2
ex +2
=
y3 y4
12
sin z
cosn z dz
=
4
3
+c
3
28
dy =
cos1 n (z)
n 1
cosec2 p
(2+cot p)4 dp
y4
7
3
+c
+ c for n 2 N
1
3(cot p+2)3
=
12
2
+c
Solution 6.3. The integrals can be performed using integration by parts.
a.
b.
c.
d.
R
R
R
R
x2 ln xdx =
x3
9
(3 ln x
1) + c
y 2 sin ydy = 2y sin y
ln(z)
z 6 dz
5 ln z+1
25z 5
=
p2 2p dp =
(y 2
2) cos y + c
+c
2p (p2 ln2 (2) p ln(4)+2)
ln3 (2)
+c
Solution 6.4. The integrals can be performed using partial fractions.
a.
b.
c.
d.
R
R
R
R
3x2 +x+2
x2 1 dx
= 3x + 3 ln(1
y 2 +3
y 3 +2y dy
=
3
2
z 3 +2
z 2 +2z 3 dz
=
1
4
ln(y)
1
4
(2(z
p5 2p4 +p3 +p+5
p3 2p2 +p 2 dp
=
x)
2 ln(x + 1) + c
ln(y 2 + 2) + c
4)z + 3 ln(1
p3
3
1
2
z) + 25 ln(z + 3)) + c
ln(p2 + 1) + 3 ln(2
p)
arctan(p) + c
Solution 6.5. The integral can be considered as the sum of two cyclical integrals.
Z
e2x sin(2x)
e3x cos(3x) dx =
e2x
( 3 sin(2x) + 2ex sin(3x) + 3 cos(2x) + ex cos(3x)) + c
12
Solution 6.6. The completed square forms are obtained as follows and any real roots follow directly.
a. x2
4x + 6 = (x
2)2 + 2 and no real roots exist
b. 2x2 + x + 5 = 2 x +
c. 5x2
d. x2
20x + 5 = 5(x
5x + 50 = x
1 2
4
+
2)2
5 2
2
+
39
8
and no real roots exist
15 leading to real roots at x = 2 ±
175
4
and no real roots exist
p
3
Solution 6.7. The completed-square forms of the integrands were obtained above and these can be
used to identify each integral as variation a on a standard form.
⇣
⌘
R
a. x2 14x+6 dx = p12 arctan xp22 + c
b.
c.
d.
R
R
R
1
2y 2 +y+5 dy
p2
39
=
1
5z 2 20z+5 dz
1
p2 5p+50 dp
=
=
arctan
4y+1
p
39
⌘
+c
p
ln( 3 + 2
1p
10 3
2
p
5 7
⇣
arctan
⇣
2pp 5
5 7
p
ln( 3
z)
⌘
2 + z) + c
+c
Solution 6.8. The integral can be performed as follows.
Z
p
2
4x 2
x2
dx =
Z
p
2
2
2)2
(x
dx = 2 arcsin
✓
x 2
p
2
◆
+c
Solution 6.9. The derivative can be ‘reversed’ by the integration
f (✓) =
Z
cos(✓) sin(✓)
d✓
sin(✓) + cos(✓)
This is determined using the substitution u = cos ✓
sin ✓ and we obtain
f (✓) = ln (sin ✓ + cos ✓) + c
where c is any real constant.
Solution 6.10. The derivative can be ‘reversed’ by the integration
g(!) =
Z
p
arcsin( !)d! =
Z
p
1 ⇥ arcsin( !)d!
This is determined by integration by parts and we obtain
g(!) =
1 ⇣p
!
2
! 2 + (2!
p ⌘
1) arcsin( ! + c
where c is any real constant.
Chapter 7 solutions
Solution 7.1. The following integrals can be determined using an appropriate change of variable.
a.
b.
c.
d.
R
R
⇡
3
sin3 xdx =
0
⇡
2
4z+6
dz
0 z 2 +3z+1
R2
1
12
1
dp
2 1+ep
3
4
cos(3x)
cos(2y)
4 sin(y) cos(y)+1 dy
0
R1
⇥
=
⇥1
4
cos x
⇤ ⇡3
5
24
=
0
ln(2 sin(2y) + 1)
⇤ ⇡2
=0
0
⇥
⇤2
= 2 ln z 2 + 3z + 1
= ln(25)
2
ln(1 + ep )]
= [p
2
2
=2
Solution 7.2. Various approaches are required to determine the following integrals.
a.
b.
c.
d.
R9
1
dx
4 1+9x2
R
⇡
2
⇡
4
R4
R1
0
=
⇥1
3
arctan(3x)
sin(3y) cos(3y)dy =
25 z
dz
4 z 2 25
p
= [2 ln(5
1
dp
1+2p p2
=
h
⇥
⇤9
4
1
12
z)
arcsin
1
6
=
cos(6y)
⇤ ⇡2
⇡
4
=
3 ln(z + 5)]
⇣
1p p
2
195
2108
arctan
⌘i1
0
=
4
4
1
12
⇡
10.986
⇡
4
Solution 7.3. In each case we can exploit the symmetry of integrand.
a.
b.
c.
d.
R2
R5
1
(y
R 12
8
R
sin(x3 )dx = 0 as the integrand is an odd function about x = 0.
2
(z
3⇡
2
⇡
2
3)3 + (y
10)4 sin(z
cos(p
3)dy = 0 as the integrand is an odd function about y = 3.
10)dz = 0 as the integrand is an odd function about z = 10.
⇡) sin (p
⇡)5
(p
⇡)3 dp = 0 as the integrand is an odd function about p = ⇡.
Solution 7.4. The integrand can be factorised as x3
9x2 + 27x
27 ⌘ (x
3)3 and immediately
identified as an odd function about x = 3. Since this is the midpoint of the integration domain, we
conclude that the integral is zero.
Solution 7.5. Each area is illustrated in Figure 7.1.
a. The functions f (x) = x2 and g(x) =
given by
x2 do not cross within x 2 (0, 3]. The area is therefore
Z
3
2x2 dx = 18
0
b. The functions f (x) = e2x and g(x) = ex do not cross within x 2 (0, 3]. The area is therefore given
by
Z
3
e3x
0
ex dx =
1 3
e
2
1
2
⇡ 182.129
c. The functions f (x) = x2
x) cross within x 2 [0, 3] at x =
2x + 1 and g(x) = x(6
1
2
4
p
14 .
The area is therefore given by
Z
1
2 (4
p
14)
1
2x
x)x + x2 dx+
(6
0
Z
3
1
2 (4
d. The functions f (x) = ex sin x and g(x) =
therefore given by
Z
p
1 + 2x + (6
x2 dx =
x)x
14)
1⇣ p
14 14
3
⌘
7 ⇡ 15.128
f (x) do not cross within x 2 (0, 3]. The area is
3
2ex sin xdx = 1 + e3 (sin(3)
0
10
400
5
300
cos(3)) ⇡ 23.719
200
0
100
5
0
10
0
0.5
1
1.5
x
2
2.5
3
0
0.5
1
1.5
x
2
2.5
3
2.5
3
(b)
(a)
10
10
8
5
6
0
4
2
5
0
0
0.5
1
1.5
x
2
2.5
3
10
0
0.5
1
1.5
x
(c)
2
(d)
Figure 7.1: Sketches for Solution 7.5
Solution 7.6. The functions g(x) and h(x) cross at x =
7⇡
4 ,
5⇡ ⇡ 3⇡
4 , 4, 4 .
The area is therefore
calculated as
Z
7⇡
4
2⇡
✓
1
p
2
◆
Z
sin x dx+
✓
◆
◆
Z ⇡4 ✓
1
1
p + sin x dx +
p
sin x dx
7⇡
5⇡
2
2
4
4
✓
◆
✓
◆
Z 3⇡
Z
2⇡
p
p
4
1
1
p + sin x dx +
p
+
sin x dx = 4 2 + 2⇡
⇡
3⇡
2
2
4
4
5⇡
4
Solution 7.7. The areas are illustrated in Figure 7.2. In each case we begin by finding where the
curves cross.
x2 cross at x = ± p12 . The area is therefore given by
a. The functions f (x) = x2 and g(x) = 1
Z
b. The functions f (x) = x2
1
p
2
1
p
2
2 and g(x) = x
by
Z
1
2
1
2
p
(1+
(1
2x2 )dx =
(1
p
c. The functions f (x) = ex and g(x) = 5
1 cross at x =
5)
1+x
5)
p
2 2
3
x
2
x2 cross at x =
1
2
1±
p
5 The area is therefore given
p
5 5
dx =
6
2.21144 and x = 1.24114 (as determined
numerically or with Wolfram Alpha). The area is therefore given by
Z
d. The functions f (x) = x3
1.24114
5
ex
x2 dx = 9.6706
2.21144
2x + 2 and g(x) = 3
2x
2x4 cross at x =
1 and x = 0.73898 (as
determined numerically or with Wolfram Alpha). The area is therefore given by
Z
0.73898
1
1
x3
2x4 dx = 1.42628
1
1
0.8
0
0.6
1
0.4
0.2
0
2
1
0.5
0
x
0.5
1
1
0.5
0
0.5
x
1
1.5
2
(b)
(a)
4
3
4
2
2
1
0
0
3
2
1
0
1
2
2
x
1.5
1
0.5
x
(c)
0
0.5
1
(d)
Figure 7.2: Sketches for Solution 7.7
Solution 7.8. The functions g(x) = 2 and f (x) =
2x2
x2 +1
do not cross within the interval x 2 [0, 1]. The
area is therefore straightforward to express and is given by
Z
1
0
✓
2
2x2
x2 + 1
◆
1
dx = [2 arctan x]0 =
⇡
2
Solution 7.9. We consider the problem in terms of functions of y. That is, we require the area bounded
between f (y) = 0 and g(y) = y 2
2y. The curves cross at y = 0 and 2 and so the area is given by
Z
2
2y
0
y 2 dy =
4
3
Solution 7.10. We consider the problem in terms of functions of y. That is, we require the area
p
2
bounded between f (y) = 0.5 and g(y) = ey . The curves cross at y = ± ln(2) and so the area is given
by
Z pln(2) ⇣
2
p
ey
2
ln(2)
⌘
dy ⇡ 1.18518
Chapter 8 solutions
Solution 8.1. The expressions are simplified as follows.
a. i
⇣
b.
i11
1
c.
1
1+i
d.
⇣
⌘
i4 =
i5
i3
⇣
i6
i4
=
1 i
1+i
i2
i
1
2
⌘3
(1
4
2i
⌘
=0
i)
=i
Solution 8.2. The modulus and principal argument are determine as follows.
a. a = 1 + i ) |a| =
p
2i ) |b| =
b. b = 1
2 and Arg(a) =
p
5 and Arg(b) =
⇡
4
⇡ 0.785
arctan(2) ⇡
1.107
p
c. c = 2 + 6i ) |c| = 2 10 and Arg(c) = arctan(3) ⇡ 1.249
d. d =
4
3
4
3i ) |d| = 5 and Arg(d) = arctan
⇡⇡
2.498
Solution 8.3. The Argand diagram is shown in Figure 8.1.
10
Im(z)
2 + 6i
5
1+i
Re(z)
-5
5
1
4
2i
3i
-5
Figure 8.1: Argand diagram for Question 8.3
Solution 8.4. The operations result in the following complex numbers.
a. a + b
b. ab
c
3d = 12 + 2i
dc =
7 + 29i
c
d
c.
a
b
+
+
33
25 i
d.
a
d
(a + bcd) =
294
25
=
21
25
+
392
25 i
Solution 8.5. An nth-order polynomial has n roots in the complex plane. In each case we identify
an obvious real root and use this to factorise the polynomial into linear and quadratic factors that are
easily solved.
a. f (x) = x2 + 2x + 2 has roots x =
b. g(x) = x3 + x
2)(x2 + 2x + 5) has roots x =
10 = (x
c. h(x) = x4 + 2x3 + x2 + 8x
d. l(x) = x5
x4 + 5x3
1±i
12 = (x
5x2 + 4x
1 ± 2i and 2
1)(x + 3)(x2 + 4) has roots x =
4 = (x
3, 1 and ±2i
1)(x2 + 4)(x2 + 1) has roots x = 1, ±i and ±2i
Solution 8.6. Some straightforward manipulation leads to the following results. Note that each Arg 2
( ⇡, ⇡) and so is the principal argument.
a. |z1 z2 | = 8 and Arg(z1 z2 ) =
3⇡
4
b. z13 = 64 and Arg(z13 ) =
c.
z1
z2
= 2 and Arg
d.
z12
z23
= 2 and Arg
⇣
⇣
z1
z2
z12
z23
⌘
⌘
⇡
8
=
3⇡
8
=
7⇡
8
Solution 8.7. We use the moduli and arguments obtained in Question 2 and write down the following.
a. a = 1 + i =
b. b = 1
p
2i =
⇡
2ei 4
p
5e
i arctan(2)
p
c. c = 2 + 6i = 2 10ei arctan(3)
d. d =
4
3i = 5ei(arctan( 4 )
3
⇡)
Solution 8.8. We begin each case by expressing the complex number in rei✓ form.
If z1 = 1 + i and z2 = 2
p
3
a. z14 =
⇡
2ei 4
3
4
1
3
d.
q
q
z1
z2
=
1
z12
z23
=
⇣q
⇣
5
3⇡
⇣p
2 i arctan(3)
5e
2
p
3
= 2 8 ei 16 , 2 8 e
b. (z1 z2 ) 3 = (3+i) 3 =
c.
i, determine all complex values of the following expressions.
e i( ⇡
5
i 5⇡
16
3
10ei arctan( 3 )
⌘ 12
1
2
arctan( 11
))
q
4
=
⌘ 12
i 13⇡
16
, 28 e
⌘ 13
=
p
6
2 i2 arctan(3)
5e
=
p
2
3
54
e 2 (⇡
i
3
and 2 8 ei
11⇡
16
10e 3 arctan( 3 ) ,
i
1
q
4
and
p
6
10e 3 arctan( 3 )+
i
1
2i⇡
3
and
p
6
10e 3 arctan( 3 )+
i
1
2 i⇡+i2 arctan(3)
5e
2
arctan( 11
))
and
p
2
2 i⇡+ 2i (⇡ arctan( 11
))
3
54
e
Solution 8.9. Each demonstration begins with the substitution of the following expressions into the
left hand side.
sin x =
1 ix
e
2i
e
ix
and
cos x =
1 ix
e +e
2
ix
Some manipulation eventually confirms the following.
4 cos x sin3 x = sin(4x)
a. 4 cos3 x sin x
b. cos4 x
6 cos2 x sin2 x + sin4 x = cos(4x)
Solution 8.10. We are required to find z = x + iy with x, y 2 R such that,
2=
1 iz
e
2i
e
iz
Using the connection between eix and the circular functions, we can equate real and imaginary parts
to obtain
ey + e
y
sin x =4
ey
y
cos x =0
e
The second expression leads to either y = 0 (and so sin x = 2 which cannot be true), or cos x =
which case (ey + e
y
) = 4 which can be true. That is
e2y
which is solved by
4ey + 1 = 0
⇣
p ⌘
y = ln 2 ± 3
⇡
2
in
4i⇡
3
We therefore see that
⇣
p ⌘
⇡
+ i ln 2 ± 3
2
sin z = 2 ) z =
Chapter 9 solutions
Solution 9.1. Each statement is a fundamental axiom of probability theory.
Solution 9.2. We have ⌦ = N+ , A = {1, 2, 3, 4} and B = {2, 4, 6, 8, . . .}.
a. A \ B = {2, 4}
b. A \ B = {1, 3}
c. B [ B c = {2, 4, 6, 8, . . .} [ {1, 3, 5, 7, . . .} = ⌦
d. A [ B c = {1, 2, 3, 4} [ {1, 3, 5, 7, . . .} = {1, 2, 3, 4, 5, 7, 9, . . .}
Solution 9.3. The following are applications of combinations and permutations.
a. Order does not matter and so we need the number of combinations of 3 from 5 distinct items.
5 C3
=
5!
= 10
3!2!
b. Order does matter and so we need the number of permutations of 3 from 15 distinct letters.
15 P3
=
15!
= 2730
12!
c. The order from each of the three sets of draws does not matter and so we need the number of
combinations in each case. The total number is then the product of the three combinations.
3 C2
⇥ 5 C4 ⇥ 4 C2 =
3!
5!
4!
⇥
⇥
= 90
2!1! 4!1! 2!2!
d. Order does matter and so we need the number of permutations of 4 from 26 distinct letters.
26 P4
=
26!
= 358, 800
22!
Solution 9.4. The answers to the above question can be used to determine the following probabilities.
a. The draw R Y B in any order is one particular combination from the 10 that are possible.
The probability is therefore
1
10
= 0.1.
b. The draw RAT is one is particular permutation from the 2730 that are possible. The probability
is therefore
1
2730 .
c. We rewrite the required draw as 1 3
4 5 7 8
B R and confirm that it is one possible
combination of the 90 calculated in Question 3b. The probability is therefore
1
90 .
d. These are three mutually exclusive results from a single draw of 4 from 26 where order matters.
We know that there are 358,800 permutations and so the probability is
3
358,800
=
1
119,600 .
Solution 9.5. We denote the five candidates by Ci for i = 1, . . . , 5 and the probability that candidate
i achieves grade A by P (Ci = A). We are given that
P (C1 = A) = 2p and P (C2 = A) = P (C3 = A) = P (C4 = A) = P (C5 = A) = p
a. We require
5
X
i=1
P (Ci ) = 1 ) 6p = 1 and so p = 16 . That is,
P (C1 = A) =
1
1
and P (C2 = A) = P (C3 = A) = P (C4 = A) = P (C5 = A) =
3
6
b. These are three mutually exclusive events and so
P (C1 = A [ C2 = A [ C3 = A) = P (C1 = A) + P (C2 = A) + P (C3 = A) =
Solution 9.6. There are 52 cards, 13 of which are |.
• the probability that the 1st card is | is
13
52
• the probability that the 2nd is also | is
12
51
• the probability that the 3rd is also | is
11
50
• the probability that the 4th is also | is
10
49
• the probability that the 5th is also | is
9
48
1 1 1
2
+ + =
3 6 6
3
The probability is therefore
P (5 ⇥ |) =
13 12 11 10
9
33
⇥
⇥
⇥
⇥
=
52 51 50 49 48
66, 640
Solution 9.7. We introduce the events F = {insect is female} and L = {insect is a locust}, and require
P (F [ L) from
P (F [ L) = P (F ) + P (L)
Note that P (F ) =
20+5
30+10
=
5
8
and P (L) =
10
40
P (F \ L)
= 14 . Furthermore P (F \ L) =
P (F [ L) =
1 5
+
4 8
5
10
=
1
2
and so
1
3
=
2
8
Solution 9.8. Standard results can be used to determine the following.
a. P (Ac ) = 1
P (A) = 0.4
b. P (A1 |A2 ) =
P (A1 \A2 )
P (A2 )
=
0.3
0.4
c. P (A2 |A1 ) =
P (A2 \A1 )
P (A1 )
=
P (A1 \A2 )
P (A1 )
= 0.75
d. P (A1 [ A2 ) = P (A1 ) + P (A2 )
=
0.3
0.6
= 0.5
P (A1 \ A2 ) = 0.6 + 0.4
0.3 = 0.7
Solution 9.9. Let {Bi } be the event that Baker i baked the pastry, where i = 1, 2, 3, and let {S}
denote the event that the pastry is substandard.
a. Using the law of total probability, we have
P (S) = P (B1 )P (S|B1 )+P (B2 )P (S|B2 )+P (B2 )P (S|B2 ) = 0.1⇥0.04+0.2⇥0.06+0.7⇥0.10 = 0.086
b. We require P (B2 |S) and obtain this from Bayes’ formula
P (B2 |S) =
P (B2 )P (S|B2 )
0.2 ⇥ 0.06
6
=
=
P (B1 )P (S|B1 ) + P (B2 )P (S|B2 ) + P (B2 )P (S|B2 )
0.086
43
Solution 9.10. Let {O} be the event that it is an oak leaf, {E} be the event that it is an elm leaf,
and {L} be the event than the leaf is longer than 4 inches. We require P (O|L), that is the probability
that the long leaf is an oak leaf.
Using Bayes’ formula, we have
P (O|L) =
P (O)P (L|O)
0.55 ⇥ 0.7
77
=
=
P (0)P (L|O) + P (E)P (L|E)
0.55 ⇥ 0.7 + 0.45 ⇥ 0.4
113
Chapter 10 solutions
Solution 10.1. The dimensions of each matrix dictate which operations can be performed.
2
8
2
6
6
6 2
6
6
a. A + B = 6
6 6
6
6
6 5
4
3
b.
3
4
7
8
2
6 0
6
3D = 6
6 3
4
6
c. A + 3B
3
6
2
3
3
7
7
3 7
7
7
3 7
7
7
7
5 7
5
11
7
7
12 7
7
5
18
14
6
6
6 3
6
6
C=6
6 10
6
6
6 6
4
4
0
4
16
11
20
d. D has di↵erent dimensions to
2
3
7
7
4 7
7
7
4 7
7
7
7
4 7
5
15
2(A + B + C) and so
2(A + B + C)
D does not exist.
Solution 10.2. The dimensions of each matrix dictate which products can be performed.
2
3
6
4
6
7
6
7
6
a. F E = 6 2 1 7
7
4
5
0
4
2
6 13
b. GE = 4
3
3
17 7
5
5
c. F G does not exist
2
6 32
d. GF = 4
2
7
0
3
38 7
5
5
Solution 10.3. All of the following statements are true.
a. (2M + 3N )T = 2M T + 3N T
b. ( M N )T =
NT MT
c. (3P (M + N ))T = 3(M T + N T )P T
(M N P )T )T = 2P T N T M T
d. (2M N P
MNP
Solution 10.4. The determinants are computed as follows.
1⇥2
a. det M1 =
2⇥1=
4 and M1 does have an inverse.
b. We expand along the second row and determine that det M2 =
1. M2 does have an inverse.
c. We expand along the fourth row and determine that det M3 =
27. M3 does have an inverse.
d. det M1T =
4 and M1T does have an inverse.
Solution 10.5. All of the following statements are true.
a. N1 1 N1 = In = N1 N1
b. N2
1
1
1
= N2
c. N2 (N1 N2 )
1
d. N1 (N1 N2
N2 N1 )
= N1
1
1
N2 = N1 N2 1 N1 N2
In
Solution 10.6. The following can be calculated using the adjoint/adjugate method.
2
6 2
a. 4
2
2
6 1
b. 4
1
3
1 7
5
0
3
1 7
5
2
1
2
1
6 0
= 12 4
2
2
6 2
=4
1
3
1 7
5
2
3
1 7
5
1
2
6
c. 4
3
4 7
5
2
3
2
2
6 0.5
d. det 4
2
1
=
1
14
3
2
2
6
4
2
3
4 7
5
3
0.5 7
5 = 0 and so the matrix has no inverse
2
Solution 10.7. The following can be calculated using the adjoint/adjugate method.
2
3 1
2
3
1
1
2
0
4
3
6
7
6
7
6
7
6
7
1 6
7
a. 6
=
0 3 7
7 6
1 7
7 6
6 1
7
4
5
4
5
1 0 4
0 1 1
2
6 2
6
b. 6
6 3
4
1
4
1
6 0
6
c. 6
6 1
4
2
6 1
6
d. 6
6 0
4
0
1 7
7
3 7
7
5
1
2
2
2
3
1
0
2
0
4
0
3
3 7
7
5 7
7
5
6
3
0 7
7
0 7
7
5
2
1
2
1
1
6 1
6
=6
6 0
4
1
2
6 10
6
1 6
= 26 4
4
2
2
6 4
6
= 14 6
6 0
4
0
0
1
0
3
3
10 7
7
3 7
7
5
8
1
2
12
6
2
3
0 7
7
0 7
7
5
2
3
5 7
7
3 7
7
5
1
Solution 10.8. Parts a, b and c can be answered by manipulating the expressions at the matrix level.
Part d requires one to work at the level of matrix entries.
2
3
2
3
1
0
4
1
6
7
6
7
a. AT 4
5 = 3I2 ) A = 4
5
1 1
0 4
b. 3A
1 T
2
6 1
=4
1
6 1
c. A 1 4
1
d. AT = I2
2
3
2
0 7
6 3
5)A=4
1
0
3
2
0 7
6 0
5 = 24
1
1
2
6 1
3A ) 14 4
0
3
3 7
5
3
3T
2
0 7
6 1
5 ) A = 24
1
1
3
0 7
5
1
3
1 7
5
1
Solution 10.9. The systems are expressed in terms of matrices and we proceed by inverting the
coefficient matrix . Cramer’s rule can be implemented to confirm the value of each unknown in both
cases.
a.
2
2
6
6
6 5
6
6
6
6 0
4
1
b.
2
1
6
6
6 1
6
6
6
6 1
4
1
1
1
1
1
0
0
1
1
1
1
1
1
1
1
x
3
2
3
3
2
x
3
2
1
3
3
2
x
3
2
3
3
76
7 6
7
6
7 6
7
76
7 6
7
6
7 6
7
6 y 7 6 10 7
6 y 7 6 2 7
1 7
76
7 6
7
6
7 6
7
76
7=6
7)6
7=6
7
76
7 6
7
6
7 6
7
2 76 z 7 6 8 7
6 z 7 6 2 7
54
5 4
5
4
5 4
5
0
a
3
a
3
2
1
32
1
1
32
x
3
2
0
76
7 6
7
6
7 6
76
7 6
7
6
7 6
6 y 7 6 2 7
6 y 7 6 1
1 7
76
7 6
7
6
7 6
76
7=6
7)6
7=6
76
7 6
7
6
7 6
1 76 z 7 6 4 7
6 z 7 6 2
54
5 4
5
4
5 4
1
a
6
a
0
7
7
7
7
7
7
7
5
Solution 10.10. The second and third equations are not independent. The determinant of the coefficient matrix will therefore be zero and will not be invertible.
Chapter 11 solutions
Solution 11.1. We implicitly di↵erentiate the expressions with respect to x and obtain the following.
a. y 2 + xy = sin(x) ) y 0 (x) =
xy
xy
b. e 20 = ln(y) ) y 0 (x) =
y 2 e 20
xy
20 xye 20
c. y 2 + 2y + 2 = x ) y 0 (x) =
d. ln(yx2
cos(x) y
2y+x
1
2(y+1)
y 2 x) = 2 ) y 0 (x) =
y(y 2x)
x(x 2y)
Solution 11.2. In each case we use the gradient function obtained above and fit a straight line of the
form y = mx + c.
a. The curve is vertical at (x, y) = (0, 0) and so x = 0 is tangent.
b. The gradient at (x, y) = (0, e) is y 0 =
e2
20
c. The gradient at (x, y) = (5, 1) is y 0 =
1
4
and the tangent line is y =
e2
20 x
and the tangent line is y = 41 (x
+e
1)
d. The gradient at (x, y) =
y = 1.01506x
0.22505
✓
q
10, 5 + 25
e2
10
◆
is approximately 1.01506 and the tangent line is
Solution 11.3. We classify the ODEs in terms of linearity, order and coefficients.
a.
df
dx
+ 2f = x is a first-order, linear ODE with constant coefficients
b. g 00 + xg = 2 is a second-order, nonlinear ODE with variable coefficients
c. h00
4h2 = ⇡ sin(x) is a second-order, nonlinear ODE with constant coefficients
mx + 4m2 = xe
d. mxxx + mxx
x
is a third-order, nonlinear ODE with constant coefficients
Solution 11.4. Note that each ODE is separable and so can be integrated directly.
a. y(x) =
1
2
b. g(z) =
20
5 ln(z+2) 5 ln(2 z)+2
c. h(y) =
d. f (x) =
arccos cos(2)
p
2 + ⇡2
p
3
x2
4x
2 cos y
2ex2 +1
2
23
Solution 11.5. Note that each ODE is linear and can be integrated after the use of an appropriate
integrating factor.
a. y(t) = 3e2t
1
b. f (x) = 12 x3 (x
c. g(y) = 5e
y4
4
1)
4
1
2 (sin x
d. h(x) = 2e
cos x)
1
2
⇣
1
4e 2 (sin x
cos x)+ 12
Solution 11.6. Note that each ODE is exact.
q
a. g(z) =
b. y(x) =
c. x = e
1
2
2f
z 2 +3
z
p
ln
e2x + 8
⇣ ⌘
ex
f
5
d. h(p) = cosec(p)
p
1 + sin p
sin p cos p
1
+3
⌘
Solution 11.7. Each ODE is solved on its own merits.
a. This is an exact ODE and we have 2x2 f (x) + x + sin(y) = 1
b. This is separable ODE and we find g(z) =
arccos
⇣
1 2z +ln(2) cos(1)
ln(2)
⌘
y
c. This ODE can be solved using an integrating factor and we find h(y) = 2e 2 y
y
2
y
y2
1
d. This ODE can be solved using an integrating factor and we find y(x) = 2earcsin x + e1+arcsin x
2
Solution 11.8. We need to solve the ODE and determine the minimum of the resulting function. The
ODE is solved by S(t) = 5.45et(16.35t
10.9)
and we determine that this has a minimum at t = 13 .
Solution 11.9. We solve the ODE subject to P (0) = P0 and find
P (t) = (P0
5) e
t
50
+5
Figure 11.1 shows the evolution of the population for various P0 2 [0, 10]. It is clear that P (t) ! 5
as t ! 1 for all P0 . That is, the populations tends to a value of 5 irrespective of the starting value.
Referring to the original ODE, we note that
P0 =
5
8
>
>
>< 0 for P > 5
>
>
>
<
P
= 0 for P = 5
50 >
>
>
>
>
>
:> 0 for P < 5
That is, the gradient always acts to point the time evolution towards P = 5.
10
8
P (t)
6
4
2
0
0
20
40
60
80
100
t
Figure 11.1: The family of solutions to Question 11.9
Solution 11.10. Part a. is separable and easily solved. We proceed to factorise parts b.–c. so the result
of a. can be generalised to combinations of exponential terms. This leads to the following.
a. f 0
2f = 0 ) f (x) = c1 e2x
b. f 00
3f 0 + 2f =
c. f 000
f 00
d. f (4)
df
dx
4f 0 + 4 =
2f 000
c2 ex + c3 e
⇣
⇣
df
dx
3f 00 + 8f 0
2x
⌘⇣
1 df
dx
1
⌘⇣
4f =
+ c4 ex
⌘
2 f = 0 ) f (x) = c1 e2x + c2 ex
df
dx
⇣
df
dx
⌘⇣
⌘
2x
2 df
+ c2 ex + c3 e
dx + 2 f = 0 ) f (x) = c1 e
⌘⇣
1 df
dx
1
⌘⇣
df
dx
2
⌘⇣
df
dx
⌘
+ 2 f = 0 ) f (x) = c1 e2x +
Chapter 12 solutions
Solution 12.1. The derivatives can be performed using standard results.
f (x, y, z) = exy sin(xz) ln(y + z)
a.
@f
@x
= exy ln(x + y) (y sin(xz) + z cos(xz)),
@f
@y
=
exy sin(xz
y+z
@f
@z
=
exy sin(xz)
y+z
b. fxy =
+ xexy sin(xz) ln(y + z),
+ xexy cos(xz) ln(y + z)
exy (z cos(xz)(1+x(y+z) ln(y+z))+(y+(1+xy)(y+z) ln(y+z)) sin(xz))
y+z
2x
c. @xz f =
d. fyz =
exy (cos(xz)(z+(1+xy)(y+z) ln(y+z))+(y xz(y+z) ln(y+z)) sin(xz))
y+z
exy (x(y+z) cos(xz)(1+x(y+z) ln(y+z))+( 1+x(y+z)) sin(xz))
(y+z)2
Solution 12.2. We determine the following stationary points.
33
2)2 has a minimum of 2, 10
a. f (x, y) = y(x
1)2
b. g(x, y) = 0.2(x
0.5(y
2)2
4 has a saddle at (1, 1)
c. h(y, z) = y 3 8z 3 2y 2 z + 4yz 2 4y + 8z has a maximum at
⇣q
⌘
⇣ p
⌘
⇣p
⌘
2
p1
and saddle points at
2, p12 and
2, p12
3,
6
d. m(p, q) = 2
⇣ q
2 p1
3, 6
⌘
, a minimum at
ep sin(pq) has a saddle point at (0, 0)
Solution 12.3. The function
f (x, y) = 1 + sin x + cos2 y
has maxima at (x, y) = 2⇡n + ⇡2 , 2⇡m and 2⇡n + ⇡2 , 2⇡m + ⇡ , and minima at 2⇡n
and 2⇡n
⇡
2 , 2⇡m
⇡
2
⇡
2 , 2⇡m
+
⇡
2
for integer n and m.
Solution 12.4. The method of Lagrange multipliers can be used in each case.
a. The function x + y 2 such that x2 + xy = 2 has local maximum at (x, y) ⇡ ( 1.5579, 0.2741)
b. The function y 2 z
c. The function (x
7 2
3, 3,
2 such that y + 2z = 1 has a local maximum at (y, z) =
2)2 + (y
1)2 + z 2 such that x
y
1 1
3, 6
2z = 3 has a local minimum at (x, y) =
2
3
d. The function x2 y 2 z such that x + y = 5 and x + y + z = 5 has a local maximum such that
(x, y, z) = (2, 2, 1)
Solution 12.5. We are required to maximise
P (O1 , O2 ) = 5 + O1 O2 +
O2 (O1 + O2 )
2
such that O1 + O2 = 125. The method of Lagrange multipliers can be used to determine that this is
maximised when (O1 , O2 ) = 125
1 3
4, 4
. In this case the profit is P =
140,705
16 ,
that is $8, 794.06.
Solution 12.6. We are required to maximise R(D, M ) = 3D1.3 M subject to R+D = 100. The method
of Lagrange multipliers can be used to determine that this occurs when (D, M ) = 100
13 10
23 , 23
. This
leads to R ⇡ £24, 732.80, which is a remarkable return on £100!
Solution 12.7. The integrals are separable and can be performed as the product of independent,
single-variable integrals.
a.
b.
R ⇡/2 R 2
0
RR
1
7
3
y 2 cos xdydx =
sin(2x) cos(4y)dxdy =
D
c.
RR
e2x
3z
dxdz =
D
d.
R 2 R 1 R 10
2 0
2
R0 R1
2
1
R ⇡/2 R 1.5⇡
0
⇡
e2x e
3z
sin(2x) cos(4y)dxdy = 0
dxdz =
e4 +e6 e10 1
6e2
x2 y 2 z 3 dxdydz = 0
Solution 12.8. The integrals are not separable and are performed iteratively.
a.
b.
R5R5
4
RR
2
(xy + 2ex ) dxdy =
(y cos x + x sin y) dxdy =
D
c.
RR ⇣
2x2 +
D
d.
189
4
R2R0 R2
0
7
1
z2 9
⌘
dxdz =
yx2 + zy
2
+ 2(e5
e2 )
R ⇡/2 R 1.5⇡
0
⇡
R0 R1 ⇣
2
1
(y cos x + x sin y) dxdy = 4.9348
2x2 +
1
z2 9
⌘
dxdz =
1
3
(8
ln(5))
ze2x dzdzdy = 0
Solution 12.9. The integrals are over non-square domains and some care is required.
a.
RR
x2 ydxdy =
D
b.
RR
yex dydx =
D
c.
RR
x
e 2 dxdy =
D
d.
RR
D
x
2 dydx
=
R 1 R y2
1 y 2
R 4 R 2+x
2
1 x
R 2 R 2y
0
y
yex dydx =
x
e 2 dxdy =
R ⇡ R cos x
0
14
5
x2 ydxdy =
x
dydx
cos x 2
3e2 (7e2 3)
2
6+
=
4
e
+ 2e2
2
Solution 12.10. Despite being an integral over three variables, the process is exactly as for two
variables. Note that the order of integration indicated by dydxdz is in fact correct for this particular
domain of integration, that is we integrate with respect to y then x then z. We find that
ZZZ
V
xydydxdz =
Z
1
0
Z
where V = x, y, z|x 2 [0, ez ], y 2 [1
ez
0
Z
2x
xydydxdz =
1 x
1
288
x, 2 + x], z 2 [0, 1] .
23
36e2 + 32e3 + 27e4
Chapter 13 solutions
Solution 13.1. A plot of each function for positive values of the independent variable is shown in
Figure 13.1. From this an appropriate starting interval can be determined.
3
4
2
f (x)
g(y)
1
2
0
0
1
2
2
0
1
2
3
4
5
0
0.5
1
y
x
(a) f (x) = ln(x2
1.5
(b) g(y) = y 3
2)
e
2
y
3
0.5
2
h(z)
m(p)
0
1
0.5
0
1
0
0.2
0.4
0.6
0.8
1
1
0
1
(c) h(z) = e2z
2
3
4
5
p
z
2ez
(d) m(p) = p
1
2
ln(p + 2)
Figure 13.1: Sketches for Solution 13.1
a. f (x) = ln(x2
2) has a root in the interval x 2 [1, 2]. Using this as the starting range, we find
x ' 1.732.
b. g(y) = y 3
e
y
has a root in the interval y 2 [0, 1]. Using this as the starting range, we find
y ' 0.773.
c. h(z) = e2z
z ' 0.693.
2ez has a root in the interval z 2 [0.5, 1]. Using this as the starting range, we find
d. m(p) = p
1
2
ln(p + 2) has a root in the interval p 2 [1, 2]. Using this as the starting range,
we find p ' 1.847.
Solution 13.2. In each case, the regula falsi method over the same interval leads to the same root.
Solution 13.3. The number of iterations of the bisection method is given by
n
ln(6
4)
ln(0.0002)
= 13.29
ln 2
that is, n = 14 iterations are required. The method can be implemented to find that x ' 4.4553.
Solution 13.4. The Newton–Raphson method beginning at the upper end of each interval leads to the
same root in each case.
Solution 13.5. The secant method beginning at the ends of the interval leads to the same root in each
case.
Solution 13.6. The expression can be interpreted as an equation of value with i an annual interest
rate. The total income received is 60 and this is in return for an expenditure of 50, that is a return
of 10 (i.e. 20%) over the four years. The interest rate is therefore positive but much less than 20% per
annum. Some experimentation confirms that the value is between i = 0.05 and i = 0.10 and we use
these as the starting values of the secant method (to some appropriate error tolerance) to determine
that i ' 0.0677. The secant method is used to avoid the need for an explicit expression for f 0 (i). It is
also quicker than either interval method.
Solution 13.7. The node points are given by { 5, 4.5, . . . 4.5, 5}. The function can be evaluated at
each point and the three methods implemented.
Solution 13.8. Various sets of node points can be generated for di↵erent step-size values h, { 1, 1 +
h, 1 + 2h, . . . , 1}. The derivative function can be calculated numerically using the 2nd order centraldi↵erence approach across each and compared to the actual derivative g(x) =
2e
sin2 (x)
sin(x) cos(x)
to determine the error. The error at a fixed point, say x = 0, can then be plotted as a function of h to
confirm the quadratic behaviour. The process can be repeated at one or two other fixed values of x.
Solution 13.9. We implement the two approaches with k = 4, as required.
a. Using the trapezoidal rule we find
R2
dx
0 x+1
' 1.11667, Simpson’s rule leads to 1.1.
R2
b. Using the trapezoidal rule we find
c. Using the trapezoidal rule we find
d. Using the trapezoidal rule we find
2
R5
0
R1
1
e
(x 1)2
dx ' 1.5702, Simpson’s rule leads to 1.72567.
x sin(2x)dx ' 0.635283, Simpson’s rule leads to 3.97818.
sinh2 x2 dx ' 0.754362, Simpson’s rule leads to 0.734123.
Solution 13.10. The theoretical error of the trapezoidal rule is given by
h2 0
(f (5)
12
f 0 (3)) =
h2
12
✓
6
2
+
e9
e
◆
= 0.001 ) h ⇡ 0.128
A step size of h = 0.128 cannot be used over the interval of width 2 and we are forced to use h = 0.10
which corresponds to k = 20. This can be implemented and we find
Z
5
e
3
(x 2)2
' 0.00420647