BSc/MSci EXAMINATION
PHY-121
Mathematical Techniques 1
SOLUTIONS
Date: Summer 2010
Instructions:
Answer ALL questions in section A. Answer ONLY TWO questions from section B. Section A carries 50 marks, each question
in section B carries 25 marks. An indicative marking-scheme is
shown in square brackets [ ] after each part of a question. Show all
appropriate intermediate steps between the question and solution.
The use of numeric calculators is permitted in this examination.
Examiners: Dr A. Bevan (CO), Dr A. J. Martin (DCO)
SOLUTIONS
Section A : Answer ALL questions in this section.
A1
(a) y = cos(x), so y 0 =
sin(x).
(b) y = eax , so y 0 = aeax .
(c) y = x2 (2x + 1), so y 0 = 6x2 + 2x.
(d) y = (2x + 1)/(x2
1), so y 0 =
2(x2 1) 2x(2x+1)
(x2 1)2
=
( 2x2 2x 2)
.
(x2 1)2
(e) y = esin(x) , so let u = sin(x) and
du
= cos(x),
dx
dy
= eu ,
du
dy
dy du
=
⇥
= cos(x)esin(x) ,
dx
du dx
(f ) Determine all first and second derivatives of the function z = xy
@z
@x
@2z
@x2
@z
@y
@2z
@y 2
@2z
@y@x
= y
= x
=
3.
2xey ,
2ey ,
=
=
x2 ey
x2 ey ,
x2 ey ,
@2z
=1
@x@y
2xey .
[10]
A2
(a)
R
R
cos(x)dx = sin(x) + C.
3x + ex dx = 3x2 /2 + ex + C.
R
(c) I = x ln x xdx, so on integrating the first term by parts, let u = ln x, u0 = 1/x,
v 0 = x, and v = x2 /2, so
Z
I1 =
x ln xdx
Z
x2 ln x
x
=
dx.
2
2
x2
=
(ln x 1/2) + C1 .
2
(b)
Hence
I =
=
(d)
Z
✓
x ln x
xdx
x2
ln x
2
3x2
4
◆
+ C.
Z
2x
dx = ln |x2 + 1| + C.
x2 + 1
as we can recognize the function to integrate as something of the form f 0 (x)/f (x).
(e) Evaluate the following integral
I=
(x
Z
(x
1
dx.
1)(2x + 1)
1
A
B
A(2x + 1) + B(x 1)
=
+
=
.
1)(2x + 1)
x 1 2x + 1
(x 1)(2x + 1)
so
A(2x + 1) + B(x
1) = 1.
If we let x = 1, we find that A = 1/3. Similarly if we let x = 1/2, we find that
B = 2/3. So
Z
1
I =
dx,
(x 1)(2x + 1)
Z
1 1
1 2
=
dx,
3 x 1 3 2x + 1
1
1
1
x 1
=
ln |x 1|
ln |2x + 1| + C = ln
+ C.
3
3
3
2x + 1
[10]
A3
(a)
3
P
kxk = x + 2x2 + 3x3 .
k=1
(b) Sn =
nP1
ark is a GP, where the sum is Sn = a(1
rn )/(1
r).
k=0
(c)
f (x) = f (x = 0) + f 0 (x = 0)x +
f 00 (x = 0) 2 f 000 (x = 0) 3
x +
x + ...
2!
3!
(d)
(1 + x)n = 1 + nx +
(e)
lim
x!1

n(n 1) 2 n(n
x +
2!
 0
f (x)
f (x)
= lim
x!1 g 0 (x)
g(x)
1)(n
3!
2)
x3 + . . .
where g 0 (x) 6= 0
[10]
A4
(a) i3 =
i.
(b) (1 + i) + (1
(c) (1 + i)(1
i) = 2 + 0i.
i) = 1
i+i
i2 = 2 + 0i.
(d) (1 + i)/(1 i) = (1 + 2i + i2 )/2 = i.
p
(e) (1 + i) = 2ei⇡/4 .
(f ) z = 1, so |z| = 1, and ✓ = arg(z) = ⇡. We know that z 1/n = r1/n ei(✓+2⇡m)/n , where
m = 0, 1, . . . n 1. So
z1 = ei⇡/3 ,
z2 = ei⇡ ,
z3 = ei5⇡/3 ,
Im
These can be drawn on an argand diagram as shown in Fig 1:
1
z1
0.8
0.6
0.4
0.2
z2
0
-0.2
-0.4
-0.6
z3
-0.8
-1
-1
-0.5
0
0.5
1
Re
Figure 1: An argand diagram showing the three cubed roots of z =
1.
[10]
A5
(a) The fourier series is given by
✓
◆
✓
◆
1 
X
1
2⇡n
2⇡n
y(t) =
A0 +
An cos
t + Bn sin
t
2
T
T
n=1
where
An =
Bn =
A0
=
2
=
=
where T = t2
✓
◆
Z
2 t2
2⇡n
y(t) cos
t dt,
T t1
T
✓
◆
Z
2 t2
2⇡n
y(t) sin
t dt,
T t1
T
Z
1 2 t2
y(t)dt,
2 T t1
Z t2
1
y(t)dt,
t2 t1 t1
< y(t) >,
t1 is a period.
(b)
Z+1
(x
(x
x0 )dx = 1
1
x0 ) = 0 for x 6= x0 .
Write down the equation for a delta function centred about x = x0 , and note the
normalization condition for such a function.
R1
(c)
f (x) (x 3)dx = f (x = 3) = f (3).
1
[10]
Section B : Answer ONLY TWO questions from this section.
B1
a)
sin(x)
,
x
x cos(x) sin(x)
=
,
x2
x2 [cos(x) x sin(x)
=
y =
y0
y 00
=
When x = ⇡, y 0 =
where
cos(x)] 2x[x cos(x)
x4
2
2x cos(x) + 2x sin(x)
,
x4
x3 sin(x)
sin(x)]
,
1/⇡, and y 00 = 2/⇡ 2 . The radius of curvature is given by R,
R=
h
1+
i3/2
dy 2
dx
d2 y
dx2
.
which, for x = ⇡ is simply
[1 + 1/⇡ 2 ]
R=
2/⇡ 2
3/2
= 5.70.
b) Determine the location of the centre of curvature corresponding to the point on the
curve given by x = ⇡.
(x0 , y0 )
xc
yc
tan ✓
So ✓ =
= (⇡, 0),
= x0 R cos ✓
= y0 + R sin ✓
=
1/⇡ from the gradient.
0.308 rad ( 17.65 ).
xc = ⇡ 5.70 cos( 17.65 ) = 2.29,
yc = 0 + 5.70 sin( 17.65 ) = 1.73.
c) Determine the position of nature of any stationary points of the function z = ex +
e x y2.
@z
@x
@2z
@x2
@z
@y
@2z
@y 2
= ex
e x,
= ex + e x ,
=
2y,
=
2,
So for there to be a turning point, we need to have the first derivatives being equal
to zero. This can only occur if x = 0 and y = 0. When x = 0, the second derivative
of z with respect to x is +2, and when y = 0, the second derivative of z with respect
to y is 2, so this point (0, 0) must be a saddle point.
[25]
B2
a) Determine the centroid position x for the lamina given by the curve y = xex bounded
by the x-axis, and the lines x = 0 and x = 1.
x
Z
dA =
Z
xdA,
where dA = ydx.
Z
dA =
Z1
ydx,
=
Z1
xex dx,
0
0
=
[xex ]10
Z1
ex dx,
0
= e
[ex ]10 = 1,
and
Z
Z1
xdA =
x2 ex dx,
0
= [x2 ex ]10
2
Z1
xex dx,
0
= e
So x = e
2.
2.
b) If the lamina in part (a) is rotated about the x-axis it forms a volume in a threedimensional space (x, y, z). Compute the centroid position (x, y, z) of that volume,
using symmetry where appropriate.
By symmetry y = 0, and z = 0, and
Z
Z
x dV = xdV,
where dV = ⇡y 2 dx. Hence
R1 3 2x
R
x e dx
xdV
0
x= R
= 1
,
R
dV
2
2x
x e dx
0
So integrating the following by parts (taking u = xn , where n is integer),
Z1
2 2x
x e dx =
0
=
=

x2 e2x
2

x2 e2x
2
xe2x
2
2 2x
2x

=
[solution continuued on next page]
xe
2
e2
1
4
.
1
0
Z1
xe2x dx,
0
1
1
+
2
0
2x
xe
e
+
2
4
Z1
0
1
,
0
e2x dx,
And similarly for the numerator, we have
Z1
3 2x
x e dx =
0
=
=
=
=

x3 e2x
2

x3 e2x
2

x3 e2x
2

3 2x
xe
2
1
3
2
0
Z1
x2 e2x
dx,
2
0
3 2 2x
xe
4
1
0
3
2
x=
3 2 2x 3 2x
x e + xe
4
4
1 e2 + 3
= 0.813.
2 e2 1
xe2x dx,
0
3 2 2x 3 2x
x e + xe
4
4
1 2
(e + 3).
8
Thus
Z1
1
0
3
4
Z1
e2x dx,
0
2x
3e
8
1
,
0
[25]
B3
a) Find the limiting value of the following when x ! 1
x5 + 7x2 + 1
1 + 7/x3 + 1/x5
=
.
3x5 + 2
3 + 2/x5
So when x ! 1, the quotient approaches the limit 1/3.
b) Using the first four terms of a binomial series expansion, estimate
4
p
0.8.
n(n 1) 2 n(n 1)(n 2) 3
x +
x + ...,
2!
3!
1
1 3 ( 0.2)2 1 3 7 ( 0.2)3
= 1 + ( 0.2) +
+
,
4
4 4
2!
4 4 4
3!
= 1 0.05 0.00375 0.0004375,
= 0.94581
(1 + x)2 = 1 + nx +
(1
0.2)1/4
c) Compute the first four non-zero terms of the series expansion for the function x cosh(x)
when expanding about x = 0. First of all determine the Maclaurin series for cosh(x).
y1 = cosh(x),
y10 = sinh(x),
y100 = cosh(x),
y1000 = sinh(x),
y10000 = cosh(x),
y100000 = sinh(x),
y1000000 = cosh(x).
So
y1 (0) = 1,
y10 (0) = 0,
y100 (0) = 1,
y1000 (0) = 0,
y10000 (0) = 1,
y100000 (0) = 0,
y1000000 (0) = 1.
Hence,
x2 x4 x6
+
+ ,
2!
4!
6!
x3 x5 x7
x cosh(x) ' x +
+
+ .
2!
4!
6!
cosh(x) ' 1 +
d) Given two complex numbers A1 = aei✓ , and A2 = bei , for the sum A = A1 + A2
compute |A|2 = AA⇤ .
|A|2 = (aei✓ + bei )(aei✓ + bei )⇤ ,
= (aei✓ + bei )(ae i✓ + be i ),
= a2 + b2 + abei(✓ ) + abei( ✓) .
e) Simplify your result from B3 part (d) for the special case when ✓ = .
|A|2 = a2 + b2 + abei(✓
= a2 + b2 + 2ab.
)
+ abei(
✓)
,
[25]
B4
a) Consider the function determined by y = h when 0  x + nT  1, y = 0 elsewhere,
where T = 2 is the periodicity of the function. Determine the Fourier series expansion for y. By inspection the constant term A0 /2 = h/2, and by symmetry about
the origin all coefficiencts An are zero. The period T = 2
Z
2 1
Bn =
y(t) sin(!nt)dt,
T 0
Z 1
=
h sin(!nt)dt,
0
h
=
[ cos(!nt)]10 ,
!n
h
=
[ cos(⇡nt)]10 ,
⇡n
h
=
[1 cos(⇡n)].
⇡n
where we use the fact that ! = 2⇡/T . So
1
h X h
y(t) = +
[1
2 i=1 ⇡n
cos(⇡n)] sin(⇡nt),
where the sum is only over odd terms as Bn = 0 for even values of n, and hence
Bn=odd = 2.
b) Sketch the function in B4 part (a), and overlay the distributions corresponding to
the sum of the constant and first harmonic of y on the same plot. This is shown in
Figure 2.
1.4
1.2
1
y(t)
0.8
0.6
0.4
0.2
0
-0.2
-2
-1.5
-1
-0.5
0
0.5
1
1.5
2
t
Figure 2: The function y(t) and Fourier series of the constant and first harmonic terms.
c) Compute the Fourier transform of the function given by y(x) = 1
and y(x) = 0 elsewhere.
Y (u) =
Z1
y(x)e
2i⇡ux
x for 0  x  1,
dx,
1
=
Z1
(1
x)e
2i⇡ux
dx,
0
which can be integrated by parts taking u = 1
v = e 2i⇡ux /2i⇡u:
Y (u) =
 ✓
1 x
2⇡iu
◆
1
e
2i⇡ux
⇥
1
1
=
+
e
2⇡iu (2⇡iu)2
⇥
1
1
=
+
e
2⇡iu (2⇡iu)2
0
x, so u0 =
Z1
2i⇡ux
2i⇡ux
, and
dx,
0
⇤
2i⇡ux 1
0
2i⇡u
1
e
2⇡iu
1, v 0 = e
,
⇤
1 .
[25]
End of Examination Paper