University of Hong Kong
Fall 2011
Math 1813
Mathematical Methods for Actuarial Science
Instructor: Siye Wu
Solution 4
I. In cylindrical polar coordinates:
Z
=
{
(
r
cos
θ, r
sin
θ, z
)
T
∈
R
3
: 0
6
θ <
2
π, r
= 2
, z
∈
R
}
S
=
{
(
r
cos
θ, r
sin
θ, z
)
T
∈
R
3
: 0
6
θ <
2
π, r
>
0
, z
∈
R
, r
2
+
z
2
= 9
}
C
=
{
(
r
cos
θ, r
sin
θ, z
)
T
∈
R
3
: 0
6
θ <
2
π, r
>
0
, z
∈
R
, r
2
=
z
2
}
In spherical polar coordinates:
Z
=
{
(
r
sin
θ
cos
φ, r
sin
θ
sin
φ, r
cos
θ
)
T
∈
R
3
: 0
< θ < π,
0
6
φ <
2
π, r
sin
θ
= 2
}
S
=
{
(
r
sin
θ
cos
φ, r
sin
θ
sin
φ, r
cos
θ
)
T
∈
R
3
: 0
6
θ
6
π,
0
6
φ <
2
π, r
= 3
}
C
=
{
(
r
sin
θ
cos
φ, r
sin
θ
sin
φ, r
cos
θ
)
T
∈
R
3
:
θ
=
π
4
or
3
π
4
,
0
6
φ <
2
π, r
>
0
}
II. We have
J
f
(
x, y
) =

y
sin(
xy
)

x
sin(
xy
)
cos(
x
+
y
)
cos(
x
+
y
)
¶
, so
J
f
(0
, π
) =
0
0

1

1
¶
.
(a) One has
J
F
(2
,
3) =
J
f
(0
, π
)
J
g
(2
,
3) =
0
0

1

1
¶
3
7

2
0
¶
=
0
0

1

7
¶
.
(b) Note that
F
(2
,
3) =
f
(0
, π
) = (1
,
0)
T
.
The linear approximation
L
(
s, t
) at point (
a, b
)
T
is
L
(
s, t
) =
F
(
a, b
) +
J
F
(
a, b
)
s

a
t

b
¶
. So
L
(2
.
01
,
2
.
98) =
F
(2
,
3) +
J
F
(2
,
3)
0
.
01

0
.
02
¶
=(1
,
0)
T
+
0
0

1

7
¶
0
.
01

0
.
02
¶
= (1
,
0)
T
+ (0
,
0
.
13)
T
= (1
,
0
.
13)
T
.