Instructor(s): Dr David Ridout and Dr Bryan Wang
Mathematical Sciences Institute
Second Semester, 2013
Australian National University
MATH1113
Mathematical Foundations for Actuarial Studies
Tutorial 8 Solutions
Question 1. Linear algebra – Determinants and Areas
Find the area of the region in the plane defined by 13x2 + 16xy + 5y 2 ≤ 4. (Hint:
Write 13x2 + 16xy + 5y 2 as a sum of two perfect squares. Then apply a linear
transformation to turn R into a disk of radius 2).
Solution. Let us denote the region by R. We have
13x2 + 16xy + 5y 2 = (2x + y)2 + (3x + 2y)2 .
Let T : R2 → R2 be the linear transformation with standard matrix
2 1
A=
,
3 2
u
2 1
x
that means, if
=
, u = 2x + y, v = 3x + 2y, then T maps R in
v
3 2
y
R2 (the (x, y)-plane) onto the disk D given by u2 + v 2 ≤ 4 in R2 (the (u, v)-plane),
which is the disk of radius 2 centered at the origin. Therefore, we have
{area of D} = {area of T (R)} = | det A| · {area of R},
which yields
4π = 1 · {area of R},
and therefore {area of R} = 4π.
Question 2. Multivariable Calculus
(a) Draw four level curves of the function
f (x, y) = x +
9y 2
.
x
What do you notice for x = 0? Can you explain it?
MATH1113, Tutorial 8 Solutions
2
(b) Determine the domain D of the function
g (x, y) =
x2 − y −2
.
xy − 1
Show that the limit of g (x, y) exists as xy tends to 1.
Solution. x
(a) The level curves take the form f (x, y) = k 0 for some constant k 0 ∈ R. A little
algebra gives
2
k0
k02
2
2
0
2
x + 9y = k x
⇒
x−
+ 9y =
.
2
4
Letting k = k 0 /2 for convenience, we arrive at the equations of the level curves:
(x − k)2 + (3y)2 = k 2 .
This is of course the equation of an ellipse centered at (k, 0). We plot the level
curves for k = −3, −2, −1, 1, 2, 3 below.
1.0
0.5
-6
-4
-2
2
4
6
- 0.5
- 1.0
We see that all the level curves intersect at the origin. This is ridiculous —
the function can’t take more than one value at any point. In fact, the function
is undefined when x = 0, so the level curves should not include the origin.
The fact that they intersect tells us that the limit of f (x, y) doesn’t exist as
(x, y) → (0, 0) — if it did, then we’d have to get the same limit along every
curve through the origin, even these elliptical ones.
MATH1113, Tutorial 8 Solutions
3
(b) The function is not defined only when y = 0 or xy = 1. Thus, the domain is
1
2
(x, y) ∈ R : y 6= 0 and y 6=
.
x
To compute the required limit, we merely factorise the numerator:
1 x2 y 2 − 1
xy + 1
2
= lim
= 2.
2
2
xy→1 y
xy→1
xy − 1
y
y
lim g (x, y) = lim
xy→1
Of course, when xy = 1, we have y = 1/x, so we could equally well say that
the limit is 2x2 .