SOLUTIONS FOR ASSIGNMENT 6
Problem Set 5: A1 (iii): a). Find and classify the critical points of
f
(
x, y
) = (
x
2
+
y
2
−
1)
y
b). Sketch the level curves of
f
in a neighbourhood of each critical point. The Taylor polynomial
P
2
,va
(
vx
)
at
each critical point will be helpful. Use symmetry to reduce the number of cases, if appropriate.
a). To find the critical points of
f
we need to solve
∇
f
(
x, y
) = (0
,
0)
,
i.e.
we need to solve the system of equations:
∂ f
∂x
(
x, y
) = 2
xy
= 0
(1)
∂ f
∂y
(
x, y
) =
x
2
+ 3
y
2
−
1 = 0
.
(2)
From equation (1) we have:
either
x
= 0
and thus equation (2) gives
y
=
±
1
/
√
3
,
or
y
= 0
and thus equation (2) gives
x
=
±
1
.
Therefore, the critical points of
f
are:
(0
,
−
1
/
√
3)
,
(0
,
1
/
√
3)
,
(
−
1
,
0)
,
(1
,
0)
.
In order to classify these critical points we will try to make use of the Second Derivative Test. The Hessian matrix of
f
is:
Hf
(
x, y
) =
p
2
y
2
x
2
x
6
y
P
.
Then:
Hf
(0
,
1
/
√
3) =
p
2
/
√
3
0
0
6
/
√
3
P
which is positive definite (since its quadratic form
Q
(
u, V
) =
2
√
3
(
u
2
+ 3
V
2
)
>
0
for all
(
u, V
)
n
= (0
,
0)
). Thus, by the
Second Derivative Test the critical point
(0
,
1
/
√
3)
is a local minimum of
f
.
Hf
(0
,
−
1
/
√
3) =
p
−
2
/
√
3
0
0
−
6
/
√
3
P
is negative definite (since its quadratic form
Q
(
u, V
) =
−
2
√
3
u
2
−
6
√
3
V
2
<
0
for all
(
u, V
)
n
= (0
,
0)
). Thus, by the
Second Derivative Test the critical point
(0
,
−
1
/
√
3)
is a local maximum of
f
.
Finally,
Hf
(
±
1
,
0) =
p
0
±
2
±
2
0
P
,
which are indefinite matrices (since their quadratic forms
Q
(
u, V
) =
±
4
uV
are both positive and negative at points
(
u, V
)
n
= (0
,
0)
). Thus by the Second Derivative Test the critical points
(
±
1
,
0)
are saddle points of
f
.
1