SOLUTION OF QUIZ 3
MINGFENG ZHAO
March 03, 2013


x3 − xy 2


 2
, if (x, y) 6= (0, 0)
∂f ∂f
x + y2
1. [5 Points] Prove that f (x, y) =
is continuous on R2 , and
,

∂x ∂y


 0, if (x, y) = (0, 0)
exist everywhere, but f is not differentiable at (0, 0).
Proof. Claim I: f is continuous on R2 .
By the definition of f , it is clear that f is continuous at (x, y) ∈ R2 \{(0, 0)}. Now it suffices to
show that f is continuous at (0, 0). In fact, for all (x, y) ∈ R2 \{(0, 0)}, we have
3
2
2
2
2
x − xy 2 = |x| · |x − y | ≤ |x| · x + y = |x|.
|f (x, y) − 0| = |f (x, y)| = 2
x + y2
x2 + y 2
x2 + y 2
So we know that f (x, y) → 0, as (x, y) → (0, 0), that is, f is continuous at (0, 0). Therefore, we
know that f is continuous on R2 .
Claim II:
∂f
exist everywhere on R2 .
∂x
By the definition of f , it is clear that
∂f
(x, y)
∂x
Now it suffices to show that
∂f
(x, y) exists for all (x, y) ∈ R2 \{(0, 0)}, and
∂x
=
(3x2 − y 2 )(x2 + y 2 ) − (x3 − xy 2 ) · 2x
[x2 + y 2 ]2
=
3x4 + 3x2 y 2 − x2 y 2 − y 4 − 2x4 + 2x2 y 2
[x2 + y 2 ]2
=
x4 + 4x2 y 2 − y 4
[x2 + y 2 ]2
∂f
x3 − 0
exists at (0, 0). In fact, notice that f (x, 0) = 2
= x for all
∂x
x +0
x 6= 0, then for all x 6= 0, we have
f (x, 0) − f (0, 0)
x − 0
− 1 = − 1 = 0.
x−0
x−0
1
2
MINGFENG ZHAO
f (x, 0) − f (0, 0)
∂f
∂f
So we know that − 1, as x → 0, that is,
(0, 0) exists and
(0, 0) = 1.
x−0
∂x
∂x
∂f
exist everywhere on R2 .
Therefore, we know that
∂x
∂f
Claim III:
exist everywhere on R2 .
∂y
∂f
By the definition of f , it is clear that
(x, y) exists for all (x, y) ∈ R2 \{(0, 0)} and
∂y
∂f
(x, y)
∂y
=
−2xy · (x2 + y 2 ) − (x3 − xy 2 ) · 2y
[x2 + y 2 ]2
=
−2x3 y − 2xy 3 − 2x3 y + 2xy 3
[x2 + y 2 ]2
=
−4x3 y
[x2 + y 2 ]2
0−0
∂f
exists at (0, 0). In fact, notice that f (0, y) =
= 0 for all
Now it suffices to show that
∂y
0 + y2
f (0, y) − f (0, 0)
∂f
∂f
y 6= 0, so we know that − 0, as y → 0, that is,
(0, 0) exists and
(0, 0) = 0.
y−0
∂y
∂y
∂f
exist everywhere on R2 .
Therefore, we know that
∂y
Claim IV: f is not differentiable at (0, 0).
Assume f is differentiable at (0, 0), by the results of Claim II and Claim III, we know that
1 and
(1)
∂f
(0, 0) = 0, then we know that
∂y
f (x, y) − f (0, 0) − ∂x
∂x (0, 0)x −
p
lim
(x,y)→0
x2 + y 2
∂f
∂y (0, 0)y =
lim
(x,y)→0
|f (x, y) − x|
p
= 0.
x2 + y 2
On the other hand, for f (x, y) − x, we know that for all (x, y) 6= (0, 0), we have
3
3
2
x − xy 2
x − xy 2 − x3 − xy 2 =
= 2|x|y .
−
x
|f (x, y) − x| = 2
2
2
2
2
x +y
x +y
x + y2
Then for all (x, y) 6= (0, 0), we have
2|x|y 2
|f (x, y) − x|
p
=
3 .
[x2 + y 2 ] 2
x2 + y 2
Let x = 0 and y 6= 0, we have
|f (x, y) − x|
0
p
=
3 = 0.
2
2
[0 + y 2 ] 2
x +y
∂f
(0, 0) =
∂x
SOLUTION OF QUIZ 3
3
Let x = y = t > 0, we have
2|x|y 2
[x2
So we know that
lim
(x,y)→0
+
y2 ]
3
2
=
2t3
3
[2t2 ] 2
=
2
2
3
2
1
=√ =
6 0.
2
|f (x, y) − x|
p
does not exist, which contradicts with (1). Therefore, we
x2 + y 2
know that f is not differentiable at (0, 0).
2. [5 Points] The equation z = x2 + y 2 defined a paraboloid in R3 . First sketch the paraboloid. Then
find all points on the paraboloid where the tangent plane is parallel to the plane x − y − z = 0 and
write down the equation of the tangent plane.
Proof. The graph of z = x2 + y 2 looks like the following:
Figure 1. z = x2 + y 2
Let F (x, y, z) = x2 + y 2 − z, then
∇F (x, y, z) = (2x, 2y, −1).
1
So we know that ∇F (x, y, z)//(1, −1, −1) if and only if (2x, 2y, −1) = (1, −1, −1), that is, x =
2
2 2
1
1
1
1
and y = − , which implies that z =
+ −
= . So the equation of the tangent plane at
2
2
2
2
4
MINGFENG ZHAO
1
1 1
2, −2, 2
is given by:
1
1
1
−1· y+
−1· z−
= 0.
1· x−
2
2
2
That is,
x−y−z =
1
.
2
Department of Mathematics, University of Connecticut, 196 Auditorium Road, Unit 3009, Storrs, CT
06269-3009
E-mail address: mingfeng.zhao@uconn.edu