Man Kit CHAN
MATH 2011 (Fall 20142015)
WeBWork Homework3 due 11/03/2020 at 05:00pm HKT
You may need to give 4 or 5 significant digits for some (floating point) numerical answers in order to have them accepted by the
computer.
1.
(2 points) Consider the function
f
(
x
,
y
) =
(
2
xy
(
x
2
+
y
2
)
2
,
(
x
,
y
)
6
= (
0
,
0
)
0
,
(
x
,
y
) = (
0
,
0
)
(a)
Use a computer to draw a contour diagram for
f
. Which
of the following is the contour diagram?
•
?
•
figure 1
•
figure 2
•
figure 3
•
figure 4
1.
2.
Is
f
differentiable at
(
0
,
0
)
?
•
?
•
yes
•
no
•
it is not possible to tell
(e)
Find the partial derivative
f
y
at
(
0
,
0
)
by calculating it di
rectly with a limit:
f
y
=
lim
h
→
0
1
h
(
f
(
,
)

f
(
,
))
=
Do the partial derivatives
f
x
and
f
y
exist and are they contin
uous at
(
0
,
0
)
?
(Hint: to test continuity, you may want to use a
similar calculation as you used to test the continuity of f)
•
?
•
yes
•
no
•
it is not possible to tell
•
they exist but are not continuous
(Be sure that you can justify all of your answers in this prob
lem.)
SOLUTION
(b)
Is
f
differentiable at all points
(
x
,
y
)
6
= (
0
,
0
)
?
(c)
Calculate the partial derivatives of
f
for
(
x
,
y
)
6
= (
0
,
0
)
:
f
x
=
f
y
=
Do the partial derivatives
f
x
and
f
y
exist and are they continuous
at all points
(
x
,
y
)
6
= (
0
,
0
)
?
•
?
The function
f
is differentiable at all points
(
x
,
y
)
6
= (
0
,
0
)
,
as it is a rational fraction with denominator
(
x
2
+
y
2
)
2
=
0 only
when
(
x
,
y
) = (
0
,
0
)
.
(c)
The partial derivatives of
f
at points
(
x
,
y
)
6
= (
0
,
0
)
are
given by
f
x
(
x
,
y
) =
2
y
(
x
2
+
y
2
)
2

8
x
2
y
(
x
2
+
y
2
)
(d)
A first test for whether
f
is differentiable at
(
0
,
0
)
is to see
if it is continuous there. Calculate each of the following limits
x
2
+
y
2
)
4
.
(b)
The function
f
is differentiable at all points
(
x
,
y
)
6
= (
0
,
0
)
,
as it is a rational fraction with denominator
(
x
2
+
y
2
)
2
=
0 only
when
(
x
,
y
) = (
0
,
0
)
.
(c)
The partial derivatives of
f
at points
(
x
,
y
)
6
= (
0
,
0
)
are
given by
f
x
(
x
,
y
) =
2
y
(
x
2
+
y
2
)
2

8
x
2
y
(
x
2
+
y
2
)
(
x
2
+
y
2
)
4
.
to determine if
f
is continuous at
(
0
,
0
)
:
lim
h
→
0
→
0
f
(
h
,
0
) =
(d)
The function
f
is not continuous at
(
0
,
0
)
. To see this,
we can calculate the indicated limits, which should all have the
same value if the function is continuous:
lim
h
→
0
f
(
h
,
h
) =
(In each case, enter
DNE
if the limit does not exist.)
lim
h
→
0