Sample Problem Solutions for Exam One
1. Show that lim
(
x,y
)
→
(0
,
0)
(
x
3
y
2
+
y
6
)
/
(
x
6
+
y
4
) does not exist.
We know that if the limit exists, then it is the same along every path approaching
(0,0). First try
y
= 0, where the limit is clearly 0. Next try
y
=
x
3
/
2
, where we
obtain
lim
x
→
0
x
6
+
x
9
2
x
6
= lim
x
→
0
1+
x
3
2
=
1
2
.
Thus the limit does not exist.
2. Show that lim
(
x,y
)
→
(0
,
0)
(
x
4

2
x
3
y
+
y
4
)
/
(
x
2
+
y
2
) = 0.
Using the polar coordinates
r
=
x
2
+
y
2
and
θ
, we obtain
F
(
x, y
) =
r
2
(
cos
4
θ

2
cos
3
θ sinθ
+
sin
4
θ
),
so that
F
(
x, y
)
≤
4
r
2
. Since
lim
(
x,y
)
→
(0
,
0)
r
= 0, it follows that the limit is 0.
3. Show that if
F
(
x, y
) is differentiable, then the first partials exist.
We have
A
and
B
such that
F
(
x, y
)

F
(
x
0
, y
0
) =
A
(
x

x
0
) +
B
(
y

y
0
) +
η
(
P

P
0
),
where
η
(
P
) =
η
x
η
y
is a vector functon with lim
P
→
P
0
η
= 0.
Thus
F
(
x, y
0
)

F
(
x
0
, y
0
) =
A
(
x

x
0
) +
B
(
y
0

y
0
) +
η
x
(
x

x
0
) +
η
y
(
y
0

y
0
),
so
F
(
x,y
0
)

F
(
x
0
,y
0
)
x

x
0
=
A
+
η
x
.
Then
F
x
(
x
0
, y
0
) = lim
x
→
x
0
F
(
x,y
0
)

F
(
x
0
,y
0
)
x

x
0
= lim
x
→
x
0
A
+
η
x
=
A
.
Similarly
F
y
(
x
0
, y
0
) exists.
4. Show that any differentiable function is continuous.
We have
F
(
P
)

F
(
P
0
) =
J
F
(
P

P
0
) +
η
F
(
P

P
0
), so that
lim
P
→
P
0
F
(
P
) =
F
(
P
0
) + (
J
F
+
η
F
)(
P

P
0
) =
F
(
P
0
) +
J
F
(0) =
F
(
P
0
).
5. Let
f
(
x, y
) =
xy/
(
x
2
+
y
2
) for (
x, y
) = (0
,
0) and f(0,0)=0. Show that
f
x
and
f
y
both exist at (0,0) but that f is not continuous there.
f
x
(0
,
0) =
lim
h
→
0
f
(
h,
0)

f
(0
,
0)
h
=
lim
h
→
0
0 = 0 and similarly
f
y
(0
,
0) = 0.
Substituting
y
=
x
,
lim
(
x,y
)
→
(0
,
0)
f
(
x
) =
lim
x
→
0
x
2
2
x
2
=
1
2
=
f
(0
,
0),
so that
f
is not continuous at (0,0).
6.
Let
f
(
x, y
) =
x
4
x
2
+
y
2
for (
x, y
) = (0
,
0) and
f
(0
,
0) = 0.
Show that
f
is
differentiable at (0,0).
f
x
(0
,
0) =
lim
h
→
0
f
(
h,
0)

f
(0
,
0)
h
=
lim
h
→
0
h
= 0 and
f
y
(0
,
0) =
lim
h
→
0
f
(0
,h
)

f
(0
,
0)
h
=
lim
h
→
0
0 = 0. Thus the linear approximation
L
(
x, y
) =
f
(0
,
0) +
Ax
+
By
= 0 near
(0,0).
Now
lim
(
x,y
)
→
(0
,
0)
f
(
x,y
)

L
(
x,y
)
√
x
2
+
y
2
=
lim
(
x,y
)
→
(0
,
0)
x
4
(
x
2
+
y
2
)
3
/
2
=
lim
r
→
0
rcos
4
θ
= 0,
so that
f
is differentiable at (0,0).
7. Define the directional derivative
D
v
F
of a function
F
at a point
P
in direction
v
and show that if
F
is differentiable, then
D
v
F
=
F
(
P
)
v
.
D
v
F
= lim
h
→
0
F
(
P
+
hv
)

F
(
P
)
h
.
Given that
F
is differentiable, we have
F
(
P
+
hv
) =
F
(
P
) +
F
(
P
)(
hv
) +
η
(
P
+
hv
)(
hv
)
where
lim
Q
→
P
η
(
Q
) = 0.
Thus
D
v
(
F
) = lim
h
→
0
F
(
P
)
v
+
η
(
P
+
hv
)(
hv
}
/h
) =
F
(
P
)
v
,
since
hv /h
=
±
1 is bounded.
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8. Let
F
x
y
=
x
2
+
y
2
2
xy
2
x
+
y
.
Find
F
1
3
and give the linear approximation
L
x
y
about the point.
Find the directional derivative
D
v
F
at
1
3
in the direction
v
=
4
/
5
3
/
5
.
F
(1
,
3) = (10
,
6
,
5) and
F
=
2
x
2
y
2
y
2
x
2
1
=
2
6
6
2
2
1
So
L
(
x, y
) =
F
(1
,
3) +
F
(1
,
3)
x

1
y

3
=
2
x
+ 6
y

10
6
x
+ 2
y

6
2
x
+
y
.
9.
Let
G
:
R
n
→
R
m
and
F
:
R
m
→
R
k
be differentiable functions and let
H
=
F
(
G
(
P
)). Show that
H
is differentiable and furthermore,
J
H
=
J
F
J
G
.
Let
Q
=
G
(
P
) and
Q
0
=
G
(
P
0
). We have
G
(
P
)

G
(
P
0
) =
J
G
(
P
)(
P

P
0
) +
η
G
(
P
)(
P

P
0
) and
F
(
Q
)

F
(
Q
0
) =
J
F
(
Q
)(
Q

Q
0
) +
η
F
(
Q
)(
Q

Q
0
),
where
lim
P
→
P
0
η
G
(
P
) = 0 and
lim
Q
→
Q
0
η
F
(
Q
) = 0.
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 Fall '08
 Staff
 Sin, Trigraph, P0

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