Math 23
B. Dodson
Week 6 Homework:
14.2 limits
14.3 partial derivatives, 2nd order deriv
Week 6 Homework:
14.2 limits
Problem 14.2.6:
Find the limit
lim
(
x,y
)
→
(6
,
3)
(
xy
cos (
x

2
y
))
.
Solution:
We see that the function
f
(
x, y
) =
xy
cos (
x

2
y
)
is continuous. (why?)
Since
xy
and
x

2
y
are polynomials;
cos
x

2
y
is a composite of continuous functions;
and then
f
is a product of continuous functions
(that is, of
xy
and cos
x

2
y
). So the
limit is
f
(6
,
3) = 6
·
3 cos 6

6 = 18
·
1 = 18
.
Problem 14.2.13:
Show that the limit
lim
(
x,y
)
→
(0
,
0)
2
x
2
y
x
4
+
y
2
does not exist.
2
.
Solution:
Along the
x
axis (
x,
0) the limit is
lim
(
x,
0)
→
(0
,
0)
2
x
2
y
x
4
+
y
2
=
lim
x
→
0
0
x
4
= 0
,
recalling that
x
= 0 while
x
→
0
.
But along
the parabola
y
=
x
2
, so (
x, x
2
),
lim
(
x,x
2
)
→
(0
,
0)
2
x
2
y
x
4
+
y
2
=
lim
x
→
0
2
x
4
2
x
4
= 1
,
again
using that
x
= 0 while
x
→
0
.
Since we have
two different limits among points with (
x, y
)
→
(0
,
0)
,
the limit does not exist. Note that the limit along
lines (0
, y
) (the
y
axis) gives
0
y
2
→
0
and (
x, mx
) (with
y
=
mx, m
= 0) gives
2
mx
3
x
4
+
m
2
x
2
→
2
mx
x
2
+
m
2
→
0
m
2
= 0
,
(since
m
= 0)
.
So the limits along all lines approach 0,
and we needed the parabola to get a nonzero value and
show that the limit doesn’t exist.
Problem 14.3.15:
Find the partial derivatives of
the function
z
=
f
(
x, y
) =
xe
3
y
.
Find
f
x
(2
,
1)
.
3
Solution:
For
f
x
we (temporarily) hold
y
constant,
so
e
3
y
constant, giving
f
as constant
·
x
.
We then take the derivative in
x
, with
(constant
·
x
)’ = constant, or
f
x
=
e
3
y
.
Likewise, with
x
constant,
f
y
=
x
∂
∂y
(
e
3
y
)
=
x
d
dy
(
e
3
y
) = 3
xe
3
y
.
Finally,
f
x
(2
,
1) =
e
3
gives the rate of change
of
f
at (2
,
1) with respect to
x
; which we
may constrast with
f
y
(2
,
1) = 6
e
3
,
the rate of
change of
f
at (2
,
1) with respect to
y
.
For example,
f
is growing six times more rapidly
in
y
than in
x
. We also solved #49, 14.3, and
in particular verified that the second partials
z
xy
=
z
yx
.