x,y
(
)
$
0,0
(
)
x
!
f
(x,0)=0/
x
2
=0
x
%
0
f
(x,y)
$
0
f
(x,x)=2
x
2
/ 3
x
2
(
)
=
2
3
x,y
(
)
$
0,0
(
)
x
=
y
f
(x,y)
$
2
3
T
x
(6,4)=lim
h
$
0
T
(6+h,4)
!
T
(6,4)
h
T
x
(6,4)
h
=
&
2
T
x
(6,4)
’
T
(8,4)
!
T
(6,4)
2
=
86
!
80
2
=3
T
x
(6,4)
’
T
(4,4)
!
T
(6,4)
!
2
=
72
!
80
!
2
=4
T
x
(6,4)
3.5
(
C/m.
T
y
6,4
(
)
=lim
h
$
0
T
(6,4+
h
)
!
T
(6,4)
h
h
=
&
2
T
y
(6,4)
’
T
(6,6)
!
T
(6,4)
2
=
75
!
80
2
=
!
2.5
T
y
(6,4)
’
T
(6,2)
!
T
(6,4)
!
2
=
87
!
80
!
2
=
!
3.5
T
y
(6,4)
!
3.0
(
C/m.
u
=
1
2
,
1
2
D
u
T
(6,4)=
)
T
(6,4)
*
u
=
T
x
(6,4)
1
2
+
T
y
(6,4)
1
2
D
u
T
(6,4)
’
(3.5)
1
2
+(
!
3.0)
1
2
=
1
2
2
’
0.35
(6,4)
u
0.35
(
C/m.
D
u
T
(6,4)=lim
h
$
0
T
6+
h
1
2
,4+
h
1
2
!
T
(6,4)
h
h
=
&
2
2
.
D
u
T
(6,4)
’
T
(8,6)
!
T
(6,4)
2
2
=
80
!
80
2
2
=0
D
u
T
(6,4)
’
T
(4,2)
!
T
(6,4)
!
2
2
=
74
!
80
!
2
2
=
3
2
D
u
T
(6,4)
’
3
2
2
’
1.1
(
C/m.
T
xy
(x,y)=
+
+
y
T
x
(x,y)
=lim
h
$
0
T
x
(x,y+
h
)
!
T
x
(x,y)
h
T
xy
(6,4)=lim
h
$
0
T
x
(6,4+
h
)
!
T
x
(6,4)
h
10. As
along the
axis,
for
, so
along this line. But
, so as
along the line
,
. Thus the limit doesn't
exist.
11.
(a)
, so we can approximate
by considering
and
using the values given in the table:
,
. Averaging these values, we estimate
to be
approximately
Similarly,
, which we can approximate
with
:
,
. Averaging these values, we estimate
to be approximately
(b)
Here
, so by Equation 15.6.9 [ET 14.6.9],
. Using our estimates from part (a), we have
. This means that as we move through the point
in the direction of
, the temperature increases at a rate of approximately
Alternatively, we can use Definition 15.6.2 [ET 14.6.2]:
, which we can estimate with
Then
,
. Averaging
these values, we have
(c)
, so
which
3
Stewart Calculus ET 5e 0534393217;14. Partial Derivatives; Review: Exercises