14.1 SOLUTIONS
951
CHAPTER FOURTEEN
Solutions for Section 14.1
Exercises
1.
If
h
is small, then
f
x
(3
,
2)
≈
f
(3 +
h,
2)

f
(3
,
2)
h
.
With
h
= 0
.
01
, we Fnd
f
x
(3
,
2)
≈
f
(3
.
01
,
2)

f
(3
,
2)
0
.
01
=
3
.
01
2
(2+1)

3
2
(2+1)
0
.
01
= 2
.
00333
.
With
h
= 0
.
0001
, we get
f
x
(3
,
2)
≈
f
(3
.
0001
,
2)

f
(3
,
2)
0
.
0001
=
3
.
0001
2
(2+1)

3
2
(2+1)
0
.
0001
= 2
.
0000333
.
Since the difference quotient seems to be approaching
2
as
h
gets smaller, we conclude
f
x
(3
,
2)
≈
2
.
To estimate
f
y
(3
,
2)
, we use
f
y
(3
,
2)
≈
f
(3
,
2 +
h
)

f
(3
,
2)
h
.
With
h
= 0
.
01
, we get
f
y
(3
,
2)
≈
f
(3
,
2
.
01)

f
(3
,
2)
0
.
01
=
3
2
(2
.
01+1)

3
2
(2+1)
0
.
01
=

0
.
99668
.
With
h
= 0
.
0001
, we get
f
y
(3
,
2)
≈
f
(3
,
2
.
0001)

f
(3
,
2)
0
.
0001
=
3
2
(2
.
0001+1)

3
2
(2+1)
0
.
0001
=

0
.
9999667
.
Thus, it seems that the difference quotient is approaching

1
, so we estimate
f
y
(3
,
2)
≈ 
1
.
2.
Using Frst
Δ
x
= 0
.
1
and
Δ
y
= 0
.
1
, we have the estimates:
f
x
(1
,
3)
≈
f
(1
.
1
,
3)

f
(1
,
3)
0
.
1
=
0
.
0470

0
.
0519
0
.
1
=

0
.
0493
,
and
f
y
(1
,
3)
≈
f
(1
,
3
.
1)

f
(1
,
3)
0
.
1
=
0
.
0153

0
.
0519
0
.
1
=

0
.
3660
.
Now, using
Δ
x
= 0
.
01
and
Δ
y
= 0
.
01
, we have the estimates:
f
x
(1
,
3)
≈
f
(1
.
01
,
3)

f
(1
,
3)
0
.
01
=
0
.
0514

0
.
0519
0
.
01
=

0
.
0501
,
and
f
y
(1
,
3)
≈
f
(1
,
3
.
01)

f
(1
,
3)
0
.
01
=
0
.
0483

0
.
0519
0
.
01
=

0
.
3629
.