Garcia, Ilse – Homework 9 – Due: Oct 24 2007, 3:00 am – Inst: Fonken
1
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printout
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have
20
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before answering.
The due time is Central
time.
001
(part 1 of 1) 10 points
Find the linearization of
f
(
x
) =
1
√
7 +
x
at
x
= 0.
1.
L
(
x
) =
1
√
7
‡
1 +
1
14
x
·
2.
L
(
x
) =
1
√
7

1
7
x
3.
L
(
x
) =
1
7
‡
1 +
1
14
x
·
4.
L
(
x
) =
1
7
‡
1

1
7
x
·
5.
L
(
x
) =
1
√
7
+
1
7
x
6.
L
(
x
) =
1
√
7
‡
1

1
14
x
·
correct
Explanation:
The linearization of
f
is the function
L
(
x
) =
f
(0) +
f
0
(0)
x .
But for the function
f
(
x
) =
1
√
7 +
x
= (7 +
x
)

1
/
2
,
the Chain Rule ensures that
f
0
(
x
) =

1
2
(7 +
x
)

3
/
2
.
Consequently,
f
(0) =
1
√
7
,
f
0
(0) =

1
14
√
7
,
and so
L
(
x
) =
1
√
7
‡
1

1
14
x
·
.
keywords: linearization, square root function,
differentials
002
(part 1 of 1) 10 points
Use linear approximation with
a
= 4 to
estimate the number
√
4
.
5 as a fraction.
1.
√
4
.
5
≈
2
3
20
2.
√
4
.
5
≈
2
3
40
3.
√
4
.
5
≈
2
1
8
correct
4.
√
4
.
5
≈
2
7
40
5.
√
4
.
5
≈
2
1
10
Explanation:
For a general function
f
, its linear approxi
mation at
x
=
a
is defined by
L
(
x
) =
f
(
a
) +
f
0
(
a
)(
x

a
)
and for values of
x
near
a
f
(
x
)
≈
L
(
x
) =
f
(
a
) +
f
0
(
a
)(
x

a
)
provides a reasonable approximation for
f
(
x
).
Now set
f
(
x
) =
√
x,
f
0
(
x
) =
1
2
√
x
.
Then, if we can calculate
√
a
easily, the linear
approximation
√
a
+
h
≈
√
a
+
h
2
√
a
provides a very simple method via calculus
for computing a good estimate of the value of
√
a
+
h
for small values of
h
.
In the given example we can thus set
a
= 4
,
h
=
5
10
.
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Garcia, Ilse – Homework 9 – Due: Oct 24 2007, 3:00 am – Inst: Fonken
2
For then
√
4
.
5
≈
2
1
8
.
keywords: linear approximation, square roots
003
(part 1 of 1) 10 points
A cube with sides 6 inches long is covered
with a fiberglass coating
.
04 inches thick. Es
timate the volume of the fiberglass shell.
1.
fiberglass vol
≈
431
50
cu.ins.
2.
fiberglass vol
≈
216
25
cu.ins.
correct
3.
fiberglass vol
≈
43
5
cu.ins.
4.
fiberglass vol
≈
214
25
cu.ins.
5.
fiberglass vol
≈
429
50
cu.ins.
Explanation:
A cube with side length
x
has volume
V
(
x
) =
x
3
.
If the length of each side is
changed by an amount Δ
x
, then the approxi
mate change, Δ
V
, in volume is given by
Δ
V
≈
V
0
(
x
)Δ
x
= 3
x
2
Δ
x .
Now a
.
04 inch thick fiberglass coating on each
face will increase the side length by
2
25
= (2
×
.
04) inches
.
When the side length of the cube is 6 inches,
therefore, the volume of this fiberglass shell
will be approximately
Δ
V
= fiberglass vol
≈
216
25
cu. ins.
.
keywords: cube, volume, estimate
004
(part 1 of 1) 10 points
The radius of a circle is estimated to be
14 inches, with a maximum error in measure
ment of
±
0
.
06 inches.
Use differentials to
estimate the maximum error in calculating
the area of the circle using this estimate.
1.
Max error
≈ ±
5
.
2778 sq.ins
correct
2.
Max error
≈ ±
5
.
2938 sq.ins
3.
Max error
≈ ±
5
.
3018 sq.ins
4.
Max error
≈ ±
5
.
2698 sq.ins
5.
Max error
≈ ±
5
.
2858 sq.ins
Explanation:
The area of a circle having radius
x
is given
by
A
=
πx
2
. By differentials, therefore,
Δ
A
≈
Δ
x
dA
dx
= 2
πx
Δ
x.
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 Spring '08
 schultz
 Calculus, Derivative, Continuous function, abs. max. value

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