husain (aih243) – HW06 – Gilbert – (56215)
1
This printout should have 26 questions.
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beFore answering.
001
10.0 points
Richard removes a turkey From the oven
aFter it reaches a temperature oF 166
◦
F
and
places on a table in his dining room where
the temperature is 85
◦
F
. AFter 10 minutes
the temperature oF the turkey is 158
◦
F
, while
aFter 20 minutes it is 152
◦
F
. Use linear ap
proximation to predict the temperature oF the
turkey aFter halF an hour.
1.
temp
≈
143
◦
F
2.
temp
≈
147
◦
F
3.
temp
≈
144
◦
F
4.
temp
≈
146
◦
F
correct
5.
temp
≈
145
◦
F
Explanation:
IF
T
(
t
) is the temperature oF the turkey
aFter
t
minutes, then the given data tell us
that
T
(0) = 166
, T
(10) = 158
, T
(20) = 152
,
while the Fact that the room temperature is
85
◦
F
means that
T
(
t
) gets closer to 85
◦
F
as
t
increases. Now by linear approximation
T
(
t
)
≈
T
(20) +
T
′
(20)(
t
−
20)
,
(
t >
20)
,
with
T
′
(20) = lim
t
→
20
T
(
t
)
−
T
(20)
t
−
20
.
But
T
′
(20)
≈
T
(10)
−
T
(20)
10
−
20
=
158
−
152
−
10
=
−
3
5
◦
degrees/min
.
Consequently,
temp =
T
(30)
≈
146
◦
F
.
002
10.0 points
±ind the linearization oF
f
(
x
) =
1
√
3 +
x
at
x
= 0.
1.
L
(
x
) =
1
3
p
1 +
1
6
x
P
2.
L
(
x
) =
1
3
p
1
−
1
3
x
P
3.
L
(
x
) =
1
√
3
+
1
3
x
4.
L
(
x
) =
1
√
3
p
1
−
1
6
x
P
correct
5.
L
(
x
) =
1
√
3
p
1 +
1
6
x
P
6.
L
(
x
) =
1
√
3
−
1
3
x
Explanation:
The linearization oF
f
is the Function
L
(
x
) =
f
(0) +
f
′
(0)
x .
But For the Function
f
(
x
) =
1
√
3 +
x
= (3 +
x
)
−
1
/
2
,
the Chain Rule ensures that
f
′
(
x
) =
−
1
2
(3 +
x
)
−
3
/
2
.
Consequently,
f
(0) =
1
√
3
,
f
′
(0) =
−
1
6
√
3
,
and so
L
(
x
) =
1
√
3
p
1
−
1
6
x
P
.
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2
003
10.0 points
Use linear approximation with
a
= 16 to
estimate the number
√
15
.
4 as a fraction.
1.
√
15
.
4
≈
3
39
40
2.
√
15
.
4
≈
3
15
16
3.
√
15
.
4
≈
3
77
80
4.
√
15
.
4
≈
3
37
40
correct
5.
√
15
.
4
≈
3
19
20
Explanation:
For a general function
f
, its linear approxi
mation at
x
=
a
is de±ned by
L
(
x
) =
f
(
a
) +
f
′
(
a
)(
x
−
a
)
and for values of
x
near
a
f
(
x
)
≈
L
(
x
) =
f
(
a
) +
f
′
(
a
)(
x
−
a
)
provides a reasonable approximation for
f
(
x
).
Now set
f
(
x
) =
√
x,
f
′
(
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
= 16
,
h
=
−
6
10
.
For then
√
15
.
4
≈
3
37
40
.
004
10.0 points
Estimate the value of 17
1
/
4
using di²eren
tials.
1.
17
1
/
4
≈
63
32
2.
17
1
/
4
≈
31
16
3.
17
1
/
4
≈
2
4.
17
1
/
4
≈
33
16
5.
17
1
/
4
≈
65
32
correct
Explanation:
Set
f
(
x
) =
x
1
/
4
. Then
df
dx
=
1
4
x
3
/
4
.
By di²erentials, therefore, we see that
f
(
a
+ Δ
x
)
−
f
(
a
)
≈
dx
v
v
v
x
=
a
Δ
x
=
Δ
x
4
a
3
/
4
.
Thus, with
a
= 16 and Δ
x
= 1,
17
1
/
4
−
2
≈
1
32
.
Consequently,
17
1
/
4
≈
65
32
.
005
10.0 points
At the HEB stores throughout Texas the
daily demand (in pounds) for candy at $
x
per
pound is given by
D
= 8000
−
100
x
2
,
1
≤
x
≤
7
.
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 Spring '06
 McAdam
 Critical Point, Derivative, Husain

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