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Exercises 3.3
where
M
(
t
) = 32 is the temperature of ice. This equation is linear and is rewritten in the
standard form as
dT
dt
+
KT
(
t
)=32
K.
We Fnd that the integrating factor is
e
Kt
. Multiplying both sides by
e
Kt
and integrating gives
e
Kt
dT
dt
+
e
Kt
KT
(
t
)=32
Ke
Kt
⇒
e
Kt
T
(
t
)=
Z
32
Ke
Kt
dt
⇒
e
Kt
T
(
t
)=32
e
Kt
+
C
⇒
T
(
t
)=32+
Ce
−
Kt
.
By setting
t
= 0 and using the initial temperature 70
◦
±, we Fnd the constant
C
.
70 = 32 +
C
⇒
C
=38
.
Knowing that it takes 15 minutes for the wine to chill to 60
◦
±, we can Fnd the constant,
K
:
60 = 32 + 38
e
−
K
(15)
.
Solving for
K
yields
K
=
−
1
15
ln
±
60
−
32
38
²
≈
0
.
02035
.
Therefore,
T
(
t
)=32+38
e
−
0
.
02035
t
.
We can now determine how long it will take for the wine to reach 56
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This note was uploaded on 03/29/2010 for the course MATH 54257 taught by Professor Hald,oh during the Spring '10 term at University of California, Berkeley.
 Spring '10
 Hald,OH
 Math, Differential Equations, Linear Algebra, Algebra, Equations

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