Between 8 a.m and 12 p.m on a hot summer day, the temperature rose at a rate of 10 ° F per hour from an initial temperature of 65 ° F . At 9 a.m, the temperature of an object was measured to be 35 ° F and was, at that time, increasing at a rate of 5 ° F per hour. Show that the temperature of the object at time t was T ( t ) = 10 t − 15 + 40 e ( 1 − t ) / 8 , 0 ≤ t ≤ 4.
Between 8 a.m and 12 p.m on a hot summer day, the temperature rose at a rate of 10 ° F per hour from an initial temperature of 65 ° F . At 9 a.m, the temperature of an object was measured to be 35 ° F and was, at that time, increasing at a rate of 5 ° F per hour. Show that the temperature of the object at time t was T ( t ) = 10 t − 15 + 40 e ( 1 − t ) / 8 , 0 ≤ t ≤ 4.
Solution Summary: The author explains that the temperature of the object at time t is given by T(t)=10t-15+40e
Between 8 a.m and 12 p.m on a hot summer day, the temperature rose at a rate of
10
°
F
per hour from an initial temperature of
65
°
F
. At 9 a.m, the temperature of an object was measured to be
35
°
F
and was, at that time, increasing at a rate of
5
°
F
per hour. Show that the temperature of the object at time t was
T
(
t
)
=
10
t
−
15
+
40
e
(
1
−
t
)
/
8
,
0
≤
t
≤
4.
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