The differential equation below models the temperature of a 94°C cup of coffee in a 24°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 74°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in °C, and let t be the time in minutes, with t = 0 corresponding to the time when the temperature was 94°C.) 1 dy (y- 24) dt 50 Step 1 dy We have dt (y-24), 50 Which separates to become dy = -dt Step 2 Ignoring the constant of integration, integrating the left side gives (remember to use In Ju| where appropriat S dy = y-24 Step 3 Ignoring the constant of integration, integrating the right side gives Sat = dt3D 50 Step 4 We now have Inly - 24| = -t + C.and so ly – 24| = 50 Step 5 Consequently, we write y-K +24, where K- te Step 6 Finally, using y (0) -94 allows us to solve for K- Step 7 The temperature of the coffee at the time is described by the following equation.

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Chapter6: Exponential And Logarithmic Functions
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The differential equation below models the temperature of a 94°C cup of coffee in a 24°C
room, where it is known that the coffee cools at a rate of 1°C per minute when its
temperature is 74°C. Solve the differential equation to find an expression for the
temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in °C,
and let t be the time in minutes, with
t = 0
corresponding to the time when the temperature was 94°C.)
1
dy
(y- 24)
dt
50
Step 1
dy
We have
dt
(y-24),
50
Which separates to become
dy = -dt
Step 2
Ignoring the constant of integration, integrating the left side gives (remember to use In
Ju| where appropriat
S dy =
y-24
Step 3
Ignoring the constant of integration, integrating the right side gives
Sat =
dt3D
50
Step 4
We now have Inly - 24| = -t + C.and so ly – 24| =
50
Step 5
Consequently, we write y-K
+24, where K- te
Step 6
Finally, using y (0) -94 allows us to solve for
K-
Step 7
The temperature of the coffee at the time is described by the following equation.
Transcribed Image Text:The differential equation below models the temperature of a 94°C cup of coffee in a 24°C room, where it is known that the coffee cools at a rate of 1°C per minute when its temperature is 74°C. Solve the differential equation to find an expression for the temperature of the coffee at time t. (Let y be the temperature of the cup of coffee in °C, and let t be the time in minutes, with t = 0 corresponding to the time when the temperature was 94°C.) 1 dy (y- 24) dt 50 Step 1 dy We have dt (y-24), 50 Which separates to become dy = -dt Step 2 Ignoring the constant of integration, integrating the left side gives (remember to use In Ju| where appropriat S dy = y-24 Step 3 Ignoring the constant of integration, integrating the right side gives Sat = dt3D 50 Step 4 We now have Inly - 24| = -t + C.and so ly – 24| = 50 Step 5 Consequently, we write y-K +24, where K- te Step 6 Finally, using y (0) -94 allows us to solve for K- Step 7 The temperature of the coffee at the time is described by the following equation.
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