Help Please EXPONENTIAL REGRESSION Data A cup of hot coffee

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EXPONENTIAL REGRESSION

Data: A cup of hot coffee was placed in a room maintained at a constant temperature of 69 degrees, and the coffee temperature was recorded periodically, in Table 1.

t = Time Elapsed

(minutes)

C = Coffee

Temperature (degrees F.)

0

166.0

10

140.5

20

125.2

30

110.3

40

104.5

50

98.4

60

93.9

TABLE 1

REMARKS:

Common sense tells us that the coffee will be cooling off and its temperature will decrease and approach the ambient temperature of the room, 69 degrees.

So, the temperature difference between the coffee temperature and the room temperature will decrease to 0.

We will fit the temperature difference data (Table 2) to an exponential curve of the form y = A e^-bt.  

Notice that as t gets large, y will get closer and closer to 0, which is what the temperature difference will do.  

So, we want to analyze the data where t = time elapsed and y = C - 69, the temperature difference between the coffee temperature and the room temperature.

TABLE 2

t = Time Elapsed (minutes)

y = C - 69 Temperature

Difference

(degrees F.)

0

97.0

10

71.5

20

56.2

30

41.3

40

35.5

50

29.4

60

24.9

Exponential Function of Best Fit (using the data in Table 2):

      y = 89.976 e ^-0.023t where t = Time Elapsed (minutes) and y = Temperature Difference (in degrees)

(a) Use the exponential function to estimate the temperature difference y when 25 minutes have elapsed. Report your estimated temperature difference to the nearest tenth of a degree. (explanation/work optional)

(b) Since y = C - 69, we have coffee temperature C = y + 69. Take your difference estimate from part (a) and add 69 degrees. Interpret the result by filling in the blank:

            When 25 minutes have elapsed, the estimated coffee temperature is ________   degrees.

(c) Suppose the coffee temperature C is 100 degrees. Then y = C - 69 = ____ degrees is the temperature difference between the coffee and room temperatures.

(d) Consider the equation   _____ = 89.976 e ^{- 0.023t} where the ____ is filled in with your answer from part (c).

Show algebraic work to solve this part (d) equation for t, to the nearest tenth. Interpret your results clearly in the context of the coffee application. [Use additional paper if needed]

t = Time Elapsed (minutes) C = Coffee Temperature (degrees F.) 0 166.0 10 140.5 20 125.2 30 110.3 40 104.5 50 98.4 60 93.9 TABLE 1

REMARKS: Common sense tells us that the coffee will be cooling off and its temperature will decrease and approach the ambient temperature of the room, 69 degrees. So, the temperature difference between the coffee temperature and the room temperature will decrease to 0. We will fit the temperature difference data (Table 2) to an exponential curve of the form y = A e-bt.   Notice that as t gets large, y will get closer and closer to 0, which is what the temperature difference will do.   So, we want to analyze the data where t = time elapsed and y = C - 69, the temperature difference between the coffee temperature and the room temperature.

TABLE 2 t = Time Elapsed (minutes) y = C - 69 Temperature Difference (degrees F.) 0 97.0 10 71.5 20 56.2 30 41.3 40 35.5 50 29.4 60 24.9

Solution

a. When t = 25,

y = 89.976*e^{-0.023 t} = 89.976*e^{-0.023 *25} = 50.6 degrees

b. C = 50.6 + 69 = 119.6

When 25 minutes have elapsed, the estimated coffee temperature is   119.6 degrees.

c. When C = 100,

y = C - 69 = 31

d. Now,

31 = 89.976 e ^{- 0.023t}

or,

e^{0.023 t} = 2.9024

Taking log on both sides,

0.023*t = ln(2.9024) = 1.0656

t = 1.0656 / 0.023 = 46.3 minutes

So Coffee temperature is 100 degrees after 46.3 minutes have elapsed.