Question

NO LINKS!!

A cup of water at an initial temperature of 81°C is placed in a room at a constant temperature of 24°C. The temperature of the water is measured every 5 minutes during a half-hour period. The results are recorded as ordered pairs of the form (t, T), where t is the time (in minutes ) and T is the temperature (in degrees Celsius).

t T
0 81
5 69
10 60.5
15 54.2
20 49.3
25 45.4
30 42.6

a. Subtract the room temp. from each of the temp. in the ordered pairs. Use a graphing utility to plot the data points (t, T) and (t, T-24). An exponential model for the data (t, T-24) is T-24 = 54.4(0.964)^t. Solve for T.
T=
Graph the model. Compare the result with the plot of the original data.

b. Use a graphing utility to plot the points (t, ln(T-24)) and observe that the points appear to be linear. Use the regression feature of the graphing utility to fit a line to these data. This resulting line has the form ln(T-24) = at + b, which is equivalent to e^(ln(T-24)) = e^(at+b). Solve for T, and verify that the result is equivalent to the model in part (b). Round all numerical values to 3 decimal places).
T=

c. Fit a rational model to the data. Take the reciprocals of the y-coordinates of the revised data points to generate the points. (t, 1/(T-24))
Use a graphing utility to graph these points and observe that they appear to be linear. Use the regression feature of a graphing utility to fit a line to these data. The resulting line has the form 1/(T - 24) = at + b.
Solve for T. (Round all numerical values to 4 decimal places). Use a graphing utility the rational function and the original data points.
T=

Answer 1

Part (a)

Given data table.

`\begin{array} \cline{1-2}t & T\\\cline{1-2}0 & 81\\\cline{1-2}5 & 69\\\cline{1-2}10 & 60.5\\\cline{1-2}15 & 54.2\\\cline{1-2}20 & 49.3\\\cline{1-2}25 & 45.4\\\cline{1-2}30 & 42.6\\\cline{1-2}\end{array}`

where,

• t = time in minutes
• T = temperature in Celsius

It's unfortunate that T is used twice for these variables. It might cause some confusion.

This is what happens when we subtract 24 from each temperature.

`\begin{array}c \cline{1-2}t & T-24\\\cline{1-2}0 & 57\\\cline{1-2}5 & 45\\\cline{1-2}10 & 36.5\\\cline{1-2}15 & 30.2\\\cline{1-2}20 & 25.3\\\cline{1-2}25 & 21.4\\\cline{1-2}30 & 18.6\\\cline{1-2}\end{array}`

The graph of each set of points is shown in figure A. I used GeoGebra to make the graph.

The first set of data points are in red. The second set in blue. Notice the set of blue points is the result of shifting the red points 24 units down.

The green exponential model is also shown as the curve through the set of red points.

The green regression exponential model doesn't go through each point perfectly. Rather it tries to get as close as possible.

To go from

T-24 = 54.4(0.964)^t

to

T = 54.4(0.964)^t + 24

we simply add 24 to both sides. This means there isn't much to show in terms of steps when solving for T.

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Part (b)

Use spreadsheet technology or something like GeoGebra to generate this data table.

`\begin{array}c \cline{1-2}t & ln(T-24)\\\cline{1-2}0 & 4.043051\\\cline{1-2}5 & 3.806662\\\cline{1-2}10 & 3.597312\\\cline{1-2}15 & 3.407842\\\cline{1-2}20 & 3.230804\\\cline{1-2}25 & 3.063391\\\cline{1-2}30 & 2.923162\\\cline{1-2}\end{array}`

The values in the 2nd column are approximate.

Then plot each of those points on the same xy grid. Use technology of your choice to determine the regression line is roughly

y = -0.03723x + 3.99739

So we can say

ln(T-24) = -0.03723t + 3.99739

the right hand side fits the format at + b where

• a = -0.03723
• b = 3.99739

both of which are approximate.

Then,

e^(ln(T-24)) = e^(at+b)

T-24 = e^(at+b)

T-24 = e^(-0.03723t + 3.99739)

T-24 = e^(-0.03723t)*e^(3.99739)

T-24 = e^(-0.03723t)*54.45583

T-24 = 54.45583*e^(-0.03723t)

T = 54.45583*e^(-0.03723t) + 24

T = 54.456*e^(-0.037t) + 24

Let's verify that this is equivalent to the model found in part (a).

T = 54.456*e^(-0.037t) + 24

T = 54.456*(e^(-0.037))^t + 24

T = 54.456*(0.963676)^t + 24

T = 54.5*(0.964)^t + 24

which is very close to the model mentioned in part (a). There's some slight rounding error. That's to be expected in problems like this.

-----------------------------

Answer: T = 54.456*e^(-0.037t) + 24

See figure B for the graph. Like with the other regression curve, this straight line doesn't go through all of the points. It tries to get as close as possible.

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Part (c)

Refer to the 2nd table mentioned in part (a).

We'll apply the reciprocal to each item in the column labeled T-24.

For instance, the item 57 becomes 1/57 = 0.017544 approximately.

This is what the data table looks like:

`\begin{array}c \cline{1-2}t & 1/(T-24)\\\cline{1-2}0 & 0.017544\\\cline{1-2}5 & 0.022222\\\cline{1-2}10 & 0.027397\\\cline{1-2}15 & 0.033113\\\cline{1-2}20 & 0.039526\\\cline{1-2}25 & 0.046729\\\cline{1-2}30 & 0.053763\\\cline{1-2}\end{array}`

The values in the 2nd column are approximate.

The set of points and regression line are shown in figure C1. This line does not go through all of the points. Like with the others, it tries to get as close as possible to each point.

Use technology to determine the equation of the regression line is approximately:

y = 0.00121x + 0.01613

Which tells us that

1/(T-24) = 0.00121t + 0.01613

Let's solve for uppercase T.

1/(T-24) = 0.00121t + 0.01613

T-24 = 1/(0.00121t + 0.01613)

T = 1/(0.00121t + 0.01613) + 24

T = 1/(0.0012t + 0.0161) + 24

which is the approximate final answer.

Figure C2 shown below represents the original data points (t, T) in red. The blue curve is y = 1/(0.0012x+0.0161)+24, which is the rational regression curve we just found. This curve tries its best to get as close as possible to each red point.

Compare the regression curves of figure A and figure C2. We have a very close tie in terms of which curve does a better job of fitting the original data.