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Appendix C C-37<br />

PROJECT<br />

Coffee Temperature<br />

Suppose that a cup of hot coffee at a temperature of T 0 is set down to cool in a<br />

room where the temperature is kept at T 1 . Then the temperature of the coffee as<br />

it cools can be modeled by the function<br />

f(t) (T 0 T 1 )e kt T 1<br />

(1)<br />

Here, f(t) is the temperature of the coffee after t minutes; t 0 corresponds to<br />

the initial instant when the temperature of the hot coffee is T 0 ; and k is a (negative)<br />

constant that depends, among other factors, on the dimensions of the cup<br />

and the material from which it is constructed. [This model is derived in calculus.<br />

It is based on Newton’s law of cooling: The rate of change of temperature<br />

of a cooling object is proportional to the difference between the temperature of<br />

the object and the surrounding temperature. (Note: y is proportional to x means<br />

y kx for some constant k.)]<br />

After studying the following example, solve Problem A below by applying<br />

equation (1) and using the technique shown in the example. Then try your hand<br />

at Problem B. You won’t be able to complete Problem B using the techniques<br />

developed in this text up to now. In complete sentences, explain exactly at<br />

which point you get stuck. What would you have to know how to do to complete<br />

the solution? (In the next section we discuss logarithms, which will allow<br />

us to complete Problem B.) For now, use a graphing utility to obtain an approximate<br />

answer for Problem B.<br />

EXAMPLE<br />

Coffee Temperature Project<br />

Given that the graph of the function y ae kx passes through the points (0, 2)<br />

and (3, 5), find y when x 7.<br />

SOLUTION<br />

Substituting x 0 and y 2 in the given equation yields 2 ae 0 , and therefore<br />

a 2. Next, substituting x 3 and y 5 in the equation y 2e kx yields<br />

5 2e k(3) , or e 3k 2.5. This last equation can be rewritten (e k ) 3 2.5. Taking<br />

the cube root of both sides gives us<br />

e k 2.5 13<br />

(2)<br />

The original function can now be written as follows<br />

y 2e kx 2(e k ) x<br />

2(2.5 13 ) x<br />

using equation (2)

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