Page 175 - How To Solve Word Problems In Calculus
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EXAMPLE 9
                               According to Newton’s law of cooling, the temperature of an
                               object changes at a rate proportional to the difference in tem-
                               perature between the object and the outside medium. If an ob-
                                                           ◦
                               ject whose temperature is 70 Fahrenheit is placed in a medium
                                                        ◦
                               whose temperature is 20 and is found to be 40 after 3 min-
                                                                               ◦
                               utes, what will its temperature be after 6 minutes?
                                   Solution
                                   If u(t) represents the temperature of the object at time t,
                               du
                                  represents its rate of change. Newton’s law of cooling may
                               dt
                                          du
                               be written    = k(u − 20).
                                          dt
                                                         1   du
                                                                = k
                                                      u − 20 dt
                                                   d
                                                      ln(u − 20) = k
                                                   dt

                                                      ln(u − 20) = kt + C

                               We can solve for C by observing that u(0) = 70

                                                  ln(70 − 20) = k · 0 + C

                                                            C = ln 50

                               Next we determine u(t).

                                                      ln(u − 20) = kt + ln 50

                                              ln(u − 20) − ln 50 = kt

                                                        u − 20
                                                    ln           = kt
                                                          50
                                                         u − 20     kt
                                                                 = e
                                                           50
                                                             u(t) = 20 + 50e kt

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