Page 165 - How To Solve Word Problems In Calculus
P. 165

ln y = kt + ln y 0
                                                  ln y − ln y 0 = kt

                                                           y
                                                        ln    = kt
                                                           y 0
                                                           y
                                                              = e  kt
                                                           y 0

                                                            y = y 0 e kt

                               If the value of y 0 is known and the value of k can be deter-
                               mined, we can represent the amount of the substance as a
                               function of time.

                               EXAMPLE 1

                               A bacteria culture has an initial population of 500. After
                               4 hours the population has grown to 1000. Assuming the
                               culture grows at a rate proportional to the size of the popula-
                               tion, find a function representing the population size after t
                               hours and determine the size of the population after 6 hours.

                                   Solution
                                   Let y(t) represent the size of the bacteria population after
                               t hours. Since the rate of growth of the population is pro-
                               portional to the population size, it follows from the above
                               discussion that
                                                       y(t) = y 0 e  kt

                                                       y(t) = 500e  kt

                               Since y(4) = 1000, it follows that


                                            1000 = 500e 4 k

                                                2 = e 4 k
                                                                      Recall that y = ln x ⇔ x = e  y
                                              4k = ln 2
                                                    1
                                                k =   ln 2
                                                    4
                               152
   160   161   162   163   164   165   166   167   168   169   170