Page 164 - How To Solve Word Problems In Calculus
P. 164

fundamental role in problems involving exponential growth
                                                                x
                                and decay. Observe that lim e = 0 and lim e     −x  = 0.
                                                         x→−∞             x→∞
                                             20                              20

                                             15                              15

                                             10                              10
                                             5                                5


                                −3  −2   −1        1    2    3  −3   −2   −1       1    2    3
                                     f (x) = e x  −3 ≤ x ≤ 3         f (x) = e −x  −3 ≤ x ≤ 3

                                Exponential-Growth and Decay
                                If a substance grows or decays at a rate proportional to its size,
                                its growth may be described by the equation
                                                           dy
                                                              = ky
                                                           dt
                                where y is the amount of the substance and k is the constant
                                of proportionality. Positive values of k correspond to growth
                                and negative values correspond to decay.
                                    To solve for y, we rewrite the equation in the form
                                                           1 dy
                                                                = k
                                                           y dt

                                from which it follows by the chain rule that

                                                          d
                                                            ln y = k
                                                          dt

                                Antidifferentiating, we get

                                                        ln y = kt + C


                                If y = y 0 when t = 0, it follows that ln y 0 = C and we may write




                                                                                        151
   159   160   161   162   163   164   165   166   167   168   169