Page 167 - Calculus Workbook For Dummies
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151
                                                    Chapter 8: Using Differentiation to Solve Your Problems



                    m    s t =  t +  t -  t
                               4
                                  2
                          ^ h
                                                            3
                         a. Find the zeros of v t ^ h: v t = l ^h  s t =  t 4 +  t 2 -  1
                                                       h
                                                ^
                           You’ll need your calculator for this:
                                      3
                           Graph  y =  4 x +  2 x - and locate the x-intercepts. There’s just one: x .  .385. That’s the only
                                            1
                                  l ^
                           zero of s t = ^h  v th.
                           Don’t forget that a zero of a derivative can be a horizontal inflection as well as a local
                           extremum. You get a turnaround point only at the local extrema.
                                          1
                           Because v 0 = - (a leftward velocity) and v 1 =  5 (a rightward velocity), s ^ .385h must be
                                    ^ h
                                                                  ^ h
                           a turnaround point (and it’s also a local min on the position graph). Does the first derivative
                           test ring a bell?
                           Thus, the sloth is going left from t = 0 sec to t = .385 sec and right from .385 to 5 sec. He
                                                                                   4
                                                                                          2
                                                                             h
                           turns around, obviously, at .385 sec when he is at s ^ .385 =  .385 +  .385 -  .385 or –.215
                           meters — that’s .215 meters to the left of the trunk. I presume you figured out that there
                           must be another branch on the tree on the other side of the trunk to allow the sloth to go left
                           to a negative position.
                         b. There are two legs of the sloth’s trip. He goes left from t = 0 till t = .385, then right from t = .385
                           till t = 5. Just add up the positive lengths of the two legs.
                           length leg 1 =  s ^ .385 - ^h  s 0h
                                  = - .215 -  0
                                  =  .215  meters
                           length leg 2 =  s 5 - ^ .385h
                                          s
                                      ^ h
                                         2
                                      4
                                           5
                                  =  5 +  5 - - - .215h
                                              ^
                                  =  645 .215  meters
                           The total distance is thus .215 + 645.215, or 645.43 meters. That’s one big tree! The branch
                           is over 2000 feet long.
                           His average speed is 645.43 / 5, or about 129.1 meters/second. That’s one fast sloth!
                           Almost 300 miles/hour!
                                                ^
                         c. Total displacement is s 5 - ^h  s 0h, that’s 645 – 0 = 645 meters. Lastly, his average velocity
                           is simply total displacement divided by total time — that’s 645/5, or 129 meters/second.
                    n    s t =  t +  1
                          ^ h
                               2
                              t +  4
                         a. Find the zeros of v t ^ h:
                                          1 l
                                              2
                                                           2
                                                        1 _
                                       ^ t + h  _ t + i  t + h  t +  4i l
                                                4 - ^
                            l ^
                                    h
                           s t = ^h  v t =            2
                                                  2
                                                _ t +  4i
                                        2       2
                                          4
                                       t + - _  t 2 +  t 2 i
                                     =          2
                                            2
                                           _ t +  4i
                                         2
                                        t - -  t 2 +  4
                                     =        2
                                          2
                                        _ t +  4i
                           Set this equal to zero and solve:
                               2
                              t - -  t 2 +  4
                                    2  =  0
                               2
                              _ t +  4i
                               2
                                     4
                              t +  t 2 - =  0
                                        - 2 !  4 - - 16h
                                                 ^
                                     t =
                                              2
                                      . -  . 3 236 or  . 1 236
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