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Sullivan  04 apcalc4e 45342 ch02 166 233 5pp  August 7, 2023  12:54



               210     Chapter 2 • The Derivative and Its Properties

                                                    EXAMPLE 8 Finding Higher-Order Derivatives

                                                                                              x
                                                                                           2
                                                   Find the second and third derivatives of y = (1 + x )e .
                                                   Solution
                                                                                            x
                                                                                                2
                                                                                                           x
                                                                                     x
                                                                                   2
                                                   In Example 1, we found that y = (1 + x )e + 2xe = (x + 2x + 1)e . To find y , use
                                                                                                                    ′′
                                                                           ′
                                                   the Product Rule with y .
                                                                      ′
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                                                            d                                d       d
                                                         ′′      2         x     2             x         2         x
                                                        y =   [(x + 2x + 1)e ] = (x + 2x + 1)  e  +    (x + 2x + 1) e
                                                            dx               ↑              dx      dx
                                                                         Product Rule
                                                              2
                                                                                       2
                                                                                  x
                                                                        x
                                                          = (x + 2x + 1)e + (2x + 2)e = (x + 4x + 3)e x
                                                            d                             d        d
                                                        ′′′      2         x     2           x       2          x
                                                       y =    [(x + 4x + 3)e ] = (x + 4x + 3)  e +  (x + 4x + 3) e
                                                            dx               ↑            dx     dx
                                                                         Product Rule
                                                              2
                                                                                       2
                                                                                  x
                                                                        x
                                                          = (x + 4x + 3)e + (2x + 4)e = (x + 6x + 7)e x
                                                                             R
                                                    NOW WORK   Problem 45 and AP Practice Problems 9, 10, and 11.
                                                    4 Find the Acceleration of an Object Moving on a Line
                                                   For an object moving on a line whose signed distance s from the origin at time t
                                                   is the position function s = s(t), the derivative s (t) has a physical interpretation as the
                                                                                          ′
                                                   velocity of the object. The second derivative s , which is the rate of change of the
                                                                                          ′′
                                                   velocity, is called acceleration.
                                                    DEFINITION Acceleration
                                                    For an object moving on a line, its signed distance s from the origin at time t is given
                                                                                             ds
                                                    by a position function s = s(t). The first derivative  is the velocity v = v(t) of the
                                                                                             dt
                                                    object at time t.
                                                        The acceleration a = a(t) of an object at time t is defined as the rate of change
                                                    of velocity with respect to time. That is,
                                                                                                    2
                                                                               dv   d    d    ds     d s
                IN WORDS Acceleration is the second                  a = a(t) =  =    v =        =   2
                                                                               dt  dt    dt  dt    dt
                 derivative of a position function with respect
                 to time.

                                                    EXAMPLE 9 Analyzing Vertical Motion

                                                   A ball is propelled vertically upward from the ground with an initial velocity
                                                   of 29.4 m/s. The height s (in meters) of the ball above the ground is
                                                                            2
                                                   approximately s = s(t) = −4.9t + 29.4t, where t is the number of seconds that elapse
                                                   from the moment the ball is released.

                                                   (a) What is the velocity of the ball at time t? What is its velocity at t = 1 s?
                                                   (b) When will the ball reach its maximum height?
                                                   (c) What is the maximum height the ball reaches?
                                                   (d) What is the acceleration of the ball at any time t?
                                                   (e) How long is the ball in the air?
                                                   (f) What is the velocity of the ball upon impact with the ground?
                                                      What is its speed?
                                                   (g) What is the total distance traveled by the ball?
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