AP calculus BC scoring guidelines from the 2019 exam administration

AP Calculus BC Scoring Guidelines from the 2019 Exam Administration AP ® Calculus BC Scoring Guidelines 2019 © 2019 The College Board College Board, Advanced Placement, AP, AP Central, and the acorn l[.]

2019 AP Calculus BC đ Scoring Guidelines â 2019 The College Board College Board, Advanced Placement, AP, AP Central, and the acorn logo are registered trademarks of the College Board Visit the College Board on the web: collegeboard.org AP Central is the official online home for the AP Program: apcentral.collegeboard.org AP® CALCULUS AB/CALCULUS BC 2019 SCORING GUIDELINES Question (a) 2:  : integral : answer 2:  : integral : answer 0 E  t  dt  153.457690 To the nearest whole number, 153 fish enter the lake from midnight to A.M (b) L t  dt  6.059038  50 The average number of fish that leave the lake per hour from midnight to A.M is 6.059 fish per hour (c) The rate of change in the number of fish in the lake at time t is given by E  t   L t  E  t   L t    t  6.20356  : sets E  t   L t    :  : answer  : justification E  t   L t   for  t  6.20356, and E  t   L t   for 6.20356  t  Therefore the greatest number of fish in the lake is at time t  6.204 (or 6.203) — OR — Let A t  be the change in the number of fish in the lake from midnight to t hours after midnight A t   t 0  E  s   L s   ds A t   E  t   L t    t  C  6.20356 t C A t  135.01492 80.91998 Therefore the greatest number of fish in the lake is at time t  6.204 (or 6.203) (d) E    L   10.7228  Because E    L   0, the rate of change in the number of fish is decreasing at time t   : considers E   and L  2:  : answer with explanation © 2019 The College Board Visit the College Board on the web: collegeboard.org AP® CALCULUS BC 2019 SCORING GUIDELINES Question (a) 0   r   2 d  3.534292 2:  : integral : answer 2:  : integral : answer The area of S is 3.534 (b)  r   d  1.579933   0 The average distance from the origin to a point on the curve r  r   for     is 1.580 (or 1.579) (c) tan   tan 0 1 y  m    tan 1 m x m   r   2 d     r   2 d  2  (d) As k  , the circle r  k cos  grows to enclose all points to the right of the y -axis   r   2 d    sin     : equates polar areas  : inverse trigonometric function  3:  applied to m as limit of  integration   : equation 2:  : limits of integration : answer with integral lim A k   k     d  3.324 © 2019 The College Board Visit the College Board on the web: collegeboard.org AP® CALCULUS AB/CALCULUS BC 2019 SCORING GUIDELINES Question (a) 2  7  (b) 6 f  x  dx  6 f  x  dx  2 f  x  dx  f  x  dx     2 6 f  x  dx   11  2 9   9 9  4 4  : f  x  dx  6   3:   : 2 f  x  dx  : answer  3  f  x    dx  23 f  x  dx  3 dx   f    f  3     3 2:  2 6 f  x  dx  2 f  x  dx : Fundamental Theorem of Calculus : answer    3      3      — OR — x 5 3  f  x    dx   f  x   x x 3   f    20    f  3  12      20        12  22 (c) g  x   f  x    x  1, x  , x  g x x 2 1 1  9 11   : g  x   f  x   :  : identifies x  1 as a candidate  : answer with justification On the interval 2  x  5, the absolute maximum value 9 of g is g    11  (d) lim x 1 10 x  f  x  101  f 1  f  x   arctan x f 1  arctan 10      arctan  1 : answer © 2019 The College Board Visit the College Board on the web: collegeboard.org AP® CALCULUS AB/CALCULUS BC 2019 SCORING GUIDELINES Question (a) V   r h   12 h   h dV dh    cubic feet per second    dt h  dt h  10     (b) d h    dh     h  10 200 20 h dt 20 h dt 2 d h Because   for h  0, the rate of change of the 200 dt height is increasing when the height of the water is feet (c) dh   dt 10 h  dh    dt  h  10 h   tC 10   0  C  C  10 h  t2 10 h t    t  20    : dV   dh 2:  dt dt  : answer with units   1: d  h    dh 10 20 h  3:  d h dh 1:     dt h 20 dt   : answer with explanation  : separation of variables  : antiderivatives  :  : constant of integration  and uses initial condition   : h t  Note: if no separation of variables Note: max [1-1-0-0] if no constant of integration © 2019 The College Board Visit the College Board on the web: collegeboard.org AP® CALCULUS BC 2019 SCORING GUIDELINES Question (a) f  x   f    (b)    2x  2 x  2x  k   : denominator of f  x   :  : f  x   : answer 2 1   k2   k  3 k 1 A B    x x  2 x x      x  2x   : partial fraction  decomposition 3:   : antiderivatives  : answer   A x  2  B  x  4 1  A , B 6 0 1    f  x  dx    dx    0  x  x   x 1 1   ln x   ln x     x  1 1 ln  ln  ln  ln   ln  6 6     2 1 1 dx   dx   dx   dx    2 0 x  x  0  x  1 0  x  1 1  x  12  (c)   lim    b 1  Because lim   b 1 b 1 dx  lim  dx  b 1 b  x  1 0  x  12   x b  x2   lim    lim     b 1  x  x   b 1  x  x  b  1  lim    lim 1    b b  1 b 1 b 1      does not exist, the integral diverges b 1 © 2019 The College Board Visit the College Board on the web: collegeboard.org  : improper integral  :  : antiderivative  : answer with reason AP® CALCULUS BC 2019 SCORING GUIDELINES Question (a) f    and f    2 2: The third-degree Taylor polynomial for f about x  is 23  3 23 3  2x  x  x   x  x2  x 2! 3! 12 (b) The first three nonzero terms of the Maclaurin series for e x are 1  x  x2 2!  : two terms : remaining terms  : three terms for e x 2:   : three terms for e x f  x  The second-degree Taylor polynomial for e x f  x  about x  is 3  x  x  x 1  x   x 1 2! 3   3  2 x  2 x 2       x  x2 (c) h1  0 f  t  dt     2t  t  23 t dt  12 0 2:  : antiderivative : answer 23  t 1  3t  t  t  t  48  t  23 97  1   48 48 (d) The alternating series error bound is the absolute value of the first omitted term of the series for h1   t 1   54 t dt   t    20  t  20 0 4! Error   : uses fourth-degree term  of Maclaurin series for f  :  : uses first omitted term  of series for h1   : error bound  0.45 20 © 2019 The College Board Visit the College Board on the web: collegeboard.org  ... lim A k   k     d  3.324 © 2019 The College Board Visit the College Board on the web: collegeboard.org AP? ? CALCULUS AB /CALCULUS BC 2019 SCORING GUIDELINES Question (a) 2  7  (b)... 10      arctan  1 : answer © 2019 The College Board Visit the College Board on the web: collegeboard.org AP? ? CALCULUS AB /CALCULUS BC 2019 SCORING GUIDELINES Question (a) V   r h  ... and L  2:  : answer with explanation © 2019 The College Board Visit the College Board on the web: collegeboard.org AP? ? CALCULUS BC 2019 SCORING GUIDELINES Question (a) 0   r   2 d