AP calculus AB scoring guidelines from the 2019 exam administration

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

2019 AP Calculus AB 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 AB 2019 SCORING GUIDELINES Question  : vP  2.8   vP  0.3   :  : justification, using  Mean Value Theorem (a) vP is differentiable  vP is continuous on 0.3  t  2.8 vP  2.8   vP  0.3 55  55  0 2.8  0.3 2.5 By the Mean Value Theorem, there is a value c, 0.3  c  2.8, such that vP  c   — OR — — OR — vP is differentiable  vP is continuous on 0.3  t  2.8  : vP  0.3  vP 1.7   and vP 1.7   vP  2.8  2:   : justification, using  Extreme Value Theorem By the Extreme Value Theorem, v p has a minimum on  0.3, 2.8 vP  0.3  55  29  vP 1.7  and vP 1.7   29  55  vP  2.8  Thus vP has a minimum on the interval  0.3, 2.8  Because vP is differentiable, vP  t  must equal at this minimum (b) 2.8 0 v    vP  0.3   vP  0.3  vP 1.7   vP  t  dt  0.3  P   1.4   2     vP 1.7   vP  2.8     1.1     29   55  29  55  55  0.3  1.4    1.1 2    40.75     (c) vQ  t   60  t  A  1.866181 or t  B  3.519174 vQ  t   60 for A  t  B B A : answer, using trapezoidal sum  : interval  :  : definite integral  : distance vQ  t  dt  106.108754 The distance traveled by particle Q during the interval A  t  B is 106.109 (or 106.108) meters (d) From part (b), the position of particle P at time t  2.8 is xP  2.8   2.8 0 vP  t  dt  40.75 xQ  2.8   xQ    2.8 0 vQ  t  dt  90  135.937653  45.937653 Therefore at time t  2.8, particles P and Q are approximately 45.937653  40.75  5.188 (or 5.187) meters apart © 2019 The College Board Visit the College Board on the web: collegeboard.org  : 2.8 v  t  dt 0 Q  :  : position of particle Q   : answer 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 AB 2019 SCORING GUIDELINES Question (a) 0    x  1    2  3cos  x    dx       x   x  1    2 x  sin  x       2   12                   3   h  x   g  x   dx   2 x2 x0  12  (b) The area of R is 44 2 0 A x  dx   0 x   : integrand    : antiderivative of 3cos x  4:  : antiderivative of  remaining terms   : answer   2 44 4  3 dx 2: x2   ln  x  3 x   ln  ln  : integral : answer The volume of the solid is ln  ln   (c)     g  x  2    h x  2 dx  : limits and constant  :  : form of integrand  : integrand © 2019 The College Board Visit the College Board on the web: collegeboard.org AP® CALCULUS AB 2019 SCORING GUIDELINES Question (a) h   : answer (b) a x   x h x   3x3h x   : form of product rule  :  : a x   : a  a    22 h    23 h   36   24   160 (c) Because h is differentiable, h is continuous, so lim h x   h   x2 Also, lim h x   lim x2  2 x 4 x 4 , so lim  x    f  x  3  f  x  x 2     Because lim x   0, we must also have lim   f  x  3  x2 x2 Thus lim f  x    x2  4  : xlim    f  x  3  :  : f  2  : L’Hospital’s Rule   : f   x2 Because f is differentiable, f is continuous, so f    lim f  x   x 2 Also, because f is twice differentiable, f  is continuous, so lim f  x   f   exists x2 Using L’Hospital’s Rule, x2  2x lim  lim   x    f  x  3 x  3  f  x   f  x  3 1  f   Thus f     (d) Because g and h are differentiable, g and h are continuous, so lim g  x   g    and lim h x   h   x2 : continuous with justification x2 Because g  x   k  x   h x  for  x  3, it follows from the squeeze theorem that lim k  x   x2 Also,  g    k    h   4, so k    Thus k is continuous at x  © 2019 The College Board Visit the College Board on the web: collegeboard.org  ... 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 AB 2019 SCORING GUIDELINES Question  : vP  2.8   vP  0.3.. .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