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AP@CALCULUS AB 2003 SCORING GUIDELINES (Form B) Question 4 A particle moves along the x-axis with velocity at time t >0 APR8S SOLUTIONS given by v ( t ) = -1 + elpt. (a) Find the acceleration of the particle at time t = 3. (b) Is the speed of thc particle increasing at time t = 3 ? Give a reason for your answer. (c) Find all values of t at which the particle changes direction. Justify your answer. (d) Find the total distance traveled by the particle over the time interval 0 5 t 5 3. (a) a(t) = v l ( t ) = -el-' a(3) = -eV2 1 : answer with reason (b) 4 3 ) < 0 v(3) = -1 + ep2 <0 Speed is increasing since v(3) < 0 and a(3) < 0 . (c) v ( t ) = 0 when 1 = e l p t , so t = 1. 1 : solves v ( t ) = 0 to get t = 1 v ( t ) > 0 for t < 1 and v ( t ) < 0 for t > 1 . 1 : justifies change in Thercfore, the particle changes dircction at t = 1. (d) Distance = iI3 1 v ( t )1 dt direction at t = 1 1 : limits 1 : integrand 1 : antidifferentiation 1 : evaluation x ( t ) = -t 1 : any antiderivative - el-" 1 : evaluates x ( t ) when x(0) = -e t = 0, 1, 3 x(1) = -2 x(3) = -eP2 - 1 : evaluates distance 3 Distance = ( x ( 1 )- x ( 0 ) ) = (-2 + e ) = e+ep2 + (x(1) - x(3)) between points + (1 + e - 2 ) APR8S Page 1 of 3 1 : evaluates total distance -1 Copyright O 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 5 AP@CALCULUS AB 2003 SCORING GUIDELINES (Form B) APR8S SOLUTIONS Question 5 Let f be a function defined on the closed interval [0,7]. The graph of f, consisting of four line segments, is shown above. Let g be the furiction given by g(i) = ST 2 f(t)dt. (a) Find g ( 3 ) , g1(3), and g1'(3). (b) Find the averagc rate of change of g on thc interval 0 5 x 5 3. (c) For how many values c, where 0 < c < 3, is gl(c) equal to the average rate found in part (b)? Explain your reasoning. Graph off (d) Find the %coordinate of each point of inflection of the graph of g on the interval 0 < z < 7. Justify your answer. (a) g(3) = J23 f ( t ) d t = -21( 4 + 2 ) = 3 g1(3) = f (3) = 2 0 -4 911(3) = f1(3) = -= -2 4 -2 2: ( 1 : g(3) - g(0) = Xf (t)dt 1 : answer [ 1 : answer of 2 (c) There are two values of c. 7 : W e need - = gl(c) = f(c) 3 7 The graph of f intersects the line y = - a t two 3 places between 0 and 3. 2: decreasing a t x = 2, and from decreasing t o increasing a t x = 5. 1 : reason Note: 112 if answer is 1 by MVT (d) x = 2 and x = 5 because g1 = f changes from increasing t o 1 1 1 : x = 2 and x = 5 only 1 : justification (ignore discussion a t x = 4) APR8S Page 2 of 3 Copyright O 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com. 6 AP@CALCULUS AB 2003 SCORING GUIDELINES (Form B) APR8S SOLUTIONS Question 6 Let f be the function satisfying f l ( z ) = z m for all real numbers z,where f ( 3 ) = 25 (a) Find f " ( 3 ) . dy = (b) Write an expression for y = f ( z ) by solving the differential equation dx 2J3 with the initial condition f ( 3 ) = 25. (a) fN(z)= m + z . - x2 2 : f ll(x) < -2 > product or chain rule error 1 : value at x = 3 1 (b) - d y = x d z : a 1 : separates variables 1 : antiderivative of dy term 1 : antiderivative of dx term 1 : constant of integration 1 : uses initial condition f(3) = 25 1 : solves for y Note: max 316 [l-1-1-0-0-01 if no constant of integration Note: 016 if no separation of variables APR8S Page 3 of 3 Copyright O 2003 by College Entrance Examination Board. All rights reserved. Available at apcentral.collegeboard.com.