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AP CALCULU AB 5 CORING GUIDELINE (Form B) Question 1 Let f and g be the functions given b f ( ) = 1+ sin( ) and g( ) = e. Let R be the shaded region in the first quadrant enclosed b the graphs of f and g as shown in the figure above. (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the -ais. (c) The region R is the base of a solid. For this solid, the cross sections perpendicular to the -ais are semicircles with diameters etending from = f( ) to = g( ). Find the volume of this solid. The graphs of f and g intersect in the first quadrant at (, T ) = ( 1.1569, 1.76446 ). 1 : correct limits in an integral in (a), (b), or (c) (a) Area = ( f( ) g( ) ) d = ( 1+ sin( ) e ) d =.49 1 : integrand : ( ) (b) Volume = π ( f( ) ) ( g( ) ) d = π (( 1+ sin( ) ) ( e ) ) d = 4.66 or 4.67 : integrand 1 each error Note: if integral not of form : b c ( R ( ) r ( ) ) d a (c) Volume π f( ) g( ) = π 1+ sin( ) e = =.77 or.78 d d : integrand : Copright 5 b College Board. All rights reserved.
AP CALCULU AB 5 CORING GUIDELINE (Form B) Question A water tank at Camp Newton holds 1 gallons of water at time t =. During the time interval t 18 hours, water is pumped into the tank at the rate t () 95 tsin ( ) W t = gallons per hour. 6 During the same time interval, water is removed from the tank at the rate t () 75sin ( ) Rt = gallons per hour. (a) Is the amount of water in the tank increasing at time t = 15? Wh or wh not? (b) To the nearest whole number, how man gallons of water are in the tank at time t = 18? (c) At what time t, for t 18, is the amount of water in the tank at an absolute minimum? how the work that leads to our conclusion. (d) For t > 18, no water is pumped into the tank, but water continues to be removed at the rate R() t until the tank becomes empt. Let k be the time at which the tank becomes empt. Write, but do not solve, an equation involving an integral epression that can be used to find the value of k. (a) No; the amount of water is not increasing at t = 15 since W( 15) R( 15) = 11.9 <. 1 : answer with reason 18 (b) 1 + ( W() t R() t ) dt = 19.788 11 gallons : 1 : limits 1 : integrand (c) W() t R() t = t =, 6.4948, 1.9748 t (hours) gallons of water 1 6.495 55 1.975 1697 18 11 1 : interior critical points 1 : amount of water is least at : t = 6.494 or 6.495 1 : analsis for absolute minimum The values at the endpoints and the critical points show that the absolute minimum occurs when t = 6.494 or 6.495. k (d) Rt () dt= 11 18 1 : limits : 1 : equation Copright 5 b College Board. All rights reserved.
AP CALCULU AB 5 CORING GUIDELINE (Form B) Question A particle moves along the -ais so that its velocit v at time t, for t 5, is given b () ( ) vt = ln t t +. The particle is at position = 8 at time t =. (a) Find the acceleration of the particle at time t = 4. (b) Find all times t in the open interval < t < 5 at which the particle changes direction. During which time intervals, for t 5, does the particle travel to the left? (c) Find the position of the particle at time t =. (d) Find the average speed of the particle over the interval t. 5 (a) a( 4) = v ( 4) = 1 : answer 7 (b) vt () = t t + = 1 t t + = ( t ) ( t 1) = t = 1, 1 : sets vt () = : 1 : direction change at t = 1, 1 : interval with reason vt () > for < t < 1 vt () < for 1 < t < vt () > for < t < 5 The particle changes direction when t = 1 and t =. The particle travels to the left when 1 < t <. t (c) () = ( ) + ln( + ) ( ) = 8 + ln( + ) s t s u u du s u u du = 8.68 or 8.69 : ( ) 1 : ln u u + du 1 : handles initial condition 1 (d) () vt dt=.7 or.71 1 : integral : Copright 5 b College Board. All rights reserved. 4
AP CALCULU AB 5 CORING GUIDELINE (Form B) Question 4 The graph of the function f above consists of three line segments. (a) Let g be the function given b g( ) = f( t) dt. 4 For each of g( 1, ) g ( 1, ) and g ( 1, ) find the value or state that it does not eist. (b) For the function g defined in part (a), find the -coordinate of each point of inflection of the graph of g on the open interval 4 < <. Eplain our reasoning. (c) Let h be the function given b h ( ) = f( tdt ). Find all values of in the closed interval 4 for which h ( ) =. (d) For the function h defined in part (c), find all intervals on which h is decreasing. Eplain our reasoning. 1 1 15 (a) g( 1) = f( t) dt = ( )( 5) = 4 g ( 1) = f( 1) = g ( 1) does not eist because f is not differentiable at = 1. : 1 : g( 1) 1 : g ( 1) 1 : g ( 1) (b) = 1 g = f changes from increasing to decreasing at = 1. 1 : = 1 (onl) : 1 : reason (c) = 1, 1, : correct values 1 each missing or etra value (d) h is decreasing on [, ] h = f < when f > 1 : interval : 1 : reason Copright 5 b College Board. All rights reserved. 5
AP CALCULU AB 5 CORING GUIDELINE (Form B) Question 5 Consider the curve given b = +. d (a) how that =. d (b) Find all points (, ) on the curve where the line tangent to the curve has slope 1. (c) how that there are no points (, ) on the curve where the line tangent to the curve is horizontal. (d) Let and be functions of time t that are related b the equation d value of is and = 6. Find the value of d at time t = 5. dt dt = +. At time t = 5, the (a) = + ( ) = = 1 : implicit differentiation : 1 : solves for (b) 1 = = = =± (, ), (, ) 1 1 : = : (c) = = The curve has no horizontal tangent since + for an. 1 : = : 1 : eplanation (d) When =, dt t = 5 = + so = d d d d = = dt d dt dt d 9 d At t = 5, 6 = = 7 dt 11 dt 6 d = 7. : 1 : solves for 1 : chain rule Copright 5 b College Board. All rights reserved. 6
AP CALCULU AB 5 CORING GUIDELINE (Form B) Question 6 d Consider the differential equation =. Let d = f( ) be the particular solution to this differential equation with the initial condition f ( 1) =. (a) On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the test booklet.) (b) Write an equation for the line tangent to the graph of f at = 1. (c) Find the solution = f( ) to the given differential equation with the initial condition f ( 1) =. (a) 1 : zero slopes : 1 : nonzero slopes ( 14 ) (b) lope = = = ( + 1) 1 : equation (c) 1 d = d 1 = + C 4 1 1 1 = + C; C = 4 4 1 4 = = 1 + 1 + 4 4 6 : 1 : separates variables : antiderivatives 1 : constant of integration 1 : uses initial condition 1 : solves for Note: ma 6 [1----] if no constant of integration Note: 6 if no separation of variables Copright 5 b College Board. All rights reserved. 7