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SCORING GUIDELINES Question On a certain workday, the rate, in tons per hour, at which unprocessed gravel arrives at a gravel processing plant t is modeled by Gt () = 9 + 45cos, where t is measured in hours and. t At the beginning of the workday ( t =, ) the plant has 5 tons of unprocessed gravel. During the hours of operation, t, the plant processes gravel at a constant rate of tons per hour. (a) Find G ( 5. ) Using correct units, interpret your answer in the context of the problem. (b) Find the total amount of unprocessed gravel that arrives at the plant during the hours of operation on this workday. (c) Is the amount of unprocessed gravel at the plant increasing or decreasing at time t = 5 hours? Show the work that leads to your answer. (d) What is the maximum amount of unprocessed gravel at the plant during the hours of operation on this workday? Justify your answer. (a) G ( 5) = 4.5 (or 4. 57) The rate at which gravel is arriving is decreasing by 4.5 (or 4.57) tons per hour per hour at time t = 5 hours. : G ( 5) : : interpretation with units tons : { : integral (b) G() t dt = 5.55 : answer (c) G ( 5) = 9.474 < At time t = 5, the rate at which unprocessed gravel is arriving is less than the rate at which it is being processed. Therefore, the amount of unprocessed gravel at the plant is decreasing at time t = 5. (d) The amount of unprocessed gravel at time t is given by t At () = 5 + ( Gs ( ) ) ds. A () t = G() t = t = 4.94 : compares G( 5 ) to : : conclusion : : considers A () t = : answer : justification t At () 5 4.94 5.7 55.559 The maximum amount of unprocessed gravel at the plant during this workday is 5.7 tons. The College Board.
SCORING GUIDELINES Question The graphs of the polar curves r = and r = 4 sin are shown in the figure above. The curves intersect when = and (a) Let S be the shaded region that is inside the graph of r = and also inside the graph of r = 4 sin. Find the area of S. = (b) A particle moves along the polar curve r = 4 sin so that at time t 5. seconds, = t. Find the time t in the interval t for which the x-coordinate of the particle s position is. (c) For the particle described in part (b), find the position vector in terms of t. Find the velocity vector at time t =.5. 5 (a) Area = + ( 4 sin ) d = 4.79 (or 4.7) : : integrand : limits and constant : answer (b) x = r cos x( ) = ( 4 sin ) cos : x( ) or x( t) xt () = ( 4 sin ( t )) cos( t ) : : x( ) = or x( t) = x() t = when t =.4 (or.47) : answer (c) y = r sin y( ) = ( 4 sin ) sin () = ( 4 sin ( t )) sin ( t ) yt : position vector : { : velocity vector Position vector = xt (), yt () ( 4 sin ( t )) cos( t ),( 4 sin ( t )) sin ( t ) = v(.5) = x (.5 ), y (.5) =.7,.7 ( or.7,.7 ) The College Board.
SCORING GUIDELINES Question t (minutes) Ct () (ounces) 4 5 5..... 4.5 Hot water is dripping through a coffeemaker, filling a large cup with coffee. The amount of coffee in the cup at time t, t, is given by a differentiable function C, where t is measured in minutes. Selected values of Ct (), measured in ounces, are given in the table above. (a) Use the data in the table to approximate C (.5 ). Show the computations that lead to your answer, and indicate units of measure. (b) Is there a time t, t 4, at which C () t =? Justify your answer. (c) Use a midpoint sum with three subintervals of equal length indicated by the data in the table to approximate the value of C() t dt. Using correct units, explain the meaning of () C t dt in the context of the problem..4t (d) The amount of coffee in the cup, in ounces, is modeled by Bt () = e. Using this model, find the rate at which the amount of coffee in the cup is changing when t = 5. C( 4) C( ).. (a) C (.5) = 4 =. ounces min : { : approximation : units (b) C is differentiable C is continuous (on the closed interval) C( 4) C( ).. = = 4 Therefore, by the Mean Value Theorem, there is at least one time t, < t < 4, for which C () t =. C( 4) C( ) : : 4 : conclusion, using MVT (c) () () ( ) ( 5) C t dt C + C + C = ( 5. +. +. ) = (. ) =. ou n ce s () C t dt is the average amount of coffee in the cup, in ounces, over the time interval t minutes. t (d) B () t = (.4) e =.4e.4.4t.4( 5).4 B ( 5) =.4 e = ounces min e : : : midpoint sum : approximation : interpretation : B () t : B ( 5) The College Board.
SCORING GUIDELINES Question 4 The figure above shows the graph of f, the derivative of a twice-differentiable function f, on the closed interval x. The graph of f has horizontal tangent lines at x =, x =, and x = 5. The areas of the regions between the graph of f and the x-axis are labeled in the figure. The function f is defined for all real numbers and satisfies f ( ) = 4. (a) Find all values of x on the open interval < x < for which the function f has a local minimum. Justify your answer. (b) Determine the absolute minimum value of f on the closed interval x. Justify your answer. (c) On what open intervals contained in < x < is the graph of f both concave down and increasing? Explain your reasoning. (d) The function g is defined by ( ) ( ( )) 5 g x = f x. If f ( ) =, find the slope of the line tangent to the graph of g at x =. (a) x = is the only critical point at which f changes sign from negative to positive. Therefore, f has a local minimum at x =. (b) From part (a), the absolute minimum occurs either at x = or at an endpoint. f( ) = f( ) + f = f ( ) f = 4 = f( ) = f( ) + f f ( ) = 4 = f ( ) f = 4 7 = The absolute minimum value of f on the closed interval, is. (c) The graph of f is concave down and increasing on < x < and < x < 4, because f is decreasing and positive on these intervals. (d) g ( x ) = f f 5 ( ) ( ) g ( ) = f ( ) f = 4 = 75 : answer with justification : : considers x = and x = : answer : justification : { : answer : explanation : g : : answer The College Board.
SCORING GUIDELINES Question 5 dy Consider the differential equation y ( x ). = + Let y = f be the particular solution to the differential equation with initial condition f ( ) =. (a) Find f + lim. Show the work that leads to your answer. x sin x (b) Use Euler s method, starting at. x = with two steps of equal size, to approximate f ( ) (c) Find y = f, the particular solution to the differential equation with initial condition f ( ) =. (a) lim ( f + ) = + = and lim sin x = x x Using L Hospital s Rule, f + f f ( ) ( ) lim = lim = = = x sin x x cos x cos : L Hospital s Rule : { : answer (b) f( ) f( ) + f ( 4 )( 4) = + ( )( ) = 4 : Euler s method : { : answer ( ) ( ) + 4 ( 4 )( 4) ( ) ( )( ) f f f = + + = 4 4 dy (c) = y ( x + ) dy = ( x + ) y dy = ( x + ) y = x + x + C y = + + C C = = x + x + y y = = x + x + ( x + ) 5 : : separation of variables : antiderivatives : constant of integration : uses initial condition : solves for y Note: max 5 [----] if no constant of integration Note: 5 if no separation of variables The College Board.
Note: This solution is valid for x. AP CALCULUS BC SCORING GUIDELINES Question 5 The College Board.
SCORING GUIDELINES Question A function f has derivatives of all orders at x =. Let Pn ( x ) denote the nth-degree Taylor polynomial for f about x =. (a) It is known that ( ) 4 f = and that ( ) P =. Show that f ( ) =. (b) It is known that f ( ) = and f ( ) =. Find P ( x ) (c) The function h has first derivative given by h = f( x). It is known that h ( ) = 7. Find the third-degree Taylor polynomial for h about x =.. (a) P = f( ) + f ( ) x = 4 + f ( ) x ( ) f ( ) P = 4 + = f ( ) = f ( ) = : : uses P : verifies f ( ) = x x 4 +!! (b) P = + x + ( ) = 4 + x x + x : : first two terms : third term : fourth term (c) Let Q ( x ) denote the Taylor polynomial of degree n for h about n x =. h = f Q ( ) ( ) ( ) x = 4 + x x x 4 x Q = 4x + 4 + C; C = Q( ) = h( ) = 7 4 Q = 7 4x + x x 9 4 : : applies h = f : constant term : remaining terms OR h = f( x), h = f ( x), h = 4f h ( ) = f( ) = 4, h ( ) = f ( ) = 4, h ( ) = 4f ( ) = x x 4 Q = 7 4x + 4 = 7 4x + x x!! 9 The College Board.