Mark Howell Gonzaga High School, Washington, D.C.
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1 Be Prepared for the Calculus Exam Mark Howell Gonzaga High School, Washington, D.C. Martha Montgomery Fremont City Schools, Fremont, Ohio Practice exam contributors: Benita Albert Oak Ridge High School, Oak Ridge, Tennessee Thomas Dick Oregon State University Joe Milliet St. Mark's School of Texas, Dallas, Texas Skylight Publishing Andover, Massachusetts
2 Copyright 5- by Skylight Publishing Chapter. Annotated Solutions to Past Free-Response Questions This material is provided to you as a supplement to the book Be Prepared for the AP Calculus Exam (ISBN ). You are not authorized to publish or distribute it in any form without our permission. However, you may print out one copy of this chapter for personal use and for face-to-face teaching for each copy of the Be Prepared book that you own or receive from your school. Skylight Publishing 9 Bartlet Street, Suite 7 Andover, MA 8 web: sales@skylit.com support@skylit.com
3 AB AP Calculus Free-Response Solutions and Notes Question AB- 6 (a) f () t dt (b) f ( 8) g( 8) = ft /hr., for t < 6 ht = 5 t 6, for 6 t< , for 7 t 9 (c) () ( ) ( t ) 9 (d) f () t dt h( 9) ( ) = Students should expect that Part A of the free-response section of the exam (the calculator part) will continue to involve questions with a real-world context.
4 4 FREE-RESPONSE SOLUTIONS ~ AB Question AB- (a) The rate at which entries were being deposited (in hundreds of entries per hour) is, E( 7) E( 5) 8 approximately, = = = (b) The approximation is This represents the average number of hundreds of entries in the box over the 8 hours (c) E(8) P() t dt 8 = hundred entries. (d) The question asks for the maximum value of P(t). P ( t) = at t = 9.85 and t =.86. The maximum value of P(t) must occur at one of these, or at the endpoints, t = 8 or t =. P(9.85) = 5.89, P(.86) =.9, P(8) =, and P() = 8. The maximum is at t =.. No need to evaluate. For the curious, the value is ( 85.5 ) = The candidate test is the simplest way to find the maximum.
5 FREE-RESPONSE SOLUTIONS ~ AB 5 Question AB- (a) The number of people who arrived at the ride between t = and t = is r() t dt = + =. (b) For t < people arrive at the rate greater than 8 people/hour and are removed from the line at the constant rate of 8 people/hour. The rate of arrival is greater than the rate of removal, so the number of people in the line is increasing. (c) Let P() t be the number of people in the line at time t. Then P () t = r() t 8. P () t > for < t <, and P ( t) < for < t < 8. Therefore, P() t reaches its maximum at t =. The line is longest at t = ; at that time there are 7 + r() t dt 8 = = 5 people in the line. t (d) r( u) ( ) du =. Question AB-4 (a) Area = / 4 6 xdx = 6x x = =. ( 7 ) 9 (b) Volume = ( x ) π dx. 6 6 (c) Volume = ( ) 4 y x x dy = dy 6.. This is the second consecutive year that an area-volume problem has appeared in Part B, the closed calculator part, of the free response. Concern over calculator programs that can unfairly assist students in the solution of area-volume problems is one possible reason for this.. Not 9 6 ( x ) dx. The sections are perpendicular to the y-axis, not the x-axis.
6 6 FREE-RESPONSE SOLUTIONS ~ AB Question AB-5 g x = + g t dt, so g( ) = 5+ g ( t) dt 5 π = + + and g = 5+ g t dt = 5 π. x (a) ( ) 5 ( ) ( ) ( ) (b) x = x =, and vice-versa at these points. x =, because g ( x) (c) At a critical point, ( ) ( ) ( ) changes from increasing to decreasing or h x = g x x= g x = x. The line y = x intersects the semicircle where x > and x + y = x = 4 h x changes sign from positive to negative, so h(x) has a relative maximum there. x =. At that point, ( ) The line y = x also intersects the graph of y = g ( x) at x =. h ( x) does not change sign at x =, so there is neither a minimum nor a maximum there.. Use geometry to calculate the areas, not the Fundamental Theorem!. The integral from to is negative.
7 FREE-RESPONSE SOLUTIONS ~ AB 7 Question AB-6 (a) At the point ( ) dy,, 8 dx =. An equation of the tangent line is y 8 ( x ) d y =. (b) The approximation is y = + 8.=.8. Since > on the open interval dx (,.), the graph of f is concave up there, and the tangent line lies below the graph of y = f( x) for x. f.. (c) dy = x dx y. Thus the approximation is less than ( ) y = +. Substituting ( ) x C, we get / 5 5 C =. Solving for y, x 8 y = y = x or 4 Since y () = >, the particular solution we are looking for is / 5 y = x =. 4 5 x 4 8 = + C 5 y = x 4 /.
8 8 FREE-RESPONSE SOLUTIONS ~ AB
9 BC AP Calculus Free-Response Solutions and Notes Question BC- See AB Question. Question BC- See AB Question. 9
10 FREE-RESPONSE SOLUTIONS ~ BC Question BC- (a) dx dx t 4 dt dt t= dy dt t = =. ( te ) + = 8. t= = =. The speed at t = is t= 4 t te dt.588. t (b) Total distance traveled = ( 4) + ( ) dy (c) The tangent to the path is horizontal when =. Solving gives.8 dt t. At dx that time, dt >, so the particle is moving to the right. (d) (i) x() t = 5 t 4t+ 8= 5 t = or t =. dy dy dx (ii) The slope is dx = dt dt =. At t =, dy / e = = dx e dy At t =, dx = =.. t (iii) ( ) ( ) ( t y = y + te ) dt = + + ( te ) dt 4. e. Alternatively, we could calculate y().. The integral can be evaluated precisely using integration by parts:., t t t t t ( te ) dt te dt te e dt = = = ( t ) e =. e
11 FREE-RESPONSE SOLUTIONS ~ BC Question BC-4 See AB Question 4. Question BC-5 (a) dy The step in Euler s method is. At (, ), the slope y dx = =, so y new = + =. At,, the slope is =, so 5 y new = + = =. 4 4 (b) Since f is continuous, f ( x) f ( ) ( ) ( ) lim = =, so we can use l Hôpital s Rule: x f x f x lim = lim =. x x x x (c) dy = dx y ln y x C = +. Using the initial condition, we get C = ln y = x y = e x. We are told that y = f( x) < for x x this solution, so y = y y e = y = e.
12 FREE-RESPONSE SOLUTIONS ~ BC Question BC-6 4 n x x n x cos = ! 4!! 4 n cos x x x n x f( x) = = x! 4! 6! n! (a) x ( ) ( n ) (b) f ( ) =. ( ) ( ) f () = f () >. Therefore, by the Second Derivative Test,! 4! f has a relative minimum at x =. (c) Integrating the first three terms of the series for f, we get 5 x x P5 ( x) = x+ + C. Since g() =, C =! 4! 5 6! 5 x x P5 ( x) = x+.! 4! 5 6! (d) g () = ! 5 6! + and P () = +. The alternating series 4! estimate error g() P () does not exceed the absolute value of the first omitted term in the series, which is. Therefore, g() P () < <. 56! 56! 6!
13 AB (Form B) AP Calculus Free-Response Solutions and Notes Question AB- (Form B) ( ) (a) Area = 6 4ln( x) dx (( ) ) (b) Volume = π 8 4ln( x) dx 68.8 ( ) (c) 6 4ln( x) dx Question AB- (Form B) (a) The graph of g has a horizontal tangent where g ( x) (b) The graph of g is concave down where g ( x) subinterval,.9 < x <... (c) (.) ( ) ( ) g g g x dx (.) = = : x =.6 and x =.59. <. This happens on one. The slope at x =. is g.47. An equation of the tangent line is ( x ) y.546 =.47.. > for.5 < x <, the graph of g is concave up there, so the tangent line at x =. lies under the graph of g. (d) Since g ( x) < on (.9,.), we could say the graph of g is concave down on the closed interval [.9,.].. Since g ( x)
14 4 FREE-RESPONSE SOLUTIONS ~ AB (FORM B) Question AB- (Form B) (a) The midpoint Riemann sum is 4 ( ) = 66 cubic feet. (b) The amount of water leaked = R () t dt cubic feet. (c) Volume = = cubic feet. (d) Let V () t be the volume of water in the pool at time t. Then ( 8) ( 8) ( 8) V = P R = 4.4 cubic feet per hour. dh dh V t = =, so V () t = 44π =. Therefore, dt dt 44π feet per hour. 44π V π r h 44π h dh dt t= 8. Or leave the answer as a fraction with π.. Don t forget the units. ( )
15 FREE-RESPONSE SOLUTIONS ~ AB (FORM B) 5 Question AB-4 (Form B) (a) The squirrel changes direction at t = 9 and at t = 5, because its velocity changes sign at each of these points in time. (b) The area of the trapezoid extending from t = to t = 9 is = 4, so the squirrel moves 4 units towards B during that time. The area of the trapezoid from t = 9 to t = 5 is = 5, so the squirrel moves back 5 units towards A during that time. The area of the trapezoid from t = 5 to t = 8 is + = 5, so the squirrel moves 5 units towards B during that time. Therefore, the distances from A are 4 at t = 9, 9 at t = 5, and 5 at t = 8. The maximum is 4 at t = 9. (c) Based on the calculations in Part (b), the total distance traveled by the squirrel is = 5. (d) a() t = = ; v( t) = ( t 9) = t+ 9 ; 7 x() t = v() t dt = 5t + 9t+ C. From Part (b), x (9) = C = 4 C = 65 x( t) = 5t + 9t 65.. The graph of velocity on the interval 7< t < is a straight line. Its slope is equal to the acceleration. The equation of the line is written in the point-slope form for the v t = t 7. point (9, ). We could use the point (7, ) instead: ( ) ( ). Be careful not to confuse v(9) = on the graph, which applies to the equation for the velocity, with x(9) = 4, which we use to find the equation for the distance.
16 6 FREE-RESPONSE SOLUTIONS ~ AB (FORM B) Question AB-5 (Form B) (a) dy (b) dx = when x + =, that is, y = x and y. y ydy x dx, so (c) = ( + ) y x x = y x = + x + C. y () = C = y x x = Since the initial condition has y <, we choose y x x = We are asked to sketch the solution on < x <. Without this restriction, the domain of this solution would be (, ). A particular solution to a differential equation is always defined on a connected interval. This differential equation is not defined when y =, so no solution can cross or touch the x-axis.
17 FREE-RESPONSE SOLUTIONS ~ AB (FORM B) 7 Question AB-6 (Form B) (a) Particle R moves to the right when r ( t) = t t+ 9> ( t )( t ) t < and < t 6. > π π (b) Particle P moves to the right when p () t = sin t > 4 π π < 4 t < π 4< t 6. π sin 4 t < R P t The particles travel in opposite directions for < t < and < t < 4. (c) The acceleration is p ( ) π π π π = cos = = The velocity 6 π π π at t = is p ( ) = sin = <. Since the velocity and acceleration 4 4 have opposite signs, the particle is slowing down. (d) The distance between the particles is p ( t) r( t) interval t is () () p t r t dt.. A chart is not required but might be helpful. and the average distance on the
18 8 FREE-RESPONSE SOLUTIONS ~ AB (FORM B)
19 Question BC- (Form B) See AB Question. BC (Form B) AP Calculus Free-Response Solutions and Notes Question BC- (Form B) dx (a) The tangent line is vertical when and t.5. (b) Let the position of the particle be ( ( ), ( )) dt = and dy dt. This occurs at t.45 x t y t. Then () = ( ) + 4cos( ) sin( t ) () = ( ) + + sin( ) x x t e dt y y t dt dy dy dx = dt dt dx t = t = + = and ( x ) y 4.6 = =. At t =,.86. An equation for the tangent line is (c) The speed is dx dy + = dt dt t= 4.5. (d) The acceleration vector is the derivative of the velocity vector. At t =, this is 8.45,.6. ( ). Enter the given functions dx dt of the solution. and dy dt into your calculator and save them for the rest 9
20 FREE-RESPONSE SOLUTIONS ~ BC (FORM B) Question BC- (Form B) See AB Question. Question BC-4 (Form B) See AB Question 4. Question BC-5 (Form B) (a) g ( x) ( + x ) x x x ( + 4x ) ( + 4x ) = =. For = when x >, g ( x) x =. For < x <, g ( x) > and g is increasing; for x >, g ( x) < and g is decreasing. The absolute maximum value of g therefore occurs at x =. It is g = =. g has no minimum on (, ) (b) The area is given by the improper integral 4x dx = x + 4x b lim ln x ln ( + 4x ) = lim ln b ln ( + 4b ) + ln 5 = b b b lim ln + ln 5 = lim ln + ln 5 = ln + ln 5. b 4 b + b + 4 b. Leave it at this to save time and avoid mistakes.
21 FREE-RESPONSE SOLUTIONS ~ BC (FORM B) Question BC-6 (Form B) (a) By the ratio test, the series converges when gives x < or < x <. At ( x) ( x) n+ lim n <. Evaluating the limit n n n x =, the series is n ( ) ( ) = n n. n= n= This is the harmonic series, which diverges. At x =, the series is. This n= n is an alternating series with terms decreasing by absolute value and approaching zero. Therefore, it converges by the Alternating Series Test. The interval of convergence is < x. n ( ) n (b) y f ( x) n n ( ) n( x) n ( ) n ( x) = =, so xy = and n= n n= n n n n n ( ) n ( x) ( ) ( x) xy y =. Factoring gives n= n n n n ( ) ( x) ( n ) n xy y = = ( x) n= n n= 4x first term 4x and the common ratio x, so it converges to + x that is, for < x <.. This is a geometric series with the n when x <,
Mark Howell Gonzaga High School, Washington, D.C.
Be Prepared for the Calculus Exam Mark Howell Gonzaga High School, Washington, D.C. Martha Montgomery Fremont City Schools, Fremont, Ohio Practice exam contributors: Benita Albert Oak Ridge High School,
More informationMark Howell Gonzaga High School, Washington, D.C.
Be Prepared for the Sylight Publishing Calculus Exam Mar Howell Gonzaga High School, Washington, D.C. Martha Montgomery Fremont City Schools, Fremont, Ohio Practice exam contributors: Benita lbert Oa Ridge
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