Mark Howell Gonzaga High School, Washington, D.C.

Transcription

1 Be Prepared for the Calculus Eam Mark Howell Gonzaga High School, Washington, D.C. Martha Montgomery Fremont City Schools, Fremont, Ohio Practice eam contributors: Benita Albert Oak Ridge High School, Oak Ridge, Tennessee Thomas Dick Oregon State University Joe Milliet St. Mark's School of Teas, Dallas, Teas Skylight Publishing Andover, Massachusetts

2 Copyright 5-8 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 Eam (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 8 AB AP Calculus Free-Response Solutions and Notes Question AB- The graphs intersect at the points with -coordinates of and. 3 (a) Area of R = sin( π ) ( 4 ) d 4.. (b) The line y = intersects the graph of 3 y 4 = at =.539 and again at ( ) b 3 =.675. Let a =.539 and b =.675. Area = ( 4 ) ( ) (c) Volume = 3 sin ( π ) ( 4 ) d ( sin π 4 ) ( 3 ) 3 (d) Volume = ( ) ( ) d 8.37 a d. Even though it is easy to antidifferentiate the integrand in this question, it is probably safer and faster to use your calculator to evaluate the integral.. No need to evaluate the integral.. 3

4 4 FREE-RESPONSE SOLUTIONS ~ 8 AB Question AB- (a) Rate at t = 5.5 ( ) L( ) L = = people per hour. 4 (b) Average value of L(t) is () 4 Ltdt. Using a trapezoidal sum, this is approimately = people. (c) L () t must be at least three times. L is continuous and so, by the Intermediate Value Theorem, there must be a time t between t = and t =, another time t between t = 3 and t = 4, another time t 3 between t = 4 and t = 7, and another time t 4 between t = 7 and t = 8 when Lt ( ) = 45. By the Mean Value Theorem, there must be a time between t and t, another time between t and t 3, and a third time between t 3 and t 4 when L () t =. 3 3 (d) r () t dt It is OK not to calculate the sum.. Or Rolle s theorem.. There were 973 tickets sold. 3. Alternative justification: L() > L() and L() > L(4), so L has a local maimum on the interval (, 4); L(4) < L(3) and L(4) < L(7), so L has a local minimum on the interval (3, 7); L(7) > L(4) and L(7) > L(8), so L has a local maimum on the interval (4, 8). L () t must be at these three local etrema.

5 FREE-RESPONSE SOLUTIONS ~ 8 AB 5 Question AB-3 dv dv dh dr (a) We are given =. Since V = π r h, = πr + πrh. At the instant dt dt dt dt dr in question, we have r =, h =.5, and.5 dt =. Substituting, dh dh 5π = π + π.5.5. Solving gives =.39 dt dt π centimeters per minute. dv dv (b) Once the recovery device starts working, = 4 t. > for t < 5 dt dt dv and < for t > 5. Thus, the maimum volume occurs at t = 5 minutes after the dt device starts removing oil. 5 (c) + ( ) 6 4 t dt. The numeric answer is optional.

6 6 FREE-RESPONSE SOLUTIONS ~ 8 AB Question AB-4 t (a) () t () = + v t dt. From the graph, ( t) < for < t < 3 and for 5 < t < 6. () t > for 3 < t < 5. So a minimum could occur at t = 3 or at t = 6. ( 3) = 8= and ( ) position of at t = 3. (b) The position () t is a continuous function. 6 = 8+ 3 = 9. The particle is at its leftmost t (t) 7 9 Reason Given From Part(a) + 3 = 7 7 = 9 By the Intermediate Value Theorem, ( t ) = 8 for some t, < t < 3, again for some t, 3 < t < 5, and again for some t, 5 < t < 6. So there are three times when () t = 8. (c) On the interval < t < 3, v(t) < and v(t) is increasing. Therefore, the speed is decreasing. (d) Since a() t v () t =, the acceleration is negative when the velocity is decreasing. < < and 4< t < 6. From the graph, this occurs for t

7 FREE-RESPONSE SOLUTIONS ~ 8 AB 7 Question AB-5 (a) dy d dy d (b) Separating variables, =, so = y y. Thus, ln y = + C. From the initial condition, ln = + C C = y = e. Since the initial condition has y <, y e f = e. (c) ( ) lim f = e = e.. Or: ( ) f = ee. = and ( )

8 8 FREE-RESPONSE SOLUTIONS ~ 8 AB Question AB-6 ln ( e ) (a) f ( e ) = e = e and ( f e ) line is y = ( e ). e e 4 ( e ) (b) f ( ) = at = e. f ( ) > for < e and f ( ) local maimum at = e. (c) f ( ) ln = =. An equation for the tangent 4 4 e e < for > e. Therefore, f has a ( ln)( ) ( + ln) ln 3 = = =. Since the second derivative changes sign at = e, the graph of f has a point of inflection there. (d) The numerator of f approaches and the denominator approaches, so lim f does not eist. + ( ). Or: lim f ( ) + =.

9 8 BC AP Calculus Free-Response Solutions and Notes Question BC- See AB Question. Question BC- See AB Question. Question BC-3 (a) P ( ) = P() + P ()( ) = ( ). h(.9) P (.9) = (.) = 67.. Since h ( ).9, this approimation is less than h (.9) = h(.9) P3 (.9) = (.) + (.) + (.). 6 8 (b) P ( ) 8 8 ( ) ( ) ( ) > for all in the interval (c) Since the fourth derivative is increasing, its maimum on the interval.9 occurs at =. It is 584. Therefore, the Lagrange error bound is ( ) < 3. 4!. The graph of h is concave up

10 FREE-RESPONSE SOLUTIONS ~ 8 BC Question BC-4 See AB Question 4. Question BC-5 (a) Since = 3 is a critical point and f ( ) < for < 3 and f ( ) a local minimum at = 3. > for > 3, f has (b) f is decreasing for < 3 since f ( ) < there. f ( ) ( 3) e e e ( ) = + =. It is positive when >. So f is decreasing and concave up for < < 3. 3 (c) ( ) = + ( ) f e d. Using integration by parts with u = 3 and dv = e d, we have du = d and v= e. So, ( ) f (3) = f() + ( 3) e d= 7 + ( 3) e e d= 7 ( ) ( ) e e e e = + e e.. Or 3.

11 FREE-RESPONSE SOLUTIONS ~ 8 BC Question BC-6 (a) dy (b) At (, 8), y new y = 8+ = 7. At 7 7 = 7+ = dt =. ( ) new,7, dy 7 dt = 8. d y 3 3 = y dy + dy = dy y + y = dy y. At t =, dt 8 dt 8 dt dt dt 4 4 d y = =. P () t = 8 t+ t f () P () = 8 +. dt (c) ( 6 y) (d) 6< y 8.. = 7.5.

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13 8 AB (Form B) AP Calculus Free-Response Solutions and Notes Question AB- (Form B) The graphs intersect at the points (, ) and (9, 3). (a) Area 9 A R = d =. (b) ( y) 3 y; ( y) y = =. Using washers, 3 y y dy 3.6. Volume = π ( 3 + ) ( + ) 3 (c) Volume = ( ) 3y y dy 8.. 3

14 4 FREE-RESPONSE SOLUTIONS ~ 8 AB (FORM B) Question AB- (Form B) (a) Distance = r () t dt 6.37 km. d d d () = g( ). From Part (a), at t = the car has traveled 6.37 km. dt d dt g liters/km. At time t =, the speed of the car is (b) g( t ) ( ) r ( ). km/hour. So, the rate of change of the number of liters of gasoline with respect to time is liters km.5 = 6 liters/hour. km hour (c) Solving r() t = 8 gives t = T =.3345 hours. At this time, the car has T traveled S = r() t dt km. The amount of fuel consumed is g( S).537 liters.. Or, if you use symbolic differentiation, g ( ) =.5 e + e g () =.5.. Store this value as T in your calculator. 3. Store this value, too, as S and use your calculator to evaluate g(s).

15 FREE-RESPONSE SOLUTIONS ~ 8 AB (FORM B) 5 Question AB-3 (Form B) (a) The area is approimately = 5 ft. Let A = 5. (b) Using the approimation from Part (a), volumetric flow F is approimated by F = a v() t. The average value of F over [, ] is () Avt dt 87.7 ft 3 /min. 4 (c) Area = f ( ) d.3 ft. Let B =.3. 6 (d) Average volumetric flow over [4, 6] is B vtdt () ft 3 /min. Since this is greater than, the water must be diverted.. Store in your calculator. 8.93

16 6 FREE-RESPONSE SOLUTIONS ~ 8 AB (FORM B) Question AB-4 (Form B) = +. (a) f ( ) = and g ( ) 3cos( ) 4 9sin ( ) (b) The slope is g ( π) = 3cos( π) 4+ 9sin ( π) = 6. g( ) f ( ) line is y = 6( π ). = at (c) g ( ) π g has a maimum at =. Since g ( ) π π = =. Tangent π π > for < and g ( ) < for < π, π 3 g = f = 4 + t dt. =. (). You could also use the candidate test, evaluate g ( ) = and g ( π ) =, and note that g π >, so it s the maimum.

17 FREE-RESPONSE SOLUTIONS ~ 8 AB (FORM B) 7 Question AB-5 (Form B) (a) The graph of y g( ) = has points of inflection at = and = 4, since its derivative has local etrema at those points. (b) From the graph of g, we see that g decreases from = 3 to = and again from = to = 6 since g ( ) <. From = to = and from = 6 to = 7, g is increasing since g ( ). > Therefore, g has a local maimum at = where g changes sign from positive to negative. g also has maima at the endpoints Evaluating, we get g( 3) = 5+ g ( t) dt = 5 + 4= and 7 3 g( 7 ) = 5 + g ( t) dt = =. Therefore, the absolute maimum value of g is 5. (c) (d) g = =. 5 ( ) g( ) ( 7) g ( 3) =. Since g ( ) g is not differentiable at = (nor at =, nor at = 4, but just one point is sufficient), the Mean Value Theorem does not apply to g on [ 3, 7]. ( ). Even though g ( c) = for any c, such that < c <.

18 8 FREE-RESPONSE SOLUTIONS ~ 8 AB (FORM B) Question AB-6 (Form B) (a) ( ) dy y y + y = y y + y = y = = = d y + y + ( ) 3 3. (b) The slope at (,) is 4 = The tangent line is y ( ) (c) 3 For a vertical tangent, we need y + = or y =, while. Substituting into + + 4= 5 + 8= + 4 =. the equation for the curve, ( )( ) So the points are ( 4, ) and (, ). dy (d) We would need (, ) d = where y =. So ( + ) = and =. Substituting into the equation of the curve, we get = 5, which is never true. So there is no point where the curve intersects the -ais and has a horizontal tangent.

19 Question BC- (Form B) 8 BC (Form B) AP Calculus Free-Response Solutions and Notes d dt d y dt (a) At t = 4 the acceleration vector is a =, = (.433,.87) (b) y( ) 5 4 dy dt dt. = +.6 t= 4. (c) Solving ( t ) t 3.5 = 3 + 3cos gives t.6. 4 (d) Distance = ( t) t 3 + 3cos dt 3.8. Question BC- (Form B) See AB Question. Question BC-3 (Form B) See AB Question 3. 9

20 FREE-RESPONSE SOLUTIONS ~ 8 BC (FORM B) Question BC-4 (Form B) 3 (a) ( ) k = k = = or = k. 3 Area = ( ) k k k k d = = k k = k = k = (b) Volume = ( ) k π k d. k (c) Perimeter = ( ) ( ) k ( 3 ). k+ + f d= k+ + k d Question BC-5 (Form B) See AB Question 5.

21 FREE-RESPONSE SOLUTIONS ~ 8 BC (FORM B) Question BC-6 (Form B) (a) Use the geometric series to generate the series for f : 3 4 n = Replace by : n... ( )... + = Multiple by : n n+ = ( ) ( n ) (b) At =, the series is + + Since lim ( ) n-th term test, the series does not converge. n does not eist, by the (c) Antidifferentiating the series for ln ( + ) = we get the series for ln ( ) + : (d) 5 When =, + =. The series from Part (c) converges at = because it is 4 an alternating series with descreasing terms, and the magnitude of the terms 5 approaches. we can use it to approimate ln 4 : 5 ln = By the alternating series error bound, ln <. Therefore, A ln <, where 4 A =. 4 3