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

Transcription

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 by Skylight Publishing Chapter 0. 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 70 Andover, MA 080 web: sales@skylit.com support@skylit.com

3 006 AB AP Calculus Free-Response Solutions and Notes Question Solving ln x= x, the graphs intersect at x = and x =.469. Let a = and b =.469. b x.949. a A = ln x ( x ) d ( ) ( ) b Vy lnx x dx. = = π + + a 4.99 b y (c) By washers : Vx= 0 = π ( y+ ) ( e ) ln a ln ( ) by shells : Vx= 0 = π x ln x ( x ) dx. b a dy;. Store the intersection points in calculator variables and use those variables when calculating the integrals. See Be Prepared, page 56.. Use your calculator to evaluate the integrals; don t bother trying to antidifferentiate.

4 4 FREE-RESPONSE SOLUTIONS ~ 006 AB Question 8 0 Ltdt ( ) cars. Solving Lt ( ) = 50 gives t.48 and t Average value of t L on the closed interval [ t, t] is Lt () dt t t cars per hour. t 6 (c) The number of cars turning left for 4 t 6 hours is L () t dt cars = 06 > So, yes, a traffic light is needed.. Store t and t in calculator variables and use those variables when calculating the integrals.. Be attentive to the units. Question 4 g(4) = f( t) dt = + 4 = ; g (4) = f(4) = 0 ; 0 g (4) = f (4) = =. 5 By the Fundamental Theorem of Calculus, g ( x) = f( x). From the graph of f, we see that g ( x) changes sign from negative to positive at x =. Therefore, g has a relative minimum at x = ( ) () () () = (c) g 0 = f t dt = f t dt+ f t dt = ( ) () () 0 0 ( ) ( ) ( ) ( ) g08 = f t dt+ f t dt = + = 44. g 08 = f 08 = f 5 + = f =. The tangent line equation is y 44 ( x 08) =.

5 FREE-RESPONSE SOLUTIONS ~ 006 AB 5 Question 4 Average acceleration = 49 5 = feet per second per second vt () dt t = 70 seconds. is the net change in height of the rocket in feet from t = 0 seconds to 70 0 () v t dt = 00 feet. (c) For Rocket A, v (80) = 49 ft/sec. For rocket B, at () = v () t and v (0) = so v(80) = + a( t) dt = + ( ) t dt = + t + = + = 0 ft/sec + 0 Rocket B is traveling faster.. Or ft/sec. Question 5 y O x dy + y dy dx = = ln + y = ln x + C dx x + y x A y C = e. ( ) we have to choose the negative sign and y+ = x y = x. Since is x < 0. + =± Ax, where y = A = dy dx has a discontinuity at x = 0, the domain

6 6 FREE-RESPONSE SOLUTIONS ~ 006 AB Question 6 ax g ( x) = ae + f ( x) g (0) = a+ f (0) = a 4. ax g ( x) = a e + f ( x) g (0) = a + f (0) = a +. h(0) = cos(0) f(0) =. h ( x) = ksin( kx) f( x) + cos( kx) f ( x) h (0) = ksin(0) f(0) + cos(0) f (0) = f (0) = 4. An equation for the tangent line to the graph of h at x = 0 is y = 4x.

7 006 BC AP Calculus Free-Response Solutions and Notes Question See AB Question. Question See AB Question. 7

8 8 FREE-RESPONSE SOLUTIONS ~ 006 BC Question d x d y a() t =, a ( ) ( 0.96, 0.74) dt dt v 0.874, Speed at t = is ( ) ( ). If the tangent line is vertical, we must have dx t 0 arcsin e = 0 t = ln. dt = ( ) (c) 4t mt () = + t arcsin t ( e ) lim mt ( ) = 0. t dy 4t (d) c = y() + dt = + dt dt + t.. No need to do symbolic differentiation use your calculator.. Or use equation solver on your calculator: t Since we are told a horizontal asymptote exists, it is sufficient to just write the answer. In general, if we were asked to show that a horizontal asymptote exists, we would need to show that lim xt ( ) = and lim y( t) = c (or lim xt ( ) = and lim y( t) = c for some T). t T t t t T Question 4 See AB Question 4.

9 FREE-RESPONSE SOLUTIONS ~ 004 BC 9 Question 5 dy 6 = 5 = 6. dx 6 d y (, 4) dy = 0x+ 6 ( y ) dx dx d y dx (, 4) = = 9. 6 No. On the x-axis, y = 0, so dy must have 0 dx =. dy dx = 5x + > 0. But if y = 0 is a tangent, then we 9 T, f = 4+ 6( x+ ) x+. (c) ( ) (d) At x =, y = 4, and slope = 6 x =, y = =. Here, slope = =. y = + =. Thus, f (0) Or draw and fill a table: Point 0 Point Point x y = m= 5x = y = 4 8

10 0 FREE-RESPONSE SOLUTIONS ~ 006 BC Question 6 Ratio test: ( n+ ) n+ x n+ ( ) lim n + n+ x n + = lim = x <. n n n n nx n+ nx n + At x = the series is series diverges because n n, which diverges because lim 0. At x = the n n + n= n + ( ) n n lim 0. The interval of convergence is < x <. n n y (0) = f (0) g (0) = = 0. y (0) = f (0) g (0) = =. y has a relative minimum at x = 0 since y (0) = 0 and y (0) > 0.. Don t forget the limit and the absolute values.. The endpoints must be tested since we are asked for the interval of convergence, not just the radius.

11 006 AB (Form B) AP Calculus Free-Response Solutions and Notes Question Solving (c) A R f( x) = 0 x =.7. Let a =.7 0 = f( x) dx.90. a 0 ( ) V = y π f( x) dx. = + = 59.6 a f (0) = and f (0) = an equation for the tangent line at x = 0 is y = x+. The graph of f intersects this tangent line at x = 0, and again at b x = AS = x+ f( x) 0 dx, where b = Store the intersection points in calculator variables and use those variables when calculating the integrals. See Be Prepared, page 56.. Use your calculator to evaluate the integrals; don t bother trying to antidifferentiate..

12 FREE-RESPONSE SOLUTIONS ~ 006 AB (FORM B) Question The graph of f is concave down on the interval.7 < x <.9 since f is decreasing on that interval. By the Fundamental Theorem of Calculus, x f ( x) = 5 + f ( t) dt. f can have an absolute maximum either at one of the endpoints or where f ( x) changes from positive to negative, at x f (0) = 5 ; f () 5.579; f (.7745) f has an absolute maximum at x = (c) f () = 5.6 and f ().459. An equation for the tangent line at x = is y = ( x ). Question f ( x) = ax. f(4) = 6a = and f (4) = 8a = 6a = 8a, which is impossible for a 0. (c) (d) x g ( x) = cx. g(4) = 64c = c=. Substituting c in g gives 8 4 g ( 4) = 4 =. 8 x x g ( x) = x = x g ( x) < 0 for 4 0 < x < g is not increasing between x = 0 and x = 4. n n n n 4 n nx n4 n4 h(4) = = k = 4. h ( x) = h (4) = = = n= 4. n k k k x 4x x k = 4 = 56. Thus hx ( ) = and h ( x) = =. Property (i): h (0) = 0 and h (0) = 0. Property (iii): Since h ( x) > 0 for x > 0, h is increasing between x = 0 and x = 4.

13 FREE-RESPONSE SOLUTIONS ~ 006 AB (FORM B) Question 4 5 f () = = 4 0 calories per minute per minute. 5 9 For 4 < t < 4, the greatest rate of increase of f is =. For 0< t < 4, 6 f () t = t + t and f () t = t+. 4 f () = 0 and f changes sign from positive to negative there, so f has a maximum at t =. f () = > f is increasing at its greatest rate at t =. (c) Total number of calories burned from t = 6 to t = 8 is f () t dt = f() t dt+ f() t dt+ f() t dt = = =. (d) The existing setting burns an average of = calories per minute. To increase that to 5 calories per minute requires c = 4.. Or calories/minute.. You could leave the answer in this form for Part (c) no need to simplify but you need that number for Part (d).

14 4 FREE-RESPONSE SOLUTIONS ~ 006 AB (FORM B) Question 5 y O x c =. dy dx (c) ( y ) = cos( π x) ( ) cos( π ) ( y ) = sin ( π x) + C. () 0 π ( y ) = sin ( π x) + π y dy = x dx 0 = sin + C C =. Thus, π ( ) y = ( π ) y = sin π f( x) =. sin + π ( π x) + ( π x)

15 FREE-RESPONSE SOLUTIONS ~ 006 AB (FORM B) 5 Question 6 60 vt () dtis the total distance in feet traveled by the car from t = 0 seconds to 0 t = 60 seconds. The trapezoidal approximation to this integral is feet = = 85 feet. 0 0 at dt= v t dt= v v = 0 = 0 () () (0) (0) 4 ( 0) 6 ft/sec. This is the net change in velocity in feet per second of the car from t = 0 seconds to t = 0 seconds. (c) Yes. v is continuous with v (5) = 0 and v (50) = 0. By the Intermediate Value Theorem, there must be a time t between 5 and 50 when vt () = 5. (d) v(5) v(0) Yes. v is differentiable (its derivative is a(t) ), and = 0. By the Mean 5 0 Value Theorem, there must be a time between t = 0 and t = 5 when v () t = a() t = 0.. You can leave the answer in this form no need to simplify.

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17 006 BC (Form B) AP Calculus Free-Response Solutions and Notes Question See AB Question. Question At t =, dy dx ( e ) ( e ) sec =.78. The tangent line equation is y = +.78( x ). tan d x d y a() t =, a ( ) ( 0.4, 0.5). dt dt vt () sec e tan = + e.9. Speed at t = is ( ) ( ) (c) Total distance traveled = dx dy + dt dt.059. ( ) t t t (d) No. xt () = + tan e dt x(0).4. Since 0< e < for t > 0, dx t = tan ( e ) > 0 for t > 0 xt () is increasing for t > 0 dt xt () > x(0).4> 0for t > 0.. No need to do symbolic differentiation use your calculator.. Use your calculator to evaluate the integral; don t bother trying to antidifferentiate. Question See AB Question. 7

18 8 FREE-RESPONSE SOLUTIONS ~ 006 BC (FORM B) Question 4 See AB Question 4. Question 5 dy dy x dx y = ( 6 x) ydx = ( 6 ) ln y = 6x x + C. y (4) = 6x x 8 = + C C = y = e. Since y > 0 at the initial point, x x 8 f( x) = e lim gx ( ) = and lim g ( x) = 0. x x (c) The graph of g has a point of inflection at y =, because dy, as a function of y, is dx a parabola, which reaches its maximum halfway between its zeroes, y = 0 and y =. dy 9 = =. dx = y. lim g( x) x is the carrying capacity, and lim g ( x) is 0 for any logistic curve. x. You can stop here.

19 FREE-RESPONSE SOLUTIONS ~ 006 BC (FORM B) 9 Question n n x x x n x... ( ) The given series is the Maclaurin series for f ( x) at series converges to 4 6 f = = n+ x x x n x n + (c) x ( ) x =. f ( x) = x ( + x ). The (d) / 45 f() t dt + =. When x =, the series in Part (c) is alternating with decreasing absolute values of terms, and the n-th term approaches 0 as n. By the alternating series error bound, the magnitude of the error does not exceed the absolute value of the first omitted term, which is 0 0 = 0 04 < You can leave it like this no need to calculate.