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

Transcription

1 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 High School, Oa Ridge, Tennessee Thomas Dic Oregon State University Joe Milliet St. Mar's School of Texas, Dallas, Texas Sylight Publishing ndover, Massachusetts

2 Copyright - by Sylight Publishing Chapter. nnotated Solutions to Past Free-Response Questions This material is provided to you as a supplement to the boo Be Prepared for the P Calculus Exam. 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 boo that you own or receive from your school. Sylight Publishing 9 Bartlet Street, Suite 7 ndover, M 8 web: sales@sylit.com support@sylit.com

3 B P Calculus Free-Response Solutions and Notes Question B- W() W(9) =.7 degrees/minute. The average 9 9 rate of change of the temperature of the water in degrees Fahrenheit per minute from t = 9 to t = minutes is.7. This approximates W, the (a) W ( ) instantaneous rate of change of the temperature at t = minutes, indicating that the temperature is increasing at approximately.7 degrees Fahrenheit per minute at that time. () ( ) ( ) (b) W t dt = W W = 7.. = 6. degrees. This is the net change ( ) in temperature of the water in degrees Fahrenheit from to minutes (c) W () t dt ( ) 6.79 =. This value underestimates the average temperature of the water, because a left Riemann sum for an integral of an increasing function underestimates the integral. (d) ( ) () ( ) W = W + W t dt = 7.4 degrees Fahrenheit.. It is important to mention the units both for the variable t (minutes) and the temperature (degrees Fahrenheit).

4 4 FREE-RESPONSE SOLUTIONS ~ B Question B- (a) ln x = x at x = Let =.6944 rea = x dx + ( x d. ln x ) (b) Volume = ( ln ) + ( ) (c) ln x = at x dx x dx.. x = e and x = at x = ln x dx+ ) e ( ) ( = ln + ( ) x d. ( ). x xdx x dx. Store this number in your calculator, and use the stored value in the subsequent calculations.. Or: ( ) ( ) e. ln x dx+ x dx=.988 simpler equation results from integrating with respect to y: ( y e B y) dy y ( ) y e dy B y y = or ( y e ) dy = ( y e ) dy where B =.,

5 FREE-RESPONSE SOLUTIONS ~ B Question B- = f t dt = (negative area of a right triangle with legs and f()); 4 = f t dt = f () t dt π =. (a) g( ) ( ) g ( ) ( ) ( ) = ( ) = g ( ) f ( ) (b) g f ; = =. (c) The graph of g has horizontal tangents where g ( x) = f( x) =, namely at x = and x =. gx ( ) has a relative maximum at x = because g ( x) changes sign from positive to negative there. gx ( ) has neither a relative maximum nor a relative minimum at g x does not change sign there. x = because ( ) (d) The graph of g has three points of inflection, at x =, x =, and x =, because g ( x) changes from increasing to decreasing at x = and at x =, and from decreasing to increasing at x =.. When evaluating definite integrals of functions defined by geometric figures, it might be easier to wor from left to right.. n alternative justification involves relating the sign change in g ( x) from positive to negative at x = and at x = and from negative to positive at x =.

6 6 FREE-RESPONSE SOLUTIONS ~ B Question B-4 x = = x / (a) f ( x) ( x ) ( x). 4 = +. 4 (b) f ( ) = 9 = 4; f ( ) =. The tangent line is y 4 ( x ) (c) By definition, g is continuous at ( ) ( ) x ( ) lim g x = lim f x = f ( ) 4 x lim g x x x = if lim g( x) = g( ). x = and lim g( x) lim ( x ) + + x x = + 7 = 4. Therefore, exists and is equal to 4 = f( ) = g( ). So g is continuous at x =. ( ) x = ( ) =. (d) / ( ) / x x dx= ( x ) ( x) dx=. Or use more formal u-substitution u = x du, x dx =, du xdx =.

7 FREE-RESPONSE SOLUTIONS ~ B 7 Question B- db (a) When B = 4, 6 dt = =, and when B = 7, db = = 6. Therefore, the dt bird is gaining weight faster when it weighs 4 grams than when it weighs 7 grams. d B d db dt dt dt. For db < B <, <, so the graph of B must be concave down, but the graph in the dt picture is not. (b) = ( B) = = ( B) = ( B) (c) (c) get e db = dt B ln B = t+ C. Substituting (, ) we ln ( 8) = C ln B = t ln(8) ln B = ln(8) t B ln(8) = e B 8e t = t B = 8e. ln t. Since B >, B = B.

8 8 FREE-RESPONSE SOLUTIONS ~ B Question B-6 (a) The particle is moving to the left when v( t ) <. This taes place when cos π t π π t π < < < 6 6 < t < 9. 6 (b) The total distance traveled is equal to v () t dt. π πt a t = v t = 6 6. π π π v( 4) = cos < and a ( 4) = sin <. Since the velocity and 6 acceleration have the same signs at t = 4, speed of the particle is increasing at that time. (c) () () sin 4 6 x 4 x cos π π = + t dt sin t = + = 6 π 6 (d) ( ) ( ) 4 6 4π + sin sin() π 6.. No need to simplify further. If you insist, = +. π

9 BC P Calculus Free-Response Solutions and Notes Question BC- See B Question. Question BC- dx (a) = >, so the particle is moving to the right at t =. The slope of the path of dt t= e dy dy sin the particle at t = is = dt =.. dx dx /e dt x = 4dx 4 t + x 4 = x + dt = + dt. t dt e (b) ( ) ( ). (c) Speed = a ( ) dx dy + dt dt d x d y 4 =, dt dt t = 4 t = 4.7. (.4,.989). 4 (d) Total distance traveled = t + + ( sin () t ) d t e t.6. Question BC- See B Question. 9

10 FREE-RESPONSE SOLUTIONS ~ BC Question BC-4 (a) The tangent line is y = 8( x ). t x =.4, 8(.4) f (.4) 8. (b) The midpoint sum is. (.4) 4.6 y = + = 8., and. f + = = 4.6. This approximates f ( ) f (.4 ), so (c) f (.) + 8. = 6.6 and f (.4) = 9. T ( x) = + 8 x + x. (d) The second-degree Taylor polynomial is ( ) ( ) ( ) ( ) T (.4) = Can stop here.. It would be incorrect to write: ( ) =. Thus, ( ) f f.4 = 8., with an equal sign such an error is usually penalized.. You could leave it as f.4 + (. +. ). Either way, not f (.4) =... ( ) with an equal sign. 4. Can stop here; either way not f (.4) =.... gain, do not use the equal sign here. Question BC- See B Question.

11 FREE-RESPONSE SOLUTIONS ~ BC Question BC-6 (a) (b) n+ x n+ lim n + x n+ = lim = x < n n+ n n x + x n+ < x <. Both at x = and at n + x = the series is an alternating series whose terms decrease in absolute value to, so both these series converge by the alternating series test. Therefore, the interval of convergence is x. The alternating series error bound guarantees that the sum of the first two terms differs from g by the next non-zero term of the series, which is = < n n n+ x n + (c) The series is ( ) ( ) x x