AP CALCULUS AB 2017 SCORING GUIDELINES

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1 AP CALCULUS AB 07 SCORING GUIDELINES

2 Consider the differential equation AP CALCULUS AB 06 SCORING GUIDELINES y =. Question 4 On the aes provided, sketch a slope field for the given differential equation at the si points indicated. (b) Let y = f ( ) be the particular solution to the given differential equation with the initial condition f ( ) = 3. Write an equation for the line tangent to the graph of y = f ( ) at =. Use your equation to approimate f (. ). (c) Find the particular solution y = f ( ) to the given differential equation with the initial condition f ( ) = 3. { : zero slopes : : nonzero slopes (b) = (, y) (, 3) 3 = = 9 { : tangent line equation : : approimation An equation for the tangent line is y = 9( ) + 3. f (. ) 9(. ) + 3 = 3. 9 (c) = y = y = ln + C y = ln + C C = 3 3 = ln y 3 y = ln( ) 3 3 Note: This solution is valid for < < + e. : separation of variables : antiderivatives : : constant of integration and uses initial condition : solves for y Note: ma 3 [--0-0] if no constant of integration Note: 0 if no separation of variables 06 The College Board. Visit the College Board on the Web:

3 Consider the differential equation y. = AP CALUCLUS AB/CALCULUS BC 0 SCORING GUIDELINES Question 4 On the aes provided, sketch a slope field for the given differential equation at the si points indicated. d y (b) Find in terms of and y. Determine the concavity of all solution curves for the given differential equation in Quadrant II. Give a reason for your answer. (c) Let y = f( ) be the particular solution to the differential equation with the initial condition f ( ) = 3. Does f have a relative minimum, a relative maimum, or neither at =? Justify your answer. (d) Find the values of the constants m and b for which y = m + b is a solution to the differential equation. : { : slopes where = 0 : slopes whe re = (b) (c) d y = = ( y) = + y In Quadrant II, < 0 and y > 0, so + y > 0. Therefore, all solution curves are concave up in Quadrant II. = = ( ) 3= =/ 0 (, y) (, 3) Therefore, f has neither a relative minimum nor a relative maimum at =. d y : : : concave up with reason : : considers (, y) = (, 3) : conclusion with justification (d) y = m + b d = ( m + b ) = m y = m ( m + b) = m ( m) ( m + b) = 0 m = 0 m = b = m b = d : ( m + b ) = m 3: : y = m : answer Therefore, m = and b =. 0 The College Board. Visit the College Board on the Web:

4 04 SCORING GUIDELINES Question 6 Consider the differential equation ( 3 y) cos. = Let y = f( ) be the particular solution to the differential equation with the initial condition f ( 0) =. The function f is defined for all real numbers. A portion of the slope field of the differential equation is given below. Sketch the solution curve through the point ( 0, ). (b) Write an equation for the line tangent to the solution curve in part at the point ( 0, ). Use the equation to approimate f ( 0. ). (c) Find y = f( ), the particular solution to the differential equation with the initial condition f ( 0) =. : solution curve (b) (, y) = ( 0, ) = cos 0 = An equation for the tangent line is y = +. f =.4 ( ) ( ) : tangent line equation : { : approimation (c) = ( 3 y) cos = cos 3 y ln 3 y = sin + C ln = sin 0 + C C = ln ln 3 y = sin ln Because y ( 0) =, y < 3, so 3 y = 3 y sin 3 y = e sin y = 3 e Note: this solution is valid for all real numbers. 04 The College Board. Visit the College Board on the Web: : separation of variables : antiderivatives 6 : : constant of integration : uses initial condition : solves for y Note: ma 3 6 [ ] if no constant of integration Note: 0 6 if no separation of variables

5 y Consider the differential equation e ( ) AP CALCULUS AB 03 SCORING GUIDELINES Question = Let y = f( ) be the particular solution to the differential equation that passes through (, 0 ). Write an equation for the line tangent to the graph of f at the point (, 0 ). Use the tangent line to approimate f (. ). (b) Find y = f( ), the particular solution to the differential equation that passes through (, 0 ). ( ) 0 = e 3 6 = 3 (, y) = (, 0) An equation for the tangent line is y = 3(. ) f (. ) 3(. ) = 0.6 : at the point (, y) = (, 0 ) 3 : : tangent line equation : approimation (b) = ( 3 6 ) y e = ( 3 6 y ) e y 3 e = 3 + C 0 3 e = 3 + C C = y 3 e = 3 + y 3 e = + 3 ( 3 ) ( 3 ) y = ln + 3 y = ln : : separation of variables : antiderivatives : constant of integration : uses initial condition : solves for y Note: ma 3 6 [ ] if no constant of integration Note: 0 6 if no separation of variables Note: This solution is valid on an interval containing 3 = for which + 3 > The College Board. Visit the College Board on the Web:

6 0 SCORING GUIDELINES Question The rate at which a baby bird gains weight is proportional to the difference between its adult weight and its current weight. At time t = 0, when the bird is first weighed, its weight is 0 grams. If B() t is the weight of the bird, in grams, at time t days after it is first weighed, then db = ( 00 B). dt Let y = B() t be the solution to the differential equation above with initial condition B ( 0) = 0. Is the bird gaining weight faster when it weighs 40 grams or when it weighs 70 grams? Eplain your reasoning. d B d B (b) Find in terms of B. Use to eplain why the graph of B dt dt cannot resemble the following graph. (c) Use separation of variables to find y = B(), t the particular solution to the differential equation with initial condition B ( 0) = 0. db = ( 60 ) = dt B= 40 db dt B= 70 = ( 30 ) = 6 : db : uses dt : answer with reason db Because dt > db, dt the bird is gaining B= 40 B= 70 weight faster when it weighs 40 grams. d B db (b) ( 00 ) ( 00 ) = = B = B dt dt Therefore, the graph of B is concave down for 0 B < 00. A portion of the given graph is concave up. db (c) = ( 00 B) dt db = dt 00 B ln 00 B = t + C Because 0 B < 00, 00 B = 00 B. ln ( 00 0) = ( 0) + C ln ( 80) = C t 00 B = 80e t Bt () = e, t 0 : : d B : in terms of B dt : eplanation : separation of variables : antiderivatives : constant of integration : uses initial condition : solves for B Note: ma [ ] if no constant of integration Note: 0 if no separation of variables 0 The College Board. Visit the College Board on the Web:

7 0 SCORING GUIDELINES Question At the beginning of 00, a landfill contained 400 tons of solid waste. The increasing function W models the total amount of solid waste stored at the landfill. Planners estimate that W will satisfy the differential dw equation = ( W 300) for the net 0 years. W is measured in tons, and t is measured in years from dt the start of 00. Use the line tangent to the graph of W at t = 0 to approimate the amount of solid waste that the landfill contains at the end of the first 3 months of 00 (time t = ). 4 dw dw (b) Find in terms of W. Use to determine whether your answer in part is an underestimate or dt dt an overestimate of the amount of solid waste that the landfill contains at time t =. 4 dw (c) Find the particular solution W = W( t) to the differential equation = ( W 300) with initial dt condition W ( 0) = 400. dw = ( W ( 0) 300) = ( ) = 44 dt t= 0 The tangent line is y = t. ( ) ( ) W = 4 tons 4 4 : at 0 : dw t dt = : answer dw dw (b) = = ( W 300) and W 400 dt dt 6 dw Therefore > 0 on the interval 0 t. dt 4 The answer in part is an underestimate. : dw : dt : answer with reason dw (c) = ( W 300) dt dw = dt W 300 ln W 300 = t + C ln ( ) = ( 0) + C ln ( 00) = C W 300 = 00e t t W() t = e, 0 t 0 : : separation of variables : antiderivatives : constant of integration : uses initial condition : solves for W Note: ma [ ] if no constant of integration Note: 0 if no separation of variables 0 The College Board. Visit the College Board on the Web:

8 00 SCORING GUIDELINES (Form B) Consider the differential equation = +. y Question On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated, and for < <, sketch the solution curve that passes through the point ( 0, ). (Note: Use the aes provided in the eam booklet.) (b) While the slope field in part is drawn at only twelve points, it is defined at every point in the y-plane for which y 0. Describe all points in the y-plane, y 0, for which. = (c) Find the particular solution y = f( ) to the given differential equation with the initial condition f ( 0) =. 3 : : zero slopes : nonzero slopes : solution curve through ( 0, ) (b) + = y = y = for all (, y ) with y = and y 0 : description (c) y= ( + ) y = + + C ( ) 0 = C C = y = Since the solution goes through ( 0, ), y must be negative. Therefore y = : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y Note: ma [ ] if no constant of integration Note: 0 if no separation of variables 00 The College Board. Visit the College Board on the Web:

9 Solutions to the differential equation particular solution to the differential equation AP CALCULUS AB 00 SCORING GUIDELINES Question 6 3 y = also satisfy d y 3 y ( y ) 3 = y with () = + 3. Let y = f( ) be a f =. Write an equation for the line tangent to the graph of y = f( ) at =. (b) Use the tangent line equation from part to approimate f (. ). Given that f( ) > 0 for < <., is the approimation for f (.) greater than or less than f (. )? Eplain your reasoning. (c) Find the particular solution y = f( ) with initial condition f () =. () f = = 8 (, ) An equation of the tangent line is y = + 8( ). : f () : : answer (b) f (.).8 Since y = f( ) > 0 on the interval <., ( y ) d y 3 = y + 3 > 0 on this interval. : { : approimation : conclusion with eplanation Therefore on the interval < <., the line tangent to the graph of y = f( ) at = lies below the curve and the approimation.8 is less than f (. ). (c) 3 y = = 3 y = + C y = + C C = 8 y = 4 f( ) =, < < 4 : : separation of variables : antiderivatives : constant of integration : uses initial condition : solves for y Note: ma [ ] if no constant of integration Note: 0 if no separation of variables 00 The College Board. Visit the College Board on the Web:

10 008 SCORING GUIDELINES Question y Consider the differential equation =, where 0. On the aes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the aes provided in the eam booklet.) (b) Find the particular solution y = f( ) to the differential equation with the initial condition f ( ) = 0. (c) For the particular solution y = f( ) described in part (b), find lim f ( ). : zero slopes : { : all other slopes (b) = y ln y = + C y = e + C y = e e y = ke, where k = ± e = ke k = e C ( ) ( ) f = e, > 0 C 6 : : separates variables : antidifferentiates : includes constant of integration : uses initial condition : solves for y Note: ma 3 6 [ ] if no constant of integration Note: 0 6 if no separation of variables ( ) (c) lim e e = : limit 008 The College Board. All rights reserved. Visit the College Board on the Web:

11 Consider the differential equation AP CALCULUS AB 007 SCORING GUIDELINES (Form B) y. = + Question On the aes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the aes provided in the eam booklet.) d y (b) Find in terms of and y. Describe the region in the y-plane in which all solution curves to the differential equation are concave up. (c) Let y = f( ) be a particular solution to the differential equation with the initial condition f ( 0) =. Does f have a relative minimum, a relative maimum, or neither at = 0? Justify your answer. (d) Find the values of the constants m and b, for which y = m + b is a solution to the differential equation. : Sign of slope at each point and relative steepness of slope lines in rows and columns. (b) (c) d y = + = + y Solution curves will be concave up on the half-plane above the line y = +. ( 0, ) d y = 0 + = 0 and ( 0, ) Thus, f has a relative minimum at ( 0, ). = 0 + > 0 3 : d y : : description : answer : { : justification (d) Substituting y = m + b into the differential equation: m = + ( m + b) = ( m + ) + ( b ) Then 0 = m + and m = b : m = and b =. : value for m : { : value for b 007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for students and parents).

12 006 SCORING GUIDELINES (Form B) Question Consider the differential equation = ( y ) cos ( π ). On the aes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the aes provided in the eam booklet.) (b) There is a horizontal line with equation y = c that satisfies this differential equation. Find the value of c. (c) Find the particular solution y = f ( ) to the differential equation with the initial condition f () = 0. : zero slopes : { : all other slopes (b) The line y = satisfies the differential equation, so c =. : c = (c) = cos( π ) ( y ) ( y ) = sin( π ) + C π = sin ( π ) + C y π = sin( π ) + C = C π = sin ( π ) + y π π y = sin( π) + π π y = for sin ( π) + π < < 6 : : separates variables : antiderivatives : constant of integration : uses initial condition : answer Note: ma 3 6 [ ] if no constant of integration Note: 0 6 if no separation of variables 006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6

13 006 SCORING GUIDELINES Question + y Consider the differential equation =, where 0. On the aes provided, sketch a slope field for the given differential equation at the eight points indicated. (Note: Use the aes provided in the pink eam booklet.) (b) Find the particular solution y = f ( ) to the differential equation with the initial condition f ( ) = and state its domain. : sign of slope at each point and relative steepness of slope lines in rows and columns (b) = + y ln + y = ln + K ln + y = e + y = C = C + y = + K y = and < 0 or y = and < 0 7 : : separates variables : antiderivatives 6 : : constant of integration : uses initial condition : solves for y Note: ma 3 6 [ ] if no constant of integration Note: 0 6 if no separation of variables : domain 006 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6

14 00 SCORING GUIDELINES (Form B) Question 6 y Consider the differential equation =. Let y = f( ) be the particular solution to this differential equation with the initial condition f ( ) =. On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the test booklet.) (b) Write an equation for the line tangent to the graph of f at =. (c) Find the solution y = f( ) to the given differential equation with the initial condition f ( ) =. : zero slopes : : nonzero slopes ( 4 ) (b) Slope = = y = ( + ) : equation (c) y = = + C y 4 = + C; C = y = = : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y Note: ma 3 6 [ ] if no constant of integration Note: 0 6 if no separation of variables Copyright 00 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 7

15 Consider the differential equation AP CALCULUS AB 00 SCORING GUIDELINES =. y Question 6 On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the pink test booklet.) (b) Let y = f ( ) be the particular solution to the differential equation with the initial condition (). f = Write an equation for the line tangent to the graph of f at (, ) and use it to approimate f (. ). (c) Find the particular solution y = f ( ) to the given differential equation with the initial condition f () =. : zero slopes : { : nonzero slopes (b) The line tangent to f at (, ) is y + = ( ). Thus, f (.) is approimately 0.8. : : equation of the tangent line : approimation for f (.) (c) = y y= y = + C = + C; C = 3 y = + 3 Since the particular solution goes through (, ), y must be negative. Thus the particular solution is y = 3. : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y Note: ma [ ] if no constant of integration Note: 0 if no separation of variables Copyright 00 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 7

16 004 SCORING GUIDELINES (Form B) Question 4 Consider the differential equation ( y. ) = On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the test booklet.) (b) While the slope field in part is drawn at only twelve points, it is defined at every point in the y-plane. Describe all points in the y-plane for which the slopes are negative. (c) Find the particular solution y = f( ) to the given differential equation with the initial condition f ( 0) = 0. : : zero slope at each point (, y) where = 0 or y = positive slope at each point (, y) where 0 and y > : negative slope at each point (, y) where 0 and y < (b) Slopes are negative at points (, y ) where 0 and y <. : description (c) 4 = y ln y = + C y = e e 0 y = Ke, K = ± e = Ke = K y = e C C 6 : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y 0 if y is not eponential Note: ma 3 6 [ ] if no constant of integration Note: 0 6 if no separation of variables Copyright 004 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 6

17 004 SCORING GUIDELINES Question 6 Consider the differential equation ( y. ) = On the aes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the aes provided in the pink test booklet.) (b) While the slope field in part is drawn at only twelve points, it is defined at every point in the y-plane. Describe all points in the y-plane for which the slopes are positive. (c) Find the particular solution y = f( ) to the given differential equation with the initial condition f ( 0) = 3. : : zero slope at each point (, y) where = 0 or y = positive slope at each point (, y) where 0 and y > : negative slope at each point (, y) where 0 and y < (b) Slopes are positive at points (, y ) where 0 and y >. : description (c) = y 3 ln y = + C 3 y = e e 0 C y = Ke, K = ± e = Ke = K y = + e C 6 : : separates variables : antiderivatives : constant of integration : uses initial condition : solves for y 0 if y is not eponential Note: ma 3 6 [ ] if no constant of integration Note: 0 6 if no separation of variables Copyright 004 by College Entrance Eamination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and (for AP students and parents). 7

18 003 SCORING GUIDELINES (Form B) Question 6 Let f be the function satisfying f ( ) = f( ) for all real numbers, where f ( 3) =. Find f ( 3 ). (b) Write an epression for y = f() by solving the differential equation = y with the initial condition f ( 3) =. f ( ) = f( ) f( ) + = f( ) + f ( ) f (3) = = 3 : : f( ) < > product or chain rule error : value at = 3 (b) = y y = + C = (3) + C ; C = 6 : : separates variables : antiderivative of term : antiderivative of term : constant of integration : uses initial condition f (3) = : solves for y y = ( ) ( ) y = + = Note: ma 3/6 [ ] if no constant of integration Note: 0/6 if no separation of variables Copyright 003 by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com. 7

19 AP CALCULUS AB 003 SCORING GUIDELINES Question A coffeepot has the shape of a cylinder with radius inches, as shown in the figure above. Let h be the depth of the coffee in the pot, measured in inches, where h is a function of time t, measured in seconds. The volume V of coffee in the pot is changing at the rate of h cubic inches per second. (The volume V of a cylinder with radius r and height h is V = r h. ) dh h Show that. dt = dh h (b) Given that h = 7 at time t = 0, solve the differential equation = for dt h as a function of t. (c) At what time t is the coffeepot empty? V = h dv dh = = h dt dt dh h h = = dt 3 : dv : = h dt dv : computes dt : shows result (b) dh dt = h dh = dt h h = t + C 7 = 0 + C ( ) t 7 h = + 0 : separates variables : antiderivatives : constant of integration : : uses initial condition h = 7 when t = 0 : solves for h Note: ma / [ ] if no constant of integration Note: 0/ if no separation of variables (c) ( ) t + = : answer t = 0 7 Copyright 003 by College Entrance Eamination Board. All rights reserved. Available at apcentral.collegeboard.com. 6

20 00 SCORING GUIDELINES (Form B) Г N O = AJO BN >AJDAF=HJE?K=HIKJEJJDACELA@EBBAHAJE=AGK=JEBH N # IK?DJD=JJDAEA O Г EIJ=CAJJJDACH=FDBB.E@JDAN?H@E=JABJDAFEJB J=CA?O=@@AJAHEAMDAJDAHBD=I=?==NEK?=EEKHAEJDAH=JJDEI FEJKIJEBOOKH=IMAH > AJO CN >AJDAF=HJE?K=HIKJEJJDACELA@EBBAHAJE=AGK=JEBH Г N & MEJDJDAEEJE=?@EJE C$ Г".E@O ГO ГO=! Г O! Г IBD=I=?=EEK=JJDEIFEJ H *A?=KIABEI?JEKKIBH N # JDAHA EIFIEJELAJJDA HECDJBN!6DAHABHABD=I=?= EEK=JN! > O@O! O! N Г N + & & Г& + +& O $ N ГN $ O Г $ N ГN $ N!!?=EEK KIJEBE?=JE IAF=H=JAIL=HE=>AI =JE@AHEL=JELAB@OJAH =JE@AHEL=JELAB@N JAH $?IJ=JBEJACH=JE KIAIEEJE=?@EJEC$ Г" ILAIBHO JA=N!$EB?IJ=J BEJACH=JE JA$EBIAF=H=JEBL=HE=>AI Copyright 00 by College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board. 6

21 !0Å#ALCULUSÅ!"n DY X #ONSIDERÅTHEÅDIFFERENTIALÅEQUATIONÅ Y DX E A &INDÅAÅSOLUTIONÅY F X ÅTOÅTHEÅDIFFERENTIALÅEQUATIONÅSATISFYINGÅ F B &INDÅTHEÅDOMAINÅANDÅRANGEÅOFÅTHEÅFUNCTIONÅFÅFOUNDÅINÅPARTÅA Y ÅA E DY X DX Y E X # Y E X # LN Y X # LN # ÅÅ # E LN Y X E ÅSEPARATESÅVARIABLES ÅANTIDERIVATIVEÅOFÅDYÅTERM ÅANTIDERIVATIVEÅOFÅDX ÅTERM ÅCONSTANTÅOFÅINTEGRATION Å ÅUSESÅINITIALÅCONDITIONÅF ÅSOLVESÅFORÅY Å.OTEÅÅIFÅY ÅISÅNOTÅAÅLOGARITHMICÅFUNCTIONÅOFÅX.OTEÅMAXÅÅ;=ÅIFÅNOÅCONSTANTÅOF INTEGRATION.OTEÅÅIFÅNOÅSEPARATIONÅOFÅVARIABLES B $OMAINÅX E X E X E E ANGEÅd Y d ÅX E ÅDOMAIN Å ÅÅÅÅÅÅ.OTEÅÅIFÅÅISÅNOTÅINÅTHEÅDOMAIN ÅRANGE.OTEÅÅIFÅYÅISÅNOTÅAÅLOGARITHMICÅFUNCTIONÅOFÅX Copyright 000 by College Entrance Eamination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the College Entrance Eamination Board.

22 998 AP Calculus AB Scoring Guidelines 4. Let f be a function with f() = 4 such that for all points (, y) on the graph of f the slope is given by 3 +. y Find the slope of the graph of f at the point where =. (b) Write an equation for the line tangent to the graph of f at = and use it to approimate f(.). (c) Find f() by solving the separable differential equation = 3 + with the initial y condition f() = 4. (d) Use your solution from part (c) to find f(.). = 3 + y = y = 4 = = 4 8 = : answer (b) y 4 = ( ) f(.) 4 (. ) { : equation of tangent line : uses equation to approimate f(.) (c) f(.) = 4. y = (3 + ) y = (3 + ) y = C 4 = + + C 4 = C y = y = is branch with point (, 4) f() = : separates variables : antiderivative of term : antiderivative of term : uses y = 4 when = to pick one function out of a family of functions : solves for y 0/ if solving a linear equation in y 0/ if no constant of integration Note: ma 0/ if no separation of variables Note: ma / [ ] if substitutes value(s) for, y, or / before antidifferentiation (d) f(.) = : answer, from student s solution to the given differential equation in (c) Copyright 998 College Entrance Eamination Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Eamination Board.