AP Calculus AB Free Response Notebook Table of Contents

Area and Volume ...... 1-25

Charts with Riemann Sums, MVT, Ave. Rates/Values ...... 41-53

Analyzing Graph of f’ ...... 54-59

Slope Fields with differential Equations ...... 60-70

Related Rates ...... 71-77

Accumulation Functions...... 78-91

Implicit Differentiation ...... 92-97

Particle Motion ...... 98-108

Charts of f, f’, f’’ ...... 109-113

Functions/Misc...... 114-128 1998 AP Calculus AB Scoring Guidelines 1. Let R be the region bounded by the x–axis, the graph of y = √x, and the line x = 4. (a) Find the area of the region R. (b) Find the value of h such that the vertical line x = h divides the region R into two regions of equal area. (c) Find the volume of the solid generated when R is revolved about the x–axis. (d) The vertical line x = k divides the region R into two regions such that when these two regions are revolved about the x–axis, they generate solids with equal volumes. Find the value of k.

(a) y 3 4 y = √x  1: A = Z √x dx 2  0 2  1: answer  1 R

O 1 2 3 4 5 x

4 4 2 3/2 16 A = Z √x dx = x = or 5.333 3 3 0 0

h 8 h 4 (b) Z √x dx = Z √x dx = Z √x dx 1: equation in h 0 3 0 h 2 ( –or– 1: answer 2 8 2 16 2 h3/2 = h3/2 = h3/2 3 3 3 3 − 3

h = √3 16 or 2.520 or 2.519

4 4 x2 2 (c) V = π Z (√x) dx = π = 8π 1: limits and constant 0 2  0  3  1: integrand or 25.133 or 25.132  1: answer 

k k 4 2 2 2 (d) π Z (√x) dx = 4π π Z (√x) dx = π Z (√x) dx 1: equation in k 0 0 k 2 ( –or– 1: answer k2 k2 k2 π = 4π π = 8π π 2 2 − 2

k = √8 or 2.828

AB{2 / BC{2 1999

The shaded region, R , is b ounded by the graph of y = x and the line 2. y

y x 2

y =4,asshown in the gure ab ove.

(a) Find the area of R .

y 4

(b) Find the volume of the solid generated by revolving R ab out the

x{axis.

There exists a number k , k>4, such that when R is revolved ab out

y = k , the resulting solid has the same volume as the solid in

the line O

part (b). Write, but do not solve, an equation involving an integral

expression that can b e used to nd the value of k .

(

1: integral

(4 x ) dx (a) Area =

1: answer

= 2 (4 x ) dx

= 2 4x

= =10:666 or 10:667

2 2 2

1: limits and constant

4 (x ) dx (b) Volume =

1: integrand

1: answer

= 2 (16 x ) dx

= 2 16x

256

= = 160:849 or 160:850

256

2 2 2

1: limits and constant

dx = (c) (k x ) (k 4)

2: integrand

< 1 > each error

1: equation

2 !0Å#ALCULUSÅ!"nÅÅ"#n

y ,ETÅ2ÅBEÅTHEÅSHADEDÅREGIONÅINÅTHEÅFIRSTÅQUADRANTÅENCLOSEDÅBYÅTHEÅGRAPHSÅOF YE X ÅYXCOS ÅANDÅTHEÅY AXIS ÅASÅSHOWNÅINÅTHEÅFIGUREÅABOVE 1.5 A &INDÅTHEÅAREAÅOFÅTHEÅREGIONÅ2 1 2 ye= – x B &INDÅTHEÅVOLUMEÅOFÅTHEÅSOLIDÅGENERATEDÅWHENÅTHEÅREGIONÅ2ÅISÅREVOLVED ABOUTÅTHEÅX AXIS 0.5 R yx= 1 – cos

C 4HEÅREGIONÅ2ÅISÅTHEÅBASEÅOFÅAÅSOLIDÅ&ORÅTHISÅSOLID ÅEACHÅCROSSÅSECTION x O 0.5 1 1.5 PERPENDICULARÅTOÅTHEÅX AXISÅISÅAÅSQUAREÅ&INDÅTHEÅVOLUMEÅOFÅTHISÅSOLID

2EGIONÅ2 Å #ORRECTÅLIMITSÅINÅANÅINTEGRALÅINÅA ÅB EXX COSÅATÅXÅÅÅÅ! ORÅC

! X ¦£ ÅINTEGRAND A !REA ¨ COSEXDX Å¤¦ ¦ ÅANSWER ¥¦ ÅÅORÅ

! £ ÅINTEGRANDÅANDÅCONSTANT B 6OLUME Å Q ¨ EXDXX COS ¦ ¦ Å¤¦ÅÅÅÅÅ ÅEACHÅERROR ¦ ¦ ÅANSWER Å Q ÅÅÅORÅ ¥¦

! ¦£ ÅINTEGRAND X ¦ C 6OLUME ¨ EXDX COS ¦ ¦ÅÅÅÅÅ ÅEACHÅERROR ¦ ¦ÅÅÅÅÅÅ.OTEÅÅIFÅNOTÅOFÅTHEÅFORM Å Å¤¦ ¦ ¦ D ¦ÅÅÅÅÅÅÅÅÅÅÅÅÅKFXGXDX¨ ¦ C ¦ ÅANSWER ¥¦

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Question 1 Let R and S be the regions in the first quadrant shown in the figure above. The region R is bounded by the x-axis and the graphs of yx2 3 and yx tan . The region S is bounded by the y-axis and the graphs of yx2 3 and yx tan . (a) Find the area of R. (b) Find the area of S. (c) Find the volume of the solid generated when S is revolved about the x-axis.

Point of intersection 2tanxx3 at (AB , ) (0.902155,1.265751)

A 3 2 ¦£ 1 : limits (a) Area R = ¨¨tanxdx 2 x3 dx = 0.729 ¦ 0 A ¦ 3 : ¤ 1 : integrand or ¦ ¦ 1 : answer B ¥¦ Area R = ¨ (2yydy )1/3 tan 1 = 0.729 0 or 3 2 A Area R = ¨¨22tanxdx33 x xdx = 0.729 00

A £ 3 ¦ 1 : limits (b) Area S = ¨ 2tanxxdx = 1.160 or 1.161 ¦ 0 ¦ 3 : ¤ 1 : integrand or ¦ ¦ B 2 ¥¦ 1 : answer Area S = ¨¨tan11/3ydy (2 y ) dy = 1.160 or 1.161 0 B or Area S 2 B = ¨¨(2ydy )1/3 (2 y ) 1/3 tan 1 ydy 00 = 1.160 or 1.161

A £ 322 ¦ 1 : limits and constant (c) Volume = 3¨ 2tanxxdx ¦ 0 ¦ 3 : ¤ 1 : integrand = 2.6523 or 8.331 or 8.332 ¦ ¦ ¥¦ 1 : answer

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Question 1

Let 4 be the region bounded by the O-axis and the graphs of N ! O and ON" , as shown in the figure above. N (a) Find the area of 4. (b) Find the volume of the solid generated when 4 is revolved about the N-axis. (c) The region 4 is the base of a solid. For this solid, each cross section perpendicular to the N-axis is a square. Find the volume of this solid.

Region 4 1 : Correct limits in an integral in (a), (b), or (c). N ! " N at N = 1.487664 = ) N

) ¬! £ 1 : integrand N ¦ (a) Area = ¨ " N@N 2 ¤¦ ® N ¦ 1 : answer ¥¦ = 3.214 or 3.215

(b) Volume

¦£ 2 : integrand and constant ) ¬¬N ! ¦ = 3 (4 2N@N ) ¦ ¨ ¦ ® ®1 N 3 ¤ < 1 > each error ¦ ¦ 1 : answer = 31.884 or 31.885 or 10.1493 ¥¦

) ¬N ! ¦£ 2 : integrand (c) Volume = " N@N ¦ ¨ ¦ ® N ¦ 1 each error ¦ ¦ note: 0/2 if not of the form = 8.997 3 ¤¦ ¦ ¦ @ ¦ ¦ kfxgxdx¨ ( ( ) ( )) ¦ ? ¦ 1 : answer ¥¦

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Question 1

Let B and C be the functions given by BN AN and CN( )ln. N (a) Find the area of the region enclosed by the graphs of B and C between N and N 1.

(b) Find the volume of the solid generated when the region enclosed by the graphs of B and C between N and N is revolved about the line O 4.

(c) Let D be the function given by DN( ) BN( ) CN( ). Find the absolute minimum value of DN on the 1 closed interval >>N 1, and find the absolute maximum value of DN on the closed interval 2 1 >>N 1. Show the analysis that leads to your answers. 2

£ 1 : integral (a) Area = AN@NN ln = 1.222 or 1.223 ¦ ¨ 2 ¤ ¦ 1 : answer ¥¦

¦£ 1 : limits and constant (b) Volume = 3 4lnNA@N 4 N ¦ ¨ ¦ ¦ 2 : integrand ¦ = 7.5153 or 23.609 ¦ 1 each error ¦ 4 ¤¦ ¦ Note: 0 / 2 if not of the form ¦ ¦ > ¦ kRxrxdx ( ) ( ) ¦ ¨ ¦ = ¦ 1 : answer ¥¦

£ 1: considers DNa ( ) 0 (c) DNaaa B N CN AN ¦ N ¦ ¦ 1 : identifies critical point ¦ N 0.567143 3 ¤ ¦ and endpoints as candidates ¦ ¦ 1: answers Absolute minimum value and absolute ¥¦ maximum value occur at the critical point or at the endpoints. Note: Errors in computation come off the third point. D(0.567143) 2.330 D(0.5) 2.3418 D(1) 2.718

The absolute minimum is 2.330. The absolute maximum is 2.718.

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Question 1

Let R be the shaded region bounded by the graphs of yx= and ye= 3x and the vertical line x = 1, as shown in the figure above. (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the horizontal line y = 1. (c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a rectangle whose height is 5 times the length of its base in region R. Find the volume of this solid.

Point of intersection 1: Correct limits in an integral in ex3x = at (T, S) = (0.238734, 0.488604) (a), (b), or (c)

1 £ 1 : integrand (a) Area = ()xe 3x dx ¦ ¨T 2 : ¤ ¦ 1 : answer = 0.442 or 0.443 ¥

1 3x 2 2 £ 2 : integrand (b) Volume = Q ()11exdx () ¦ ¨T ()¦ ¦ < 1> reversal = 0.453 Q or 1.423 or 1.424 ¦ ¦ < 1> error with constant 3 : ¤¦ ¦ < 1> omits 1 in one radius ¦ ¦ < 2> other errors ¦ ¦ 1 : answer ¥¦

(c) Length = xe 3x ¦£ 2 : integrand ¦ Height = 5()xe 3x ¦ < 1 > incorrect but has ¦ ¦ 3x 3 : ¤ xe ¦ 1 ¦ 3x 2 ¦ as a factor Volume = ¨ 5()xe dx = 1.554 ¦ T ¦ 1 : answer ¥¦

2 AP® CALCULUS AB 2004 SCORING GUIDELINES (Form B)

Question 1

Let R be the region enclosed by the graph of yx=−1, the vertical line x = 10, and the x-axis. (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the horizontal line y = 3. (c) Find the volume of the solid generated when R is revolved about the vertical line x = 10.

10  1 : limits (a) Area =−=xdx118  ∫1 3 :  1 : integrand  1 : answer

10 2  1 : limits and constant (b) Volume =−−−π ()93()x 1dx  ∫1 3 :  1 : integrand = 212.057 or 212.058  1 : answer

3 2 2  1 : limits and constant (c) Volume =−+π ()10()ydy 1  ∫0 3 :  1 : integrand = 407.150  1 : answer

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2 AP® CALCULUS AB 2004 SCORING GUIDELINES

Question 2

Let f and g be the functions given by f ()xxx=−21 ( ) and g()xxx=−31 ( ) for 01.≤≤x The graphs of f and g are shown in the figure above. (a) Find the area of the shaded region enclosed by the graphs of f and g. (b) Find the volume of the solid generated when the shaded region enclosed by the graphs of f and g is revolved about the horizontal line y = 2.

(c) Let h be the function given by hx()=− kx (1 x ) for 0≤≤x 1. For each k > 0, the region (not shown) enclosed by the graphs of h and g is the base of a solid with square cross sections perpendicular to the x-axis. There is a value of k for which the volume of this solid is equal to 15. Write, but do not solve, an equation involving an integral expression that could be used to find the value of k.

1 1 : integral (a) Area =−()fx() gx () dx 2 : { ∫0 1 : answer 1 =−−−=()21xx()() 3 x 1 xdx 1.133 ∫0

1 1 : limits and constant (b) Volume =−−−π ()()22g()xfxdx22()()  ∫0  2 : integrand  1 2 2 − =−−−−−π ()()23()x 1xxxdx() 221()  1 each error ∫0  4 :  Note: 0 2 if integral not of form = 16.179  b  cRxrxdx()22()− () ∫a    1 : answer

1 2 2 : integrand (c) Volume =−()hx() gx () dx 3 : { ∫0 1 : answer 1 2 ()kx()()131−− x x − x dx = 15 ∫0

Question 1

Let f and g be the functions given by f ()xx=+1sin2 ( ) and gx()= ex 2. Let R be the shaded region in the first quadrant enclosed by the graphs of f and g as shown in the figure above. (a) Find the area of R. (b) Find the volume of the solid generated when R is revolved about the x-axis. (c) The region R is the base of a solid. For this solid, the cross sections perpendicular to the x-axis are semicircles with diameters extending from yfx= () to ygx= (). Find the volume of this solid.

The graphs of f and g intersect in the first quadrant at 1 : correct limits in an integral in (a), (b), ()(ST,= 1.13569, 1.76446 ) . or (c)

S (a) Area =−()fx() gx () dx ⎧ 1 : integrand ∫0 2 : ⎨ 1 : answer S ⎩ =1sin2()+−()x edxx 2 ∫0 = 0.429

S 22 (b) Volume =−π ()()fx() () gx() dx ⎧ 2 : integrand ∫0 ⎪ − 1 each error S 2 ⎪ =+π ⎮⌠ (()1sin2()x 2 −()edxx 2 ) ⎪ Note: 0 2 if integral not of form ⌡0 3 : ⎨ b = 4.266 or 4.267 ⎪ cRxrxdx()22()− () ⎪ ∫a ⎪ ⎩ 1 : answer

S 2 ⌠ π ⎛⎞fx()− gx () (c) Volume = ⎮ ⎜⎟dx ⎧ 2 : integrand ⌡ 22⎝⎠ 3 : ⎨ 0 ⎩ 1 : answer S 2 ⌠ π ⎛⎞1sin2+−()xex 2 = dx ⎮ 22⎜⎟ ⌡0 ⎝⎠ = 0.077 or 0.078

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Question 1

1 − Let f and g be the functions given by f ()xx=+sin ()π and gx()= 4.x Let 4 R be the shaded region in the first quadrant enclosed by the y-axis and the graphs of f and g, and let S be the shaded region in the first quadrant enclosed by the graphs of f and g, as shown in the figure above.

(a) Find the area of R. (b) Find the area of S. (c) Find the volume of the solid generated when S is revolved about the horizontal line y =−1.

1 − f ()xgx= () when +=sin()π x 4 x . 4 f and g intersect when x = 0.178218 and when x = 1. Let a = 0.178218.

a (a) ()gx()−= f () x dx 0.064 or 0.065 ⎧ 1 : limits ∫0 ⎪ 3 : ⎨ 1 : integrand ⎩⎪ 1 : answer

1 (b) ()fx()−= gx () dx 0.410 ⎧ 1 : limits ∫a ⎪ 3 : ⎨ 1 : integrand ⎩⎪ 1 : answer

1 (c) π ()()fx()+−122() gx() + 1 dx = 4.558 or 4.559 2 : integrand ∫a 3 : { 1 : limits, constant, and answer

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Question 1

xxx32 Let f be the function given by f ()xx=−−+3cos . Let R 432 be the shaded region in the second quadrant bounded by the graph of f, and let S be the shaded region bounded by the graph of f and line l, the line tangent to the graph of f at x = 0, as shown above. (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the horizontal line y =−2. (c) Write, but do not evaluate, an integral expression that can be used to find the area of S.

For x < 0, fx()= 0 when x =−1.37312. Let P =−1.37312.

0 (a) Area of Rfxdx==() 2.903 1 : integral ∫ P 2 : { 1 : answer

0 2 (b) Volume =+−=π ()()fx() 24 dx 59.361 ⎧ 1 : limits and constant ∫ P ⎪ 4 : 2 : integrand ⎨ ⎩⎪ 1 : answer

1 (c) The equation of the tangent line is yx=−3. ⎧ 1 : tangent line l 2 ⎪ 3 : ⎨ 1 : integrand ⎪ 1 : limits The graph of f and line l intersect at A = 3.38987. ⎩

A ⌠ 1 Area of Sxfxdx=−−⎮ ()()3 () ⌡0 2

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Question 1

Let R be the shaded region bounded by the graph of yx= ln and the line yx=−2, as shown above. (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the horizontal line y =−3. (c) Write, but do not evaluate, an integral expression that can be used to find the volume of the solid generated when R is rotated about the y-axis.

ln()xx=− 2 when x = 0.15859 and 3.14619.

Let S = 0.15859 and T = 3.14619

T (a) Area of Rxxdx=−−=()ln() ( 2 ) 1.949 ⎧ 1 : integrand ∫S ⎪ 3 : 1 : limits ⎨ ⎪ ⎩ 1 : answer

T (b) Volume =+−−+π ()()ln()x 32 ()xdx 2 3 2 2 : integrand ∫S 3 : { 1 : limits, constant, and answer = 34.198 or 34.199

− T 2 2 (c) Volume =+−π ⎮⌠ ((2)yedy2 ()y ) 2 : integrand ⌡S −2 3 : { 1 : limits and constant

Question 1

2 Let R be the region bounded by the graph of ye= 2x− x and the horizontal line y = 2, and let S be the region bounded by the graph of 2 ye= 2x− x and the horizontal lines y = 1 and y = 2, as shown above. (a) Find the area of R. (b) Find the area of S. (c) Write, but do not evaluate, an integral expression that gives the volume of the solid generated when R is rotated about the horizontal line y = 1.

2 e2xx− = 2 when x = 0.446057, 1.553943 Let P = 0.446057 and Q = 1.553943

Q ⎧ 1 : integrand 2xx− 2 ⎪ (a) Area of Re=−=⌠ 20.5dx14 3 : ⎨ 1 : limits ⌡P ( ) ⎪ 1 : answer ⎩

2 (b) e2xx− = 1 when x = 0, 2 ⎧ 1 : integrand ⎪ 3 : ⎨ 1 : limits 2 2 ⎪ Area of Se=−−⌠ 2xx− 1 dxArea of R ⎩ 1 : answer ⌡0 ( ) =2.06016 −Area of R = 1.546 OR

P 222 ⌠⌠edxQPed22xx−−−+−⋅+11() xx −1x ⌡⌡0 ( ) Q ( ) =++=0.219064 1.107886 0.219064 1.546

Q 2 ⌠ ⎛⎞2xx− 2 2 2 : integrand (c) Volume =−π ⎮ ed121−−()x 3 : ⎜⎟( ) { 1 : constant and limits ⌡P ⎝⎠

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Question 1

20 Let R be the region in the first and second quadrants bounded above by the graph of y = and 1 + x2 below by the horizontal line y = 2. (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the x-axis. (c) The region R is the base of a solid. For this solid, the cross sections perpendicular to the x-axis are semicircles. Find the volume of this solid.

20 = 2 when x =±3 1 : correct limits in an integral in + 2 1 x (a), (b), or (c)

3 ⌠ ⎛⎞20 1 : integrand (a) Area =−=⎮ ⎜⎟2dx 37.961 or 37.962 2 : { 2 ⌡−3 ⎝⎠1 + x 1 : answer

3 2 ⌠ ⎛⎞⎛⎞20 2 2 : integrand (b) Volume =−=π ⎜⎟⎜⎟2dx 1871.190 3 : { ⎮ ⎝⎠+ 2 1 : answer ⌡−3 ⎝⎠1 x

3 2 π ⌠ 120 2 : integrand (c) Volume =−⎛⎛2 ⎞⎞dx 3 : { 22⎮ ⎜⎜+ 2 ⎟⎟ 1 : answer ⌡−3 ⎝⎝1 x ⎠⎠ 3 2 π ⌠ 20 =−=⎛⎞2dx 174.268 8 ⎮ ⎜⎟+ 2 ⌡−3 ⎝⎠1 x

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Question 1

x Let R be the region in the first quadrant bounded by the graphs of yx= and y = . 3 (a) Find the area of R. (b) Find the volume of the solid generated when R is rotated about the vertical line x =−1. (c) The region R is the base of a solid. For this solid, the cross sections perpendicular to the y-axis are squares. Find the volume of this solid.

x The graphs of yx= and y = intersect at the points 3 ()0, 0 and ()9, 3 .

9 x (a) ⎮⌠ ()xdx−=4.5 ⎧ 1 : limits ⌡0 3 ⎪ 3 : ⎨ 1 : integrand OR ⎩⎪ 1 : answer 3 ()34.5yydy−=2 ∫ 0

3 2 (b) π ⎮⌠ (()31yydy+−2 ()2 + 1) ⎧ 1 : constant and limits ⌡0 ⎪ 4 : ⎨ 2 : integrand 207π ==130.061 or 130.062 ⎪ 1 : answer 5 ⎩

3 2 1 : integrand (c) ()38.1yy−=2 dy 2 : { ∫ 0 1 : limits and answer

Question 1

Let R be the region bounded by the graphs of yx= sin()π and yx=−3 4, x as shown in the figure above. (a) Find the area of R. (b) The horizontal line y =−2 splits the region R into two parts. Write, but do not evaluate, an integral expression for the area of the part of R that is below this horizontal line. (c) The region R is the base of a solid. For this solid, each cross section perpendicular to the x-axis is a square. Find the volume of this solid. (d) The region R models the surface of a small pond. At all points in R at a distance x from the y-axis, the depth of the water is given by hx()=−3. x Find the volume of water in the pond.

3 (a) sin()π x =−xx 4 at x = 0 and x = 2 ⎧ 1 : limits ⎪ 2 3 : 1 : integrand Area =−−=()sin()π xx()3 4 xdx 4 ⎨ ∫ 0 ⎩⎪ 1 : answer

(b) xx3 −=−42 at r = 0.5391889 and s = 1.6751309 1 : limits 2 : { s The area of the stated region is ()−−24()x3 −xdx 1 : integrand ∫ r

2 2 1 : integrand (c) Volume =−−=()sin()π xx()3 4 xdx 9.978 2 : { ∫ 0 1 : answer

2 1 : integrand (d) Volume =−()()3xxxxdx() sinπ −−()3 4 = 8.369 or 8.370 2 : { ∫ 0 1 : answer

Consider the curve defined by the equation y + cosy = x +1 for 0 ≤ y ≤ 2π .

dy (a) Find in terms of y . dx

(b) Write an equation for each vertical tangent to the curve.

d 2y (c) Find in terms of y . dx2

dy dy (a) −=sin y 1 dx dx dy ()1s−=iny 1 dx

dy = 1 − dx 1siny

dy (b) undefined when sin y =1 dx π y = 2 π +=01x + 2 π x =−1 2

1 d  2 − (c) dy= 1siny 2 dx dx  −− dy cos y = dx 2 ()1s− iny

1 cos y  1s− iny = ()− 2 1siny = cos y ()1s− iny 3

Let R be the region between the graphs of y =1 + sin(πx) and y = x 2 from x = 0 to x = 1.

(a) Find the area of R .

(b) Set up, but do not integrate an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the x-axis.

(c) Set up, but do not integrate an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the y-axis.

1 (a) Ax=+∫ 1sin()π −x2 dx 0 1 11 =−xxcos()π −x3 π 3 0 111 =−11()−− −0−−0 ππ3  =+22 3 π

1 2 (b) Vx=+ππ∫ ()1sin()−x4 dx 0 or 2 1 ⌠ 2 22ππ∫ yd3/2 y+− y1arcsin()y−1dy 0 π ⌡1 

1 (c) Vx=+21ππ∫ ()sin()x−x2 dx 0 or

2 22 1 ⌠ 11 ππ∫ ydy+− 1 arcsin ()y−1 −arcsin ()y−1 dy 0 ππ ⌡1

Let R be the region enclosed by the graphs of y = ex , y =(x −1) 2 , and the line x =1.

(a) Find the area of R .

(b) Find the volume of the solid generated when R is revolved about the x-axis.

(c) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the y-axis.

22 1990 AB3 Solution

1 (a) Ae=−∫ x ()x−1 2 dx 0 1 =−∫ exx 2 +21x−dx 0 1 =−x  1 1 −3  ex ()1  0 3 0 14 =−()ee1 −=− 33

1 (b) Ve=−π ∫ 2x ()x−1 4 dx 0 1 2x 1 e 1 5 =−ππ ()x −1 25 0 0 ee2211 7 =ππ−−=− 22 5 210

1 e Vy=−21ππ⌠ ()1−ydy+2∫ y()1−lnydy ⌡ 0 1 e 211  11 =⋅22ππyy5/2 + 2 −y2 lny−y2  52 24 0 1 2 412 3e 7 =+ππ2 e −=π− 544210

1 (c) Vx=−21π ∫ ex ()x−2 dx 0  or 1 2 e Vy=−ππ⌠ 11()−dy+∫ 1−()lny2 dy ⌡ 0 1

23 1989 AB2

Let R be the region in the first quadrant enclosed by the graph of y = 6x + 4 , the line y = 2x , and the y-axis.

(a) Find the area of R .

(b) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the x-axis.

(c) Set up, but do not integrate, an integral expression in terms of a single variable for the volume of the solid generated when R is revolved about the y-axis.

2 (a) Area =+∫ 6x 4 −2xdx 0 2 12 3/2 =⋅()64x + −x2 63 0 64 8 20 =−4 −= 999

(b) Volume about x-axis

2 V = π ∫ ()64x +−4xd2 x 0

2 32π V = π ∫ ()64xd+x− 0 3

2 V = 26π ∫ x()xx+−42dx 0 or 4 4 2 ⌠ yy2 ⌠ 2 − 4 V = ππ dy −  dy ⌡ 26 0 ⌡ 2

Question 2

The rate at which people enter an amusement park on a given day is modeled by the function E defined by 15600 Et . tt 24 160 The rate at which people leave the same amusement park on the same day is modeled by the function L defined by 9890 Lt . tt 38 370

Both -J and Lt are measured in people per hour and time t is measured in hours after midnight. These functions are valid for 923,>>J the hours during which the park is open. At time J 9, there are no people in the park. (a) How many people have entered the park by 5:00 P.M. (J % )? Round answer to the nearest whole number. (b) The price of admission to the park is $15 until 5:00 P.M. (J % ). After 5:00 P.M., the price of admission to the park is $11. How many dollars are collected from admissions to the park on the given day? Round your answer to the nearest whole number. J (c) Let Ht Ex Lxdx for 923.>>J The value of 0 % to the nearest whole number is 3725. ¨' Find the value of 0 = % and explain the meaning of 0 % and 0 = % in the context of the park. (d) At what time t, for 923,>>J does the model predict that the number of people in the park is a maximum?

% £ (a) -J@J( ) 6004.270 ¦ 1 : limits ¨' ¦ ¦ 6004 people entered the park by 5 pm. 3 ¤ 1: integrand ¦ ¦ 1: answer ¥¦ % ! (b) 15¨¨-J@J( )11 -J@J( ) 104048.165 ' % 1 : setup The amount collected was $104,048. or ! -J@J( ) 1271.283 ¨ % 1271 people entered the park between 5 pm and 11 pm, so the amount collected was $15 ¸¸(6004) $11 (1271) $104, 041.

(c) HELa(17) (17) (17) 380.281 ¦£ 1 : value of 0 a (17) ¦ ¦ 2 : meanings There were 3725 people in the park at t = 17. ¦ ¦ The number of people in the park was decreasing 3 ¤¦ 1 : meaning of 0 (17) ¦ at the rate of approximately 380 people/hr at ¦ 1 : meaning of 0 a (17) ¦ time t = 17. ¦ 1 if no reference to J 17 ¦¥

£1: Et( ) Lt( )0 (d) Hta Et Lt ¦ 2 ¤ ¦ 1 : answer t = 15.794 or 15.795 ¥¦ Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered 26trademarks of the College Entrance Examination Board. 3 AP® CALCULUS AB 2003 SCORING GUIDELINES (Form B)

Question 2

A tank contains 125 gallons of heating oil at time t = 0. During the time interval 012bbt hours, heating oil is pumped into the tank at the rate 10 Ht( ) =+2 gallons per hour. ()1ln1++()t During the same time interval, heating oil is removed from the tank at the rate ¬t2 Rt()= 12 sin gallons per hour. ® 47 (a) How many gallons of heating oil are pumped into the tank during the time interval 0bbt 12 hours? (b) Is the level of heating oil in the tank rising or falling at time t = 6 hours? Give a reason for your answer. (c) How many gallons of heating oil are in the tank at time t = 12 hours? (d) At what time t, for 0bbt 12, is the volume of heating oil in the tank the least? Show the analysis that leads to your conclusion.

12 £ 1 : integral (a) Htdt() = 70.570 or 70.571 ¦ ¨ 2 : ¤¦ 0 ¦ 1 : answer ¥¦

(b) HR(6)(6) 2.924,= 1 : answer with reason so the level of heating oil is falling at t = 6.

12 + ()Ht Rt dt ¦£ 1 : limits (c) 125¨ ( )( ) = 122.025 or 122.026 ¦ 0 ¦ 3 : ¤ 1 : integrand ¦ ¦ 1 : answer ¥¦

(d) The absolute minimum occurs at a critical point ¦£ 1 : Htsets Rt = ( )( )0 ¦ or an endpoint. ¦ 1 : volume is least at ¦ Ht Rt = t = t = ¦ ()()0 when 4.790 and 11.318. 3 : ¤¦ t = 11.318 ¦ ¦ ¦ 1 : analysis for absolute The volume increases until t = 4.790, then ¦ ¦ minimum decreases until t = 11.318, then increases, so the ¥¦ absolute minimum will be at t = 0 or at t = 11.318.

11.318 125+ ()Ht ()() Rt dt = 120.738 ¨0

Since the volume is 125 at t = 0, the volume is least at t = 11.318.

3 AP® CALCULUS AB 2004 SCORING GUIDELINES (Form B)

Question 2

For 031,≤≤t the rate of change of the number of mosquitoes on Tropical Island at time t days is t modeled by Rt()= 5cos t () mosquitoes per day. There are 1000 mosquitoes on Tropical Island at 5 time t = 0. (a) Show that the number of mosquitoes is increasing at time t = 6. (b) At time t = 6, is the number of mosquitoes increasing at an increasing rate, or is the number of mosquitoes increasing at a decreasing rate? Give a reason for your answer. (c) According to the model, how many mosquitoes will be on the island at time t = 31? Round your answer to the nearest whole number. (d) To the nearest whole number, what is the maximum number of mosquitoes for 0≤≤t 31? Show the analysis that leads to your conclusion.

(a) Since R()6=> 4.438 0, the number of mosquitoes is 1 : shows that R()60> increasing at t = 6.

(b) R′()61.913=−  1 : considers R′() 6 2 :  Since R′()60,< the number of mosquitoes is  1 : answer with reason increasing at a decreasing rate at t = 6.

+=31 ()  1 : integral (c) 1000∫ Rt dt 964.335 2 :  0  1 : answer To the nearest whole number, there are 964 mosquitoes.

(d) Rt()= 0 when t = 0 , t = 2.5π , or t = 7.5π  2 : absolute maximum value Rt()> 0 on 0< 0 on 7.5π <

There are 964 mosquitoes at t = 31, so the maximum number of mosquitoes is 1039, to the nearest whole number.

3 AP® CALCULUS AB 2004 SCORING GUIDELINES

Question 1

Traffic flow is defined as the rate at which cars pass through an intersection, measured in cars per minute. The traffic flow at a particular intersection is modeled by the function F defined by t Ft()=+82 4sin() for 030,≤≤t 2 where F()t is measured in cars per minute and t is measured in minutes. (a) To the nearest whole number, how many cars pass through the intersection over the 30-minute period? (b) Is the traffic flow increasing or decreasing at t = 7? Give a reason for your answer. (c) What is the average value of the traffic flow over the time interval 10≤≤t 15 ? Indicate units of measure. (d) What is the average rate of change of the traffic flow over the time interval 10≤≤t 15 ? Indicate units of measure.

30  1 : limits (a) Ft() dt= 2474 cars  ∫0 3 :  1 : integrand  1 : answer

(b) F′()7=− 1.872 or − 1.873 1 : answer with reason Since F′()70,< the traffic flow is decreasing at t = 7.

1 15  1 : limits (c) Ft() dt= 81.899 cars min  5 ∫10 3 :  1 : integrand  1 : answer

FF()15− () 10 1 : answer (d) = 1.517 or 1.518 cars min2 15− 10

Units of cars min in (c) and cars min2 in (d) 1 : units in (c) and (d)

2 AP® CALCULUS AB 2005 SCORING GUIDELINES (Form B)

Question 2

A water tank at Camp Newton holds 1200 gallons of water at time t = 0. During the time interval 018≤≤t hours, water is pumped into the tank at the rate t Wt()= 95 t sin2 () gallons per hour. 6 During the same time interval, water is removed from the tank at the rate t Rt()= 275sin2 () gallons per hour. 3 (a) Is the amount of water in the tank increasing at time t = 15 ? Why or why not? (b) To the nearest whole number, how many gallons of water are in the tank at time t = 18 ? (c) At what time t, for 018,≤≤t is the amount of water in the tank at an absolute minimum? Show the work that leads to your conclusion. (d) For t > 18, no water is pumped into the tank, but water continues to be removed at the rate R()t until the tank becomes empty. Let k be the time at which the tank becomes empty. Write, but do not solve, an equation involving an integral expression that can be used to find the value of k.

(a) No; the amount of water is not increasing at t = 15 1 : answer with reason since WR()15−=−< () 15 121.09 0.

18 (b) 1200+−=()Wt() Rt () dt 1309.788 ⎧ 1 : limits ∫0 ⎪ 3 : 1 : integrand 1310 gallons ⎨ ⎪ ⎩ 1 : answer

(c) Wt()−= Rt () 0 ⎧ 1 : interior critical points t = 0, 6.4948, 12.9748 ⎪ 1 : amount of water is least at 3 : ⎨ = t (hours) gallons of water ⎪ t 6.494 or 6.495 0 1200 ⎩⎪ 1 : analysis for absolute minimum 6.495 525 12.975 1697 18 1310

The values at the endpoints and the critical points show that the absolute minimum occurs when t = 6.494 or 6.495.

k (d) Rt() dt= 1310 ⎧ 1 : limits ∫18 2 : ⎨ ⎩ 1 : equation

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Question 2

The tide removes sand from Sandy Point Beach at a rate modeled by the function R, given by 4πt Rt()=+25sin( ) . 25 A pumping station adds sand to the beach at a rate modeled by the function S, given by 15t St()= . 13+ t Both R()t and St() have units of cubic yards per hour and t is measured in hours for 0≤≤t 6. At time t = 0, the beach contains 2500 cubic yards of sand. (a) How much sand will the tide remove from the beach during this 6-hour period? Indicate units of measure. (b) Write an expression for Yt(), the total number of cubic yards of sand on the beach at time t. (c) Find the rate at which the total amount of sand on the beach is changing at time t = 4. (d) For 0≤≤t 6, at what time t is the amount of sand on the beach a minimum? What is the minimum value? Justify your answers.

6 (a) Rt() dt= 31.815 or 31.816 yd3 1 : integral ∫0 2 : { 1 : answer with units

t (b) Yt()=+2500 () Sx () − Rx () dx ⎧ 1 : integrand ∫0 ⎪ 3 : 1 : limits ⎨ ⎩⎪ 1 : answer

(d) Yt′()= 0 when St()−= Rt () 0. ⎧ 1 : sets Yt′()= 0 ⎪ The only value in [0, 6] to satisfy St()= Rt () 3 : ⎨ 1 : critical t -value is a = 5.117865. ⎩⎪ 1 : answer with justification

t Yt()

0 2500 a 2492.3694 6 2493.2766

The amount of sand is a minimum when t = 5.117 or 5.118 hours. The minimum value is 2492.369 cubic yards.

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Question 2

At an intersection in Thomasville, Oregon, cars turn t left at the rate Lt()= 60 t sin2 ( ) cars per hour 3 over the time interval 0≤≤t 18 hours. The graph of yLt= () is shown above. (a) To the nearest whole number, find the total number of cars turning left at the intersection over the time interval 0≤≤t 18 hours. (b) Traffic engineers will consider turn restrictions when Lt()≥ 150 cars per hour. Find all values of t for which Lt()≥ 150 and compute the average value of L over this time interval. Indicate units of measure. (c) Traffic engineers will install a signal if there is any two-hour time interval during which the product of the total number of cars turning left and the total number of oncoming cars traveling straight through the intersection is greater than 200,000. In every two-hour time interval, 500 oncoming cars travel straight through the intersection. Does this intersection require a traffic signal? Explain the reasoning that leads to your conclusion.

18 (a) Lt() dt≈ 1658 cars 1 : setup ∫0 2 : { 1 : answer

(b) Lt()= 150 when t = 12.42831, 16.12166 ⎧ 1 : tLt -interval when ()≥ 150 ⎪ Let R = 12.42831 and S = 16.12166 3 : ⎨ 1 : average value integral ≥ Lt() 150 for t in the interval [R, S] ⎩⎪ 1 : answer with units 1 S Lt() dt= 199.426 cars per hour SR− ∫R

(c) For the product to exceed 200,000, the number of cars ⎧ 1 : considers 400 cars turning left in a two-hour interval must be greater than 400. ⎪ 1 : valid interval []hh ,+ 2 ⎪ 4 : ⎨ h+2 15 1 : value of Lt() dt Lt() dt=>431.931 400 ⎪ ∫h ∫13 ⎪ 1 : answer and explanation ⎩

OR OR

The number of cars turning left will be greater than 400 ⎧ 1 : considers 200 cars per hour on a two-hour interval if Lt()≥ 200 on that interval. ⎪ 1 : solves Lt()≥ 200 ()≥ 4 : ⎨ Lt 200 on any two-hour subinterval of ⎪ 1 : discusses 2 hour interval [13.25304, 15.32386] . ⎩⎪ 1 : answer and explanation

Yes, a traffic signal is required.

Question 2

The amount of water in a storage tank, in gallons, is modeled by a continuous function on the time interval 07,≤≤t where t is measured in hours. In this model, rates are given as follows: (i) The rate at which water enters the tank is f ()ttt= 1002 sin() gallons per hour for 07.≤≤t (ii) The rate at which water leaves the tank is ⎧ 250 for 0≤

7 1 : integral (a) ftdt() ≈ 8264 gallons 2 : { ∫ 0 1 : answer

(b) The amount of water in the tank is decreasing on the 1 : intervals 2 : { intervals 0≤≤t 1.617 and 3≤≤t 5.076 because 1 : reason ()< () ≤< << f tgt for 0t 1.617 and 3t 5.076.

(c) Since f ()tgt− () changes sign from positive to negative ⎧ 1 : identifies t = 3 as a candidate only at t = 3, the candidates for the absolute maximum are ⎪ 1 : integrand ⎪ at t = 0, 3, and 7. 5 : ⎨ 1 : amount of water at t = 3 ⎪ 1 : amount of water at t = 7 ⎪ t (hours) gallons of water ⎩ 1 : conclusion 0 5000 3 3 5000+−=ftdt() 250 () 3 5126.591 ∫ 0

7 7 5126.591+−=ftdt() 2000 ( 4 ) 4513.807 ∫ 3

The amount of water in the tank is greatest at 3 hours. At that time, the amount of water in the tank, rounded to the nearest gallon, is 5127 gallons.

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Question 2

− 2 For time t ≥ 0 hours, let rt()=−120( 1 e10t ) represent the speed, in kilometers per hour, at which a car travels along a straight road. The number of liters of gasoline used by the car to travel x kilometers is − modeled by gx()=−0.05 x() 1 ex 2 .

(a) How many kilometers does the car travel during the first 2 hours? (b) Find the rate of change with respect to time of the number of liters of gasoline used by the car when t = 2 hours. Indicate units of measure. (c) How many liters of gasoline have been used by the car when it reaches a speed of 80 kilometers per hour?

2 1 : integral (a) rt() dt= 206.370 kilometers 2 : { ∫ 0 1 : answer

dg dg dx dx (b) =⋅; = rt() 2 : uses chain rule dt dx dt dt 3 : { 1 : answer with units dg dg =⋅r()2 dttx==2 dx 206.370 ==()()0.050 120 6 liters hour

(c) Let T be the time at which the car’s speed reaches ⎧ 1 : equation rt()= 80 ⎪ 80 kilometers per hour. 4 : ⎨ 2 : distance integral

⎩⎪ 1 : answer Then, rT()= 80 or T = 0.331453 hours.

At time T, the car has gone T xT()== r () t dt 10.794097 kilometers ∫ 0 and has consumed gxT()()= 0.537 liters of gasoline.

Question 3

Oil is leaking from a pipeline on the surface of a lake and forms an oil slick whose volume increases at a constant rate of 2000 cubic centimeters per minute. The oil slick takes the form of a right circular cylinder with both its radius and height changing with time. (Note: The volume V of a right circular cylinder with radius r and height h is given by Vrh= π 2 . ) (a) At the instant when the radius of the oil slick is 100 centimeters and the height is 0.5 centimeter, the radius is increasing at the rate of 2.5 centimeters per minute. At this instant, what is the rate of change of the height of the oil slick with respect to time, in centimeters per minute? (b) A recovery device arrives on the scene and begins removing oil. The rate at which oil is removed is R()tt= 400 cubic centimeters per minute, where t is the time in minutes since the device began working. Oil continues to leak at the rate of 2000 cubic centimeters per minute. Find the time t when the oil slick reaches its maximum volume. Justify your answer. (c) By the time the recovery device began removing oil, 60,000 cubic centimeters of oil had already leaked. Write, but do not evaluate, an expression involving an integral that gives the volume of oil at the time found in part (b).

dV dV dr (a) When r = 100 cm and h = 0.5 cm, = 2000 cm3 min ⎧ 1 : == 2000 and 2.5 dt ⎪ dt dt ⎪ dr 4 : ⎨ dV and = 2.5 cm min. 2 : expression for dt ⎪ dt ⎩⎪ 1 : answer dV dr dh =+2ππrh r2 dt dt dt 2 dh 2000=+ 2ππ()()() 100 2.5 0.5 () 100 dt dh = 0.038 or 0.039 cm min dt

dV dV (b) =−2000R()t , so = 0 when Rt()= 2000. ⎧ 1 : Rt()= 2000 dt dt ⎪ 3 : This occurs when t = 25 minutes. ⎨ 1 : answer ⎪ dV dV ⎩ 1 : justification Since > 0 for 025< 25, dt dt the oil slick reaches its maximum volume 25 minutes after the device begins working.

(c) The volume of oil, in cm3 , in the slick at time t = 25 minutes 1 : limits and initial condition 2 : { 25 1 : integrand is given by 60,000+−() 2000R()tdt . ∫ 0

Oil is being pumped continuously from a certain oil well at a rate proportional to the dy amount of oil left in the well; that is, = ky, where y is the amount of oil left in the dt well at any time t . Initially there were 1,000,000 gallons of oil in the well, and 6 years later there were 500,000 gallons remaining. It will no longer be profitable to pump oil when there are fewer than 50,000 gallons remaining.

(a) Write an equation for y , the amount of oil remaining in the well at any time t .

(b) At what rate is the amount of oil in the well decreasing when there are 600,000 gallons of oil remaining?

dy  = kdt dy y = ky  =+ (a) dt or ln yktC1 kt  = kt+C yCe ye= 1 

=⇒ = 66= tC010,C1 ln10 ∴=ye106 kt 1 te=⇒6 =6k 2 ln 2 ∴=k − 6 −−tt ln 2 ye==10666610 ⋅2

dy ln 2 (b) ==ky − ⋅61⋅05 dt 6 =−105 ln 2 Decreasing at 105 ln 2 gal/year

5. The temperature outside a house during a 24-hour period is given by

πt F (t) = 80 10 cos , 0 t 24, − 12 ≤ ≤ where F (t) is measured in degrees Fahrenheit and t is measured in hours. (a) Sketch the graph of F on the grid below. (b) Find the average temperature, to the nearest degree Fahrenheit, between t = 6 and t = 14. (c) An air conditioner cooled the house whenever the outside temperature was at or above 78 degrees Fahrenheit. For what values of t was the air conditioner cooling the house? (d) The cost of cooling the house accumulates at the rate of $0.05 per hour for each degree the outside temperature exceeds 78 degrees Fahrenheit. What was the total cost, to the nearest cent, to cool the house for this 24–hour period?

(a) 100 1: bell–shaped graph

t minimum 70 at t = 0, t = 24 only ¢ i

e 90 h

n maximum 90 at t = 12 only e r h a F

80 s e e r

g £ e

D 70

¡ ¡ 60 0 6 12 18 24 Time in Hours 1 14 πt 2: integral (b) Avg. = Z 80 10 cos dt  14 6 12  6 −  1: limits and 1/(14 6) −  −  1: integrand 1  = (697.2957795)  8 3  1: answer = 87.162 or 87.161  0/1 if integral not of the form   b  1 87◦ F  Z F (t) dt ≈  b a a  − πt (c) 80 10 cos 78 0 1: inequality or equation − 12 − ≥ 2 ( πt 1: solutions with interval 2 10 cos 0 − 12 ≥ 5.230 18.769   or  t  or 5.231 ≤ ≤ 18.770  

18.770 or 2: integral 18.769 πt   (d) C = 0.05 Z 80 10 cos 78 dt  1: limits and 0.05 5.231 − 12 −  or  1: integrand  5.230  3  1: answer = 0.05(101.92741) = 5.096 $5.10  0/1 if integral not of the form ≈   b   k Z (F (t) 78) dt  a −  Copyright ©1998 College Entrance Examination38 Board. All rights reserved. Advanced Placement Program and AP are registered trademarks of the College Entrance Examination Board. !0Å#ALCULUSÅ!"n

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Question 2

The number of gallons, 2J , of a pollutant in a lake changes at the rate 2Ja ! A0.2 J gallons per day, where J is measured in days. There are 50 gallons of the pollutant in the lake at time J = 0. The lake is considered to be safe when it contains 40 gallons or less of pollutant. (a) Is the amount of pollutant increasing at time J = 9 ? Why or why not? (b) For what value of J will the number of gallons of pollutant be at its minimum? Justify your answer. (c) Is the lake safe when the number of gallons of pollutant is at its minimum? Justify your answer. (d) An investigator uses the tangent line approximation to 2J at J = 0 as a model for the amount of pollutant in the lake. At what time J does this model predict that the lake becomes safe?

1 : answer with reason (a) 2Aa(9) 1 30.6 0.646 < 0 so the amount is not increasing at this time.

(b) 2Ja ! A0.2 J ¦a£ 1 : sets 2J( )0 ¦ ¦ J = 5ln3 = 30.174 3 ¤ 1 : solves for J ¦ 2J= is negative for 0 < J < 5ln3 and positive ¦ 1 : justification ¥¦ for J > 5ln3 . Therefore there is a minimum at J = 5ln3 .

30.174 0.2 J £ 1 : integrand (c) 2A@J(30.174) 50 1 3 ¦ ¨0 ¦ ¦ 1 : limits ¦ = 35.104 < 40, so the lake is safe. 3 ¤¦ ¦ 1 : conclusion with reason ¦ ¦ ¦ based on integral of 2Ja( ) ¥¦

(d) 2 a ! . The lake will become safe ¦£ 1 : slope of tangent line 2 ¤¦ ¦ 1 : answer when the amount decreases by 10. A linear model ¥¦ predicts this will happen when J = 5.

AB{3 / BC{3 1999

t 3. The rate at whichwater ows out of a pip e, in gallons p er hour, is R (t)

given by a di erentiable function R of time t. The table ab ove

(hours) (gallons p er hour)

shows the rate as measured every 3 hours for a 24{hour p erio d.

0 9.6

(a) Use a midp oint Riemann sum with 4 sub divisions of equal

3 10.4

10.8 6

length to approximate R (t) dt. Using correct units, explain

9 11.2

the meaning of your answer in terms of water ow.

11.4 12

(b) Is there some time t,0

15 11.3

your answer.

10.7 18

21 10.2

. Use Q(t)toapproximate the Q(t)= 768 + 23t t

24 9.6

average rate of water ow during the 24{hour time p erio d.

Indicate units of measure.

1: R (3) + R (9) + R (15) + R (21)

R (t) dt 6[R (3) + R (9) + R (15) + R (21)] (a)

1: answer

=6[10:4+ 11:2+ 11:3+ 10:2] 3

= 258.6 gallons

1: explanation

This is an approximation to the total owin

gallons of water from the pip e in the 24{hour

period.

(

1: answer

(b) Yes;

1: MVT or equivalent

Since R (0) = R (24) = 9:6, the Mean Value

Theorem guarantees that there is a t,0

suchthat R (t)=0.

1: limits and average value constant

average value of Q(t)

3 1: Q(t)asintegrand

1: answer

1 1

(768 + 23t t ) dt =

24 79

=10:785 gal/hr or 10.784 gal/hr

(units) Gallons in part (a) and gallons/hr in 1: units

part (c), or equivalent.

41 1998 Calculus AB Scoring Guidelines v(t) t v(t) 90 (seconds) (feet per second) 80 0 0 70 5 12 60 10 20 50 15 30 40 20 55 Velocity 30 25 70 30 78

(feet per second) 20 35 81 10 40 75 t O 5 10 15 20 25 30 35 40 45 50 45 60 50 72 Time (seconds) 3. The graph of the velocity v(t), in ft/sec, of a car traveling on a straight road, for 0 t 50, is shown above. A table of values for v(t), at 5 second intervals of time t, is shown to the right of the graph.≤ ≤ (a) During what intervals of time is the acceleration of the car positive? Give a reason for your answer. (b) Find the average acceleration of the car, in ft/sec2, over the interval 0 t 50. ≤ ≤ (c) Find one approximation for the acceleration of the car, in ft/sec2, at t = 40. Show the computations you used to arrive at your answer. 50 (d) Approximate Z v(t) dt with a Riemann sum, using the midpoints of ﬁve subintervals of equal length. 0 Using correct units, explain the meaning of this integral.

(a) Acceleration is positive on (0, 35) and (45, 50) because 1: (0, 35) the velocity v(t) is increasing on [0, 35] and [45, 50]  3  1: (45, 50) 1: reason  Note: ignore inclusion of endpoints

v(50) v(0) 72 0 72 (b) Avg. Acc. = − = − = 1: answer 50 0 50 50 − or 1.44 ft/sec2

(c) Diﬀerence quotient; e.g. 1: method ( v(45) v(40) 60 75 2 − = − = 3 ft/sec2 or 1: answer 5 5 − Note: 0/2 if ﬁrst point not earned v(40) v(35) 75 81 6 − = − = ft/sec2 or 5 5 −5 v(45) v(35) 60 81 21 − = − = ft/sec2 10 10 −10 –or– Slope of tangent line, e.g. 90 75 through (35, 90) and (40, 75): − = 3 ft/sec2 35 40 − − 50 (d) Z v(t) dt 1: midpoint Riemann sum 0   10[v(5) + v(15) + v(25) + v(35) + v(45)] 3  1: answer ≈ = 10(12 + 30 + 70 + 81 + 60) 1: meaning of integral  = 2530 feet  This integral is the total distance traveled in feet over the time 0 to 50 seconds.

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Question 2

The temperature, in degrees Celsius (°C), of the water in a pond is a t Wt() differentiable function W of time t. The table above shows the water (days) (°C) 0 20 temperature as recorded every 3 days over a 15-day period. 3 31 (a) Use data from the table to find an approximation for W =(12) . Show the 6 28 9 24 computations that lead to your answer. Indicate units of measure. 12 22 (b) Approximate the average temperature, in degrees Celsius, of the water 15 21

over the time interval 015>>t days by using a trapezoidal approximation with subintervals of length %t 3 days. (c) A student proposes the function P, given by Pt( ) 20 10 te(/3)t , as a model for the temperature of the water in the pond at time t, where t is measured in days and Pt() is measured in degrees Celsius. Find P=(12) . Using appropriate units, explain the meaning of your answer in terms of water temperature. (d) Use the function P defined in part (c) to find the average value, in degrees Celsius, of Pt() over the time interval 015>>t days.

(a) Difference quotient; e.g. £ ¦ 1 : difference quotient a xWW(15) (12) 1 ° ¤ W (12) C/day or 2 : ¦ 15 12 3 ¥¦ 1 : answer (with units) WW(12) (9) 2 W a(12) x °C/day or 12 9 3 WW(15) (9) 1 W a(12) x °C/day 15 9 2 3 ¦£ 1 : trapezoidal method (b) 20 2(31) 2(28) 2(24) 2(22) 21 376.5 ¦ 2 2 : ¤ ¦ 1 : answer 1 ¥ Average temperature x(376.5) 25.1 °C 15

a tt/310 /3 £ a (c) Pete(12) 10 ¦ 1 : P (12) (with or without units) 3 t12 2 : ¤ ¦ 1 : interpretation 30e 4 0.549 °C/day ¥¦ This means that the temperature is decreasing at the rate of 0.549 °C/day when t = 12 days.

15 1 t /3 ¦£ 1 : integrand (d) ¨ 20 10te dt 25.757 °C ¦ 15 0 ¦ ¦ 1 : limits and 3 : ¤ ¦ average value constant ¦ ¦ ¥¦ 1 : answer

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Question 3

A blood vessel is 360 millimeters (mm) long Distance with circular cross sections of varying diameter. x (mm) 0 60 120 180 240 300 360 Diameter The table above gives the measurements of the B(x) (mm) 24 30 28 30 26 24 26 diameter of the blood vessel at selected points along the length of the blood vessel, where x represents the distance from one end of the blood vessel and Bx() is a twice-differentiable function that represents the diameter at that point. (a) Write an integral expression in terms of Bx() that represents the average radius, in mm, of the blood vessel between x = 0 and x = 360. (b) Approximate the value of your answer from part (a) using the data from the table and a midpoint Riemann sum with three subintervals of equal length. Show the computations that lead to your answer. 275 Bx 2 ¬() dx (c) Using correct units, explain the meaning of Q in terms of the blood vessel. ¨125 ® 2 (d) Explain why there must be at least one value x, for 0360,<

360 Bx 1 : limits and constant 1 ()dx ¦£ (a) 2 : ¤¦ 360¨0 2 ¦ 1 : integrand ¥¦

1 BB(60) (180) B (300) ¬¯ ¦£ 1 : BB (60)++ (180) B (300) (b) ¡°120 ++ = 2 : ¦ 360¡°® 2 2 2 ¤ ¢±¦ 1 : answer 1 ¥ []60() 30++ 30 24 = 14 360

Bx() ¬Bx()2 ¦£ 1 : volume in mm3 (c) is the radius, so Q is the area of ¦ 2 ® 2 ¦ 2 : ¤ 1 : between x = 125 and the cross section at x. The expression is the ¦ ¦ x = 275 ¥¦ volume in mm3 of the blood vessel between 125 mm and 275 mm from the end of the vessel.

(d) By the MVT, Bca()= 0 for some c in £ 1 1 ¦ 2 : explains why there are two ¦ (60, 180) and Bca()= 0 for some c in ¦ values of xBx where a ( ) has 2 2 ¦ ¦ (240, 360). The MVT applied to Bxa() shows 3 : ¤ the same value ¦ that Bxaa()= 0 for some x in the interval ¦ 1 : explains why that means ¦ cc ¦ Bxaa ( )=<< 0 for 0 x 360 ()12,. ¥¦

Note: max 1/3 if only explains why Bxa()= 0 at some x in (0, 360).

4 AP® CALCULUS AB 2003 SCORING GUIDELINES

Question 3

The rate of fuel consumption, in gallons per minute, recorded during an airplane flight is given by a twice-differentiable and strictly increasing function R of time t. The graph of R and a table of selected values of Rt(), for the time interval 090bbt minutes, are shown above. (a) Use data from the table to find an approximation for Ra()45 . Show the computations that lead to your answer. Indicate units of measure. (b) The rate of fuel consumption is increasing fastest at time t = 45 minutes. What is the value of Raa()45 ? Explain your reasoning.

90 (c) Approximate the value of Rt() dt using a left Riemann sum with the five subintervals indicated ¨0 90 by the data in the table. Is this numerical approximation less than the value of Rt() dt ? ¨0 Explain your reasoning. b (d) For 090< b b minutes, explain the meaning of Rtdt() in terms of fuel consumption for the ¨0 1 b plane. Explain the meaning of Rtdt() in terms of fuel consumption for the plane. Indicate b ¨0 units of measure in both answers.

RR(50) (40) 55 40 1 : a difference quotient using (a) Ra(45) x = ¦£ 50 40 10 ¦ ¦ numbers from table and 2 ¦ = 1.5 gal/min 2 : ¤¦ ¦ interval that contains 45 ¦ ¦ 1 : 1.5 gal/min2 ¥¦ Raa = Rta £ 1 : Raa (45)= 0 (b) (45) 0 since () has a maximum at ¦ 2 : ¤ t = 45 . ¦ 1 : reason ¥¦ 90 £ 1 : value of left Riemann sum (c) Rt( ) dt x (30)(20)++ (10)(30) (10)(40) ¦ ¨0 2 : ¤ ¦ 1 : “less” with reason ++(20)(55) (20)(65) = 3700 ¥ Yes, this approximation is less because the graph of R is increasing on the interval.

b £ 2 : meanings (d) Rt() dt is the total amount of fuel in ¦ ¨0 ¦ b ¦ Rt dt gallons consumed for the first b minutes. ¦ 1 : meaning of ( ) ¦ ¨0 1 b ¦ 1 b Rt() dt is the average value of the rate of 3 : ¤ Rt dt b ¨ ¦ 1 : meaning of b ¨ ( ) 0 ¦ 0 fuel consumption in gallons/min during the ¦ < 1 > if no reference to time b ¦ b ¦ 1 : units in both answers first minutes. ¥¦

Question 3

A test plane flies in a straight line with t (min) 0 5 10 15 20 25 30 35 40 positive velocity vt(), in miles per vt() (mpm) 7.0 9.2 9.5 7.0 4.5 2.4 2.4 4.3 7.3 minute at time t minutes, where v is a differentiable function of t. Selected values of vt() for 040≤≤t are shown in the table above. (a) Use a midpoint Riemann sum with four subintervals of equal length and values from the table to

40 approximate vt() dt. Show the computations that lead to your answer. Using correct units, ∫0 40 explain the meaning of vt() dt in terms of the plane’s flight. ∫0 (b) Based on the values in the table, what is the smallest number of instances at which the acceleration of the plane could equal zero on the open interval 0<

(a) Midpoint Riemann sum is  1 : vv() 5+++ ( 15 ) v ( 25 ) v ( 35 ) 10⋅+[vv() 5 ( 15 ) + v ( 25 ) + v ( 35 )]  3 :  1 : answer =⋅[] + + + = 10 9.2 7.0 2.4 4.3 229  1 : meaning with units The integral gives the total distance in miles that the plane flies during the 40 minutes.

(b) By the Mean Value Theorem, vt′()= 0 somewhere in  1 : two instances 2 :  the interval ()0, 15 and somewhere in the interval  1 : justification ()25, 30 . Therefore the acceleration will equal 0 for at least two values of t.

1 40  1 : limits (d) Average velocity = ftdt()  40 ∫0 3 :  1 : integrand = 5.916 miles per minute  1 : answer

4 AP® CALCULUS AB 2005 SCORING GUIDELINES

Question 3

Distance 0 1 5 6 8 x (cm) Temperature

Tx() ()°C 100 93 70 62 55

A metal wire of length 8 centimeters (cm) is heated at one end. The table above gives selected values of the temperature Tx(), in degrees Celsius ()°C, of the wire x cm from the heated end. The function T is decreasing and twice differentiable. (a) Estimate T ′()7. Show the work that leads to your answer. Indicate units of measure. (b) Write an integral expression in terms of Tx() for the average temperature of the wire. Estimate the average temperature of the wire using a trapezoidal sum with the four subintervals indicated by the data in the table. Indicate units of measure. 8 8 (c) Find Txdx′() , and indicate units of measure. Explain the meaning of Txdx′() in terms of the temperature of the ∫0 ∫0 wire. (d) Are the data in the table consistent with the assertion that Tx′′()> 0 for every x in the interval 08?<

TT()86− () 55− 62 7 (a) ==−°Ccm 1 : answer 86− 2 2

1 8 1 8 (b) Txdx() ⎧ 1 : Txdx() 8 ∫0 ⎪ 8 ∫0 3 : 8 ⎨ 1 : trapezoidal sum Trapezoidal approximation for Txdx() : ⎪ ∫0 ⎩⎪ 1 : answer 10093937070626255++ ++ A =⋅+⋅+⋅+⋅1412 2222 1 Average temperature ≈=A 75.6875 ° C 8

8 (c) TxdxT′()=−=−=−° ()805510045C T () 1 : value ∫0 2 : { The temperature drops 45° C from the heated end of the wire to the 1 : meaning other end of the wire.

70− 93 (d) Average rate of change of temperature on [1, 5] is =−5.75. 1 : two slopes of secant lines 51− 2 : { 62− 70 1 : answer with explanation Average rate of change of temperature on [5, 6] is =−8. 65− ′()=− () No. By the MVT, Tc1 5.75 for some c1 in the interval 1, 5 ′()=− () and Tc2 8 for some c2 in the interval 5, 6 . It follows that ′ () ′′ T must decrease somewhere in the interval cc12,. Therefore T is not positive for every x in [0, 8] .

Units of °Ccm in (a), and ° C in (b) and (c) 1 : units in (a), (b), and (c)

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Question 6

t 0 15 25 30 35 50 60 (sec) vt() –20 –30 –20 –14 –10 0 10 ()ft sec at() 1 5 2 1 2 4 2 ()ft sec2

A car travels on a straight track. During the time interval 060≤≤t seconds, the car’s velocity v, measured in feet per second, and acceleration a, measured in feet per second per second, are continuous functions. The table above shows selected values of these functions.

60 (a) Using appropriate units, explain the meaning of vt() dt in terms of the car’s motion. Approximate ∫30 60 vt() dt using a trapezoidal approximation with the three subintervals determined by the table. ∫30 30 (b) Using appropriate units, explain the meaning of at() dt in terms of the car’s motion. Find the exact value ∫0 30 of at() dt. ∫0 (c) For 060,<

60 (a) vt() dt is the distance in feet that the car travels 1 : explanation ∫30 2 : { 1 : value from t = 30 sec to t = 60 sec. 60 Trapezoidal approximation for vt() dt: ∫30 111 A =++()14105()() 1015 +()() 1010185 = ft 222

30 (b) at() dt is the car’s change in velocity in ft/sec from 1 : explanation ∫0 2 : { 1 : value t = 0 sec to t = 30 sec.

30 30 a() t dt==− v′ () t dt v (30 ) v ( 0 ) ∫∫00 =−14 −() − 20 = 6 ft/sec

(c) Yes. Since vv()35=− 10 <− 5 < 0 = () 50 , the IVT ⎧ 1 : vv() 35<− 5 < () 50 2 : ⎨ guarantees a t in ()35, 50 so that vt()=−5. ⎩ 1 : Yes; refers to IVT or hypotheses

(d) Yes. Since vv()025,= ( ) the MVT guarantees a t in ⎧ 1 : vv() 0= ( 25 ) 2 : ⎨ ()0, 25 so that at()== v′ () t 0. ⎩ 1 : Yes; refers to MVT or hypotheses Units of ft in (a) and ft/sec in (b) 1 : units in (a) and (b)

Question 4

t 0 10 20 30 40 50 60 70 80 (seconds) vt() 5 14 22 29 35 40 44 47 49 (feet per second)

Rocket A has positive velocity vt() after being launched upward from an initial height of 0 feet at time t = 0 seconds. The velocity of the rocket is recorded for selected values of t over the interval 0≤≤t 80 seconds, as shown in the table above. (a) Find the average acceleration of rocket A over the time interval 0≤≤t 80 seconds. Indicate units of measure. 70 (b) Using correct units, explain the meaning of vt() dt in terms of the rocket’s flight. Use a midpoint ∫10 70 Riemann sum with 3 subintervals of equal length to approximate vt() dt. ∫10 3 (c) Rocket B is launched upward with an acceleration of at()= feet per second per second. At time t + 1 t = 0 seconds, the initial height of the rocket is 0 feet, and the initial velocity is 2 feet per second. Which of the two rockets is traveling faster at time t = 80 seconds? Explain your answer.

(a) Average acceleration of rocket A is 1 : answer

vv()80− () 0 49− 5 11 == ft sec2 80− 0 80 20

70 (b) Since the velocity is positive, vt() dt represents the ⎧ 1 : explanation ∫10 ⎪ ()()() = 3 : ⎨ 1 : uses vvv 20 , 40 , 60 distance, in feet, traveled by rocket A from t 10 seconds ⎪ to t = 70 seconds. ⎩ 1 : value

A midpoint Riemann sum is 20[vvv() 20++ () 40 () 60 ]

=++=20[] 22 35 44 2020 ft

() (c) Let vtB be the velocity of rocket B at time t. ⎧ 1 : 6t + 1 ⎪ ()==++⌠ 3 1 : constant of integration vtB ⎮ dt61 t C ⎪ ⌡ t + 1 4 : ⎨ 1 : uses initial condition 206==+vC() ⎪ () () B 1 : finds vvB 80 , compares to 80 , ()=+− ⎪ vtB 614 t ⎩⎪ and draws a conclusion ()=>= () vvB 80 50 49 80

Rocket B is traveling faster at time t = 80 seconds.

Units of ft sec2 in (a) and ft in (b) 1 : units in (a) and (b)

Question 5

t 0 2 5 7 11 12 (minutes) rt′() 5.7 4.0 2.0 1.2 0.6 0.5 (feet per minute)

The volume of a spherical hot air balloon expands as the air inside the balloon is heated. The radius of the balloon, in feet, is modeled by a twice-differentiable function r of time t, where t is measured in minutes. For 012,<

dV4 dr dV (b) = 3()π r2 ⎪⎧ 2 : dt3 dt 3 : ⎨ dt dV 2 ⎩⎪ 1 : answer ==4ππ() 30 2 7200 ft3 min dt t =5

12 (c) rtdt′()≈++++2 () 4.0 3 () 2.0 2 () 1.2 4 () 0.6 1 () 0.5 1 : approximation ∫ 0 2 : { 1 : explanation = 19.3 ft 12 rtdt′() is the change in the radius, in feet, from ∫ 0 t = 0 to t = 12 minutes.

(d) Since r is concave down, r′ is decreasing on 0<

Units of ft3 min in part (b) and ft in part (c) 1 : units in (b) and (c) © 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 50 AP® CALCULUS AB 2008 SCORING GUIDELINES (Form B)

Question 3

Distance from the 0 8 14 22 24 river’s edge (feet) Depth of the water (feet) 0 7 8 2 0

A scientist measures the depth of the Doe River at Picnic Point. The river is 24 feet wide at this location. The measurements are taken in a straight line perpendicular to the edge of the river. The data are shown in the table above. The velocity of the water at Picnic Point, in feet per minute, is modeled by vt()=+16 2sin() t + 10 for 0120≤≤t minutes. (a) Use a trapezoidal sum with the four subintervals indicated by the data in the table to approximate the area of the cross section of the river at Picnic Point, in square feet. Show the computations that lead to your answer. (b) The volumetric flow at a location along the river is the product of the cross-sectional area and the velocity of the water at that location. Use your approximation from part (a) to estimate the average value of the volumetric flow at Picnic Point, in cubic feet per minute, from t = 0 to t = 120 minutes. π x (c) The scientist proposes the function f, given by fx()= 8sin() , as a model for the depth of the 24 water, in feet, at Picnic Point x feet from the river’s edge. Find the area of the cross section of the river at Picnic Point based on this model. (d) Recall that the volumetric flow is the product of the cross-sectional area and the velocity of the water at a location. To prevent flooding, water must be diverted if the average value of the volumetric flow at Picnic Point exceeds 2100 cubic feet per minute for a 20-minute period. Using your answer from part (c), find the average value of the volumetric flow during the time interval 40≤≤t 60 minutes. Does this value indicate that the water must be diverted?

()07++++() 78() 82() 20 (a) ⋅+8682 ⋅+ ⋅+ ⋅ 1 : trapezoidal approximation 2222 = 115 ft2

1 120 (b) 115vt() dt ⎧ 1 : limits and average value 120 ∫ 0 ⎪ constant = 1807.169 or 1807.170 ft3 min 3 : ⎨ 1 : integrand ⎪ ⎩⎪ 1 : answer

24 π x = 2 1 : integra1 (c) 8sin()dx 122.230 or 122.231 ft 2 : { ∫ 0 24 1 : answer

(d) Let C be the cross-sectional area approximation from ⎧ 1 : volumetric flow integral ⎪ part (c). The average volumetric flow is 3 : ⎨ 1 : average volumetric flow 1 60 Cvtdt⋅=() 2181.912 or 2181.913 ft3 min. ⎩⎪ 1 : answer with reason 20 ∫ 40

Yes, water must be diverted since the average volumetric flow for this 20-minute period exceeds 2100 ft3 min.

Question 2

t (hours) 0 1 3 4 7 8 9

Lt() (people) 120 156 176 126 150 80 0

Concert tickets went on sale at noon ()t = 0 and were sold out within 9 hours. The number of people waiting in line to purchase tickets at time t is modeled by a twice-differentiable function L for 0≤≤t 9. Values of Lt() at various times t are shown in the table above. (a) Use the data in the table to estimate the rate at which the number of people waiting in line was changing at 5:30 P.M. ()t = 5.5 . Show the computations that lead to your answer. Indicate units of measure. (b) Use a trapezoidal sum with three subintervals to estimate the average number of people waiting in line during the first 4 hours that tickets were on sale. (c) For 09,≤≤t what is the fewest number of times at which Lt′() must equal 0 ? Give a reason for your answer. − (d) The rate at which tickets were sold for 09≤≤t is modeled by rt()= 550 tet 2 tickets per hour. Based on the model, how many tickets were sold by 3 P.M. ()t = 3, to the nearest whole number?

LL()74− () 150− 126 1 : estimate (a) L′()5.5≈== 8 people per hour 2 : { 74− 3 1 : units (b) The average number of people waiting in line during the first 4 hours is 1 : trapezoidal sum 2 : { approximately 1 : answer 1 ⎛⎞LL()01++ () LL() 13 () LL() 34 + () ⎜⎟()10−+ (31) −+() 43 − 42⎝⎠ 2 2 = 155.25 people (c) L is differentiable on [0, 9] so the Mean Value Theorem implies ⎧ 1 : considers change in Lt′()> 0 for some t in ()1, 3 and some t in ()4, 7 . Similarly, ⎪ sign of L′ 3 : ′ < ′ ⎨ Lt() 0 for some t in ()3, 4 and some t in ()7, 8 . Then, since L is ⎪ 1 : analysis continuous on [0, 9] , the Intermediate Value Theorem implies that ⎩⎪ 1 : conclusion Lt′()= 0 for at least three values of t in [0, 9] .

OR OR

The continuity of L on [1, 4] implies that L attains a maximum value ⎧ 1 : considers relative extrema there. Since LL()31> () and LL()34,> () this maximum occurs on ⎪ of L on () 0, 9 3 : ⎨ ()1, 4 . Similarly, L attains a minimum on ()3, 7 and a maximum on ⎪ 1 : analysis ()4, 8 . L is differentiable, so Lt′()= 0 at each relative extreme point ⎩⎪ 1 : conclusion on ()0, 9 . Therefore Lt′()= 0 for at least three values of t in [0, 9] .

[Note: There is a function L that satisfies the given conditions with Lt′()= 0 for exactly three values of t.]

3 (d) rt() dt= 972.784 1 : integrand ∫ 0 2 : { 1 : limits and answer There were approximately 973 tickets sold by 3 P.M.

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Question 4

The figure above shows the graph of f ′, the derivative of the function f, on the closed interval −≤15.x ≤ The graph of f ′ has horizontal tangent lines at x = 1 and x = 3. The function f is twice differentiable with f ()26.= (a) Find the x-coordinate of each of the points of inflection of the graph of f. Give a reason for your answer. (b) At what value of x does f attain its absolute minimum value on the closed interval −≤15?x ≤ At what value of x does f attain its absolute maximum value on the closed interval −≤ 1x ≤ 5 ? Show the analysis that leads to your answers. (c) Let g be the function defined by gx()= xfx (). Find an equation for the line tangent to the graph of g at x = 2.

(a) x = 1 and x = 3 because the graph of f ′ changes from  1 : xx== 1, 3 2 :  increasing to decreasing at x = 1, and changes from  1 : reason decreasing to increasing at x = 3.

(b) The function f decreases from x =− 1 to x = 4, then  1 : indicates f decreases then increases = = increases from x 4 to x 5.  1 : eliminates x = 5 for maximum = 4 :  Therefore, the absolute minimum value for f is at x 4. 1 : absolute minimum at x = 4 The absolute maximum value must occur at x =− 1 or   1 : absolute maximum at x =− 1 at x = 5.  5 ff()51−−= ( ) ftdt′ () < 0 ∫−1 Since ff()51,<− ( ) the absolute maximum value occurs at x =−1.

Tangent line is yx=−+4212()

545

AP® CALCULUS AB 2006 SCORING GUIDELINES (Form B)

Question 2

Let f be the function defined for x ≥ 0 with f ()05= and f ′ , the ()− first derivative of f, given by f ′()xe= x 4 sin() x2 . The graph of yfx= ′() is shown above.

(a) Use the graph of f ′ to determine whether the graph of f is concave up, concave down, or neither on the interval 1.7<

(a) On the interval 1.7<

′ (b) fx()= 0 when x = 0,πππ , 2 , 3 , K ⎧ 1 : identifies π and 3 as candidates On [0, 3] f ′ changes from positive to negative ⎪ - or - ⎪ only at π . The absolute maximum must occur at ⎪ indicates that the graph of f 3 : = π ⎨ x or at an endpoint. ⎪ increases, decreases, then increases ⎪ 1 : justifies ff()π > () 3 f ()05= ⎪ ⎩ 1 : answer π ff()π =+()0 fxdx′ () = 5.67911 ∫0 3 ff()3=+ () 0 fxdx′ () = 5.57893 ∫0

This shows that f has an absolute maximum at x = π .

2 (c) ff()2=+ () 0∫ fxdx′ () = 5.62342 ⎧ 2 : f () 2 expression 0 ⎪ − 1 : integral fe′()2==−0.5 sin () 4 0.45902 ⎪ 4 : ⎨ 1 : including f () 0 term

⎪ 1 : f ′() 2 yx−=−−5.623()() 0.459 2 ⎪ 1 : equation ⎩

Question 4

A particle moves along the x-axis so that its velocity at time t, for 06,≤≤t is given by a differentiable function v whose graph is shown above. The velocity is 0 at t = 0, t = 3, and t = 5, and the graph has horizontal tangents at t = 1 and t = 4. The areas of the regions bounded by the t-axis and the graph of v on the intervals [0, 3] , [3, 5] , and [5, 6] are 8, 3, and 2, respectively. At time t = 0, the particle is at x =− 2. (a) For 0≤≤t 6, find both the time and the position of the particle when the particle is farthest to the left. Justify your answer. (b) For how many values of t, where 06,≤≤t is the particle at x =− 8 ? Explain your reasoning. (c) On the interval 2<

(a) Since vt()< 0 for 0< 0 ⎧ 1 : identifies t = 3 as a candidate for 35,<

(b) The particle moves continuously and monotonically from ⎧ 1 : positions at tt== 3, 5, x()02=− to x()310.=− Similarly, the particle moves ⎪ and t = 6 3 : x()310=− ⎨ continuously and monotonically from to ⎪ 1 : description of motion x()57=− and also from x()57=− to x()69.=− ⎩⎪ 1 : conclusion

By the Intermediate Value Theorem, there are three values of t for which the particle is at xt()=−8. (c) The speed is decreasing on the interval 2<

The figure above shows the graph of f ′ , the derivative of a function f . The domain of f is the set of all real numbers x such that −10 ≤ x ≤ 10.

(a) For what values of x does the graph of f have a horizontal tangent?

(b) For what values of x in the interval (1− 0,10) does f have a relative maximum? Justify your answer.

(a) horizontal tangent ⇔=fx′() 0 x =−7, −1, 4, 8

(b) Relative maxima at x =−1, 8 because f ′ changes from positive to negative at these points

Question 5

dy Consider the differential equation =−xy4 ()2. dx (a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the axes provided in the test booklet.) (b) While the slope field in part (a) is drawn at only twelve points, it is defined at every point in the xy-plane. Describe all points in the xy-plane for which the slopes are negative. (c) Find the particular solution yfx= () to the given differential equation with the initial condition f ()00.=

(a)  1 : zero slope at each point ()xy ,  where xy== 0 or 2    ()   positive slope at each point x , y 2 :   where xy≠> 0 and 2    1 :    negative slope at each point ()x , y   ≠<   where xy 0 and 2

(b) Slopes are negative at points ()x, y 1 : description ≠ < where x 0 and y 2.

1 4  1 : separates variables (c) dy= x dx y − 2  2 : antiderivatives  1 lnyxC−= 2 5 +  1 : constant of integration 6 :  5 1 : uses initial condition 1  x5 yee−=2 C 5  1 : solves for y  1 5  0 1 if y is not exponential x C yKeKe−=2,5 =± −=2 Ke0 = K Note: max 3 6 [1-2-0-0-0] if no 1 constant of integration x5 ye=−225 Note: 0 6 if no separation of variables

6 AP® CALCULUS AB 2004 SCORING GUIDELINES

Question 6

dy Consider the differential equation =−xy2 ()1. dx (a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the axes provided in the pink test booklet.) (b) While the slope field in part (a) is drawn at only twelve points, it is defined at every point in the xy-plane. Describe all points in the xy-plane for which the slopes are positive. (c) Find the particular solution yfx= () to the given differential equation with the initial condition f ()03.=

(a)  1 : zero slope at each point ()xy ,  where xy== 0 or 1     positive slope at each point ()x , y 2 :   where xy≠> 0 and 1    1 :    negative slope at each point ()x , y     where xy≠< 0 and 1

(b) Slopes are positive at points ()x, y 1 : description where x ≠ 0 and y > 1.

1 (c) dy= x2 dx y − 1  1 : separates variables 1  2 : antiderivatives lnyxC−= 1 3 +  3  1 : constant of integration 1 3 6 :  x 1 : uses initial condition yee−=1 C 3   1 1 : solves for y x3  yKeKe−=1,3 =±C  0 1 if y is not exponential 2 ==Ke0 K 1 Note: max 3 6 [1-2-0-0-0] if no constant of x3 ye=+123 integration Note: 0 6 if no separation of variables

7 AP® CALCULUS AB 2005 SCORING GUIDELINES (Form B)

Question 6

dy− xy2 Consider the differential equation = . Let dx 2 yfx= () be the particular solution to this differential equation with the initial condition f ()−=12. (a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the axes provided in the test booklet.) (b) Write an equation for the line tangent to the graph of f at x =− 1. (c) Find the solution yfx= () to the given differential equation with the initial condition f ()−=12.

(a) ⎧ 1 : zero slopes 2 : ⎨ ⎩ 1 : nonzero slopes

−−()14 (b) Slope ==2 1 : equation 2 yx−=22() + 1

1 x (c) dy=− dx ⎧ 1 : separates variables y2 2 ⎪ 2 : antiderivatives 1 x2 ⎪ −=−+C 6 : ⎨ 1 : constant of integration y 4 ⎪ 1 : uses initial condition 11 1 ⎪ −=−+CC; =− ⎩⎪ 1 : solves for y 24 4 14 y == Note: max 3 6 [1-2-0-0-0] if no 22+ xx+ 11 constant of integration 44 Note: 0 6 if no separation of variables

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Question 6

dy 2x Consider the differential equation =− . dx y

(a) On the axes provided, sketch a slope field for the given differential equation at the twelve points indicated. (Note: Use the axes provided in the pink test booklet.) (b) Let yfx= () be the particular solution to the differential equation with the initial condition f ()11.=− Write an equation for the line tangent to the graph of f at ()1,− 1 and use it to approximate f ()1.1 . (c) Find the particular solution yfx= () to the given differential equation with the initial condition f ()11.=−

(a) 1 : zero slopes 2 : { 1 : nonzero slopes

(b) The line tangent to f at ()1,− 1 is yx+=12() − 1. ⎧ 1 : equation of the tangent line 2 : ⎨ Thus, f ()1.1 is approximately − 0.8. ⎩ 1 : approximation for f () 1.1

dy 2x (c) =− 1 : separates variables dx y ⎧ ⎪ 1 : antiderivatives ydy=−2 xdx ⎪ 5 : ⎨ 1 : constant of integration y2 =−x2 +C ⎪ 1 : uses initial condition 2 ⎪ 1 3 ⎩ 1 : solves for y =−1; +C C = 2 2 yx22=−23 + Note: max 2 5 [1-1-0-0-0] if no constant of integration Since the particular solution goes through ()1,− 1 , Note: 0 5 if no separation of variables y must be negative.

Thus the particular solution is yx=−32. − 2

Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudent62 s (for AP students and parents). 7 AP® CALCULUS AB 2006 SCORING GUIDELINES (Form B)

Question 5

dy Consider the differential equation =−()()yx1cos2 π . dx (a) On the axes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the axes provided in the exam booklet.)

(b) There is a horizontal line with equation yc= that satisfies this differential equation. Find the value of c. (c) Find the particular solution yfx= () to the differential equation with the initial condition f ()10.=

(a) 1 : zero slopes 2 : { 1 : all other slopes

(b) The line y = 1 satisfies the differential equation, so 1 : c = 1 c = 1.

1 (c) dy= cos()π x dx ⎧ 1 : separates variables ()− 2 y 1 ⎪ 2 : antiderivatives ⎪ −−()−1 =1 ()π + 6 : 1 : constant of integration yxC1sinπ ⎨ ⎪ 1 : uses initial condition 11 ⎪ =+sin()π x C 1 : answer 1 − y π ⎩

=+=1 ()π 1sinπ CC Note: max 3 6 [1-2-0-0-0] if no 11 constant of integration =+sin()π x 1 1 − y π Note: 06 if no separation of variables π =+sin()ππx 1 − y π y =−1 for −<<∞∞x sin()ππx +

Question 5

dy1 + y Consider the differential equation = , where x ≠ 0. dx x (a) On the axes provided, sketch a slope field for the given differential equation at the eight points indicated. (Note: Use the axes provided in the pink exam booklet.)

(b) Find the particular solution yfx= () to the differential equation with the initial condition f ()−=11 and state its domain.

(a) 2 : sign of slope at each point and relative steepness of slope lines in rows and columns

11 (b) dy= dx 1 : separates variables 1 + yx ⎧ ⎧ ⎪ ⎪ 2 : antiderivatives ⎪ ⎪ ln 1+=yxK ln + ⎪ 6 : ⎨ 1 : constant of integration ⎪ ⎪ 1 : uses initial condition ⎪ ⎪ +=ln x + K ⎪ ⎩ 1 : solves for y 1 ye 7 : ⎨ Note: max 3 6 [] 1-2-0-0-0 if no 1 +=yCx ⎪ = ⎪ constant of integration 2 C ⎪ 12+=yx ⎪ Note: 0 6 if no separation of variables yx=−21 and x < 0 ⎪ ⎪ or ⎩ 1 : domain yx=−21 − and x < 0

Question 5

dy 1 Consider the differential equation =+−xy1. dx 2 (a) On the axes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the axes provided in the exam booklet.) dy2 (b) Find in terms of x and y. Describe the region in the xy-plane in dx2 which all solution curves to the differential equation are concave up. (c) Let yfx= ( ) be a particular solution to the differential equation with the

initial condition f (01) = . Does f have a relative minimum, a relative maximum, or neither at x = 0 ? Justify your answer. (d) Find the values of the constants m and b, for which ymxb= + is a solution to the differential equation.

(a) 2 : Sign of slope at each point and relative steepness of slope lines in rows and columns.

dy2 11 dy 1 ⎧ dy2 (b) =+ =xy +− ⎪ 2 : dx2 22dx 2 3 : ⎨ dx2 Solution curves will be concave up on the half-plane above the line ⎩⎪ 1 : description 11 yx=− + . 22

dy dy2 1 (c) =+−=0110 and = 01+− > 0 1 : answer dx 2 2 2 : ()0, 1 dx ()0, 1 { 1 : justification Thus, f has a relative minimum at (0, 1) .

(d) Substituting ym=+xb into the differential equation: 1 : value for m 2 : 11 { 1 : value for b mxmxb= +() +−=+11 m xb +−() 22( ) 1 1 1 Then 0 =+m and mb=−1: m =− and b = . 2 2 2

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Question 5

dy y − 1 Consider the differential equation = , where x ≠ 0. dx x2 (a) On the axes provided, sketch a slope field for the given differential equation at the nine points indicated. (Note: Use the axes provided in the exam booklet.) (b) Find the particular solution yfx= () to the differential equation with the initial condition f ()20.=

(a) 1 : zero slopes 2 : { 1 : all other slopes

11 (b) dy= dx ⎧ 1 : separates variables y − 1 x2 ⎪ 2 : antidifferentiates 1 ⎪ lnyC−=−+ 1 x 6 : ⎨ 1 : includes constant of integration 1 ⎪ −+C 1 : uses initial condition −= x ⎪ ye1 ⎩ 1 : solves for y − 1 yee−=1 C x Note: max 36 [1-2-0-0-0] if no constant − 1 ykeke−=1,x where =±C of integration Note: 0 6 if no separation of variables − 1 −=1 ke 2 1 ke=− 2 ()11− fx()=−1,0 e2 x x >

()11− (c) lim 1−=−ee2 x 1 1 : limit →∞ x

© 2008 The College Board. All rights reserved. Visit the College Board on the66 Web: www.collegeboard.com. 1998 AP Calculus AB Scoring Guidelines 4. Let f be a function with f(1) = 4 such that for all points (x, y) on the graph of f the slope is 3x2 + 1 given by . 2y (a) Find the slope of the graph of f at the point where x = 1. (b) Write an equation for the line tangent to the graph of f at x = 1 and use it to approximate f(1.2). dy 3x2 + 1 (c) Find f(x) by solving the separable diﬀerential equation = with the initial dx 2y condition f(1) = 4. (d) Use your solution from part (c) to ﬁnd f(1.2).

dy 3x2 + 1 (a) = 1: answer dx 2y

dy 3 + 1 4 1 = = = dx x = 1 2 4 8 2

y = 4 · 1 (b) y 4 = (x 1) 1: equation of tangent line − 2 − 2 ( 1: uses equation to approximate f(1.2) 1 f(1.2) 4 (1.2 1) − ≈ 2 − f(1.2) 0.1 + 4 = 4.1 ≈

(c) 2y dy = (3x2 + 1) dx 1: separates variables   2  1: antiderivative of dy term Z 2y dy = Z (3x + 1) dx    1: antiderivative of dx term  2 3  y = x + x + C  1: uses y = 4 when x = 1 to pick one 5  42 = 1 + 1 + C function out of a family of functions   1: solves for y 14 = C    0/1 if solving a linear equation in y y2 = x3 + x + 14   0/1 if no constant of integration  y = x3 + x + 14 is branch with point (1, 4)  p Note: max 0/5 if no separation of variables f(x) = x3 + x + 14 p Note: max 1/5 [1-0-0-0-0] if substitutes value(s) for x, y, or dy/dx before antidiﬀerentiation

3 1: answer, from student’s solution to (d) f(1.2) = p1.2 + 1.2 + 14 4.114 ≈ the given diﬀerential equation in (c)

DY X #ONSIDERÅTHEÅDIFFERENTIALÅEQUATIONÅ DX EY A &INDÅAÅSOLUTIONÅYFX ÅTOÅTHEÅDIFFERENTIALÅEQUATIONÅSATISFYINGÅ F B &INDÅTHEÅDOMAINÅANDÅRANGEÅOFÅTHEÅFUNCTIONÅFÅFOUNDÅINÅPARTÅA

ÅA EDYXDXY ¦£ ÅSEPARATESÅVARIABLES ¦ ¦ ÅANTIDERIVATIVEÅOFÅDY ÅTERM ¦ EX#Y ¦ ¦ ÅANTIDERIVATIVEÅOFÅDX ÅTERM ¦ ¦ ÅCONSTANTÅOFÅINTEGRATION Å¦¤ EX#Y ¦ ¦ ¦ ÅUSESÅINITIALÅCONDITIONÅF ¦ YX# LN ¦ ¦ ÅSOLVESÅFORÅY ¦ ¦Å.OTEÅÅIFÅY ÅISÅNOTÅAÅLOGARITHMICÅFUNCTIONÅOFÅX ¥¦ LN##E ÅÅ ¦ .OTEÅMAXÅÅ; =ÅIFÅNOÅCONSTANTÅOF YXE LN INTEGRATION .OTEÅÅIFÅNOÅSEPARATIONÅOFÅVARIABLES

B $OMAINÅXE £ ¦ ÅXE ¦ ¦ ÅDOMAIN XE ¦ Å¤ ¦ÅÅÅÅÅÅ.OTEÅÅIFÅÅISÅNOTÅINÅTHEÅDOMAIN ¦ XE E ¦ ¦ ÅRANGE ¥¦ 2ANGEÅd Y d .OTEÅÅIFÅYÅISÅNOTÅAÅLOGARITHMICÅFUNCTIONÅOFÅX

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Question 5

@O! N Consider the differential equation . @N O (a) Let OBN be the particular solution to the given differential equation for #N such that the line O is tangent to the graph of B. Find the N-coordinate of the point of tangency, and determine whether B has a local maximum, local minimum, or neither at this point. Justify your answer. (b) Let OCN be the particular solution to the given differential equation for N &, with the initial condition C$ " . Find OCN .

@O (a) when N = 3 @N £ ¦ 1 : N 3 2 ¦ @O OOa! N ¦ , 3 ¤ 1 : local minimum @N22 O ¦ (3, 2) (3, 2) ¦ ¦ 1 : justification so B has a local minimum at this point. ¥¦ or Because B is continuous for #N , there is an interval containing N = 3 on which @O O< 0. On this interval, is negative to @N @O the left of N = 3 and is positive to the @N right of N = 3. Therefore B has a local minimum at N = 3.

(b) O@O! N @N £ 1 : separates variables ¦ ¦ ¦ 1 : antiderivative of @O term ONN+! ¦ ¦ 1 : antiderivative of @N term ¦ 6 ¤ ¦ 1 : constant of integration & & &+ ; + = 8 ¦ ¦ 1 : uses initial condition C(6) 4 ¦ ¦ 1 : solves for O ONN $ $ ¥¦ Note: max 3/6 [1-1-1-0-0-0] if no constant ONN$ $ of integration Note: 0/6 if no separation of variables

6 AP® CALCULUS AB 2002 SCORING GUIDELINES (Form B)

Question 6

Ship ) is traveling due west toward Lighthouse Rock at a speed of 15 kilometers per hour (km/hr). Ship * is traveling due north away from Lighthouse Rock at a speed of 10 km/hr. Let N be the distance between Ship ) and Lighthouse Rock at time J, and let O be the distance between Ship * and Lighthouse Rock at time J, as shown in the figure above. (a) Find the distance, in kilometers, between Ship ) and Ship * when N = 4 km and O = 3 km. (b) Find the rate of change, in km/hr, of the distance between the two ships when N = 4 km and O = 3 km. (c) Let 4 be the angle shown in the figure. Find the rate of change of 4 , in radians per hour, when N = 4 km and O = 3 km.

(a) Distance = !" = 5 km 1 : answer

(b) HNO £ 1 : expression for distance @H @N @O ¦ HNO ¦ @J @J @J ¦ 2 : differentiation with respect to J 4 ¤¦ or explicitly: ¦ 2 chain rule error ¦ ¦ 1 : evaluation HNO ¥¦ @H @N @O NO @J NO @J @J

At N = 4, O = 3, @H " # ! $ km/hr @J #

O £ (c) tan 4 ¦ 1 : expression for 4 in terms of NO and N ¦ ¦ 2 : differentiation with respect to J @O @N ¦ @4 NO ¦ IA? 4 @J @J ¦ 2 chain rule, quotient rule, or @J N ¦ ¦ # 4 ¤ transcendental function error At N = 4 and O = 3, IA? 4 ¦ " ¦ note: 0/2 if no trig or inverse trig ¦ @4 $ ¬ " #! ¦ ¦ function @J #® $ ¦ ¦ 1 : evaluation &# % ¦¥ = radians/hr # # Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered 70trademarks of the College Entrance Examination Board.

7 AP® CALCULUS AB 2002 SCORING GUIDELINES

Question 5 A container has the shape of an open right circular cone, as shown in the figure above. The height of the container is 10 cm and the diameter of the opening is 10 cm. Water in the container is evaporating so that its depth D is changing at the ! constant rate of cm/hr. 1 (The volume of a cone of height D and radius H is given by 8HD 3 . ) 3 (a) Find the volume 8 of water in the container when D # cm. Indicate units of measure. (b) Find the rate of change of the volume of water in the container, with respect to time, when D # cm. Indicate units of measure. (c) Show that the rate of change of the volume of water in the container due to evaporation is directly proportional to the exposed surface area of the water. What is the constant of proportionality?

# 1 5 125 1 : 8 when D = 5 (a) When D = 5, H ; 8(5)33 5 cm! 32 12 H # 1 £ 1 (b) , so HD ¦ 1 : HD in (a) or (b) D 2 ¦ 2 ¦ @8 @D ¦ 8DDD33 !; 3D ¦ !" @J" @J ¦ ¦ ¦£ 8 as a function of one variable @8 ! ¦ ¦ ! #cm ¦ ¦ 33 # hr ¦ ¦ in (a) or (b) @J D# " & ¦ ¦ ¦ 1: ¤¦ ¦ ¦ OR ¦ ¦ OR 5 ¤¦ ¦ @H ¦ ¦ ¦ ¥¦ @J @8 @D @H @H @D ¦ 3 HHD ; ¦ @J! @J @J @J @J ¦ ¦ @8 @8 1253 5 3 ¦ 2: 3 25 ¦ @J 5 ¦ @J DH5, 3 4 10 2 20 ¦ 2 chain rule or product rule error 2 ¦ ¦ 15 3 ¦ 1 : evaluation at D 5 3 cm ¦¥ 8 hr ¦

¦£ dV @8 ! @D ¦ 1 : shows ¸k area (c) 33DD ¦ dt @J"" @J ¦ !!! 2 ¤ 1 : identifies constant of 33 HH ¸=HA= ¦ " ¦ proportionality ¦ 3 ¥¦ The constant of proportionality is . 10

! cm! 1 : correct units in (a) and (b) units of cm in (a) and hr in (b)

Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered 71trademarks of the College Entrance Examination Board. 6 AP® CALCULUS AB 2003 SCORING GUIDELINES

Question 5

A coffeepot has the shape of a cylinder with radius 5 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 5Q h cubic inches per second. (The volume V of a cylinder with radius r and height h is Vrh= Q 2 . ) dh h (a) Show that = . dt 5 dh h (b) Given that h = 17 at time t = 0, solve the differential equation = for dt 5 h as a function of t. (c) At what time t is the coffeepot empty?

Vh= £ dV (a) 25Q ¦ 1 : = 5Q h ¦ dt dV dh ¦ dV ==25QQ 5 h 3 : ¤ dt dt ¦ 1 : computes dt ¦ ¦ 1 : shows result dh5Q h h ¥¦ == dt 25Q 5

dh h ¦£ 1 : separates variables (b) = ¦ dt 5 ¦ ¦ 1 : antiderivatives ¦ ¦ 1 : constant of integration 11 5 : ¦ dh= dt ¤ h = h 5 ¦ 1 : uses initial condition 17 ¦ ¦ when t = 0 ¦ 1 ¦ 1 : solves for h 2 htC= + ¥¦ 5

217=+ 0 C Note: max 2/5 [1-1-0-0-0] if no constant

of integration 1 2 ht= () + 17 10 Note: 0/5 if no separation of variables

1 2 (c) () t +=17 0 1 : answer 10

t = 10 17

6 1992 AB6

At time tt,≥ 0, the volume of a sphere is increasing at a rate proportional to the reciprocal of its radius. At t = 0, the radius of the sphere is 1 and at t =15, the radius is 4 2. (The volume Vr of a sphere with a radius is V= π r3.) 3

(a) Find the radius of the sphere as a function of t .

(b) At what time t will the volume of the sphere be 27 times its volume at t = 0?

dV k (a) = dt r dV dr = 4π r 2 dt dt kdr = 4π r 2 rdt

kdt= 4π r3 dr

kt +=C π r 4 At tr==0, 1, so C=π == +ππ= =π At tr15, 2, so 15k16 , k ππrt4 =+π

rt=+4 1

4 (b) At tr==0, 1, so V()0 =π 3 4 27V ()0 ==27ππ36 3 4 36ππ= r3 3 = r 3 4 t +=13

t = 80

The radius r of a sphere is increasing at a constant rate of 0.04 centimeters per second. 4 (Note: The volume of a sphere with radius r is V = π r3 .) 3

(a) At the time when the radius of the sphere is 10 centimeters, what is the rate of increase of its volume?

(b) At the time when the volume of the sphere is 36π cubic centimeters, what is the rate of increase of the area of a cross section through the center of the sphere?

(c) At the time when the volume and the radius of the sphere are increasing at the same numerical rate, what is the radius?

75 1990 AB4 Solution

dV 4 dr (a) =⋅3π r 2 dt 3 dt dr Therefore when r ==10, 0.04 dt dV ==41ππ023()0.04 16 cm/sec dt

4 (b) Vr=⇒36π 36 =⇒33r=27 ⇒r=3 3 Ar= π 2 dA dr = 2π r dt dt

dr Therefore when V ==36π , 0.04 dt dA 6π =⋅2ππ3()0.04 = =0.24 cm2/sec dt 25

dV dr (c) = dt dt dr dr 44ππrr22=⇒ =1 dt dt 11 Therefore rr2 =⇒= cm 4π 2 π

AB{5 / BC{5 1999

5. The graph of the function f , consisting of three line segments, is

f (t) dt. en ab ove. Let g (x)=

giv 4 (1, 4) 1

(a) Compute g (4) and g (2).

(b) Find the instantaneous rate of change of g , with resp ect to x,at

1 (2, 1)

x =1.

Find the absolute minimum value of g on the closed interval

2; 4]. Justify your answer. [ –1 (4, –1)

–2

(d) The second derivativeof g is not de ned at x = 1 and x =2.

Howmanyofthesevalues are x{co ordinates of p oints of

in ection of the graph of g ? Justify your answer.

(

1 1 5 3

(a) g (4) = +1+ = 1: g (4) f (t) dt =

2 2 2 2

1: g (2)

g (2) = f (t) dt = (12) = 6

(b) g (1) = f (1) = 4 1: answer

1: endp oint analysis

Therefore, g has absolute minimum at an 3

endp ointof[2; 4].

1: answer

Since g (2) = 6 and g (4) = ,

the absolute minimum value is 6.

1: choice of x = 1 only

(d) One; x =1

00 0

1: show(1;g(1)) is a p oint of in ection

On (2; 1), g (x)=f (x) > 0 3

00 0

1: show(2;g(2)) is not a p oint of in ection

On (1; 2), g (x)=f (x) < 0

00 0

On (2; 4), g (x)=f (x) < 0

Therefore (1;g(1)) is a p oint of in ection and

(2;g(2)) is not.

77 AP® CALCULUS AB 2001 SCORING GUIDELINES

Question 3 A car is traveling on a straight road with velocity 55 ft/sec at time t = 0. For 018>>t seconds, the car’s acceleration at(), in ft/sec2, is the piecewise linear function defined by the graph above. (a) Is the velocity of the car increasing at t = 2 seconds? Why or why not? (b) At what time in the interval 018>>t , other than t = 0, is the velocity of the car 55 ft/sec? Why? (c) On the time interval 018>>t , what is the car’s absolute maximum velocity, in ft/sec, and at what time does it occur? Justify your answer. (d) At what times in the interval 018>>t , if any, is the car’s velocity equal to zero? Justify your answer.

(a) Since vaa(2) (2) and a(2) 15 0 , the velocity is 1 : answer and reason increasing at t = 2.

£ (b) At time t = 12 because ¦ 1 : t 12 2 : ¤ 12 ¦ 1 : reason vv(12) (0)¨ atdt ( ) 0 . ¥¦ 0

£ (c) The absolute maximum velocity is 115 ft/sec at ¦ 1 : t 6 ¦ t = 6. ¦ 1 : absolute maximum velocity ¦ The absolute maximum must occur at t = 6 or ¦ 1 : identifies t 6 and ¦ ¦ at an endpoint. ¦ t 18 as candidates 4 : ¦¤ 6 ¦ or vatdt(6) 55¨ ( ) ¦ 0 ¦ ¦ indicates that v increases, 1 ¦ 55 2(15) (4)(15) 115 v (0) ¦ 2 ¦ decreases, then increases ¦ 18 ¦ 1 : eliminates t 18 ¨ at() dt 0 so vv(18) (6) ¥¦ 6

(d) The car’s velocity is never equal to 0. The absolute £ minimum occurs at t = 16 where ¦ 1 : answer 2 : ¤ 16 ¦ 1 : reason vatdt(16) 115¨ ( ) 115 105 10 0 . ¥¦ 6

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Question 4

The graph of a differentiable function B on the closed interval <>3, 15 is shown in the figure above. The graph of B has a horizontal tangent line at N = 6. Let N CN # BJ@J for >! #N > . ¨$ (a) Find C$ , C=$ , and Caa$ . (b) On what intervals is C decreasing? Justify your answer. (c) On what intervals is the graph of C concave down? Justify your answer. # (d) Find a trapezoidal approximation of ¨ BJ@J using six subintervals of length %J ! . !

$ £ (a) CBJ@J$ # ¨ # ¦ 1 : C(6) $ ¦ 3 ¦ 1 : Ca(6) ¢ ¤ CB$ ==$ ! ¦ ¦ 1 : Caa(6) CB¢¢$ == ¢$ ¥¦

(b) C is decreasing on <>3, 0 and <>12,15 since ¦£ 1 : <> 3, 0 ¦ CNa BN for N and N 12. 3 ¤¦ 1 : <> 12,15 ¦ ¦ 1 : justification ¥¦

(c) The graph of C is concave down on (6,15) since ¦£ 1 : interval 2 ¤¦ ¦ 1 : justification CB¢ = is decreasing on this interval. ¥¦

! 1 : trapezoidal method (d) !

= 12

5 AP® CALCULUS AB 2002 SCORING GUIDELINES

Question 4 The graph of the function B shown above consists of two line segments. Let C be the N function given by CN( ). B J @J ¨ (a) Find C 1, C =1, and C == 1. (b) For what values of N in the open interval 2, 2 is C increasing? Explain your reasoning. (c) For what values of N in the open interval 2, 2 is the graph of C concave down? Explain your reasoning. (d) On the axes provided, sketch the graph of C on the closed interval 2, 2¯ . ¢¡ ±°

! ¦£ 1: C(1) (a) C BJ@J BJ@J ¦ ¨¨ ¦ 3 ¤¦ 1: Ca(1) CBa ¦ ¦ 1: Caa(1) CBaa a ! ¥¦

(b) C is increasing on N because ¦£ 1 : interval 2 ¤¦ ¦ 1 : reason CNa BN on this interval. ¥¦

(c) The graph of C is concave down on N because CNaa BN a on this interval. ¦£ 1 : interval 2 ¤¦ ¦ 1 : reason or ¥¦ because CNa BN is decreasing on this interval.

(d)

¦£1 : CCC ( 2) (0) (2) 0 ¦ ¦ 1 : appropriate increasing/decreasing ¦ 2 ¤¦ ¦ and concavity behavior ¦ ¦ 1 vertical asymptote ¥¦

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Question 5

Let f be a function defined on the closed interval [0,7]. The graph of f, consisting of four line segments, is shown above. Let g be the x function given by gx()().= ftdt ¨2 (a) Find g ()3, ga()3, and gaa()3.

(b) Find the average rate of change of g on the interval 0bbx 3. (c) For how many values c, where 0<

3 1 £ g (a) gftdt(3)==+= ( )() 4 2 3 ¦ 1 : (3) ¨2 2 ¦ 3 : ¦ 1 : ga (3) gfa(3)== (3) 2 ¤ ¦ 04 ¦ 1 : gaa (3) gfaa(3)== a (3) = 2 ¥¦ 42

gg 3 3 (3) (0) 1 ftdt ¦£ gg= ftdt (b) = () ¦ 1 : (3) (0) ( ) 3 3 ¨0 2 : ¤ ¨0 11 1 7 ¦ 1 : answer = ()(2)(4)++= (4 2) ¥¦ 32 2 3

7 The graph of f intersects the line y = at two 3 Note: 1/2 if answer is 1 by MVT places between 0 and 3.

(d) x = 2 and x = 5 ¦£ 1 : xx== 2 and 5 only ¦ gf= ¦ because a changes from increasing to 2 : ¤ 1 : justification ¦ x ¦ decreasing at = 2, and from decreasing to ¦ (ignore discussion at x = 4) ¥¦ increasing at x = 5.

6 AP® CALCULUS AB 2003 SCORING GUIDELINES

Question 4

Let f be a function defined on the closed interval b34x b with f ()03.= The graph of f a, the derivative of f, consists of one line segment and a semicircle, as shown above. (a) On what intervals, if any, is f increasing? Justify your answer. (b) Find the x-coordinate of each point of inflection of the graph of f on the open interval 3<

(d) Find f ()3 and f ()4. Show the work that leads to your answers.

(a) The function f is increasing on [3,2] since ¦£ 1 : interval 2 : ¤¦ f > x < ¦ 1 : reason a 0 for b32 . ¥¦

(b) x = 0 and x = 2 f a changes from decreasing to increasing at ¦£ 1 : xx== 0 and 2 only 2 : ¤¦ x = 0 and from increasing to decreasing at ¦ ¥¦ 1 : justification x = 2

0 £ 1 (d) ff(0) ( 3) = ftdta() ¦ 1 : ± () 2 ¨3 ¦ 2 11 3 ¦ = (1)(1) (2)(2) = ¦ (difference of areas 22 2 ¦ ¦ of triangles) ¦ ¦ 39 ¦ ff(3) =+= (0) ¦ 22 ¦ 1 : answer for f ( 3) using FTC ¦ ¦ 4 : ¤¦ 4 ¦ ff= ftdt ¦ 1 (4) (0) a() ¦ 1 : ± () 8 (2)2 Q ¨0 ¦ 2 ¦ = 1 2 = + ¦ ()8(2)82QQ ¦ (area of rectangle 2 ¦ ¦ area of semicircle) ¦ ¦ ff(4)= (0) 8+= 2QQ 5+ 2 ¦ ¦ 1 : answer for f (4) using FTC ¥¦ ¦

5 AP® CALCULUS AB 2004 SCORING GUIDELINES

Question 5

The graph of the function f shown above consists of a semicircle and three line segments. Let g be the function x given by g()xftdt= () . ∫−3 (a) Find g()0 and g′()0.

(b) Find all values of x in the open interval ()−5, 4 at which g attains a relative maximum. Justify your answer. (c) Find the absolute minimum value of g on the closed interval [−5, 4] . Justify your answer.

(d) Find all values of x in the open interval ()−5, 4 at which the graph of g has a point of inflection.

0 19 1 : g() 0 (a) gftdt()0321==+= () ()( )  ∫− 2 :  3 22  1 : g′() 0 gf′()001== ()

(b) g has a relative maximum at x = 3.  1 : x = 3 ′ = 2 :  This is the only x-value where gf changes from  1 : justification positive to negative.

(c) The only x-value where f changes from negative to  1 : identifies x =− 4 as a candidate =−  positive is x 4. The other candidates for the 3 :  1 : g()−=− 4 1 location of the absolute minimum value are the  1 : justification and answer endpoints.

g()−=50 −4 gftdt()−=41 () =− ∫−3 913ππ− g()42=+() − = 222

So the absolute minimum value of g is − 1.

(d) x =−3, 1, 2 2 : correct values −1 each missing or extra value

6 AP® CALCULUS AB 2005 SCORING GUIDELINES (Form B)

Question 4

The graph of the function f above consists of three line segments.

x (a) Let g be the function given by g()xftdt= () . ∫−4 For each of g()−1, g′()−1, and g′′()−1, find the value or state that it does not exist. (b) For the function g defined in part (a), find the x-coordinate of each point of inflection of the graph of g on the open interval −< 4x < 3. Explain your reasoning.

3 (c) Let h be the function given by hx()= f () tdt. Find all values of x in the closed interval ∫x −≤43x ≤ for which hx()= 0. (d) For the function h defined in part (c), find all intervals on which h is decreasing. Explain your reasoning.

−1 115 (a) gftdt()−=135 () =−()() =− ⎧ 1 : g()− 1 ∫−4 22 ⎪ 3 : 1 : g′()− 1 gf′()−=112 () −=− ⎨ ⎪ 1 : g′′()− 1 g′′()−1 does not exist because f is not differentiable ⎩

at x =− 1.

(b) x = 1 ⎧ 1 : x = 1 (only) 2 : ⎨ gf′ = changes from increasing to decreasing ⎩ 1 : reason at x = 1.

(d) h is decreasing on [0, 2] ⎧ 1 : interval 2 : ⎨ hf′ =− <0 when f > 0 ⎩ 1 : reason

Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudent84 s (for AP students and parents). 5 AP® CALCULUS AB 2005 SCORING GUIDELINES

Question 5

A car is traveling on a straight road. For 0≤≤t 24 seconds, the car’s velocity vt(), in meters per second, is modeled by the piecewise-linear function defined by the graph above. 24 24 (a) Find vt() dt. Using correct units, explain the meaning of vt() dt. ∫0 ∫0 (b) For each of v′()4 and v′()20 , find the value or explain why it does not exist. Indicate units of measure.

24 11 (a) vt() dt=++=()()()()4 20 12 20()( 8 20 ) 360 1 : value ∫0 222 : { The car travels 360 meters in these 24 seconds. 1 : meaning with units

(b) v′()4 does not exist because ⎧ 1 : v′() 4 does not exist, with explanation ⎪ vt()−− v (44 ) vt() v ( ) 3 : ⎨ 1 : v′() 20 lim⎛⎞=≠= 5 0 lim ⎛⎞ . −+⎜⎟−− ⎜⎟ ⎪ tt→→44⎝⎠tt44 ⎝⎠ ⎩ 1 : units 20− 0 5 2 v′()20==− m sec 16− 24 2

(d) The average rate of change of v on [8, 20] is ⎧ 1 : average rate of change of v on [ 8, 20] 2 : ⎨ vv()20− () 8 5 2 ⎩ 1 : answer with explanation =− msec. 20− 8 6 No, the Mean Value Theorem does not apply to v on [8, 20] because v is not differentiable at t = 16.

Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudent85 s (for AP students and parents). 6 AP® CALCULUS AB 2006 SCORING GUIDELINES (Form B)

Question 4

The rate, in calories per minute, at which a person using an exercise machine burns calories is modeled by the function 13 f. In the figure above, ft()=− t32 + t +1 for 42 04≤≤t and f is piecewise linear for 4≤≤t 24. (a) Find f ′()22 . Indicate units of measure. (b) For the time interval 0≤≤t 24, at what time t is f increasing at its greatest rate? Show the reasoning that supports your answer. (c) Find the total number of calories burned over the time interval 618≤≤t minutes. (d) The setting on the machine is now changed so that the person burns f ()tc+ calories per minute. For this setting, find c so that an average of 15 calories per minute is burned during the time interval 6≤≤t 18.

15− 3 (a) f ′()22==− 3 calories/min/min 1 : f ′()22 and units 20− 24

(b) f is increasing on [0, 4] and on [12, 16] . ⎧ 1 : f ′ on () 0, 4 15− 9 3 ⎪ 1 : shows ft′ has a max at = 2 on () 0, 4 On ()12, 16 , ft′()== since f has 4 : 16− 12 2 ⎨ <<′′() < ( ) ⎪ 1 : shows for 12tftf 16, 2 constant slope on this interval. ⎩⎪ 1 : answer 3 On ()0, 4 , f ′()ttt=−2 +3 and 4 3 ft′′()=− t +30 = when t = 2. This is where f ′ 2 has a maximum on [0, 4] since f ′′ > 0 on ()0, 2 and f ′′ < 0 on ()2, 4 .

On [0, 24] , f is increasing at its greatest rate when 3 t = 2 because f ′()23=> . 2

18 1 (c) ftdt()=+69 ( ) ()( 4 9 ++ 15 ) 215 ( ) 1 : method ∫6 2 2 : { 1 : answer = 132 calories

1 18 (d) We want ()ft()+= cdt 15. 1 : setup 12 ∫6 2 : { 1 : value of c This means 13212+=c 15(12). So, c = 4.

OR 132 Currently, the average is = 11 calories/min. 12 Adding c to f ()t will shift the average by c. So c = 4 to get an average of 15 calories/min.

Question 3

The graph of the function f shown above consists of six line segments. Let g be the function given by x g()xftdt= () . ∫0 (a) Find g()4, g′()4, and g′′()4. (b) Does g have a relative minimum, a relative maximum, or neither at x = 1 ? Justify your answer.

(c) Suppose that f is defined for all real numbers x and is periodic with a period of length 5. The graph above shows two periods of f. Given that g()52,= find g()10 and write an equation for the line tangent to the graph of g at x = 108.

4 (a) gftdt()43== () 1 : g() 4 ∫0 ⎧ ⎪ ′() 3 : ⎨ 1 : g 4 gf′()440== () ⎩⎪ 1 : g′′() 4

gf′′()442==− ′ ()

(b) g has a relative minimum at x = 1 1 : answer 2 : { because gf′ = changes from negative to positive at 1 : reason x = 1.

(c) g()00= and the function values of g increase by 2 for ⎧ 1 : g() 10 every increase of 5 in x. ⎪ ⎪ 4 : ⎨ ⎧ 1 : g() 108 gg()10== 2 () 5 4 ⎪ ⎪ 3 : ⎨ 1 : g′() 108 ⎪ ⎪ 1 : equation of tangent line 105 108 ⎩⎩ g(108 )=+ftdt () ftdt () ∫∫0105

=+=21gg() 5 () 3 44

gff′()108=== () 108 () 3 2

An equation for the line tangent to the graph of g at x = 108 is yx−=44 2() − 108 .

Question 4

Let f be a function defined on the closed interval −55≤≤x with f (13) = . The graph of f ′, the derivative of f, consists of two semicircles and two line segments, as shown above. (a) For −<5x <5, find all values x at which f has a relative maximum. Justify your answer. (b) For −<5x <5, find all values x at which the graph of f has a point of inflection. Justify your answer. (c) Find all intervals on which the graph of f is concave up and also has positive slope. Explain your reasoning. (d) Find the absolute minimum value of f ( x) over the closed interval −5≤≤x 5. Explain your reasoning.

(a) fx′( ) = 0 at x =−3, 1, 4 1 : x -values 2 : f ′ changes from positive to negative at −3 and 4. { 1 : justification Thus, f has a relative maximum at x = −3 and at x = 4.

(b) f ′ changes from increasing to decreasing, or vice versa, at 1 : x -values 2 : x =−4, −1, and 2. Thus, the graph of f has points of { 1 : justification inflection when x =−4, −1, and 2.

(c) The graph of f is concave up with positive slope where f ′ 1 : intervals 2 : is increasing and positive: −<54x <− and 12< x < . { 1 : explanation

(d) Candidates for the absolute minimum are where f ′ ⎧ 1 : identifies x = 1 as a candidate ⎪ changes from negative to positive (at x = 1 ) and at the 3 : ⎨ 1 : considers endpoints endpoints ( x =−5, 5 ). ⎩⎪ 1 : value and explanation −5 π ff()−=+53 ′ ()xdx =−3 + 2π >3 ∫1 2 f (13) = 5 32⋅ 1 ffx()53=+′ ()dx =+3 − >3 ∫1 22 The absolute minimum value of f on [−5, 5] is f (13) = .

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Copyright © 2000 by College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the 89College Entrance Examination Board. AP® CALCULUS AB 2008 SCORING GUIDELINES (Form B)

Question 5

Let g be a continuous function with g()25.= The graph of the piecewise-linear function g′ , the derivative of g, is shown above for −≤ 3x ≤ 7. (a) Find the x-coordinate of all points of inflection of the graph of ygx= () for −< 3x < 7. Justify your answer. (b) Find the absolute maximum value of g on the interval −≤3x ≤ 7. Justify your answer. (c) Find the average rate of change of gx() on the interval −≤37.x ≤ (d) Find the average rate of change of gx′() on the interval −≤ 3x ≤ 7. Does the Mean Value Theorem applied on the interval −≤ 3x ≤ 7 guarantee a value of c, for −<37,c < such that gc′′() is equal to this average rate of change? Why or why not?

(a) g′ changes from increasing to decreasing at x = 1; 1 : x -values 2 : { g′ changes from decreasing to increasing at x = 4. 1 : justification

Points of inflection for the graph of ygx= () occur at x = 1 and x = 4.

(b) The only sign change of g′ from positive to negative in ⎧ 1 : identifies x = 2 as a candidate ⎪ the interval is at x = 2. 3 : ⎨ 1 : considers endpoints ⎩⎪ 1 : maximum value and justification −3 315 ggxdx()−=+35′ () =+−+= 5( ) 4 ∫ 2 22 g()25= 7 13 ggxdx()75=+′ () =+−+ 5 ( 4 ) = ∫ 2 22

15 The maximum value of g for −≤ 3x ≤ 7 is . 2

315− gg()73−− ( ) 3 1 : difference quotient (c) ==−22 2 : { 73−−() 10 5 1 : answer

gg′′()7314−− ( ) −−() 1 (d) == ⎧ 1 : average value of gx′() 73−−() 10 2 2 : ⎨ ⎩ 1 : answer “No” with reason No, the MVT does not guarantee the existence of a value c with the stated properties because g′ is not differentiable for at least one point in −< 3x < 7.

© 2008 The College Board. All rights reserved. Visit the College Board on the90 Web: www.collegeboard.com. 1998 AP Calculus AB Scoring Guidelines 6. Consider the curve deﬁned by 2y3 + 6x2y 12x2 + 6y = 1. dy 4x 2xy − (a) Show that = − . dx x2 + y2 + 1 (b) Write an equation of each horizontal tangent line to the curve. (c) The line through the origin with slope 1 is tangent to the curve at point P . Find the x– and y–coordinates of point P . −

dy dy dy (a) 6y2 + 6x2 + 12xy 24x + 6 = 0 1: implicit diﬀerentiation dx dx − dx  2  dy dy 1: veriﬁes expression for (6y2 + 6x2 + 6) = 24x 12xy dx dx −  dy 24x 12xy 4x 2xy = − = − dx 6x2 + 6y2 + 6 x2 + y2 + 1

dy dy (b) = 0  1: sets = 0 dx  dx   dy 4x 2xy = 2x(2 y) = 0  − −  1: solves = 0  dx x = 0 or y = 2 4   1: uses solutions for x to ﬁnd equations When x = 0, 2y3 + 6y = 1 ; y = 0.165  of horizontal tangent lines  There is no point on the curve with   1: veriﬁes which solutions for y yield  y coordinate of 2.  equations of horizontal tangent lines  y = 0.165 is the equation of the only Note: max 1/4 [1-0-0-0] if dy/dx = 0 is not horizontal tangent line. of the form g(x, y)/h(x, y) = 0 with solutions for both x and y

(c) y = x is equation of the line. 1: y = x −  − 2( x)3 + 6x2( x) 12x2 + 6( x) = 1   1: substitutes y = x into equation − − − − 3  − 8x3 12x2 6x 1 = 0 of curve − − − −   1: solves for x and y x = 1/2 , y = 1/2  −  –or– –or– dy dy = 1  1: sets = 1 dx − dx −  2 2  dy 4x 2xy = x y 1 3  1: substitutes y = x into − − − − − dx 4x + 2x2 = x2 x2 1   1: solves for x and y − − −  4x2 + 4x + 1 = 0  Note: max 2/3 [1-1-0] if importing x = 1/2 , y = 1/2 − incorrect derivative from part (a)

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Copyright © 2000 by College Entrance Examination Board and Educational Testing Service. All rights reserved. AP is a registered trademark of the 92College Entrance Examination Board. AP® CALCULUS AB 2001 SCORING GUIDELINES

Question 6

1 The function f is differentiable for all real numbers. The point 3, is on the graph of 4 dy yfx (), and the slope at each point xy, on the graph is given by yx2 62. dx dy2 1 (a) Find and evaluate it at the point 3, . dx 2 4 dy (b) Find yfx () by solving the differential equation yx2 62 with the initial dx 1 condition f(3) . 4

dy2 dy ¦£ dy2 (a) 2(62)2yxy2 ¦ 2 : dx 2 dx ¦ dx 2 ¦ 322 ¦ = 2(62)yxy 2 ¦ < 2 > product rule or ¤ 3 : ¦ ¦ chain rule error ¦ dy2 112 ¦ 1 02 ¦ 1 : value at 3, 2 1 48 ¥¦ 4 dx 3, 4

1 (b) dy(6 2 x ) dx y2 £ ¦ 1 : separates variables ¦ ¦ 1 : antiderivative of dy term 1 ¦ 6xx 2 C ¦ 1 : antiderivative of dx term y ¦ 6 : ¤ 1 : constant of integration ¦ ¦ 4189 CC 9 ¦ 1 ¦ 1 : uses initial condition f (3) ¦ 4 C 13 ¥¦ 1 : solves for y

1 y Note: max 3/6 [1-1-1-0-0-0] if no xx2 613 constant of integration Note: 0/6 if no separation of variables

Question 4

Consider the curve given by x22+=+473.yxy

dy32 y− x (a) Show that = . dx83 y− x (b) Show that there is a point P with x-coordinate 3 at which the line tangent to the curve at P is horizontal. Find the y-coordinate of P. dy2 (c) Find the value of at the point P found in part (b). Does the curve have a local maximum, a dx2 local minimum, or neither at the point P ? Justify your answer.

(a) 28x +=+yy′′ 33 y xy  1 : implicit differentiation 2 :  ()83yxyyx−=−′ 32  1 : solves for y′ 32yx− y′ = 83yx−

32yx− dy (b) =−=0; 3yx 2 0  1 : = 0 83yx−  dx 3 :  1 : shows slope is 0 at () 3, 2 When x = 3, 36y =   1 : shows () 3, 2 lies on curve = y 2

22 34225+⋅ = and 7+⋅⋅= 3 3 2 25

Therefore, P = ()3, 2 is on the curve and the slope is 0 at this point.

(c) dy2 ()()()()8332yxy−−−−−′′ 3283 yxy  dy2 =  2 : dx22()83yx− dx2  2 −− 4 :  dy2 dy ()()16 9 2 2 1 : value of at () 3, 2 At P = ()3, 2 , ==−.  2 dx22()16− 9 7  dx  1 : conclusion with justification Since y′ = 0 and y′′ < 0 at P, the curve has a local maximum at P.

Question 5

Consider the curve given by yxy2 =+2.

dy y (a) Show that = . dx2 y− x 1 (b) Find all points ()x, y on the curve where the line tangent to the curve has slope . 2

(d) Let x and y be functions of time t that are related by the equation yxy2 =+2. At time t = 5, the dy dx value of y is 3 and = 6. Find the value of at time t = 5. dt dt

(a) 2yy′′=+ y xy ⎧ 1 : implicit differentiation 2 : ⎨ ()2yxyy−=′ ⎩ 1 : solves for y′ y y′ = 2yx−

y 1 ⎧ y 1 (b) = ⎪ 1 : = 22yx− 2 : ⎨ 22yx− 22yyx=− ⎩⎪ 1 : answer x = 0 y =± 2 ()0, 2 ,() 0,− 2

y (c) = 0 ⎧ 1 : y = 0 2yx− 2 : ⎨ ⎩ 1 : explanation y = 0 The curve has no horizontal tangent since 022 ≠+⋅x 0 for any x.

7 (d) When y = 3, 3232 =+x so x = . ⎧ 1 : solves for x 3 ⎪ 3 : 1 : chain rule dy dydx y dx ⎨ =⋅= ⋅ ⎪ dt dx dt2 y− x dt ⎩ 1 : answer 39dx dx At t = 5, 6 =⋅=⋅ 711dt dt 6 − 3 dx 22 = dt t =5 3

AB{1 1999

1. A particle moves along the y {axis with velo citygiven by v (t)=t sin(t )fort 0.

(a) In which direction (up or down) is the particle moving at time t =1:5? Why?

(b) Find the acceleration of the particle at time t = 1:5. Is the velo city of the particle increasing at

t =1:5? Whyorwhynot?

(d) Find the total distance traveled by the particle from t =0 to t =2.

(a) v (1:5)=1:5 sin(1:5 )=1:167 1: answer and reason

Up, b ecause v (1:5) > 0

(

0 2 2 2

1: a(1:5)

(b) a(t)=v (t) = sin t +2t cos t

1: conclusion and reason

a(1:5) = v (1:5) = 2:048 or 2:049

No; v is decreasing at 1.5 b ecause v (1:5) < 0

1: y (t)= v (t) dt

1: y (t)= cos t + C

cos t

= t sin t dt = > + C

1: y (2)

7 1

+ C =) C = y (0) = 3 =

2 2

1 7

y (t)= cos t +

2 2

1 7

y (2) = cos 4 + =3:826 or 3:827

2 2

(d) distance = jv (t)j dt =1:173

1: limits of 0 and 2 on an integral of

v (t)orjv (t)j

or >

v (t)=t sin t =0

uses y (0) and y (2) to compute distance

t =0 or t = > 1:772

1: handles change of direction at student's

turning p oint

y (0) = 3 ; y ( )=4; y (2) = 3:826 or 3:827

1: answer

p p

) y (0)] + [y ( ) y (2)] [y (

0/1 if incorrect turning p oint

=1:173 or 1:174

96 AP® CALCULUS AB 2002 SCORING GUIDELINES (Form B)

Question 3

A particle moves along the N-axis so that its velocity L at any time J, for $bbJ , is given by LJ A2sinJ . At time J = 0, the particle is at the origin. (a) On the axes provided, sketch the graph of LJ for $bbJ . (b) During what intervals of time is the particle moving to the left? Give a reason for your answer. (c) Find the total distance traveled by the particle from J = 0 to J = 4. (d) Is there any time J, $bJ , at which the particle returns to the origin? Justify your answer.

(a) L( J) 1 : graph three humps periodic behavior J starts at origin reasonable relative max and min values

(b) Particle is moving to the left when ¦£ 2 : intervals ¦ 2sinJ ¦ LJ , i.e. A . 3 ¤ 1 each missing or incorrect interval ¦ ¦ 1 : reason (33,2 ), (3,4)33 and (5,16]3 ¥¦

" (c) ¨ LJ @J = 10.542 ¦£ 1 : limits of 0 and 4 on an integral of ¦ or ¦ LJ ( ) or LJ ( ) ¦ ¦ LJ A2sinJ ¦ or ¦ ¦ uses NN(0) and (4) to compute distance ¦ J = 0 or J 3 3 ¤¦ ¦ 1 : handles change of direction at students 3 ¦ NLJ@J3 ¨ = 10.10656 ¦ ¦ turning point " ¦ NLJ@J" ¨ = 9.67066 ¦ 1 : answer ¦ ¦ note: 0/1 if incorrect turning point NN(33) (0) NN(4) ( ) ¦¥ = 10.542

(d) There is no such time because £ 1 : no such time ¦ 6 2 ¤ LJ @J > 0 for all 6 . ¦ 1 : reason ¨ ¥

4 AP® CALCULUS AB 2002 SCORING GUIDELINES

Question 3 An object moves along the N-axis with initial position N 02. The velocity of the object at time J F is given 3 by LJ sin J . 3

(a) What is the acceleration of the object at time J 4? (b) Consider the following two statements. Statement I: For 34.5,J the velocity of the object is decreasing. Statement II: For 34.5,J the speed of the object is increasing. Are either or both of these statements correct? For each statement provide a reason why it is correct or not correct. (c) What is the total distance traveled by the object over the time interval 04?>>J (d) What is the position of the object at time J 4?

334 1 : answer (a) =L(4)a (4) cos 33 3 = or 0.523 or 0.524 $

(b) On 34.5J : ¦£ 1 : I correct, with reason 33 ¦ =J( ) La( J)cos0 J ¦ 33 3 ¤ 1 : II correct ¦ Statement I is correct since =J( )0. ¦ 1 : reason for II ¥¦ Statement II is correct since LJ and =J( )0. " £ £ (c) Distance = LJ( ) @J 2.387 ¦ ¦ limits of 0 and 4 on an integral ¨ ¦ ¦ ¦ ¦ OR ¦ ¦ of LJ ( ) or LJ ( ) 333 ¦ ¦ NJ( )cos2 J ¦ 1 : ¤ or 333 ¦ ¦ ¦ ¦ uses NN(0) and (4) to compute N ¦ ¦ ¦ ¦ 9 ¦ ¦ distance N(4) 2 3.43239 3 ¦¤ ¦¥ 23 ¦ 1 : handles change of direction at ¦ LJ when J = 3 ¦ ¦ students turning point 6 ¦ N(3) 2 3.90986 ¦ 3 ¦ 1 : answer ¦ # ¦ 0/1 if incorrect turning point or NN!"! NN = 2.387 ¦ 3 ¦ ¦ no turning point ¦¥ ¦ " ¦£ 1 : integral (d) NN(4) (0) ¨ LJ@J( )3.432 2 ¤¦ ¦ 1 : answer ¥¦ OR OR 333 £ 3 3 NJ( )cos2 J ¦ 1: NJ () cos J + 333 ¦ 3 3 9 ¦ N(4) 2 3.432 2 ¤ 1: answer 23 ¦ ¦ 0/1 if no constant of integration ¥¦

Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered 98trademarks of the College Entrance Examination Board. 4 AP® CALCULUS AB 2003 SCORING GUIDELINES (Form B)

Question 4

t A particle moves along the x-axis with velocity at time t p 0 given by vt()= 1.+ e1 (a) Find the acceleration of the particle at time t = 3. (b) Is the speed of the particle increasing at time t = 3? Give a reason for your answer. (c) Find all values of t at which the particle changes direction. Justify your answer. (d) Find the total distance traveled by the particle over the time interval 0bbt 3.

at== v t e1t £ 1 : vta ( ) (a) ()a () ¦ 2 : ¤¦ ae= 2 ¦ 1 : a (3) (3) ¥¦

(b) a(3)< 0 1 : answer with reason ve(3)= 1+<2 0 Speed is increasing since v(3)< 0 and a(3)< 0 .

(c) vt()= 0 when 1 = e1t , so t = 1. ¦£ 1 : solves vt ( )= 0 to ¦ ¦ get t = 1 ¦ 2 : ¤ vt()> 0 for t < 1 and vt()< 0 for t > 1. ¦ 1 : justifies change in ¦ t = 1. ¦ direction at t = 1 Therefore, the particle changes direction at ¥¦

3 £ 1 : limits (d) Distance = vt() dt ¦ ¨ ¦ 0 ¦ 1 : integrand 13 ¦ ()()++edt11tt edt 4 : ¤ = ¨¨11 ¦ 1 : antidifferentiation 01 ¦ tt13 ¦ = ()te11++() te ¦ 1 : evaluation 01 ¥¦ = ()11+++ee() 32 11 = ee+ 2 1 OR OR

t xt()= t e1 ¦£ 1 : any antiderivative ¦ xe(0) = ¦ 1 : evaluates xt ( ) when ¦ x(1)= 2 ¦ t = 0, 1, 3 ¦ 4 : ¤¦ xe(3)= 2 3 ¦ 1 : evaluates distance ¦ Distance = ()()xx(1)(0)(1)(3) + xx ¦ between points ¦ = (2 +++ee )() 1 2 ¦ 1 : evaluates total distance ¥¦ = ee+ 2 1

5 AP® CALCULUS AB 2003 SCORING GUIDELINES

Question 2

A particle moves along the x-axis so that its velocity at time t is given by ¬t2 vt()= () t+ 1sin . ® 2 At time t = 0, the particle is at position x = 1. (a) Find the acceleration of the particle at time t = 2. Is the speed of the particle increasing at t = 2? Why or why not? (b) Find all times t in the open interval 03<

(a) av(2)= a (2) = 1.587 or 1.588 £ 1: a(2) ¦ ¦ v(2)= 3 sin(2)< 0 2 : ¤¦ 1 : speed decreasing ¦ Speed is decreasing since a(2)> 0 and v(2)< 0 . ¦ with reason ¥¦

t2 £ 1: t = 2Q only ¦ (b) vt()= 0 when = Q 2 : ¤¦ 2 ¦ 1 : justification ¥¦ t = 2Q or 2.506 or 2.507 Since vt()< 0 for 02< 0 for 23Q <

3 ¦£ 1: limits (c) Distance = ¨ vt() dt = 4.333 or 4.334 ¦ 0 ¦ 3 : ¤ 1: integrand ¦ ¦ 1: answer ¥¦

2Q ¦£ 1 : ± (distance particle travels (d) vt( ) dt= 3.265 ¦ ¨0 ¦ 2 : ¦ while velocity is negative) 2Q ¤ ¦ xx()2Q =+ (0) vtdt ( ) = 2.265 ¦ 1: answer ¨0 ¥¦ Since the total distance from t = 0 to t = 3 is 4.334, the particle is still to the left of the origin at t = 3. Hence the greatest distance from the origin is 2.265.

3 AP® CALCULUS AB 2004 SCORING GUIDELINES

Question 3

− A particle moves along the y-axis so that its velocity v at time t ≥ 0 is given by vt()=−1tan1 () et . − At time t = 0, the particle is at y =−1. (Note: tan1 x = arctan x ) (a) Find the acceleration of the particle at time t = 2. (b) Is the speed of the particle increasing or decreasing at time t = 2 ? Give a reason for your answer. (c) Find the time t ≥ 0 at which the particle reaches its highest point. Justify your answer. (d) Find the position of the particle at time t = 2. Is the particle moving toward the origin or away from the origin at time t = 2 ? Justify your answer.

(a) av()2==−−′ () 2 0.132 or 0.133 1 : answer

(b) v()20.436=− 1 : answer with reason Speed is increasing since a()20< and v()20.<

− (c) vt()= 0 when tan1()et = 1  1 : sets vt()= 0  3 : 1 : identifies t = 0.443 as a candidate t ==ln() tan() 1 0.443 is the only critical value for y.   1 : justifies absolute maximum  vt()> 0 for 0lntan1< ln() tan() 1

yt() has an absolute maximum at t = 0.443.

2 2 (d) yvtdt()2=− 1 + () =− 1.360 or − 1.361  1 : vt() dt ∫0  ∫0  The particle is moving away from the origin since 4 :  1 : handles initial condition v()20< and y()20.<  1 : value of y() 2   1 : answer with reason

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4 AP® CALCULUS AB 2005 SCORING GUIDELINES (Form B)

Question 3

A particle moves along the x-axis so that its velocity v at time t, for 0≤≤t 5, is given by vt()=−+ln() t2 3 t 3 . The particle is at position x = 8 at time t = 0. (a) Find the acceleration of the particle at time t = 4. (b) Find all times t in the open interval 0<

5 (a) av()44==′ () 1 : answer 7

(b) vt()= 0 ⎧ 1 : sets vt()= 0 ⎪ tt2 −+=331 3 : ⎨ 1 : direction change at t = 1, 2 ⎪ 1 : interval with reason tt2 −+=320 ⎩ ()()tt−−=210 t = 1, 2

vt()> 0 for 0< 0 for 2<

The particle changes direction when t = 1 and t = 2. The particle travels to the left when 1<

t 2 (c) s()ts=+ (0ln33 ) () u2 −+ u du ⎧ 1 : ln()uu2 −+ 3 3 du ∫0 ⎪ ∫0 3 : 2 ⎨ 1 : handles initial condition s()28=+ ln33()uu2 − + du ⎪ ∫0 ⎩⎪ 1 : answer = 8.368 or 8.369

1 2 (d) vt() dt= 0.370 or 0.371 ⎧ 1 : integral 2 ∫0 2 : ⎨ ⎩ 1 : answer

Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and 102www.collegeboard.com/apstudents (for AP students and parents). 4 AP® CALCULUS AB 2007 SCORING GUIDELINES (Form B)

Question 2

A particle moves along the x-axis so that its velocity v at time t ≥ 0 is given by vt()= sin() t2 . The graph of v is shown above for 0≤≤t 5.π The position of the particle at time t is x(t) and its position at time t = 0 is x(05) = . (a) Find the acceleration of the particle at time t = 3. (b) Find the total distance traveled by the particle from time t = 0 to t = 3. (c) Find the position of the particle at time t = 3. (d) For 0≤≤t 5,π find the time t at which the particle is farthest to the right. Explain your answer.

(a) av(336cos95.4) ==′( ) =−66 or −5.467 1 : a(3)

3 (b) Distance ==vt() dt 1.702 1 : setup ∫ 0 2 : { 1 : answer OR For 03<

⎧⎧ 1 : integrand 3 ⎪ 2 ⎨ (c) xvtd()35=+ () t = 5.773 or 5.774 3 : ⎨ ⎩ 1 : uses x() 0= 5 ∫ 0 ⎩⎪1 : answer

(d) The particle’s rightmost position occurs at time t ==π 1.772. ⎧ 1 : sets vt()= 0 ⎪ The particle changes from moving right to moving left at those times t for 3 : ⎨ 1 : answer ⎪ which vt( ) = 0 with vt( ) changing from positive to negative, namely at ⎩ 1 : reason t = π ,3,5ππ (t = 1.772, 3.070, 3.963) . T Using x()Tvt=+5 ()dt, the particle’s positions at the times it ∫ 0 changes from rightward to leftward movement are: T:0πππ 3 5

xT(): 5 5.895 5.788 5.752 The particle is farthest to the right when T = π .

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Question 4

− A particle moves along the x-axis with position at time t given by x()te= t sin t for 0≤≤t 2π . (a) Find the time t at which the particle is farthest to the left. Justify your answer. (b) Find the value of the constant A for which x()t satisfies the equation Ax′′() t++= x ′ () t x () t 0 for 0<

−−tt − t (a) x′()tetetet=−sin + cos = ( cos − sin t ) ⎧ 2 : xt′() xt′()= 0 when costt= sin . Therefore, xt′()= 0 on ⎪ 1 : sets xt′()= 0 5 : ⎨ π 5π 1 : answer 02≤≤t π for t = and t = . ⎪ 4 4 ⎩⎪ 1 : justification The candidates for the absolute minimum are at ππ5 t = 0, , , and 2.π 44

t x()t 0 e0 sin() 0= 0 π π − π e 4 sin()> 0 4 4 5π 5π − 5π e 4 sin()< 0 4 4 − π 2π e 2 sin() 2π = 0

5π The particle is farthest to the left when t = . 4

−−tt (b) x′′()te=− (cos t − sin te ) + ( − sin t − cos t ) ⎧ 2 : xt′′() − ⎪ ′′ ′ =−2cosett ⎪ 1 : substitutes x ()txt , () , and xt () 4 : ⎨ ′′()++ ′ () () ⎪ into Axtxtxt Ax′′() t++ x ′ () t x () t ⎩⎪ 1 : answer −− − =−Ae()2costt te +() cossin t − te + t sin t

− =−()21cosAe + t t = 0 1 Therefore, A = . 2

A particle moves along the x-axis in such a way that its acceleration at time t for t ≥ 0 is given by a(t) = 4cos(2t) . At time t = 0 , the velocity of the particle is v(0) = 1 and its position is x(0) = 0.

(a) Write an equation for the velocity v(t) of the particle.

(b) Write an equation for the position x(t) of the particle.

(a) vt()= ∫ 4cos2tdt vt()=+2sin2t C vC()01=⇒ =1 vt()=+2sin2t 1

(b) x()tt=+∫ 2sin2 1 dt x()tt=−cos 2 +t+C xC()00=⇒ =1 xt()=−cos 2t+t+1

Question 4

x 0 01<

Let f be a function that is continuous on the interval [0, 4) . The function f is twice differentiable except at x = 2. The function f and its derivatives have the properties indicated in the table above, where DNE indicates that the derivatives of f do not exist at x = 2. (a) For 04,<

(d) For the function g defined in part (c), find all values of x, for 04,<

(a) f has a relative maximum at x = 2 because f ′ changes from 1 : relative extremum at x = 2 2 : { positive to negative at x = 2. 1 : relative maximum with justification

(b) ⎧ 1 : points at x = 0, 1, 2, 3 ⎪ and behavior at () 2, 2 2 : ⎨ ⎪ 1 : appropriate increasing/decreasing ⎩⎪ and concavity behavior

(c) gx′()== fx () 0 at x = 1, 3. ⎧ 1 : gx′()= fx () ⎪ g′ changes from negative to positive at x = 1 so g has a relative 3 : ⎨ 1 : critical points minimum at x = 1. g′ changes from positive to negative at x = 3 ⎩⎪ 1 : answer with justification so g has a relative maximum at x = 3.

(d) The graph of g has a point of inflection at x = 2 because gf′′= ′ 1 : x = 2 2 : { changes sign at x = 2. 1 : answer with justification

Copyright © 2005 by College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and 107www.collegeboard.com/apstudents (for AP students and parents). 5 AP® CALCULUS AB 2007 SCORING GUIDELINES

Question 3

x f ()x f ′()x gx() gx′()

1 6 4 2 5

2 9 2 3 1

3 10 – 4 4 2

4 –1 3 6 7

The functions f and g are differentiable for all real numbers, and g is strictly increasing. The table above gives values of the functions and their first derivatives at selected values of x. The function h is given by hx()=− f() gx () 6. (a) Explain why there must be a value r for 13<

gx() (c) Let w be the function given by wx()= f () t dt. Find the value of w′()3. ∫1 − − (d) If g 1 is the inverse function of g, write an equation for the line tangent to the graph of ygx= 1() at x = 2.

(a) hfg()11626963=−=−=−=() () f() ⎧ 1 : hh() 1 and () 3 2 : ⎨ hfgf()3=−=−=−−=−() () 36() 46167 ⎩ 1 : conclusion, using IVT Since hh()351<− < () and h is continuous, by the Intermediate Value Theorem, there exists a value r, 13,<

hh()31− () −−73 ⎧ hh()31− () (b) ==−5 ⎪ 1 : 31−− 31 2 : ⎨ 31− Since h is continuous and differentiable, by the ⎩⎪ 1 : conclusion, using MVT Mean Value Theorem, there exists a value c, 13,<

− − (d) g ()12,= so g 1 ()21.= ⎧ 1 : g 1 () 2 ⎪ ′ ()−1 ′ ()===111 3 : ⎨ ()−1 () g 2 − ′ 1 : g 2 gg′()1 ()2 g ()1 5 ⎪ ⎩⎪ 1 : tangent line equation 1 An equation of the tangent line is yx−=12.() − 5

Let f be a function that is even and continuous on the closed interval [−3,3]. The function f and its derivatives have the properties indicated in the table below.

x 00< x <111< x < 222< x < 3 f (x) 1 Positive 0 Negative −1 Negative

f ′ (x) Undefined Negative 0 Negative Undefined Positive f ′′ (x) Undefined Positive 0 Negative Undefined Negative

(a) Find the x-coordinate of each point at which f attains an absolute maximum value or an absolute minimum value. For each x-coordinate you give, state whether f attains an absolute maximum or an absolute minimum.

(b) Find the x-coordinate of each point of inflection on the graph of f . Justify your answer.

x −3 −2 −1 1 2 3

−1

−2

−3

(a) Absolute maximum at x = 0 Absolute minimum at x =±2

(b) Points of inflection at x =±1 because the sign of f ′′ (x)changes at x =1 and f is even

x −3 −2 −1 1 2 3

−1

Question 6 N 1.5 1.0 0.5 0.5 1.0 1.5 BN( ) 1 4 6 % 6 4 1 BNa % 5 3 0 3 5 7

Let B be a function that is differentiable for all real numbers. The table above gives the values of B and its derivative B a for selected points N in the closed interval bb1.5N 1.5. The second derivative of B has the property that BNaa for bb1.5N 1.5.

1.5 (a) Evaluate 3BN= ( )4 @N . Show the work that leads to your answer. ¨0 (b) Write an equation of the line tangent to the graph of B at the point where N 1. Use this line to approximate the value of B(1.2). Is this approximation greater than or less than the actual value of B(1.2)? Give a reason for your answer. (c) Find a positive real number H having the property that there must exist a value ? with 00.5? and

B?aa( ). H Give a reason for your answer. £ ¦27for0NN N (d) Let C be the function given by CN() ¤¦ ¦27for0.NN N p ¥¦ The graph of C passes through each of the points NBN, ( ) given in the table above. Is it possible that B and C are the same function? Give a reason for your answer.

1.5 1.5 1.5 £ 1 : antiderivative (a) !Baa N" @N ! B N" @N @N ¦ ¨¨¨ 2 ¤¦ 000¦ 1: answer 1.5 ¥¦ 3BN( )31"N ( 7) 4 1.5 24 0 (b) ON# " £ ¦ 1 : tangent line B(1.2)x 5 0.2 4 3 ¦ 3 ¤¦ 1 : computes ON on tangent line at 1.2 The approximation is less than B(1.2) because the ¦ ¦ 1 : answer with reason graph of B is concave up on the interval ¥¦ 11.2N . £ a (c) By the Mean Value Theorem there is a ? with ¦ 1 : reference to MVT for B (or differentiability ¦ 00.5? such that 2 ¤¦ of B a ) ¦ BBaa(0.5) (0) 30 ¦ 1 : valueHN of bb for interval 00.5 B?aa() 6 H ¥¦ 0.5 0 0.5 (d) lim CNa( )lim411 N NNll £ lim CNa( )lim41 N 1 ¦ 1 : answers no with reference to NNll ¦ 2 ¦ CCaaa or Thus C a is not continuous at N = 0, but B a is ¤ ¦ ¦ 1 : correct reason continuous at N = 0, so BCL . ¥¦ OR CNaa " for all N L , but it was shown in part (c) that B?aa $ for some ? L , so BCL .

Copyright © 2002 by College Entrance Examination Board. All rights reserved. Advanced Placement Program and AP are registered111 trademarks of the College Entrance Examination Board. 7 1998 Calculus AB Scoring Guidelines 2. Let f be the function given by f(x) = 2xe2x. (a) Find lim f(x) and lim f(x). x x →−∞ →∞ (b) Find the absolute minimum value of f. Justify that your answer is an absolute minimum. (c) What is the range of f? (d) Consider the family of functions deﬁned by y = bxebx, where b is a nonzero constant. Show that the absolute minimum value of bxebx is the same for all nonzero values of b.

(a) lim 2xe2x = 0 1: 0 as x x →−∞ 2 ( → −∞ 1: or DNE as x lim 2xe2x = or DNE ∞ → ∞ x →∞ ∞

2x 2x 2x (b) f 0(x) = 2e + 2x 2 e = 2e (1 + 2x) = 0 1: solves f 0(x) = 0 · ·  if x = 1/2  1: evaluates f at student’s critical point  −  f( 1/2) = 1/e or 0.368 or 0.367  0/1 if not local minimum from  − − − −  student’s derivative  1/e is an absolute minimum value because: 3  − 1: justiﬁes absolute minimum value  (i) f 0(x) < 0 for all x < 1/2 and  0/1 for a local argument −   f 0(x) > 0 for all x > 1/2  0/1 without explicit symbolic −   derivative –or–  Note: 0/3 if no absolute minimum based on f (x) + (ii) 0 − student’s derivative 1/2 − and x = 1/2 is the only critical number −

(c) Range of f = [ 1/e, ) 1: answer − ∞ or [ 0.367, ) − ∞ Note: must include the left–hand endpoint; or [ 0.368, ) exclude the right–hand “endpoint” − ∞

(d) y = bebx + b2xebx = bebx(1 + bx) = 0 1: sets y = bebx(1 + bx) = 0 0  0 if x = 1/b   1: solves student’s y0 = 0 − 3  At x = 1/b, y = 1/e 1: evaluates y at a critical number − −   and gets a value independent of b y has an absolute minimum value of 1/e for  all nonzero b − Note: 0/3 if only considering speciﬁc values of b

Question 5

A cubic polynomial function f is defined by

fx() 4 x32 ax bx k where a, b, and k are constants. The function f has a local minimum at x 1 , and the graph of f has a point of inflection at x 2 . (a) Find the values of a and b. 1 (b) If ¨ fxdx() = 32, what is the value of k ? 0

(a) fxa() 12 x2 2 axb ¦£ 1 : fxa ( ) ¦ aa ¦ fx() 24 x 2 a ¦ 1 : fxaa ( ) ¦ ¤¦ a 5 : ¦ 1 : f ( 1) 0 a ¦ fab(1)122 0 ¦ aa ¦ 1 : f ( 2) 0 faaa(2) 482 0 ¦ ¥¦ 1 : ab ,

a 24 ba12 2 36

1 32 ¦£ 2 : antidifferentiation (b) ¨ 42436xxxkdx ¦ 0 ¦ < 1 > each error x 1 ¦ xx43818 xkx 2 27 k 4 : ¤ x 0 ¦ 1 : expression in k ¦ ¦ ¥¦ 1 : k 27k 32

k 5

Question 4

Let h be a function defined for all x L 0 such that h(4) 3 and the derivative of h is given x 2 2 by hxa() for all x L 0 . x (a) Find all values of x for which the graph of h has a horizontal tangent, and determine whether h has a local maximum, a local minimum, or neither at each of these values. Justify your answers. (b) On what intervals, if any, is the graph of h concave up? Justify your answer. (c) Write an equation for the line tangent to the graph of h at x = 4. (d) Does the line tangent to the graph of h at x = 4 lie above or below the graph of h for x 4 ? Why?

£ o (a) hxa() 0 at x o 2 ¦ 1 : x 2 ¦ ¦ 1 : analysis ¦ = 0 und 0 4 : ¤ hx() + + ¦ 2 : conclusions ¦ 1 > not dealing with ¦ x 2 0 2 ¥¦ discontinuity at 0

Local minima at x 2 and at x 2

aa 2 ¦£ 1 : hxaa ( ) (b) hx() 1 0 for all x L 0 . Therefore, ¦ x 2 ¦ 3 : ¤ 1 : hxaa ( ) 0 the graph of h is concave up for all x L 0 . ¦ ¦ 1 : answer ¥¦

16 2 7 (c) ha(4) 42

7 yx3(4) 1 : tangent line equation 2

(d) The tangent line is below the graph because 1 : answer with reason the graph of h is concave up for x 4 .

Question 6

Let f be the function defined by

 x + 1 for 0≤≤x 3  fx()=  5 − x for 3< x ≤ 5.  (a) Is f continuous at x = 3? Explain why or why not. (b) Find the average value of f ()x on the closed interval 05.≤≤x (c) Suppose the function g is defined by kx+ 1 for 0≤≤x 3  gx()=  mx + 2 for 3< x ≤ 5,  where k and m are constants. If g is differentiable at x = 3, what are the values of k and m ?

(a) f is continuous at x = 3 because  1 : answers “yes” and equates the  limf (xfx )== lim ( ) 2.  values of the left- and right-hand − +  xx→→33 2 :  Therefore, limf (xf )== 2 (3).  limits x →3   1 : explanation involving limits 

535  35 (b) f() x dx=+ f () x dx f () x dx  1 : kfxdxkfxdx ()+ () ∫∫∫003  ∫∫03  3 5  (where k ≠ 0) 213 2  =+(1)5xxx2 +−  320 ()3 4 :   1 : antiderivative of x + 1 16 2 25 21 20  = − + − =  1 : antiderivative of 5 − x ()33() 2 2 3    1 : evaluation and answer  145 Average value: fxdx() = 53∫0

2 8 832mm=+; m = and k = 5 5

7 AP® CALCULUS AB 2007 SCORING GUIDELINES

Question 6

Let f be the function defined by f ()xkxx=−ln for x > 0, where k is a positive constant. (a) Find f ′()x and f ′′()x . (b) For what value of the constant k does f have a critical point at x = 1 ? For this value of k, determine whether f has a relative minimum, relative maximum, or neither at x = 1. Justify your answer. (c) For a certain value of the constant k, the graph of f has a point of inflection on the x-axis. Find this value of k.

k 1 (a) fx′()=− ⎧ 1 : f ′()x 2 x x 2 : ⎨ 1 : f ′′()x ⎩

1 −− f ′′()xkxx=−32 + 2 4

1 (b) fk′()1102=−=⇒ k= ⎧ 1 : sets ffx′′() 1== 0 or ( ) 0 2 ⎪ 1 : solves for k 1 ⎪ When k = 2, f ′()10= and f ′′()110.=− + > 4 : ⎨ 2 ⎪ 1 : answer f has a relative minimum value at x = 1 by the ⎩⎪ 1 : justification Second Derivative Test.

(c) At this inflection point, fx′′()= 0 and fx()= 0. ⎧ 1 : fx′′()== 0 or fx () 0 ⎪ 3 : ⎨ 1 : equation in one variable −k 14 ⎪ fx′′()= 00⇒ +=⇒ k= ⎩ 1 : answer 32 2 4x x x ln x fx()= 0ln0⇒ kx−= x⇒ k= x

4 ln x Therefore, = x x ⇒ 4ln= x ⇒ x = e4 4 ⇒ k = e2

© 2007 The College Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and www.collegeboard.com/apstudents (for students and parents). 116 AP® CALCULUS AB 2008 SCORING GUIDELINES

Question 6

ln x Let f be the function given by fx()= for all x > 0. The derivative of f is given by x 1ln− x fx′()= . x2 (a) Write an equation for the line tangent to the graph of f at x = e2. (b) Find the x-coordinate of the critical point of f. Determine whether this point is a relative minimum, a relative maximum, or neither for the function f. Justify your answer. (c) The graph of the function f has exactly one point of inflection. Find the x-coordinate of this point. (d) Find limf ()x . + x→0

ln e2 1ln− e2 ()2 ==2 ′()2 ==−1 ⎧ ()22′ () (a) fe 22, fe 24 ⎪ 1 : f efe and ee ()e2 e 2 : ⎨ ⎩⎪ 1 : answer 21 An equation for the tangent line is yxe=−() −2 . ee24

(b) fx′()= 0 when x = e. The function f has a relative maximum ⎧ 1 : xe= ⎪ at x = e because f ′()x changes from positive to negative at 3 : ⎨ 1 : relative maximum x = e. ⎩⎪ 1 : justification

1 −−−xxx2 ()1ln2 x −+32lnx (c) fx′′()== for all x > 0 ⎧ 2 : f ′′()x xx43 3 : ⎨ ⎩ 1 : answer fx′′()= 0 when −+ 3 2lnx = 0

x = e32

The graph of f has a point of inflection at x = e32 because f ′′()x changes sign at xe= 32.

ln x (d) lim =−∞ or Does Not Exist 1 : answer + x→0 x

AB{4 1999

4. Supp ose that the function f has a continuous second derivative for all x, and that f (0) = 2, f (0) = 3,

00 0 2x 0

and f (0) = 0. Let g b e a function whose derivativeisgiven by g (x)=e (3f (x)+2f (x)) for all x.

(a) Write an equation of the line tangent to the graph of f at the p oint where x =0.

(b) Is there sucient information to determine whether or not the graph of f has a p oint of in ection

when x =0? Explain your answer.

x =0.

00 2x 0 00

(d) Show that g (x)=e (6f (x) f (x)+2f (x)). Do es g have a lo cal maximum at x =0? Justify

your answer.

(a) Slop e at x =0 is f (0) = 3 1: equation

At x =0, y =2

y 2=3(x 0)

(

1: answer

(b) No. Whether f (x) changes sign at x =0 is

unknown. The only given value of f (x) is

1: explanation

f (0) = 0.

(

0 2x 0

1: g (0)

1: equation

0 0

g (0) = e (3f (0)+2f (0))

= 3(2)+2(3) = 0

y 4=0(x 0)

y =4

0 2x 0

(d) g (x)=e (3f (x)+2f (x))

2: verify derivative

0/2 pro duct or chain rule error

2x 0

g (x) = (2e )(3f (x)+2f (x))

< 1 > algebra errors

0 00

2x 0 00

1: g (0) = 0 and g (0)

+ e (3f (x)+2f (x))

1: answer and reasoning

2x 0 00

= e (6f (x) f (x)+2f (x))

00 0

g (0) = e [(6)(2) (3) + 2(0)] = 9

0 00

Since g (0) = 0 and g (0) < 0, g do es havea

lo cal maximum at x =0.

118 AP® CALCULUS AB 2003 SCORING GUIDELINES (Form B)

Question 6

Let f be the function satisfying fxa()= x fx () for all real numbers x, where f ()325.= (a) Find f aa()3. dy yfx= = xy (b) Write an expression for () by solving the differential equation dx with the initial condition f ()325.=

fxa() x2 ¦£ 2 : fxaa ( ) (a) fxaa()=+ fx () x¸ =+ fx () ¦ fx ¦ 2() 2 ¦ < 2> product or ¦ 3 : ¤ ¦ chain rule error 919 ¦ f aa(3)=+= 25 ¦ 1 : value at x = 3 22 ¥¦

1 dy= x dx ¦£ 1 : separates variables (b) y ¦ ¦ 1 : antiderivative of dy term ¦ ¦ 1 : antiderivative of dx term ¦ 1 2 6 : ¤ 2 yxC=+ ¦ 1 : constant of integration 2 ¦ ¦ f = ¦ 1 : uses initial condition (3) 25 ¦ 1 2 11 ¦ 1 : solves for y 225=+ (3) C ; C = ¥¦ 2 2

111 Note: max 3/6 [1-1-1-0-0-0] if no yx=+2 44 constant of integration Note: 0/6 if no separation of variables 2 1111 2 yx=+=()22() x +11 4416

Question 6

The twice-differentiable function f is defined for all real numbers and satisfies the following conditions: f ()02,= f ′()04,=− and f ′′()03.= (a) The function g is given by gx()=+ eax f () x for all real numbers, where a is a constant. Find g′()0 and g′′()0 in terms of a. Show the work that leads to your answers. (b) The function h is given by hx()= cos ( kxf ) () x for all real numbers, where k is a constant. Find hx′() and write an equation for the line tangent to the graph of h at x = 0.

(a) gx′′()=+ aeax f () x ⎧ 1 : gx′() ga′()04=− ⎪ 1 : g′() 0 4 : ⎨ ′′() ⎪ 1 : gx 2 ax gx′′()=+ ae f ′′ () x ⎩⎪ 1 : g′′() 0 ga′′()03=+2

(b) hx′′()=− f () xcos ( kx ) k sin ( kxfx ) () ⎧ 2 : hx′() hf′′()00cos0sin0004=− () () kff () () ==− ′ () ⎪ ⎧ 1 : h′() 0 5 : == ⎨ ⎪ () hf()0cos002 () () ⎪ 3 : ⎨ 1 : h 0 The equation of the tangent line is yx=− 4 + 2. ⎩⎩⎪ ⎪ 1 : equation of tangent line

Question 6

Let f be a twice-differentiable function such that f (2) = 5 and f (52) = . Let g be the function given by g( xffx) = ( ( )). (a) Explain why there must be a value c for 2< c < 5 such that fc′( ) = −1. (b) Show that gg′′()2= ()5. Use this result to explain why there must be a value k for 2<

(a) The Mean Value Theorem guarantees that there is a value ⎧ ff(52) − ( ) ⎪ 1 : c, with 2<

(b) g′′( xffxfx) =⋅( ( )) ′( ) ⎧ 1 : gx′( ) ⎪ gfffff′′(2225) =⋅=⋅( ( )) ′′′( ) ( ) (2) 3 : ⎨ 1 : gffg′() 2=⋅=′′ () 5 () 2 ′ () 5 gfffff′′(5552) =⋅=⋅( ( )) ′′′( ) ( ) (5) ⎩⎪ 1 : uses MVT with g′ Thus, gg′′(25) = ( ). Since f is twice-differentiable, g′ is differentiable everywhere, so the Mean Value Theorem applied to g′ on [2, 5] guarantees there is a value k, with 25< k < , such gg′′(52) − ( ) that gk′′()==0. 52−

(c) g′′( xffxfxfxffxfx) =⋅⋅+⋅ ′′( ( )) ′( ) ′( ) ′( ( )) ′′( ) ⎧ 1 : considers g′′ 2 : ⎨ If fx′′( ) = 0 for all x, then ⎩ 1 : g′′()xx= 0 for all gx′′( ) =⋅00 fx ′( ) ⋅ fx ′( ) + f ′( fx( )) ⋅=0 for all x. Thus, there is no x-value at which g′′( x) changes sign, so the graph of g has no inflection points. OR

OR If fx′′ = 0 for all x, then f is linear, so g = ffD is 1 : f is linear ( ) 2 : linear and the graph of g has no inflection points. { 1 : g is linear

(d) Let hx( ) =− f( x) x. ⎧ 1 : hh( 2) and ( 5) 2 : ⎨ hf(2) =−=−=( 2) 2523 ⎩ 1 : conclusion, using IVT

hf()55525=−=−= () −3 Since hh(20) >>( 5), the Intermediate Value Theorem guarantees that there is a value r, with 2< r < 5, such that hr( ) = 0.

AB{6 1999 y

1 1

In the gure ab ove, line ` is tangent to the graph of y = 6. y = 2 2

x x

, where w>0. Point at p oint P , with co ordinates w;

2 w

Q has co ordinates (w; 0). Line ` crosses the x{axis at the

point R , with co ordinates (k; 0).

Find the value of k when w =3.

(a) O Q R

(b) For all w>0, nd k in terms of w .

rate of change of k with resp ect to time?

(d) Supp ose that w is increasing at the constant rate of 7 units p er second. When w = 5, what is the

rate of change of the area of 4PQR with resp ect to time? Determine whether the area is increasing

or decreasing at this instant.

2 2 dy dy

= = ; (a)

dx x dx 27

x=3

x=3 2

1: answer

1 2

Line ` through 3; and (k; 0) has slop e .

9 27

2 1 2

Therefore, = or 0 = (k 3)

k 3 27 9 27

k =

1 2

1: equation relating w and k ,

(b) Line ` through w; and (k; 0) has slop e .

2 3

w w

using slop es

1: answer

2 1 2

Therefore, = or 0 = (k w )

3 2 3

k w w w w

w k =

dk dw 3 dw 3 21 dk 21

dt 2 dt 2 2 dt 2 dt

w =5

P w; 1=w

(d)

1: area in terms of w and/or k

> .

. .

> .

. .

> .

. .

> . .

. .

> .

. .

> .

. .

. dA

. .

> .

. .

implicitly . . 1:

. .

. > .

. .

dt .

. <

. .

......

R (k; 0)

Q(w; 0)

dA dw

1: using =7

> dt dt

1 1 3 1 1 1

w =5

(k w ) = w w = > A =

2 2

2 w 2 2 w 4w

1: conclusion

dA 1 dw

Note: 0/4 if A constant

dt 4w dt

1 dA

= 7=0:07

dt 100

w =5

Therefore, area is decreasing. 122 AP® CALCULUS AB 2003 SCORING GUIDELINES (Form B)

Question 1

Let f be the function given by fx()= 4 x23 x , and let A be the line yx= 18 3 , where A is tangent to the graph of f. Let R be the region bounded by the graph of f and the x-axis, and let S be the region bounded by the graph of f, the line A, and the x-axis, as shown above. (a) Show that A is tangent to the graph of yfx= () at the point x = 3. (b) Find the area of S. (c) Find the volume of the solid generated when R is revolved about the x-axis.

(a) fxa()= 8 x 3 x2 ; f a(3)= 24 27= 3 ¦£ 1 : finds ffa (3) and (3) ¦ f = = ¦ finds equation of tangent line (3) 36 27 9 ¦ ¦£ ¦ ¦ Tangent line at x = 3 is 2 : ¦ ¦ or ¤ ¦ ¦ 1 : ¤¦ yx= 3( 3)+= 9 3 x+ 18, ¦ ¦shows (3,9) is on both the ¦ ¦ ¦ ¦graph of f and line A which is the equation of line A. ¥¥¦¦¦ ¦

(b) fx()= 0 at x = 4 ¦£ 2 : integral for non-triangular region ¦ The line intersects the x-axis at x = 6. ¦ 1 : limits ¦ 1 4 ¦ Area = (3)(9)() 4xxdx23 4 : ¤ 1 : integrand 2 ¨3 ¦ ¦ 1 : area of triangular region = 7.916 or 7.917 ¦ ¦ ¦ 1 : answer OR ¥¦ 4 Area = ()()18 3xxxdx() 4 23 ¨3 1 + (2)(18 12) 2

= 7.916 or 7.917

4 ()xxdx232 ¦£ 1 : limits and constant (c) Volume = Q¨ 4 ¦ 0 ¦ 3 : ¤ 1 : integrand = 156.038 Q or 490.208 ¦ ¦ 1 : answer ¥¦

2 AP® CALCULUS AB 2004 SCORING GUIDELINES (Form B)

Question 6

Let A be the line tangent to the graph of yx= n at the point (1, 1), where n > 1, as shown above. 1 (a) Find xn dx in terms of n. ∫0 (b) Let T be the triangular region bounded by A , the x-axis, and the 1 line x = 1. Show that the area of T is . 2n (c) Let S be the region bounded by the graph of yx= n , the line A , and the x-axis. Express the area of S in terms of n and determine the value of n that maximizes the area of S.

+ 1 n 1 xn 1 1  1 : antiderivative of x (a) xdxn == 2 :  ∫0 nn++11 1 : answer 0 

(b) Let b be the length of the base of triangle T.  1 : slope of line A is n   1 1 3 : 1 : base of T is is the slope of line A , which is n  n b  1  1 : shows area is  2n 11 Area()Tb==() 1 22n

1 1 : area of Sn in terms of (c) Area()SxdxT=−n Area ()  ∫  0  1 : derivative 4 :  =−11 1 : sets derivative equal to 0 +  nn12  1 : solves for n  d 11 Area()S =− + = 0 dn (1)2nn+ 22

21nn2 =+()2

21nn=+()

1 n ==+12 21−

Copyright © 2004 by College Entrance Examination Board. All rights reserved. Visit apcentral.collegeboard.com (for AP professionals) and 124www.collegeboard.com/apstudents (for AP students and parents).

7 AP® CALCULUS AB 2006 SCORING GUIDELINES (Form B)

Question 3

The figure above is the graph of a function of x, which models the height of a skateboard ramp. The function meets the following requirements. (i) At x = 0, the value of the function is 0, and the slope of the graph of the function is 0. (ii) At x = 4, the value of the function is 1, and the slope of the graph of the function is 1. (iii) Between x = 0 and x = 4, the function is increasing.

(a) Let f ()xax= 2 , where a is a nonzero constant. Show that it is not possible to find a value for a so that f meets requirement (ii) above. x2 (b) Let gx()=− cx3 , where c is a nonzero constant. Find the value of c so that g meets requirement (ii) 16 above. Show the work that leads to your answer. (c) Using the function g and your value of c from part (b), show that g does not meet requirement (iii) above. xn (d) Let hx()= , where k is a nonzero constant and n is a positive integer. Find the values of k and n so that k h meets requirement (ii) above. Show that h also meets requirements (i) and (iii) above.

1 11 (a) f ()41= implies that a = and fa′()4241== () ⎪⎧ 1 : aa== or 16 2 : ⎨ 16 8 1 ⎪ 1 : shows a does not work implies that a = . Thus, f cannot satisfy (ii). ⎩ 8

1 (b) gc()46411=−= implies that c = . 1 : value of c 32

1 2 24() 11 When c = , gc′()434=−= () 3()() 16 −= 1 32 16 32 2

312 x (c) gx′()=−=− x x()34 x ⎧ 1 : gx′() 32 8 32 2 : ⎨ 1 : explanation 4 ⎩ gx′()< 0 for 0,<

n n 4 n ⎧ 4 (d) h()41== implies that 4.= k 1 : = 1 k ⎪ k −− ⎪ nnn44nn11 ⎪ n4n-1 h′()41==== gives n = 4 and k ==4256.4 4 : 1 : = 1 n ⎨ k 4 4 ⎪ k ⎪ 1 : values for kn and x4 ⎩⎪ 1 : verifications hx()= ⇒ h()00.= 256 4x3 hx′′()= ⇒ h()00= and hx′()> 0 for 0<

Question 4

3x The functions f and g are given by f ()xtdt=+4 2 and gx()= f (sin x ) . ∫ 0 (a) Find f ′()x and gx′(). (b) Write an equation for the line tangent to the graph of ygx= () at x = π. (c) Write, but do not evaluate, an integral expression that represents the maximum value of g on the interval 0≤≤x π . Justify your answer.

2 (a) f ′()xx=+34 ( 3 ) ⎧ 2 : f ′()x 4 : ⎨ ⎩ 2 : g′()x gx′′()=⋅ f (sin x ) cos x

=+34() 3sinx 2 ⋅ cosx

(b) g()π = 0, g′()π =−6 ⎧ 1 : gg()ππ or ′ () 2 : ⎨ Tangent line: yx=−6() −π ⎩ 1 : tangent line equation

π (c) For 0,<402 2 ∫ 0 The maximum value of g on [0, π ] is 3 4.+ tdt2 ∫ 0