calculus bc chapter 6 ... e qx′( )=⇒p(x) degree of q is n+1 differentiating the expression in...

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AP Calculus BC Chapter 6 - AP Exam Problems solutions 1. 2. 3. 4. 5. 4 2 5 3 1 2 E Expand the integrand. x + ( 1 2 2 ) ( 2 1) 3 5 dx x = + x + dx = x + x x + C + A 3 3 f () x x = sin x + 2 C , f () x = 1 x + cos x + Cx + K . Option A is the only one with this form. A sec 2 x dx d tan x ( ) = = tan x C + A Since f is linear, its second derivative is zero. The integral gives the area of a rectangle with zero height and width ( b a) . This area is zero. D 2 2 3 2 1 1 1 1 1 3 1 2 2 4 8 x dx =− x =− = . 6. D 2 2 2 7 7 2 2 2 16 ( x + k ) dx x dx = k dx 0 = 2 + ( 2) ( ) k 4k = k 4 = = + 7. C 2 2 2 2 1 2 1 1 1 1 1 2 dx x x dx = x = = 8. E 2 2 2 1 1 1 1 x dx 1 1 1 1 1 3 2 ln 1 0 2 2 2 2 2 e e x e dx x e e x x = = = x = 9. D 2 ) 2 )( dx ) ) 1 1 1 2 2 3 0 0 0 1 = 3 ( x 1 1 3 3 3 2 ( ) 1 3 33 9 ( x d = x ( x = 1 ( ) 8 1 = 10. E 1 1 1 1 2 2 2 1 0 0 0 3 3 3 2 lim + 3 2 2 lim L L L dx dx dx x x + L x x = = = which does not exist.

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AP Calculus BC

Chapter 6 - AP Exam Problems solutions

1.

2.

3.

4.

5.

4 2 5 31 2E Expand the integrand. ∫ x +( 12 2) ( 2 1)35

dx x= +∫ x + dx = x + x x+ C+

A 33

f ′( )x x= − sin x +2 C , f ( )x =1 x + cos x +Cx + K . Option A is the only one with this form.

A sec2 x dx∫ ∫ d tan x( )= = tan x C+

A Since f is linear, its second derivative is zero. The integral gives the area of a rectangle withzero height and width ( −b a) . This area is zero.

D22 3 2

1 1

1 1 1 312 2 4 8

x dx = −− −x = − −⎛ ⎞⎜ ⎟ =

⎠⎝∫ .

6. D2 2 27 72 2 2

16 (x + k)dx x dx− − −∫= k dx 0= 2+ (− 2)( )− k 4k= k⇒ 4== ∫ + ∫

7. C2 2 2 21

21 1 1

112

dxx

x− dx = x− − =∫ = ∫

8. E2

221 1

1

1 x − dx1 1 1 1 1 32 ln 1 02 2 2 2 2

eexe dx x e e

x x− ⎛ ⎞ = ⎛ −⎞ ⎛ ⎞= −= ⎜ − x ⎟ ⎜ − ⎟ ⎜ − =⎟

⎝ ⎠ ⎝ ⎠ ⎝ ⎠∫ ∫

9. D 2) 2) ( dx) )111 2 2 3

0 0 0

1= ⋅ 3( x1

−13 3 3 2 ( ) 1

3 3 3 9( x d− =x∫ ∫ ( x − = 1− ( )8−1

=

10. E111 1

22 21 0 0 0

3 3 3 2 lim+

32 2 limLL L

dx dx dx− x x → + L x x→

= = −∫ = ∫ ∫ which does not exist.

11.

13.

12.

14.

A2 22 2 2 2

11 1 1

1 x( 1+ )(x 1− ) 1(11 2 2

dxx dxx x−

= = x 1)dx− = x(1 1)− =+ +∫ ∫ ∫

C 2 3 3 30

1 218 27, so 3 3

kkx= − x⎛ ⎞⎜ ⎟ = k k⇒ = k =

⎠⎝3

A ( (k3 )2 3 3 33 3x dx = x1 1 3) ) = 1 27 0

3 3 3kk

k− −

= (− −∫ + = only when k = −3.

A The value of this integral is 2. Option A is also 2 and none of the others have a value of 2.Visualizing the graphs of = siny x and = cosy x is a useful approach to the problem.

15. D The graph is a V with vertex at x =1. The integralgives the sum of the areas of the two triangles that the V forms with the horizontal axis for x from 0 to 3. These triangles have areas of 1/2 and 2 respectively.

16. C The graph is a V with vertex at x = 3 . Theintegral gives the sum of the areas of the two triangles that the V forms with the horizontal axis for x from 1 to 4. These triangles have areas of 2 and 0.5 respectively.

17. A By the Fundamental Theorem of Calculus0 0

cc( )x∫ f ′( )x dx f= = f (c) f− (0)

18.

19.

20.

21.

B1

0 0 11 1 3

2 2e ef x( )dx x dx dx

x= ln+ e =∫ = ∫ + ∫

B500 500 500 500

1 2 1 2(13 11x x )dx− +∫ ∫ (11x 13− )x dx = (13∫ 11x x )dx− − (13∫ 11x x− )dx

2222

1 1

13 11 13 13 11 11 14.946ln13 ln11 ln13 ln11

xxx(13 11− )x dx

⎛ ⎞ − −= ⎜⎜= − ⎟⎟ = − =

⎝ ⎠∫

E ′(Q x) p(x)= ⇒degree of Q is n 1+

Differentiating the expression in (i) gives f ′′( ) = 2x a +x b

Let x = 1. Then from (ii),

a b+ = f ′(1) = 6 a b+ = f ′′2 ( =1) 18

Solving these two equations gives a =12 and b = −6 . Therefore f ′( ) 12x x2= − 6x and hence

f ( ) 4x x= −33 2x +C

Using (iii) gives

2 3 21

24 31

18 (4

)

−3x x )+C dx

Cx

C C

=

= −x x( +

(16= −8+ 2 ) (1− 1− + ) 8C = +

Hence C =10 and f x( ) 4x= −33 2x 1+ 0 .

22.

23.

24.

D1

−1 1 1

+ 5) 22 (3x 5)+2 2 2(3x 5)+2 2 (3x 5)+1 2 2= ⋅

166 3

x(3x dx 6( x d )x C−

+C = +∫ =1∫

D 1 2 ( ) ( − )1 2 20 0

cos dθ θ = 2 1+() sin 2 2 1ππ

∫ 2 1 sin+ θ( )− θ =

A Let u x 13= + . Then2

1/ 2 33

3 2 2 11

x Cx

x dx = u−1/ 2du u= +C = + ++

∫ ∫

25.

26.

D x( + ) x dx( ) = ⋅ x( )1+

1= (2 2 1311 2 2 2 3 3

0 0 0

1−

1 3 ) =2 192 2 2 22 2 3 6 6

x dx∫ ∫= x( )+

C dx = ∫1 cos(2x) (2dx) sin(2=

1cos(2x)2 2

x) C+∫

27. D3 3

00cos(3 )x1 cosπ1

− cos 0( ) = 2sin( x3 )3 3 3

dxπ π= − = −∫

28. C 3) (2dx) = − cos(212 2

sin(2x 3)dx+ =∫ ∫ sin(2x1+ x 3)+ C+

29. B3

) = − cos( )31= − ⎜

1y x( ; Let , 0 03 2 3 2

x +C x π cos⎛ ⎞π= +⎟ C⇒ C =⎝ ⎠

. y(0) = − cos 0( )3 = −1 13 3

30. A 12 2

du

−a u= sin− u⎛ ⎞ + ,C a 0>

a⎜ ⎟⎝ ⎠∫

33 1 1 10 2 0⎜ ⎟2 2 34

dx

xsin x⎛ ⎞= = sin− − 3⎛ ⎞

sin (− 0)− π=⎜ ⎟⎜ ⎟⎝ ⎠− ⎝ ⎠

31.

32.

A 1 12 2 225

sin ⎛ ⎞= ⇒− udu du dx dxa u x

sin= ⎛− x ⎞ C+a⎜ ⎟ ⎜ 5 ⎟⎝ ⎠ ⎝ ⎠−−

∫ ∫

B f x( ) x2e= + 2xx xe = xe (x x 2+ ) ;′ ′f x( ) 0< for 2− < x 0<

33. E +x ee d =x∫ ∫x xee xe( )dx . This is of the form ∫ e du, u = eu x , so ∫ +x ee dx

x xee= +C

34.

35.

A Use the technique of antiderivatives by parts2

212

x

x

u x dv e dx

du dx v e

= =

= =

2 2 2 21 1 1 12 2 2 4

x ∫e ex x− =dx xe x − e x +C

A 1 )4 4 4 111 3 3

0 0 0

1 (4x dx) = 1 14 4 4

x e dx∫ ∫ ex x= ex e(= −

36.

37.

38.

B −1 2sin( x2 )tan( x2 ) 1 ln cos(22 cos( x2 ) 2

dx dx x) +C= −∫ = − ∫

B33 3 2

22 22 2

1 2 1 ln 1 ln10 ln 5( )−1 ln= 2

2 2 2 21 1dxx x dx

x x== x( )+

1=

+ +∫ ∫

E1

1 ln1 lnx dx xxx u

x du dx⎞⎛ dx⎛ ⎞= =∫ ∫⎜ ⎟ ⎜ x ⎟⎝ ⎠ ⎝ ⎠∫ ∫ .This is ∫u du with = lnu x , so the value is

ln( 2)2x

+C

39.

40.

41.

42.

C9

1(9) −F F (1)

3(ln t) dt = 5.827t∫= using a calculator. Since F (1) = 0 F, (9) = 5.827.

Or solve the differential equation with an initial condition by finding an antiderivative for (ln x)3

x. This is of the form 3u du where = lnu x . Hence F x( ) = 1 (ln 4

4x) +C and since

4F (1) = 0 , C = 0. Therefore 1 (ln 9)4 5.827=F (9) =

B ( 2) (ln 8 ln 3− )22 2 2

2 21 1 1

21 1 1 1ln2 2 2+x x2 2

+x dxx dx+= = x 2x+ =

x x+∫ ∫

E Since F is an antiderivative of f, F (2 )x =1 1 F (6) F (2− )( )3 3

11 2 2∫ (2f x) dx =

A ; when x 2,= =u 1 and when x 4,= =u 22 2x

= ,u du 1= dx

2

2 24 2 2

12 1

11 12 2

2

x

du⋅u duudxux u

−⎛ ⎞⎜ ⎟ −−⎝ ⎠ = ∫∫ = ∫

43.

44.

45.

46.

B Separate the variables. y−2 1 1;2

dy = 2dx C ; yy +x C

−− = 2x + = . Substitute the point (1,−1)

1to find the value of C. Then −1 1,2

CC

−= ⇒ = −

+so 1

1 2y

x=

−. When =x 2,

3y = 1

− .

C This is the differential equation for exponential growth.2y y(0)e= =− −e2 2t t ; 1

2= e− t ; − =2 ln ⎛ ⎞

⎜ ⎟⇒t t =1 1− ln ⎛ ⎞

⎜ ⎟ =1 ln1 2

2 2 ⎝ ⎠2 2⎝ ⎠

A dy 3x= y2 = x +3dy 3x d2 lx n y ;K =y x aC nd e (0)y 8 s= o, 8= xy3 3

edx y

⇒ = ⇒

C dy sec= x dx ln y tan x k+ y⇒ Ce2 t= x . y(0) 5= yan ⇒ 5e= xtan

y⇒ =

47. C31

2 3 31

1, ln3

xdy x dx yy= = x C , y+ Ce= . Only C is of this form.

48.

(a)dy

dx=

3x2 + 12y

dy∣∣∣

dx ∣ x = 1y = 4

=3 + 12 · 4

=48

=12

1: answer

(b) y − 4 =12

(x− 1)

f(1.2)− 4 ≈ 1(1.2− 1)

2

f(1.2) ≈ 0.1 + 4 = 4.1

2

{1: equation of tangent line

1: uses equation to approximate f(1.2)

(c) 2y dy = (3x2 + 1) dx

2y dy =∫ ∫

(3x2 + 1) dx

y2 = x3 + x+ C

42 = 1 + 1 + C

14 = C

y2 = x3 + x+ 14

y =√x3 + x+ 14 is branch with point (1, 4)

f(x) =√x3 + x+ 14

5

1: separates variables

1: antiderivative of dy term

1: antiderivative of dx term

1: uses y = 4 when x = 1 to pick onefunction out of a family of functions

1: solves for y0/1 if solving a linear equation in y0/1 if no constant of integration

Note: max 0/5 if no separation of variablesNote: max 1/5 [1-0-0-0-0] if substitutes

value(s) for x, y, or dy/dx beforeantidifferentiation

(d) f(1.2) =√

1.23 + 1.2 + 14 ≈ 4.114 1: answer, from student’s solution tothe given differential equation in (c)

49.(a) ln y dy = −x dx

y

2

22

2 = −x +2

2

2 2or

or

lnor

−C x

= − +x(ln )y C

(ln )y C

= ± −Cy x

=y e±

(b)

2

2 2

22

2

22

4

(ln )4

(ln )

ln 4

But x = 0, ln

x

e CC

xy

y x

y x

=y e −

0= +=

4= −

= ± −

y e= ⇒ = 4−

(c) If x = 2 , then y =1 and ln y = 0 . This causes ln

xyy

− to be undefined.

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51.

(a) 2

22

d y dy2 (6y x2 ) 2ydxdx

� � �

= 2 (6y x23 2) 2y2� �

� � 4 ��22

23,1

4

1 10 28

d ydx

� � � �

3 :

1 : value at 3, 1� �

2 2 : 2

<� 2 > product rule or

chain rule error

4

d ydx

���������������������

(b) 21 dy (6 2x)dxy

� �

1 6x x2 Cy

� � � �

�4 18 � 9 � � 9 �C C�

C � �13

216 13

yx x

�� �

6 :

1 : separates variables

1 : antiderivative of term

1 : antiderivative of term

1 : constant of integration

1 : uses initial condition f (3) � 14

1 : solves for

dy

dx

y

���������������������������

Note: max 3/6 [1-1-1-0-0-0] if no

constant of integration

Note: 0/6 if no separation of variables

52.AP® CALCULUS AB

2001 SCORING GUIDELINES

Question 6

The function f is differentiable for all real numbers. The point �3, 14 � is on the graph of

y � .�f ( )x , and the slope at each point ,x y� � on the graph is given by dy 2 �6 2y xdx

� �

(a) Find2

2 d ydx

and evaluate it at the point �3, 1 �4.

(b) Find �y f ( )x by solving the differential equation dy�

2 �6 2y x�

dx w� ith the initial

4condition f (3) � 1 .

(a) 2

22

d y dy2 (6y x2 ) 2ydxdx

� � �

= 2 (6y x23 2) 2y2� �

� � 4 ��

22

23,1

4

1 10 28

d ydx

�� � �

3 :

1 : value at 3, 1� �

2 2 : 2

<� 2 > product rule or

chain rule error

4

d ydx

���������������������

(b) 21 dy (6 2x)dxy

� �

1 6x x2 Cy

� � � �

�4 18 � 9 � � 9 �C C�

13C ��

216 13

yxx

���

6 :

1 : separates variables

1 : antiderivative of term

1 : antiderivative of term

1 : constant of integration

1 : uses initial condition f (3) � 1 4

1 : solves for

dy

dx

y

���������������������������

Note: max 3/6 [1-1-1-0-0-0] if no

constant of integration

Note: 0/6 if no separation of variables

53. E4 4

111dt ln t= = ln 4 ln− 1 ln= 4∫ t

54.

55.

56.

57

C Use the technique of antiderivatives by parts:

xe − e )(1

10

2

x

x

x x x x

=u x dv e dx

du = dx

xe e dx e

− − − − −

=

v e= −

∫+− = − =1−

E Use the technique of antiderivatives by parts: Let =u x and dv = cos x d .x

cosx x d =x sx in x( ) =2 ( ) 2200 0

sin + cos2

x dx sx in x xπ ππ π

= −∫ − ∫ 1

E Use the technique of antiderivatives by parts: Let =u x and dv = cos x d .x

cosx x d =x sx in x( ) =2 ( ) 2200 0

sin + cos2

x dx sx in x xπ ππ π

= −∫ − ∫ 1

E Use the technique of antiderivatives by parts: u x and d= =v sec2 x dx

xsec2 x dx x tan x= −∫ ∫ tan x dx x tan x= + ln cos x C+

58.

B Use the technique of antiderivative by parts, which is no longer in the AB CourseDescription. The formula is u dv uv= −∫ ∫ v du . Let =u f ( )x and dv = x dx. This leads to

x( )2 2

x f dx x f x( )= −1 1

∫ ∫ x2 2 f ′ x( ) dx .

59.

60.

61.

62.

B Use the technique of antiderivatives by parts:dv = sin x dx u f ( )x=

du f ′ x( )= dx v = cos x−

f x( )sin x dx = − f x( ) cos x +∫ ∫ f ′ x( ) cos x dx and we are given that

f ( )sinx x dx = − (f ) cx os +x∫ ∫3x cos x dx ⇒ ′(f )x = 32 2x ⇒ (f )x = x3

A Use partial fractions to rewrite−( )x x( +1 2)

11 1 1 as 3 1 2x x⎛ ⎞−

+ ⎠⎟

⎝⎜ −

1)( x x− +( )1 x( − −l xn 1 ln )1 = 1 1 1 1

+ C+ =2 1 ln2 3 1 2 3 3 2

xdx Cx x

−⎛ ⎞− dx = +⎜⎝ x − ++ ⎟

⎠∫ ∫

D Use partial fractions:

1) 1) (ln x 1− ln x− 2+ )3 3 3

22 213 1 ln 8

1 2 5dx

x⎜= ⎛ − dx⎞ = ln 2= ln 5− ln1− ln 4+ =+ ⎟(x x(− + x −⎝ ⎠∫ ∫

63.

A Use partial fractions. 21 1 1 1 1

= ⎛ ⎞= −⎠⎟x x− +6 8 −( 4x x)( 2) ⎝− −⎜2 x 4 x 2−

ln1 x 4( − ln− x )21 ln1 4

2 2 26 8xdx Cx− +xx−

= 2− C+ = +−∫

C All slopes along the diagonal = −y x appear to be 0. This is consistent only with option (C).For each of the others you can see why they do not work. Option (A) does not work because all slopes at points with the same x coordinate would have to be equal. Option (B) does not work because all slopes would have to be positive. Option (D) does not work because all slopes in the third quadrant would have to be positive. Option (E) does not work because there would only be slopes for y > 0 .

64.

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67.

(a)

2 :

,x y( )

, y( )

, y( )

1 : zero slope at each point where 0 or 1

where 0 and 1

where 0 and 1

x y

positive slope at each point xx y

negative slope at each point xx y

==

>≠ 1 :

≠ <

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

1 : description

21dy = x dx(c) y

1−

3

3

3

3

13

13

0

13

ln 1 13

12

1 2

x

x C

x

y +x C

y eCe

y Ke , K = ±eKe K

y e

− =

1− =

− =

= =

= +

6 :

1 : separates variables 2 : antiderivatives 1 : constant of integration 1 : uses initial condition 1 : solves for

0 1 if y is not exponential y

Note: max 3 6 [1-2-0-0-0] if no constant of integration

Note: 0 6 if no separation of variables

68.

B This is an example of exponential growth. We know from pre-calculus that 23.522

t

w = ⎛ ⎞⎜ ⎟⎝ ⎠

is

an exponential function that meets the two given conditions. When t = 3 , w = 4.630 . Using calculus the student may translate the statement “increasing at a rate proportional to its weight” to mean exponential growth and write the equation = 2w ekt . Using the given

conditions, 3.5 = 2e2k ; ln(1.75)

22

te⋅

ln(1.75) 2 ;k k= =ln(1.75) ; w 2= . When t = 3 , w = 4.630 .

69. A A known solution to this differential equation is y t( ) = y(0)ekt . Use the fact that thepopulation is y2 (0) when t =10. Then

2 (0)y y(0) (1e 0)= ⇒ 10k ke 2= k⇒ (0.1)= ln 2 0.069=

70. A This is an example of exponential growth, 30 2

t=B B ⋅ . Find the value of t so B = 3B0 .

3 30 0 ln3= ln 2t 3ln 33 2 3 2

3 ln 2t t

B B= ⋅ ⇒ = ⇒ ⇒ t =

71.

72.

C dN kN= ⇒ N C= ekt . (0) =1000N C⇒ =1000 . =N (7) 1200 1 ln(1.2)7

=k⇒ . Therefore

127 ln(1.2)

dt

(12) =1000N e 1367 .≈

E As lim 0dP= for a population satisfying a logistic differential equation, this means that

t→∞ dtP →10,000 .

(a) kt

dy kydty Ce

=

= or

1

1lnkt+C

k dty

k= +ty C

=y e

dy =

6 61

6

6

ln 2 66 66

0 10 , ln10

162ln 26

10 10 2

kt

k

tt

t = ⇒ =C C∴ =y e

t e

y e− −

=

= ⇒ =

k∴ = −

= = ⋅

10

(b) 5

5

5

10dt 6

= −10 ln 2Decreasing at 10 ln 2 gal/year

dy ky= = − ln 2 6⋅ ⋅

4 6

2

(c) 5 10ln 20

ln 20ln 26

ln 206 6= log 20ln 2

6 ln 20 years after startingln 2

ekt

t

10⋅ =kt∴ = −

−−

=

∴ =

73.

74.

(a)

0.138 or 0.139

5

5

0 6 6

5 12 12 6

2ln 25

kt

k

k

S t Ce

S C

S e

e

k

(b)ln 2135

3

2.6ln 2 0.6ln 2

1

13 3

3

ln 2

( )Average rate 6

billion gal/yr

(19.680 billion gal/yr)

te dt

e e

(c) S S S7

51 5 2 5.5 2 6 2 6.5 74

S t dt S S

(d) This gives the total consumption, in billions of gallons, during the years 1985 and 1986.

(a) Let N = number of bacteria present at time t. Then

1

2

lnkt

dN= kN

dtdN kdtN

N k= +t C

=N C e

=

tAt = 0, 02210000 C ek⋅= = C . Therefore =10000N ekt .

At =t t1 , 20000 10000 kt= ⇒ 2e ekt1 1=

At t t1= +10 , 100000 10000 (k t1+10)= ⇒10e ekt1= 10⋅ 2e 10k k= e

10

ln51010000

tTherefore k =

ln 5 and so =N e .

(b) At t = 20 , ln5 20( )

10000N e 10= =10000 ln 25e 250000=

(c) ln51020000 10000

te=

ln5102

t= e

ln 210

=ln 5 t

10ln 2ln 5

t =

75.

76. (a) =y A kte

5

(0) 1000

3ln 35

k

Ay

e

k

==

=

=

=y Thereforet ln 3

51000e or y 1000= ⋅3 t 5

(b) 10ln3

(10) 1000y e 5= =1000 32 ⋅Therefore the population will have increased by a factor of 9.

(c) ln 35

ln35

6000 1000

65ln 6ln 3

t

t

e

e

t

=

=

=

(a) For this logistic differential equation, the carrying capacity is 12.

P t(t→∞

If P( ) =0 3 , lim ) = 12.

P t(t→∞

If P( ) =0 20 , lim ) = 12.

2 : 1 : answer1 : answer

(b) The population is growing the fastest when P is half the carrying capacity. Therefore, P is growing the fastest when P = 6.

1 : answer

(c) Y dY1 1 1= −5 1( )2 dt = 5

1t t−( )60 dt

2ln 5 120= −t tY C+

2

5 120−t tY t( ) = KeK = 3

)Y t(2

5 1203−t t

= e

5 :

1 : separates variables 1 : antiderivatives 1 : constant of integration 1 : uses initial condition 1 : solves for

0 1 if is not exponentialY

Y

Note: max 2 5 [1-1-0-0-0] if no constant of integration

Note: 0 5 if no separation of variables

(d) lim Y t( ) = 0t→∞

1 : answer 0 1 if Y is not exponential

77.