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AP Calculus BC - Parametric equations and vectors Chapter 9- AP Exam Problems solutions 1. 2. 4. 3. A 3 5 and 3 5 dx dy dt dt dx = = dy = B At t = 1, 2 1 3 3 3 2 3 1 4 3 1 8 3 1 t dy t dy dt dx dx t dt = + = = = = C At t = 1, 3 2 1 4 4 2; the point at t 1 = is (2,3) 4 3 1 t dy t t dy dt dx dx t dt = + = = = 8 = + . = + 3 2 y x ( ) 2 = 2 1 x D At t = 3, =− =− 2 6 2 3 1 slope t te + t t t t dy e t e e e dx dx dt = = dy = dt = = 0.005 5. 6. 7. A 2 2 2 2 3 dy t = 3 3 2 3 t thus dx t 2 2 2t t 4 d d dy dt = ; 3 t dx dt dt dx dx dt dy dy dx dx dx 2, dy t = = = = = = B If x = 2 then y = 5. 6 0 5 dy x y + dx dx dt dt dt dt = 0; 2(3) 5 + = dx =− E 2 2 2 2cos(2t ) c = os(2t ) sin = (2 ) t ; 2 t t t dy = x e , y dx e e = C 3 3 3 2 2 + t 4 4 8 4 8 1, 2) 3 t t 2 8; t dy = t t x t = t dx t (3 t + t 4 y = 2 t 4 + t 2 = . Vertical tangents at t = 0, 2 3 8.

10. 11. · 2016-01-15 · dy e t dx dx ee e dt − = − = dy =dt = = 0.005 5. 6. 7. A 22 2 2 3 dy t = 3 2 3 3t thus dx t22 2t t4 d d d y dt =; 3 t dt dt dx dx dx dt ⎛⎞dy ⎛⎞dy

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AP Calculus BC - Parametric equations and vectors

Chapter 9- AP Exam Problems solutions

1.

2.

4.

3.

A 35 and 35

dx dydt dt dx

= =dy

⇒ =

B At t = 1, 2 1

3 332 3 1 4

3 1 83 1 t

dytdy dt

dx dx tdt

=

+= = = =−−

C At t = 1,3

2 1

4 4 2; the point at t 1= is (2,3)43 1 t

dyt tdy dt

dx dx tdt

=

+= = =

8=

+. = +3 2y x −( )2 = 2 −1x

D At t = 3, = − = −262

3

1 slopette− +−t

t tt

dye t

e e edx dxdt

=

−=

dy= dt = = 0.005

5.

6.

7.

A2 2

22

3dy t

=3 323t thus

dx t2 2 2t t4

dd

d y dt= ;3 t dxdt dt dxdxdt

dy⎛ ⎞dy⎛ ⎞ dx⎜ ⎟

⎝ ⎠dx⎜ ⎟dx 2 , dyt= = = ⎝ ⎠ = = =

B If x = 2 then y = 5 . 605

dyx y+dx dx

dt dt dt dt= 0; 2(3) 5+ ⋅ = ⇒

dx= −

E 22 2

2cos(2t) c=

os(2t)sin= (2 )t ;2

tt t dy

=x e , ydx e e

=

C3 3

3 22+ −t44 8 4 81,

− 2)3 −t t28 ;t d y

=t tx t= − t

dx t(3t+ t4 −

− y = 2t4 + t2 − = . Vertical tangents at t = 0, 23

8.

9.

10.

11.

12.

(a) 64

= −3y t( −t 128)

tAt = 32, (32)(323−128) ⎟

⎞ (−96) =3 14464 2

y = − = −⎛⎜⎝ ⎠

.

(b) 364 64

dydt

(3 ⎞= −⎛ ⎞ −⎛t t −⎜ +⎟ ⎜ ⎟⎝ ⎠ ⎝ ⎠

128)

tAt = 32, 3 3 92 64 2

dydt

= ⎛ ⎞ ⎛− (3 ⎞ −96) = ⎛ ⎞ ⎛ ⎞ =⎜ ⎟− + ⎜ ⎟ ⎜ ⎟− + ⎜ 2 ⎟⎝ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ ⎠3

(c) Let s be the length of the string.

s x= +2 2y

−1 ⎛21 x y2 2+( ) 2 22

ds x dx+ y dy

dt dt dt⎞= ⎟⎜⎠⎝

tAt = 32, x = 4 and y = 144.

tAt = 32, 18

dxdt

= and dydt

= 3 .

Therefore tat = 32,

22 2

1 1432 4321 1 2282 4 144

dsdt

++⎞⎛2(4) 1 2(144)(3)+= ⎟ = =⎜⎠⎝+ 16 (16 9) 4+ 1297⋅

.

6 6(d) The kite will start to lose altitude at the instant when 064

tdt

=dy

= − + . This occurs

at t = 64 seconds.

1973 BC4

A kite flies according to the parametric equations 3 (,

8 64tx y= = − t t −128)

where t is measured in seconds and < t ≤0 9 . 0

(a) How high is the kite above the ground at time t = 32 seconds?

(b) At what rate is the kite rising at t = 32 seconds?

(c) At what rate is the string being reeled out at t = 32 seconds?

(d) At what time does the kite start to lose altitude?

13.1974 BC5

Given the parametric equations = 2(θ − sinθ) and x y = 2(1− cosθ) :

(a) Find dy in terms of θ . dx

(b) Find an equation of the line tangent to the graph at θ = π .

(c) Find an equation of the line tangent to the graph at θ 2= π .

(d) Set up but do not evaluate an integral representing the length of the curve over the interval ≤ θ ≤0 2π . Express the integrand as a function of θ .

1974 BC5 Solution

(a) = −2 2cosθ, 2sin

2sin sin−2 2cosθ −1 cos

dx dyd d

dydy ddx dx

d

= θθθ

θ θθ == =θ

θ

(b) At θ = π, sin 0−1 cos

dydx

π= =

π. This means the tangent line is horizontal.

At ,θ = π y = 4, so the equation of the tangent line is y = 4.

(c) At θ = 2π, sin 2−1 cos 2

dydx

π=

π which is not defined.

2 2lim sin lim cosθ→ π 1 co− θs θ→ π sin

θθ= →∞

θ

This means that there is a vertical tangent. At θ 2= π , x = 4π , so the equation of the tangent line is x 4= π

(d) Length2 2220 0

(2− 2cos )θ (2s+ in )θ θ =d 2 2 2c− osθ dππ

∫= θ∫

14.1982 BC6

Point (P x, y) moves in the xy-plane in such a way that 11

dx=

dt t +and dy

dt= 2t for t ≥ 0.

(a) Find the coordinates of P in terms of t if, when t =1, x = ln 2 and y = 0 .

(b) Write an equation expressing y in terms of x.

(c) Find the average rate of change of y with respect to x as t varies from 0 to 4 .

(d) Find the instantaneous rate of change of y with respect to x when t =1.

1982 BC6 Solution

(a) 11

1dt C

tx = = ln(t 1)+ +

+⌠⎮⌡

1)Cx C

ln(= +txln 2 (1)= = ln 2+ , so 1 1 0=

22

2 22

2

0 (1) , s o 1

1

y t dt = t +C

Cy C

y

=

= =1+ = −

t= −

(b) xe t= +1, so y = (e −1) 1− =2 2x xe − 2ex

(c) (0) 15 ( 1)−= =

− 16(4) (0x ) ln 5− − ln1 ln 5(4) −y y

x

(d) At t = 1,

1

dy 21t

dxt

==

+

4

or

From (b), using x = ln 2 when t = 1

e2 2ln 2 2 8 4dy ln2edx

−= = − 4 =

15.1984 BC2

The path of a particle is given for time t > 0 by the parametric equations xt

= +t 2 and

= 3y t2 .

(a) Find the coordinates of each point on the path where the velocity of the particle in the x direction is zero.

(b) Find dy when tdx

=1.

(c) Find2

2 d y when ydx

=12 .

1984 BC2 Solution

(a) 22dx

dt t1= −

dxdt

= 0 when t = 2 . The corresponding point on the path is (2 2,6) .

(b)

2

621

dydy tdx dx

dt t

dt= =−

tAt =1, dy 6dx

= −

(c) 2 2 3

2 3

2

6

21

⎛d dy t⎠ ⎝ ⎠td y dt dx⎜ ⎟

⎝ ⎠= =⎜⎝ t

dxdxdt t

⎞ 1−⎛ ⎞2 4⎛ ⎞⎟ − 6 ⎜ ⎟

−⎛ ⎞⎟⎜⎠⎝

When y =12 , t = 2 and so 2

32

1 662 2412

d ydx

⎛ ⎞⎜ ⎟−

= ⎝ ⎠ = −⎛ ⎞⎜ ⎟⎝ ⎠

.

16.

1989 BC4 Solution

(a)

(t + )( )

2

2

2 2

2 2

3 12

6 6

223 12 46 6 2 2 2 1

dydt

= −tdx tdt

tdy t tdx t t t− − t t t

t= −

−− −== =( )−

y (b) x

( +x )

5, 1134

311 54

or3 294 4

34 29

dydx

y

= − +xy

+y x

= − =

= −

− = −

=

(c)

28,16)( )

( )

type

2 horizontal

0 0,0 vertical

1 ( 1, 1 )1 vertica

2 4, 16 horizontal

t ( )x, y

(−−

− −

l

1989 BC4

Consider the curve given by the parametric equations x 2= −3t t3 2 and y = t3 −12t

(a) In terms of t , find dydx

.

(b) Write an equation for the line tangent to the curve at the point where t = −1.

(c) Find the x- and y-coordinates for each critical point on the curve and identify each point as having a vertical or horizontal tangent.

17. 1993 BC 2 Solution

(a) ( ) ( )( ) ( )( )

2

2 2

22

5 10 5 50

5 10 50 2600

10 26 50.990

=′ tx t y t

x y

v

t′ =

=′ ′ =

= + =

= ≈

(b)

( t 2 )

2 4

0

5 2

0

53/ 2

0

12

32 263/ 2( ) 87.7163

+t t dt

t t dt+=

2 1= +

= 1− ≈

54 4∫

t t=t( )

( )2

2 2

22

3; 3

3

3

dx x′ t t

x t

= +xtdydx

(c) dy y′

==

t= − = x +

x= +

or

(x + )

2

3/ 23

3; 32 3y t2 y;

3 3

3

=x t t = x

dydx

− +

==

x= +

18. D2dx ⎞24 4 2

0 0L dy ⎞ dt 4 1+t dt= ⎛ ⎛+ =

dt⎜ ⎟ ⎜⎝ ⎠ ⎝ dt ⎟⎠∫ ∫

19. D cos , t y sin3 3= for t 02

x π= ≤ t .≤

2 22

0L dt

π dx⎛ ⎞ ⎛ dy+ ⎞=

dt⎜ ⎟ ⎜ dt ⎟⎝ ⎠ ⎝ ⎠∫

22 2 2 2 2 4 2 4 20 0

L (−3cos sin )t (3s+ it n cos )t t dt 9cos sint 9si+ nt cost t dtππ

∫= = ∫

20 C The length of this parametric curve is given by 2dx ⎞21 1

0 0dy ⎞ dt t( ) t dt⎛ ⎛+ = 2 2 2+

dt⎜ ⎟ ⎜⎝ ⎠ ⎝ dt ⎟⎠∫ ∫ .

1972 BC7 Solution

1 , 6 22

dx dy tdt dt

= = −

= −4(y x2 −5 +x 4) for x≤ ≤1 4 , so dy= 4(2x− −5)

x = −5 1 9

22

dx

− y for 0 y≤ ≤ 9 (left half of curve), so 194

dxdy y

=−

Each of the integrals can be written in terms of dt, dx, or dy.

(a) Length2dx ⎞2b

a

dy ⎞ dt = ⎛ ⎛+dt⎜ ⎟ ⎜

⎝ ⎠ ⎝ dt ⎟⎠

16 9( − y)

62

04 2

1

9

0

4

1 1+ 6(2x 5− )

12 1

=1 (6+ 2−

=

= +

⌠⎮⎮⌡

⌠⎮⌡

⌠⎮⌡

)t dt

dx

dy

(b) Volume b

y2a

= π∫ dx

6 20

4 2 21

9

0

( (6 −t t)) ⋅12

16( 4)

= π2 9

= π

= π +

dt

x − 5x dx

y y dy

(c) Surface area = 2b

aπ∫ y ds

6 20

4 2 21

9

0

= π2 ) 1+ (6− 2

4

2 4(x −5 + 4) 1+16(2x x −5)

1= π4 1

16(9− y)

(6−t t t) dt

dx

y dy

= π −

+⌠⎮⌡

21.

22.

1974 BC5 Solution

(a) = −2 2cosθ, 2sin

2sin sin−2 2cosθ −1 cos

dx dyd d

dydy ddx dx

d

= θθθ

θ θθ == =θ

θ

(b) At θ = π, sin 0−1 cos

dydx

π= =

π. This means the tangent line is horizontal.

At ,θ = π y = 4, so the equation of the tangent line is y = 4.

(c) At θ = 2π, sin 2−1 cos 2

dydx

π=

π which is not defined.

2 2lim sin lim cosθ→ π 1 co− θs θ→ π sin

θθ= →∞

θ

This means that there is a vertical tangent. At θ 2= π , x = 4π , so the equation of the tangent line is x 4= π

(d) Length2 2220 0

(2− 2cos )θ (2s+ in )θ θ =d 2 2 2c− osθ dππ

∫= θ∫

1974 BC5

Given the parametric equations = 2(θ − sinθ) and x y = 2(1− cosθ) :

(a) Find dy in terms of θ . dx

(b) Find an equation of the line tangent to the graph at θ = π .

(c) Find an equation of the line tangent to the graph at θ 2= π .

(d) Set up but do not evaluate an integral representing the length of the curve over the interval ≤ θ ≤0 2π . Express the integrand as a function of θ .

23. E f ( ) = −t e ,− sin( );t′ ′ (f )t = (− −t te ,′ − cos ) .t

24.

25.

D2 2

2 −4 3= t 2t ⇒dy

= 4t −3 2 22 2x( )t t= −1⇒ dx

= 2t and d x= 2; (y )t 6t and d y

=12t −12tdtdt dt dt

a t( )2 2

222

d x , d y (2, 12t 12t)− a(1) (2,0)dt dt

⎞⎛⎟⎟ =⎜⎜= ⇒ =

⎝ ⎠

E2

2 2 2;2 41, 2; ln(2t + 3) ,dt t2 3 (2t + 3)

2dx= 2 ,t d x dy d y

=x t ydt dt dt

2 + = = = = −+

26.

27.

A 2 8 4v 2 ,4 , (t ⎞ 2)v

+ 2t t⎛ t2 +⎜=

6 ,8⎛ ⎞ ⎛ ,3 8⎞=⎜ ⎟= ⎜ ⎟⎟⎝ ⎠ ⎝ ⎠⎝ ⎠

28.

E (v t) 3t= −1,6(2t 1)−2 2( ) and a(t) = 6t, 24(2t 1)−( )⇒ a(1) (= 6, 24)

1975 BC3 Solution

(a) 2 2

2 2 2 2(1 t )(+ −2t) (1− t− )(2t) 4

(1+ t )dx tdt (1 )+ t

−= =

22

2 2 2 2(1+ t )dy tdt (1 )+ t

=(1 t )(2)+ − 2t(2t) 2 2−

=

The velocity vector is 2

2 2 2 2,t4 2− 2

⎝ +⎜ (1t−⎛ ⎞

⎜ ⎟⎟) (1t t )+ ⎠

dt(b) dx 0 at t= = 0 and dy 0 a= t t

dt=1. Therefore there is no t such that dx

dt dtdy

= = 0 at

the same time. Hence the particle is never at rest.

(c) 2 2

2

2

1 1lim (x t) lim=

1 lim 11 1t

t tt

t→∞t t→∞ →∞

−−

= = −+ +

1

22lim (y t) lim=

21

lim 1t

tt t

t→∞t t→∞ →∞

= =+ +

0

Hence the particle approaches the point (−1,0) as t increases without bound.

29.

1975 BC3

A particle moves on the circle +x y2 2 =1 so that at time t ≥ 0 the position is given by

the vector2

2 21 2,1 1

t tt

⎞−⎛⎟⎜

⎝ +⎜ + t ⎟⎠

.

(a) Find the velocity vector.

(c) Give the coordinates of the point that the particle approaches as t increases without bound.

30.

1987 BC5 Solution

(a) sin( cos(2 )

cos(t),

v cos(= + ( 2− sin(2

x ),t y tdx 2sin(2 )tdt dtv (cos( −t), 2sin(2 ))t

)t i t)) j

= =dy

= = −

=or

(b) 0= ⇒ cos(v t) = 0 and − 2sin(2 ) = 0t Therefore cos(t) = 0 and 4sin( ) cos(t t)− = 0 . The only choice is cos(t) = 0 .

Therefore the particle is at rest when ,2 2

t π 3π= .

(c) 22 2 2

2

sin (t),

21

x y

y x

==

= −

cos ( )t 1 2= sin− ( )t

(d)

31.

1992 BC3 Solution

(a)

( t )( t )

( )+( )

/ 2

/ 2

sin

cos −sin// s +in cos

0 1at ,= 1

2 1 0

t

t

sine t +t te cos tdtdy cose t= −t te tdt

tdy dy dt= =e

dx dx dt e t

tdx e

π

ππ −dy e

= =

dx =

) (b) se in +tt t cose t )( +(

e( sin1+ coe s1) +( cos1 si−e n1)

22

2 2

when =1 speed is

2

coste t e sin− t t

t

e e=

spee d =

(c) dist

sine t( +t te cos t ) +(1

cost t s− ine tt )22

0

1 12 2 2

00

1

0

ance is

2

22 1

e dtt

t

e d

t dt

e

e tt2 sin cos( )+= =

= = ( )e −

⌠⌡

∫ ∫

t

1992 BC3

At time t , 0 ≤ t ≤ 2π , the position of a particle moving along a path in the xy-plane is given by the parametric equations x = et sin t and y = et cos t .

2t = π .(a) Find the slope of the path of the particle at time

(b) Find the speed of the particle when t =1.

(c) Find the distance traveled by the particle along the path from t = 0 to t =1 .

32. 1993 BC 2 Solution

(a) ( ) ( )( ) ( )( )

2

2 2

22

5 10 5 50

5 10 50 2600

10 26 50.990

=′ tx t y t

x y

v

t′ =

=′ ′ =

= + =

= ≈

(b)

( t 2 )

2 4

0

5 2

0

53/ 2

0

12

32 263/ 2( ) 87.7163

+t t dt

t t dt+=

2 1= +

= 1− ≈

54 4∫

t t=t( )

( )2

2 2

22

3; 3

3

3

dx x′ t t

x t

= +xtdydx

(c) dy y′

==

t= − = x +

x= +

or

(x + )

2

3/ 23

3; 32 3y t2 y;

3 3

3

=x t t = x

dydx

− +

==

x= +

33.

1994 BC 3

(a) x ( ) ( )( ) ( )( ) ( )( )

2

2

2

22

0 0 0; 2

x, 0

cos

′x t = t +C

x C =′x t t

x t k k

x t

y t

π

′′ t = ⇒

= ⇒′ =

t= + =π =

t= +

=( ) t π( )+

(b) dy = − t +( )

( )

( ) π( )( )( + )

2

2 2

22 2

22 2 2

t2 sin

2 2t sin

4 4 sin

dt

dx dys tdt dt

t t

t

π

π

= +

= + − +

= t t+

( ) ( )

2 2

2

when x 4, 4; 4

4 4 siπ n 4

2.324

t t

s

+ =π = −π=

= −π + 4 4 −

1994 BC 3

A particle moves along the graph of y = cos x so that the x-coordinate of acceleration is always 2. At time t = 0 , the particle is at the point π,( −1) and the velocityvector of the particle is ( ).0,0

(a) Find the x- and y-coordinates of the position of the particle in terms of t.

(b) Find the speed of the particle when its position is 4,cos4( ).

34.1995 BC1

Solution

(a) 1,2 4t ; 1, 23

33 33, ; 3 ,22 22

A A

B B

V V

VV

(b)3 22

01 2 4t dtdistance

(c) Set 2 42

3tt t4;

When t 4 , the y-coordinates for A and B are also equal. Particles collide at

(2,4) when t 4 .

(d)

7

5

7

5

Viewing Window[ 7,7] [ 5,5]

35.

1997 BC1

Solution

(a) 2.5 3cos 2.5 0

2.5 5sin 2.5 5

x

y

(b)

y

x

(c) 3

(d) t

v t

)x t( 3 sin ( )t 5 cos

3 sin ,5 cos

t y

t t

(e) distance 1.75

2 2 2 2

1.259 sin 25 cos

5.392

t dtt

36. BC{1 1999

1. A particlet

moves in the xy{plane so that its position at any time t, 0 � t � �, is given by

x(t) =2

2� ln(1 + t) and y(t) = 3 sin t.

(a) Sketch the path of the particle in the xy{plane below. Indicate the direction of motion along thepath.

(b) At what time t, 0 � t � �, does x(t) attain its minimum value? What is the position (x(t); y(t)) ofthe particle at this time?

(c) At what time t, 0 < t < �, is the particle on the y{axis? Find the speed and the acceleration vectorof the particle at this time.

(a) y

O 1

1

x

2

(1: graph

1: direction

(b) x0(t) = t� 1

1 + t= 0

t2 + t� 1 = 0

t =�1 +

p5

2or t = 0:618 in [0; �]

x(0:618) = �0:290 y(0:618) = 1:738

3

8>><>>:

1: x0(t) = 0

1: solution for t

1: position

(c) x(t) =t2

2� ln(1 + t) = 0

t = 1:285 or 1:286

x0(t) = t� 1y0(t) = 3 cos t

speed =p 1 + t

(x0(1:286))2 + (y0(1:286))2 = 1:196

1

(1 + t)2x

00(t) = 1 + y00(t) = �3 sin t

acceleration vector = <x00(1:286); y00(1:286)>

= <1:191;�2:879>

4

>>><>>>>

8>>

>:

1: x(t) = 0

1: solution for t

1: speed

1: acceleration vector

37.

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38. AP® CALCULUS BC 2001 SCORING GUIDELINES

Question 1

An object moving along a curve in the xy-plane has position x t( ),y(t)� � at time t with

dtdx

� cos t3� � and 3 sin t2� �dydt

for t� �0 3 . At time t = 2, the object is at position (4,5).

(a) Write an equation for the line tangent to the curve at (4,5).

(b) Find the speed of the object at time t = 2.

(c) Find the total distance traveled by the object over the time interval t� �0 1 .

(d) Find the position of the object at time t = 3.

(a) � �� �

2

3

3 sincos

tdydx t

� �� �

2

cos 232

3 sin 215.604

t

dydx

� �

5 15.604( � 4)xy ��

1 : tangent line

(b) Speed = cos (8) � 9 sin2 2(4) = 2.275

(c) Distance = � �1 32 2 2

0cos � 9 sint t� � dt �

1= .458

3 :

2 : distance integral

< 1 each integrand error

< 1 error in limits

� �

� �

�����������������

(d) ��3 3

2(3) 4� � � cosx t = 3.9dt 53 or 3.954

��3 2

2(3) 5� � � 3 siny t = 4.906dt

4 :

1 : definite integral for

1 : definite integral for

x

y

�����������������

39. AP® CALCULUS BC 2002 SCORING GUIDELINES

Question 3 �

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40. AP® CALCULUS BC 2002 SCORING GUIDELINES (Form B)

Question 1

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41.AP® CALCULUS BC

2003 SCORING GUIDELINES

Question 2

A particle starts at point A on the positive x-axis at time t = 0 and travels

along the curve from A to B to C to D, as shown above. The coordinates of

the particle’s position x t( ), y t( )( ) are differentiable functions of t, where

x t(1

sin26

= 9costdx t

dt

+) = and y t( )

dy

dt= is not explicitly given.

At time t = 9, the particle reaches its final position at point D on the positive x-axis.

dy

dt

dx

dt(a) At point C, is positive? At point C, is positive? Give a reason for each answer.

(b) The slope of the curve is undefined at point B. At what time t is the particle at point B ?

(c) The line tangent to the curve at the point ( )8 ,x y 8( )( ) has equation9

5=y x 2. Find the

velocity vector and the speed of the particle at this point.

(d) How far apart are points A and D, the initial and final positions, respectively, of the particle?

(a) At point C, dy

is not positive because y t( ) is dt

decreasing along the arc BD as t increases.

At point C, dx

is not positive because x t( ) is dt

decreasing along the arc BD as t increases.

2 : 1 : not positive with reason

1 : not positive with reason

dy

dt

dx

dt

(b) 0dx

dt ( )= ; cos 06

t= or

1 sin 0

2

t +=

6 2

t= or

1

2

t += ; t = 3 for both.

Particle is at point B at t = 3.

2 : 1 : sets 0

1 : 3

dx

dt

t

=

=

(c) ( )sin ( )34 9(8) 9 cos

23 2x = =

(8) 5

9(8)

y dy

dxx= =

5 5(8)

9 2y x (8)= =

The velocity vector is < 4.5, 2.5 >.

Speed 4.5= + 2.52 2 5= .147 or 5.148

3 :

1 : (8)

1 : y (8)

1 : speed

x

(d)

9

0

39.255

(9)x x(0) = (x )t dt

=

The initial and final positions are 39.255 apart.

2 : 1 : integral

42. AP® CALCULUS BC 2003 SCORING GUIDELINES (Form B)

Question 4

A particle moves in the xy-plane so that the position of the particle at any time t is given by

(t ) 3 7t tx = 2e e+ and (y t ) 3= e3 2t .t e

(a) Find the velocity vector for the particle in terms of t, and find the speed of the particle at time t = 0.

dy

dx t

dy

dx(b) Find in terms of t, and find lim .

(c) Find each value t at which the line tangent to the path of the particle is horizontal, or explain why

none exists.

(d) Find each value t at which the line tangent to the path of the particle is vertical, or explain why

none exists.

(a) ( ) = 6 73 7t tx t e e

y t( ) 9= 23 2t te e+

Velocity vector is < 6 7e e3 7t t , 9 + 2e3t e 2t >

Speed = (0)x y (0)2 2+ = ( 1) +2 112

= 122

3 :

1 : (x t)

1 : (y t)

1 : speed

(b) 3 2

73

9 2

76

tt

t t

dydy e edt

dxdx e edt

+= =

23

3 7

29 9 3lim lim

2676

tt

ttt t

dy e e

dx ee

+= = =

2 : 1 : in terms of

1 : limit

dyt

dx

(c) Need y t( ) = 0, but 9 0e e23 2t t+ > for all t, so

none exists. 2 :

1 : considers (y t) = 0

1 : explains why none exists

(d) Need x t( ) = 0 and y t( ) 0.

=6 7 7tt3e e

10 7

6te =

( )71ln

10 6t =

2 : 1 : considers (x t) = 0

1 : solution

43. AP® CALCULUS BC 2004 SCORING GUIDELINES

Question 3

An object moving along a curve in the xy-plane has position x( ) ,t y( )t( ) at time t ≥ 0 with

= +3dtdx cos t2( ). The derivative dt

dy is not explicitly given. At time t = 2, the object is at position

1, 8( ). (a) Find the x-coordinate of the position of the object at time t = 4.

(b) At time t = 2, the value of dtdy is −7. Write an equation for the line tangent to the curve at the point

( )2 ,(x y 2( ) . )(c) Find the speed of the object at time t = 2.

(d) For t ≥ 3, the line tangent to the curve at x( ) ,t y( )t( ) has a slope of 2t + 1. Find the acceleration vector of the object at time t = 4.

(a) ( )x x( ) ( )( )( )( )

4 22

4 22

24 3 + cos

1 3 + cos 7.132 or 7.133

t dt

t dt

= +

= + =

∫3 :

( )( )4 22

1 : 3 + cos

1 : handles initial condition1 : answer

t dt

(b)

22

7+3 cos 4t

t

dydy dtdx dx

dt=

=

= −= = −2.983

− =8 2− .98y x3 −( ) 1

2 : 2 1 : finds

1 : equation

dydx t=

(c) The speed of the object at time t = 2 is

2( )( )x y′ ′ 2( )( )2 2+ = 7.382 or 7.383.

(d) x′′ 4( ) = 2.303

y dt dx dt 1) 3 + cos t2( )( )′ t( ) dy= = dy dx⋅ = 2t( +

y′′ 4( ) = 24.813 or 24.814 The acceleration vector at t = 4 is

2.303, 24.813 or 2.303, 24.814 .

3 :

1 : ′′ 4( )

1 :

xdydt

44. AP® CALCULUS BC 2004 SCORING GUIDELINES (Form B)

Question 1

A particle moving along a curve in the plane has position x( ) ,t y( )t( ) at time t, where dx = t4dt + 9 and dt

dy = +2 5e e−t t

for all real values of t. At time t = 0, the particle is at the point (4, 1). (a) Find the speed of the particle and its acceleration vector at time t = 0. (b) Find an equation of the line tangent to the path of the particle at time t = 0. (c) Find the total distance traveled by the particle over the time interval 0 t≤ ≤ 3. (d) Find the x-coordinate of the position of the particle at time t = 3.

(a) At time t = 0:

Speed (0)= +x y′ ′(0)2 2 3= 2 7+ 2 58=

Acceleration vector ′′( )0 ,x y′′ 0( )= = 0, 3−

2 : 1 : speed1 : acceleration vector

(b) 0( ) 730

dy =y

dx x′′( ) =

Tangent line is 73y x= −( )4 1+

2 : 1 : slope1 : tangent line

(c) Distance ( t + ()3 2 24 9 2 50

45.226 or 45.227

+ e −t t+ e ) dt =

=

⌠⌡ 3 :

2 : distance integral1 each integrand error1 error in limits

−−

(d) 3( )3 40

9

17.930 or 17.931

x = +4 +t dt

=∫ 2 :

45. AP® CALCULUS BC 2005 SCORING GUIDELINES (Form B)

Question 1

An object moving along a curve in the xy-plane has position x( ) ,t y( )t( ) at time t ≥ 0 with

dxdt 12= − 3t t2 and dt

dy ln 1= + t 4( )− 4( ).

At time t = 0, the object is at position −13, 5( ). At time t = 2, the object is at point P with x-coordinate 3.

(a) Find the acceleration vector at time t = 2 and the speed at time t = 2.

(b) Find the y-coordinate of P.

(c) Write an equation for the line tangent to the curve at P.

(d) For what value of t, if any, is the object at rest? Explain your reasoning.

(a) 1.882′′( )2 0x y, ′′( )2= = 1732− = −

a( )2 0= −, 1.88 2

Speed 122= + ln 17( )( )2 12= .329 or 12.330

2 : 1 : acceleration vector

1 : speed⎧⎨⎩

(b) y t( ) = y(0) ∫+ (+ u −( ))40

ln 1 4t

du

2( ) 1 + ( )2 40

= +5 ∫ lny u − 4( ) = 13.d 6u 713 :

4)+ −u( )1 : ln 1(2

0⎧ 4

⎪⎩ 1 : handles initial condition1 : answer

du⎪⎪⎨⎪

(c) At t = 2, slope ln 17( )12

dydtdx= = = 0.236

dt13.671− = 0.236y x 3( )−

2 : 1 : slope1 : equation

⎧⎨⎩

(d) ′x t( ) = 0 if t = 0, 4′y t( ) = 0 if t = 4

t = 4

2 : 1 : reason1 : answer

⎧⎨⎩

46. AP® CALCULUS BC 2006 SCORING GUIDELINES

Question 3

An object moving along a curve in the xy-plane is at position x( ) ,t y( )t( ) at time t, where

dtdx 1− −t( )sin 1= − 2e and 3

41

dy tdt t

=+

tfor 0.≥ At time t = 2, the object is at the point 6, −3( ). (Note: sin−1 x = arcsin x )

(a) Find the acceleration vector and the speed of the object at time t = 2. (b) The curve has a vertical tangent line at one point. At what time t is the object at this point? (c) Let m t( ) denote the slope of the line tangent to the curve at the point x( )t y( )t( ), . Write an expression for

t→∞m t( ) in terms of t and use it to evaluate lim m t( ).

(d) The graph of the curve has a horizontal asymptote =y c. Write, but do not evaluate, an expression involving an improper integral that represents this value c.

(a) a 2( ) 0.395 or 0.396,= −0.741 or 0.740−

Speed 2( )= +x y′ ′(22 2) 1.= 207 or 1.208 2 : { 1 : acceleration

1 : speed

(b) sin 1 1 2e− −t( )− = 0

1 2e−t− = 0

t ln 2= = 0.693 and dtdy ≠ 0 when t = ln 2

⎧⎨⎩

1 : ′x t( ) = 0

(c) m t( )sin 1 − 2e− −t( )3 1

4 11

tt

⋅=+

m t( )e− )(

)

3 1

1

lim 4 1lim1 sin 1 2

10 0sin− (1

ttttt −→→ ∞∞

⎛ ⎞ ⎟⎜= ⋅

+⎜ ⎟−⎝ ⎠⎞⎛

⎜= ⎟ =⎝ ⎠

2 : 1 : m t( )1 : limit value

⎧⎨⎩

t →∞(d) Since lim x t( ) = ∞,

324lim

1tyc ( )t t dt

t→

∞∞

= 3= − ++∫ 3 :

1: integrand 1: limits 1: initial value consistent

with lower limit

⎧⎪⎪⎨⎪⎪⎩

47.AP® CALCULUS BC

2006 SCORING GUIDELINES (Form B)

Question 2

An object moving along a curve in the xy-plane is at position x( ) ,t y( )t( ) at time t, where

dtdx tan e−t( )= and sec e−t( )dt

dy =

tfor 0.≥ At time t = 1, the object is at position 2, −3( ).

(a) Write an equation for the line tangent to the curve at position 2, −3( ).

(b) Find the acceleration vector and the speed of the object at time t = 1. (c) Find the total distance traveled by the object over the time interval t≤ ≤1 2 .(d) Is there a time t ≥ 0 at which the object is on the y-axis? Explain why or why not.

(a) e( −t )

( ) sine e− −t t( )sec 1tan

dydy dtdx dx

dt

== =

2,( ) ( 1 )3

1sin

dydx e−

−= = 2.780 or 2.781

e−( ) −( )113 2

siny x+ =

2 : 2,( −3) 1 :

1 : equation of tangent line

dydx

⎧⎪⎨⎪⎩

(b) x′′ 1( ) = −0.42253, y′′ 1( ) = −0.15196

a( ) = −1 0.423 −, 0.152 or 0.422,− −0.151 .

speed sec( )( )= + tane e− −1 1( )( )2 21.138= or 1.139

2 : { 1 : acceleration vector 1 : speed

(c) 2 22

1∫ x t( )( )′ ′y ( )t( ) dt+ = 1.0592 : { 1 : integral

(d) (0) =x x 1( )1

0∫− ′( )x t dt 2= 0− .775553 0>

The particle starts to the right of the y-axis. Since ′x t( ) > 0 for all t ≥ 0, the object is always moving

to the right and thus is never on the y-axis.

3 : 0( )t( )

1 : x expression0

1 : conclusion and reason 1 : x′ >⎧⎪⎨⎪⎩

48. AP® CALCULUS BC 2007 SCORING GUIDELINES (Form B)

Question 2

An object moving along a curve in the xy-plane is at position (x( ) ,t y( )t ) at time t with

( )arctan 1dx tdt = + t and dy = ln t2 + 1( )dt

for t ≥ 0. At time t = 0, the object is at position (−3, −4). (Note: tan−1 x = arctan x )

(a) Find the speed of the object at time t = 4. (b) Find the total distance traveled by the object over the time interval ≤ t ≤0 4 .(c) Find x( )4 .

there is a point on the curve where the line tangent to the curve has slope 2. At what time t is the object at this point? Find the acceleration vector at this point.

(d) For t > 0,

(a) Speed 4( )= +x y′ ′ 4( )2 2 2.912= 1 : speed at t = 4

(b) Distance4 2 2

0

⎛ = dx ⎞ dy⎛ ⎞⎜ ⎟+ dt 6.423=dt⎜ ⎟

⎝ ⎠ dt⎝ ⎠⌠⎮⌡

2 : { 1 : integral 1 : answer

4

04(c) ( ) =x x( )0

3 2.10794 0.892

( )t+ ′x dt

= − + = −

∫3 : 3 1 : uses x 0( )

⎧ ⎧ 1 : integrand 2 :

⎪ ⎨ = −⎨ ⎩⎪⎩

(d) The slope is 2, so 2,

dydtdxdt

= or ln t2 1( )+ = 2arctan ( t ).1 t+

Since t > 0, t = 1.35766. At this time, the acceleration is (x t ) , y′′(t ) t=1.35766′′ = 0.135, 0.955 .

3 : 1 : 2

1 : t-value1 : values for x′′ and

dydtdxdt

y

=

′′

⎧⎪⎪⎪⎨⎪⎪⎪⎩

49.AP® CALCULUS BC

2008 SCORING GUIDELINES (Form B)

Question 1

A particle moving along a curve in the xy-plane has position x( ) ,t y( )t( ) at time t ≥ 0 with

dtdx = 3t and

23cos .dy

dtt⎛ ⎞= 2⎜ ⎟

⎝ ⎠The particle is at position 1, 5( ) at time t = 4.

(a) Find the acceleration vector at time t = 4. (b) Find the y-coordinate of the position of the particle at time t = 0. (c) On the interval t≤ ≤0 4, at what time does the speed of the particle first reach 3.5 ?

(d) Find the total distance traveled by the particle over the time interval t≤ ≤0 4 .

(a) 4( )a x′′ 4( ) , ′′ 4y ( )= = 0.433, 11.872−

(b) y 0( )0 2

43cos 1.600t

2⎛ ⎞⎜ ⎟ =dt5= +⎝ ⎠

⌠⎮⌡

or 1.601 3 : 1 : integrand1 : uses y 4( ) = 5

⎧⎪⎨⎪⎩

(c) Speed x t( )( )= +′ ′y ( )t( )2 2

2t +3 9co 2s 3.52

t⎛ ⎞= ⎜ ⎟ =⎝ ⎠

The particle first reaches this speed when t = 2.225 or 2.226.

3 : 1 : expression for speed

1 : equation 1 : answer

⎧⎪⎨⎪⎩

(d) 4 2

2

03 9cos 13.1822t t⎛ ⎞+ ⎜ ⎟ dt =

⎝ ⎠

⌠⎮⌡

2 : { 1 : integral1 : answer