AP® Exam Practice Questions for Chapter 10 · AP® Exam Practice Questions for Chapter 10 1 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, c opied or duplicated,

AP® Exam Practice Questions for Chapter 10 1

© 2018 Cengage Learning. All Rights Reserved. May not be scanned, copied or duplicated, or posted to a publicly accessible website, in whole or in part.

AP® Exam Practice Questions for Chapter 10

1. ( ) ( )( ) ( )

2 3 2

2

52

20 2 12

20 5 6 24

y t t t x t t ty t t x t t t

= − = −′ ′= − = −

( )( )

2

2

20 5

6 2420 5

06 24

0 20 5

5 20

4

y tdydx x t

tt t

tt t

ttt

′=

′

−=−−=−

= −==

( ) ( ) ( )3 24 2 4 12 4 64x = − = −

( ) ( ) ( )254 20 4 4 40

2y = − =

The particle is at rest when 4t = at the point ( )64, 40 .−

So, the answer is B.

2. ( )2

2 4

0 0

12

2 sin sinA d dπ π

θ θ θ θ= ⋅ =

So, the answer is C.

3. cos sin

cos sin

sin cos cos sin

x r y r

dx dyd d

θ θθ θ θ θ

θ θ θ θ θ θθ θ

= == =

= − + = +

cos sin

sin cos

dy dy ddx dx d

θ θ θ θθ θ θ θ

+= =− +

At ,2

πθ = −

cos sin

1 22 2 2

sin cos22 2 2

dydx

π π π

ππ π π π

− − + − − = = = −− + −


4. 2 3 2

2

2 3 4

4 3 3 8

x t t y t tdx dyt t tdt dt

= + = +

= + = +

At 2,t =

( ) ( )( )

22 3 2 8 23 8 28.

4 3 4 2 3 11

dy dx dt t tdx dy dt t

++= = = =+ +

At 2,t = ( ) ( ) ( )22 2 2 3 2 14x = + = and

( ) ( ) ( )3 22 2 4 2 24.y = + =

An equation of the tangent line is

( )2824 14 .

11y x− = −


5.

( )( ) cos

3 5 sin cos

3 cos 5 sin cos

x r θθ θ

θ θ θ

=

= +

= +

( ) ( )( )2 2

3 sin 5 sin sin 5 cos cos

3 sin 5 sin 5 cos

dxd

θ θ θ θ θθ

θ θ θ

= − + − +

= − − +

( )( )2

sin

3 5 sin sin

3 sin 5 sin

3 cos 10 sin cos

y r

dyd

θθ θ

θ θ

θ θ θθ

=

= +

= +

= +

At 0,θ =

( )( )

2 2

3 cos 10 sin cos

3 sin 5 sin 5 cos

3 1 0 3.

0 0 5 1 5

dydy d

dxdxd

θ θ θθθ θ θ

θ

+= =− − +

+= =

− +


2 AP® Exam Practice Questions for Chapter 10


6.

( ) ( )

( )

( )

( )

( )

4 2 2

0

4 2

0

4 2

0

4

0

4 5 5 cos 2

4 25 25 cos 2

1100 1 cos 2 2

2

1100 2 sin 2 cos 2

4

1 1100 0 0 0

4 4 2 4

25

239.270

A d

d

d

π

π

π

π

θ θ

θ θ

θ θ

θ θ θ θ

π π

π

= −

= −

= −

= − +

= − + − −

=

≈


7. At 4,t =

( )( )

( )

4

4 4

2 3

cos

2 3

4 cos 41.147.

t

dy dy dtdx dx dt

y tx t

et te

−

−

=

′=

′

+=

+=

≈ −


8. (a) At 3,t = the speed is

( ) ( )( ) ( )

2

2

2 2 2 21 2

2 23 1 2

2 3 25

2 3 25 3 12.165.

tdx dy e tdt dt

e

− +

− +

+ = − + −

= − + − ≈

(b) ( ) ( )22 2 2 25 5 1 2

0 0 2 3 25

59.725

tdx dy dt e t dtdt dt

− + + = − + −

≈

(c) ( ) ( ) 3

0

3 2

0

3 0

3 3 25

45.131

dyy y dtdt

t dt

= +

= + −

≈

Note: Be sure to write down the appropriate definite integral before numerically approximating it on your calculator.


Notes: Round each answer to at least three decimal places to receive credit on the exam.

Use rather than an equal sign in presenting these approximations from your calculator. Because these are approximations, a point may be deducted if an equal sign is used.



9. (a)

sin

2 5 cos

dy dy dtdx dx dt

tt t

=

−=−

When 4,t = ( )sin 4 sin 4

.2 4 5 cos 4 8 5 cos 4

dydx

−= = −− −

So, an equation of the tangent line is

( )

( )

sin 43 1

8 5 cos 4

sin 43 1 .

8 5 cos 4

y x

y x

− = − − − −

= − +−

(b) 2 5 cos

0 2 5 cos

1.11051

dx t tdt

t tt

= −

= −≈

Because ( )1.11051 0,y′ < the direction is down.

(c) ( ) ( )

( )

( )

0

4

0

4

4

0

0 4

3 sin

3 sin

4.6536

dyy y dtdt

t dt

t dt

= +

= + −

= − −

≈

(d) ( ) ( )2 2

4 4 2 2

0 0sin 2 5 cos

26.657

dx dy dt t t t dtdt dt

+ = − + −

≈



Notes: Be sure to round each answer to at least three decimal places to receive credit on the exam.

Use rather than an equal sign in presenting these approximations from your calculator. Because these are approximations, a point may be deducted if an equal sign is used.



10. (a)

( )

2

2

2

2

12

12

2 cos

13.338

A r d

d

π

π

π

π

θ

θ θ θ

=

= +

≈

(b)

( ) sin

2 cos sin

2 sin sin cos 1

2.93575 radians

y r θθ θ θ

θ θ θ θθ

=

= +

= + =≈

( )

( ) ( ) ( ) ( )

2

2

cos

2 cos cos

2 cos cos

2.93575 2 2.93575 cos 2.93575 cos 2.93575

4.790

x r

x

θθ θ θ

θ θ θ

=

= +

= +

= +

≈ −

(c)

( )

( )

( )

2

cos

2 cos cos

2 cos cos

2 sin 2 cos

2 cos sin

x r

dx d ddt dt dt

ddt

θθ θ θ

θ θ θθ θθ θ θ

θθ θ

=

= +

= +

= − + ⋅

+ ⋅ −

When 3

and 3,4

ddt

π θθ = =

( ) ( )

( )

3 3 32 sin 3 2 3 cos

4 4 4

3 3 2 cos sin 3

4 4

11.239.

dxdt

π π π

π π

= − +

+ −

≈ −

So, the particle is moving to the left on the rectangular coordinate system.


Notes: Be sure to write down the appropriate equation before numerically approximating its solution on your calculator.

In the intermediate step, round the value of to more than three decimal places to use to determine the value of


Use rather than an equal sign in presenting the approximations from your calculator. Because these are approximations, a point may be deducted if an equal sign is used.



11. (a) ( )2

3

13 2 2 2 cos

214.197

A dπ

ππ θ θ= + ⋅ +

≈

(b)

1

dr drdt d

dr d dr ddt dr d dr

ddt

θθ θ

θθ

=

⋅ = ⋅

=

( )

2 2 cos

2 sin 2 sin 1 2 sin

rdr ddt dt

θθθ θ θ

= +

= − = − = −

At ,3

πθ =3

2 sin 2 3 1.732.3 2

drdt

π = − = − = − ≈ −

So, the particle is moving closer to the origin of the polar coordinate system.

(c)

1

dx dxdt d

dx d dx ddt dx d dx

ddt

θθ θ

θθ

=

⋅ = ⋅

=

( )

( )( )

cos

sin cos

2 2 cos sin cos

x rdx d drrdt dt dt

d drdt dt

θθθ θ

θθ θ θ

=

= − +

= + − +

When , 1,3

ddt

π θθ = = and 3,drdt

= −

( ) ( )

( ) ( )

2 2 cos sin 1 cos 33 3 3

3 13 3 2 3 3.464

2 2

dxdt

π π π = + − + −

= − + − = − ≈ −

So, the particle is moving to the left on the rectangular coordinate system.



Use rather than an equal sign in presenting the approximations from your calculator. Because these are approximations, a point may be deducted if an equal sign is used.



12. (a) ( )6 2

0

11 2 sin

2A d

πθ θ= −

(b) ( ) cos 1 2 sin cos cos 2 sin cos x r θ θ θ θ θ θ= = − = −

( ) ( )

( )

( )

2 2

2 2

sin 2 sin sin cos cos

sin 2 sin cos

2 sin 2 cos sin

dx d ddt dt dt

d ddt dt

ddt

θ θθ θ θ θ θ

θ θθ θ θ

θθ θ θ

= − − − +

= − − − +

= − −

( )( ) 2 sin 1 2 sin sin sin 2 siny r θ θ θ θ θ= = − = −

( )cos 4 sin cos dy ddt dt

θθ θ θ= −

(c) 2 2

cos 4 sin cos

2 sin 2 cos sin

dy dy ddx dx d

θ θ θ θθ θ θ θ

−= =− −

When ,θ π=

( )

2 2

2

cos 4 sin cos

2 sin 2 cos sin

1 0

0 2 1 0

1

2

dydx

π π ππ π π

−=− −

− −=− − −

=

( )( )

cos 2 sin cos

cos 2 sin cos 1

x

x

θ θ θ θ

π π π π

= −

= − = −

( )( )

2

2

sin 2 sin

sin 2 sin 0

y

y

θ θ θ

π π π

= −

= = =

So, an equation of the tangent line is

( )10 1

21 1

.2 2

y x

y x

− = +

= +

Note: You do not need to simplify these derivatives.



13. (a) dxdt

is positive at point B because the particle is moving to the right.

dydt

is positive at point B because the particle is moving upward.

(b) Because there is a cusp at point C, dydt

is undefined.

dydt

is undefined when

( )1 33

3 1 2

1 6

27 0

27

27 .

t

t

t

− =

=

=

So, the particle reaches point C when 1 627 3.t = =

(c) ( ) ( ) ( )

( )

( )

( )

( ) ( )

3

0

23 3 21 30 3

23

1 30 3

32 3

3

0

2 33 2 3

3 0

0 2 27, 327

1 32

3 27

2 327

3 2

3 27 0 27

3

y y y t dt

t dt u t du t dtt

t dtt

t

′= +

= − = − =−

= − ⋅−

= − −

= − − − −

=

(d) 2 20

3 32

3

y x

dy dxdt dt

= − +

= −

Because ( )21.837, 1.837 .

3

dx dydt dt

≈ = −

So, the speed is ( ) ( )2 2 2

2 21.837 1.837 .

3

dx dydt dt

+ = + −

Note: These approximations do not need to be simplified.

AP® Exam Practice Questions for Chapter 10 · AP® Exam Practice Questions for Chapter 10 1 © 2018 Cengage Learning. All Rights Reserved. May not be scanned, c opied or duplicated,

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