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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
− − + − − = = = −− + −
So, the answer is B.
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− = −
So, the answer is B.
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
+= =− − +
+= =
− +
So, the answer is B.
2 AP® Exam Practice Questions for Chapter 10
© 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.
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
= −
= −
= −
= − +
= − + − −
=
≈
So, the answer is B.
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
−
−
=
′=
′
+=
+=
≈ −
So, the answer is B.
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.
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.
AP® Exam Practice Questions for Chapter 10 3
© 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.
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
+ = − + −
≈
Note: Be sure to write down the appropriate definite integral before numerically approximating it on your calculator.
Note: Be sure to write down the appropriate definite integral before numerically approximating it on your calculator.
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.
4 AP® Exam Practice Questions for Chapter 10
© 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.
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.
Note: Be sure to write down the appropriate definite integral before numerically approximating it on your calculator.
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
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 the approximations from your calculator. Because these are approximations, a point may be deducted if an equal sign is used.
AP® Exam Practice Questions for Chapter 10 5
© 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.
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.
Note: Be sure to write down the appropriate definite integral before numerically approximating it on your calculator.
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 the approximations from your calculator. Because these are approximations, a point may be deducted if an equal sign is used.
6 AP® Exam Practice Questions for Chapter 10
© 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.
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.
AP® Exam Practice Questions for Chapter 10 7
© 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.
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.
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