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### Transcript of PreCalc B Second Semester Final Exam Review PreCalc B Second Semester Final Exam Review...

• PreCalc B Second Semester Final Exam Review

I.    Linear  velocity,  angular  velocity,  and   arc  length.

1A.θ = 40π t = 1min

ω = θ t =

20 2π( ) 1min

i 1min 60sec

= 2π 3

21B. 30 3

20 cm/sec

V r πω

π

⎛ ⎞= = ⎜ ⎟⎝ ⎠ =

1C. 0 cm/sec

2A.V = 5m 20sec

= 1 4

m/sec

V = rω = 1 4

m/sec = 2ω

2B. ω = 1 8

( )

( )( )2 2 2 3. 6 2 12feet

1 1 6 2 36ft 2 2

S r

A r

θ

θ

= = =

= = =

II.    Trigonometry

1A. 2 13

13 1B. − 13

3 1C. −3

2 1D. 13

2

2. 2 2

, 2 2

⎝ ⎜

⎠ ⎟

3A. 2

2 3B. −1 3C. 2 3

3 3D. 2

4. A : none P : 3π

D :°except− 5π 2

+ 3πk R :°

5. A : 2 P : 4 D :° R : −1.5,2.5[ ]

6. A : none P : 2π

D :°except 3π 2

+πk R : (−∞,−2]∪[2,∞)

7. y = 19cos π 32

x +10( )⎛ ⎝⎜

⎞ ⎠⎟ + 31

8. y = 23cos π 9

x − 2( )⎛ ⎝⎜

⎞ ⎠⎟ −1

• 9A. cos2 x 2

⎛ ⎝⎜

⎞ ⎠⎟ − sin2 x

2 ⎛ ⎝⎜

⎞ ⎠⎟

= cos x

9B. 2sin xcos x

1+ 2cos2 x −1 = sin x

cos x

= tan x

9C. tan(2(50)) = tan100o

9D. sin(2(35)) = sin70o

9E. tan2θ +1

= sec2 x

9F. 2sinθ cosθ

= sin2θ

10A. 2 1 3

⎛ ⎝⎜

⎞ ⎠⎟ 2 2 3

⎛ ⎝⎜

⎞ ⎠⎟ = 4 2

9

10B.1− 2 1 3

⎛ ⎝⎜

⎞ ⎠⎟ 2

= 1− 2 1 9

⎛ ⎝⎜

⎞ ⎠⎟ =

7 9

10C. 1− 2 2

3

1+ 2 2 3

= 3− 2 2 3+ 2 2

11A. tan−1 −1( )= − π4

11B. arccos − 2

2

⎝ ⎜

⎠ ⎟ =

3π 4

11C. tan sin−1 5

13 ⎛ ⎝⎜

⎞ ⎠⎟ = 5

12

12. cos 150 2

⎛ ⎝⎜

⎞ ⎠⎟ =

1+ − 3 2

⎛ ⎝⎜

⎞ ⎠⎟

2

= 2 − 3 4

= 2 − 3 2

13A.1− sin2β = 1− 2 1 3

⎛ ⎝⎜

⎞ ⎠⎟ 2

= 7 9

13B. 1− cosα 2

= 1− 1 2 2

= 1 2

13C. 1− tan 2α

2 tanα = 1− − 3

1 ⎛ ⎝⎜

⎞ ⎠⎟

2

2 − 3( ) = 3 3

13D. 1− 2 2

3

1+ 2 2 3

= 3− 2 2 3+ 2 2

15A.sec2θ = 4; cosθ = ±1/ 2

θ = 60o ,120o ,240o ,300o

15B.tan2θ +5tanθ + 6 = 0 = (tanθ + 3)(tanθ + 2)

θ = 108.43o ,116.57o ,288.44o ,296.57o

15C.2sin2θ − cosθ −1= 0

θ = 60o ,180o ,300o

16A. sin2 x − sin x − 6 = 0 sin x − 3( ) sin x + 2( ) = 0

sin x = 3 or sin x = −2

NoSolution

16B. 2cot2 x + csc2 x − 2 = 0 2cot2 x +1+ cot2 x − 2 = 0

3cot2 x = 1 cot2 x = 1 3

tan2 x = 3 tan x = ± 3

x = π 3

, 2π 3

, 4π 3

, 5π 3

16C. 2cos2 x − sin x −1= 0

2 1− sin2 x( )− sin x −1= 0 2− 2sin2 x − sin x −1= 0 2sin x −1( ) sin x +1( ) = 0

sin x = 1 2

orsin x = 1

x = π 6

, 5π 6

, 3π 2

17D. sin2θ = 2 2

θ = π 8

, 3π 8

, 9π 8

,11π 8

17E. 0, π 2

, 3π 2

, π

17F. tan(3θ + π 4

) = 1

θ = 0,π 3

, 2π 3

,π , 4π 3

,5π 3

• 17G. sec2θ + tanθ = 0 tan2θ + tanθ +1= 0

tanθ = −1± i 3 2

θ = no solution

17H. sinθ + cosθ = 1

θ = 0,π 2

III.    Polar  Graphing

( )

21. 4 25 29 5tan 111.8 2

29,111.8

r r

θ θ

= + =

= → = −

o

o

5 5 22. cos 5cos 4 2 5 5 2sin 5sin 4 2

5 2 5 2, 2 2

x r

y r

πθ

πθ

= = = −

= = = −

⎛ ⎞ − −⎜ ⎟⎜ ⎟⎝ ⎠

83. 3, 7 π⎛ ⎞−⎜ ⎟⎝ ⎠

4. 2 r 2 cos2θ( )− 2 r 2 sin2θ( ) = 5r sinθ 2r 2 cos2θ − 2r 2 sin2θ = 5r sinθ

5. r 2 = r − 2r sinθ

x2 + y2 = x2 + y2 − 2y

6A.    symmetry:    x-­‐axis,  y-­‐axis , 2 π⎛ ⎞

⎜ ⎟⎝ ⎠  pole

6B.    symmetry:    y-­‐axis   2 π⎛ ⎞

⎜ ⎟⎝ ⎠

6C.    symmetry:    y-­‐axis   2 π⎛ ⎞

⎜ ⎟⎝ ⎠

6D.    symmetry:    x-­‐axis,  y-­‐axis   ,

2 π⎛ ⎞

⎜ ⎟⎝ ⎠  pole

IV.  Law  of  Sines  and  Cosines

1A. 51o 1B. 18.59 cm

1C. 14.45 cm

2A. 12.64 cm 2B. 50.91o

2C. 81.09o

3A. 27.45 in 3B. 29.71 in

3C. 28o

4A. b1=51.60m b2 =5.135

4B. B1=153.45 o B2 =2.55

o

4C. C1=14.55 o C2 =165.45

o

5A. A = 15(7)(5)(3) = 1575 = 15 7 ft2

5B. A = 1 2

sin38o 6( ) 4( ) = 7.39mi2

V.  Conics

1A. x2 + 4x + 4( )− 3 y2 − 2y +1( ) = −13+ 4− 3 x + 2( )2 − 3 y −1( )2 = −12 y −1( )2

4 −

x + 2( )2 12

= 1

C : −2,1( ) V : −2,3( ) −2,−1( ) F : −2,5( ) −2,−3( ) A : y −1( ) = ± 33 x + 2( )

• 1B. 4 x2 − 4x( ) + 9 y2 + 6y( ) = −61 4 x2 − 4x + 4( ) + 9 y2 + 6y + 9( ) = −61+16+81 4 x − 2( )2 + 9 y + 3( )2 = 36

x − 2( )2 9

+ y + 3( )2

4 = 1

C : 2,−3( ) V : −1,−3( ) 5,−3( ) F : 2 ± 5,−3( ) Major length : 6 Minor length : 4

1C. 4 x2 − 2x +1( )− y2 + 6y + 9( ) = 6+ 4− 9 4 x −1( )2 − y + 3( )2 = 1

x −1( )2 1

4 −

y + 3( )2 1

= 1

C : 1,−3( ) V : 3

2 ,−3

⎛ ⎝⎜

⎞ ⎠⎟

1 2

,− 3 ⎛ ⎝⎜

⎞ ⎠⎟

F : 1± 5 2

,−3 ⎛

⎝ ⎜

⎠ ⎟

A : y + 3( ) = ±2 x −1( )

2. x2 +8x +16( )− 4y = −8+16 x + 4( )2 − 4y = 8 x + 4( )2 = 4y +8 x + 4( )2 = 4 y + 2( )

V : −4,−2( ) F : −4,−1( ) D : y = −3

3. a = 5 c = 3 25− b2 = 9 b2 = 16

x − 2( )2 25

+ y − 4( )2

16 = 1

4.V : 5,−4( ) x − h( )2 = −4a y − k( ) x −5( )2 = −8 y + 4( )

( ) ( ) ( )2 2

5. : 4,6

4 6 1

5 9

C

x y− − + =

( ) ( )

2 2

2 2

6. 4 9 5

4 1 1

4 5

b b

x y

+ = =

− + − =

( )

2

2

2

7. 4 30 4 20 900 80 11.25

45

5, 0.56 ht=19.44ft

10, 2.22 ht=17.78ft

20, 8.89 ht=11.11ft

x ay a a a

x y

x y

x y

x y

= = − = − = −

= −

= = − →

= = − →

= = − →

VI.    Parametrics

( )

( ) ( ) ( ) ( ) ( )

( ) 4 2 cos 12

( ) 4 2 cos 15

( ) 4 2 2 cos 24 4 2 2

2( ) 4 2 2 cos 4 2 2 15

x t t

or x t t

y t t

or y t t

π

π

=

⎛ ⎞− − = ⎜ ⎟⎝ ⎠

= − + −

⎛ ⎞− − = − + −⎜ ⎟⎝ ⎠

x(t)

y(t)

•   VII.  Sequences  and  Series

1A. 3, 7, 11, 15, 19; Arithmetic

1B. −14 3

, −13 3

, − 4, −11 3

,−10 3

; Arithmetic

1C. 2 3

, 3 4

, 4 5

, 5 6

, 6 7

; Neither

2A. an = 32−5 n−1( ) 2B. a n = 11 3( )n−1

3. −333 4. 3.45×1011

5.166 = 30+ 4 n−1( ) n = 35 6A. S25 =

25 2

5+ 77( ) = 1025

S5 = 5 2

5+17( ) = 55 1025−55= 970

6B. S23 = 3 1− 323( )

1− 3 = 1.41×1011

6C. 1339 280

7. 8748

8. 1 64

= 2 1 2

⎛ ⎝⎜

⎞ ⎠⎟

n−1 1

128 = 1

2 ⎛ ⎝⎜

⎞ ⎠⎟

n−1

ln 1 128

⎛ ⎝⎜

⎞ ⎠⎟ = n−1( ) ln 12

⎛ ⎝⎜