MATH 117 Test 2 Review Solutions 1. One cycle of y =12sin(8t) ends when 8t = 2π , or at t = π / 4 . Then the middle is at t = π / 8 , and a quarter cycle is at t = π /16 . One cycle of y = 8cos(6π t) ends when 6π t = 2π , which gives t =1/ 3 . Then the middle is at t = 1/6, and a quarter cycle is at t =1/12 .

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2. (a) Use y = 60sin(Bt) . One cycle ends at t =1/ 200 seconds. So B× 1200

= 2π , and

B = 400π . Thus, y = 60sin(400π t) . Five equidistant points on the cycle are at t = 0 , 1/800, 2/800, 3/800, and 4/800.

(b) Use y = 80sin(Bt) . One cycle ends at t =1/ 800 seconds. So B× 1800

= 2π , and

B =1600π . Thus, y = 80sin(1600π t) . Five equidistant points on the cycle are at t = 0 , 1/3200, 2/3200, 3/3200, and 4/3200. --------------------------------------------------------------------------------------------------------------------- 3. (a) We can compute the inverse cosine, or arccos, of any of the values

1,

€

32

, 22

, 12

, 0, − 12

, − 22

, –

€

32

, –1 :

1

32

−32

−22

22

12

−12

0

−1 0

π4

π6

π3

π22π

33π45π

6

π

cos−1 x is the radian angle between 0 and ! whose cosine is x

cos−1(1) = 0 cos−1 32

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = π

6 cos−1 2

2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = π

4 cos−1 1

2⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = π

3 cos−1(0) = π

2

cos−1 −12

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = 2π

3 cos−1 −

22

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = 3π

4 cos−1 −

32

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = 5π

6 cos−1(−1) = π

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

(b) We can compute the inverse sine, or arcsin, of any of the values

–1, − 32

, − 22

, − 12

, 0, 12

, 22

, 32

,1 :

sin−1 x is the radian angle between −π

2 and π

2 whose sine is x .

0

32

−32

−22

22

−12

12

−1

1

0

−π6

π6

−π4

π4

−π3

π3

−π2

π2

sin−1 −1( ) = − π2

sin−1 −32

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = − π

3 sin−1 −

22

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = − π

4 sin−1 – 1

2⎛ ⎝ ⎜

⎞ ⎠ ⎟ = − π

6

sin−1 0( ) = 0 sin−1 12⎛ ⎝ ⎜ ⎞ ⎠ ⎟ = π

6 sin−1 2

2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = π

4 sin−1 3

2

⎛

⎝ ⎜ ⎜

⎞

⎠ ⎟ ⎟ = π

3 sin−1 1( ) = π

2

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - (c) We can compute the inverse tangent, or arctan, of any of the values

–∞, − 3 , −1 , − 13

, 0, 13

, 1, 3 , ∞ :

tan −1 x is the radian angle between −π

2 and π

2 whose tangent is x

0

∞

−∞

−1

1

− 3

3

−13

13

tan −1 −∞( ) = − π2

tan −1 − 3( ) = − π3

tan −1 −1( ) = − π4

tan −1 −13

⎛

⎝ ⎜

⎞

⎠ ⎟ = − π

6

tan −1 0( ) = 0 tan −1 13

⎛

⎝ ⎜

⎞

⎠ ⎟ = π

6 tan −1 1( ) = π

4 tan −1 3( ) = π

3 tan −1 ∞( ) = π

2

–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

4. tan A = 7

3, with angle A in Quadrant III → x = –3, y = –7, z = 32 + 72 = 58

So: sin A = −758

and cos A = −358

sinB = 610

, with angle B in Quadrant II → y = 6, z = 10, x = − 102 − 62 = −8

So: sin B = 610

and cos B = −810

sin(A + B) = sin A cosB + cos A sinB

=−758

×−810

+−358

×610

=56 − 1810 58

=38

10 58

sin(A − B) = sin A cosB − cos A sinB

=−758

×−810

−−358

×610

=56 + 1810 58

=74

10 58

cos(A + B) = cos A cosB − sin A sinB

=−358

×−810

−−758

×610

=24 + 4210 58

=66

10 58

cos(A − B) = cosA cosB + sin A sinB

=−358

×−810

+−758

×610

=24 − 4210 58

=−18

10 58

A + B must be in Quadrant I (sin, cos both +).

A − B must be in Quadrant II (sin is + and cos is –).

sin(2A) = 2 sinA cosA

= 2 ×−758

×−358

=4258.

cos(2A) = cos2 A − sin2 A

=958

−4958

=−4058

.

2A must be in Quadrant II (sin is + and cos is –). Because B is in Quadrant II, then 90o≤ B ≤180o; so 45o≤ B / 2 ≤ 90o. Thus B / 2 is in Quadrant I, and both its sine and cosine are positive.

sin B2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = +

1− cosB2

=1 − (−8 / 10)

2

=18 / 102

=910

=310

cos B2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = +

1+ cosB2

=1 + (−8 / 10)

2

=2 / 102

=110

=110

–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

5. secA = – 3

2 with angle A in Quadrant III → x = –2, z = 3, y = − 32 − 22 = − 5

So: sin A = − 53

and cos A = −23

cot B = − 125

, with angle B in Quadrant IV → x = 12, y = –5, z = 169 = 13.

So: sin B = −513

and cos B = 1213

sin(A + B) = sin A cosB + cos A sinB

=− 53

×1213

+−23×−513

=−12 5 + 10

39< 0.

€

sin(A − B) = sinA cosB − cosA sinB

=− 53

×1213

−−23×−513

=−12 5 −10

39< 0.

cos(A + B) = cos A cosB − sin A sinB

=−23×1213

−− 53

×−513

=−24 − 5 5

39< 0.

cos(A − B) = cosA cosB + sin A sinB

=−23×1213

+− 53

×−513

=−24 + 5 5

39< 0.

Both A + B and A − B must be in Quadrant III

because the sine and cosine of both are negative.

(Note: To see that the sine and cosine of both are negative, use 5 ≈ 2.)

sin(2A) = 2 sin A cos A

= 2 ×− 53

×−23

=4 59

cos(2A) = cos2 A − sin2 A

=49−59

= −19

2A must be in Quadrant II. Because B is in Quadrant IV, then 270o≤ B ≤ 360o; so 135o≤ B / 2 ≤180o. Thus B / 2 is in Quadrant II; so its sine is positive and its cosine is negative.

sinB2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ =

1 − cosB2

=1 −12 / 13

2

=126

=126

cosB2

⎛ ⎝ ⎜

⎞ ⎠ ⎟ = −

1+ cosB2

= −1+12 / 13

2

= −2526

= −526

–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

6. Note: v =ω r , where ω is in radians per time.

(a) v =ω r = 900 revmin

×2π (rad)rev

×3 ft× 60 minhr

×1

5280miles

ft≈ 192.78 mph

ω = 900

€

revmin

× 360ºrev

× 160minsec

= 5400 degrees per second.

(b) v =ω r = 3000 revmin

×2π (rad)rev

×0.25 ft× 60 minhr

×1

5280miles

ft≈ 53.55 mph

––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––– 7. Because v =ω r , we have

60 mileshr

=ω(rad)hour

×1.6 ft× 15280

mileft

, so

ω =

€

60 mileshr

×1

1.6ft× 5280 ft

mile=198,000 (rad)

hour . Then,

ω = 198000 (rad )hr

×1(2π )

revrad

×160

hrmin

≈ 525. 2 rpm

–––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––

8. (a) v =ω r = 2π (rad)(1 / 3)hr

×2.6 miles ≈ 49 mph Note: 20 min = (1/3) hour

(b)

€

ω =360º

(1/3) hr= 1080º per hour and

€

ω =1 rev

20 min= 0.05 rpm

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9. Earth’s radius is about 3963.2 miles and one rotation takes 23h 56m 4.1s. (a) At 40º N, the cross-sectional radius is given by 3963.2×cos(40º ) mi. But we still need the angle in between 12º 4 ʹ 2 W and 18º 3 ʹ 6 E, which is given by the sum: 12º 4 ʹ 2 + 18º 3 ʹ 6 = 31.3º .

So the direct East-West distance is

31.3º × π

180º⎛ ⎝ ⎜

⎞ ⎠ ⎟ × 3963.2 ×cos(40º ) mi ≈

1658.52 miles

12º 42’ W 18º 36’ E

40º N

(b) At 25º S, the lateral radius is 3963.2× cos(−25º ) mi. The angle between 18º 3 ʹ 6 E and 12º 4 ʹ 2 E is the difference: (18 + 36/60) – (12 + 42/60) = 5.9º

So the direct East-West distance between the points is

5.9º× π / 180º× 3963.2 × cos(−25º ) mi ≈ 369.87 miles. (c) The linear rotational velocity at 38o3 ʹ 0 N is

v =ω r = 2π (rad)(23+ 56 / 60+ 4.1 / 3600)hr

×3963.2× cos(38.5) mi les ≈ 814.23 mph.

(d)

2π (rad)

(23 + 56 / 60 + 4.1 / 3600)hr× 3963.2× cos(−52.75o)miles ≈ 629.75 mph.

(e) v = ω r = 2π(rad)(10+39 / 60+ 54 / 3600)hr

× (9.4 × 3963.2 miles) ≈ 21,947.9 mph.

And ω =360º

(10+39 / 60+ 54 / 3600)hr≈ 33.755º per hour

(f) v = ω r = 2π (rad)(16+3 / 60)hr

× (3.88 × 3963.2 miles) ≈ 6019.8 mph.

And ω =360º

(16+3 / 60)hr≈ 22.43º per hour

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10. (a) To stay in orbit, v = 1.24×1012

(3963.2+ a) = 1.24×1012

(3963.2+ 240) ≈ 17,175.95 mph.

(b) a = v2

r = C

r2 = 1.24 ×1012

(3963.2 + 240)2mileshr2

(× 5280 /36002 ) ≈ 28.595 ftsec2

, which is about

88.876% of the surface gravity of 32.174 ft/s2.

(c) ω =vr=

v(3963.2 + a)

= 17,175.95(3963.2 + 240)

(rad)hr

≈ 4.08639 (rad)hr

×180ºπ

→ 234.133º per hr

(d) The angular velocity function of the Shuttle at this altitude is 234.133 t degrees. The

angular velocity function of Earth is 360(23 + 56 / 60 + 4.1 / 3600)

t = 15.041 t degrees. For

Shuttle to lap Earth, we must find the time t such that 234.133 t = 15. 041 t + 360º. That is, the Shuttle’s angle must equal the Earth’s angle plus one revolution. Solving for t we have 219.092 t = 360, which gives t ≈ 1.643 hrs ≈ 1 hr, 38 min, 35 sec. ––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––––