math r3 soln36-97

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8/10/2019 Math R3 Soln36-97

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MATH REFRESHER 3 ENGR. FUENTES

Cebu: JRT Bldg., Imus Avenue, Cebu City Tel. 2685989 90 | 09173239235 Manila: 3rd

& 4th

Fl. CMFFI Bldg. R. Papa St. Sampaloc Tel. 7365291

( ) ( ) ( ) ( )

( ) ( )

2 22

2 2 o

BCBD 15.05

2

Cosine law:

L AB BD 2 AB BD cosB

40 15.05 2 40 15.05 cos 65

L 36.3m

= =

= +

= +

=

56. Ans: 19.94 m and 11.23 m

( )2 2 2 o

2

1 2

10 18 x 2 18 x cos30

x 18 3x 224 0

x 19.94; x 11.23

= +

+ =

= =

57. Ans: 74.33Use Napiers Circle for right sphericaltriangle:

Use Sin-Cos-Op Rule - sine of any middle partis equal to the product of the cosine of twoopposite parts.

( ) ( )o o

o

sin c cos b cos a

cos c cos b cos a

cos 80 cosbcos 50

b 74.33

=

=

=

=

58. Ans: 4 p.m.

o o

12hrs ahead x

180 E 120 E

x 8hrs ahead or 4pm

=

=

59. Ans: 2.87 hours

Use Sin-Tan-Ad Rule - sine of any middle part isequal to the product of the tangents of two

adjacent parts.

oo

o

o

sinA tanc tanb

cos A cot c tan b

tanbcosA

tanc

tan14 36'cos30

tanc

60 min 1 milec 16.74 1004.4 nautical mi.

1min1

d 1004.4 n.mit 2.87hrs

v 350 n.mi / hr

=

=

=

=

= =

= = =

60. Ans: 6046.2 nautical miles

( ) ( )

( ) ( ) ( )

o o o

o o o

o o o o

o o

o o o

a 90 14 36' 75 24'

b 90 37 48' 52 12'

C 360 121 05' 122 24' 116 31'

Oblique Spherical Triangle:

Using cosine law for sides,

cosc cosacosb sinasinbcosC

cosc cos 75 24 ' cos 52 12'

sin 75 24 ' sin 52 12 ' cos 116 31'

= =

= =

= =

= +

=

+

o

o

60min 1n.mic 100.77 6064 n.mi

1min1

= =

61. Ans: 8 2

62. Ans: 5.61 yards

( ) ( )

( )

2 2

22

o o o

o o o o

1 o

1m 3.281ft

AC 20 20 20 2

ACAE EC 10 2

2

FE 3.281 10 2 14.52

180 65 115

180 35 115 30

tan 3.281 10 2 13.06

=

= + =

= = =

= + =

= =

= =

= =

o o

10 2 GE

sin30 sin35=

( ) ( ) ( ) ( )( )

o o o

2 2 2 o

1

1

T

GE 16.2 ft

FEG 65 13.06 51.94

L FE GE 2 FE GE cos51.94

Subst. values for FE and GE,

L 13.54 ft

L 3.281 13.54 16.82ft

1 yard16.82ft 5.61 yards

3ft

=

= =

= +

=

= + =

= =

63. Ans: 22 sq. units

( ) ( )

1 8 5 4 7 8A

2 2 6 1 4 2

148 5 16 14 10 24 7 32

2

A 22 sq.units

=

= + + + + +

=

64. Ans: 11.91 cm

( ) ( ) ( )

( ) ( ) ( )

( ) ( ) ( ) ( )

ABC

2

ABC

ABC ABD DBC

A s s a s b s c

45 45 36 45 30 45 24

A 357.18 cm

a b c 36 30 24s 45

2 2

A A A

1 1

357.18 r 36 r 242 2

r 11.91 cm

=

=

=

+ + + += = =

= +

= +

=

65. Ans: 6 m

( )

ins idewheels outside wheels

inside outside

inside outside

t t t

v 0.5vd d

0.5t t

2 r 1.5 2 r0.5

t t

2 r 3 r

r 3

2 r 6 Lenth or circumference of the track.

= =

=

=

=

=

=

=

66. Ans: 5.33 sq. units

2

2

22

y 4x eq.1

x 4y eq.2

Point of intersection:

x4x

4

=

=

=

18

xo30

1010

ab

A Bc

where :

A 90 A

B 90 B

c 90 c

=

=

=

o121 05'E

o14 36 'N

ob 14 36 '=

A

aC

o90

c

030

ab

A Bc

M

o121 05'E

o14 36 'N

o37 48 'N

o122 24'W

ba

C

c

d

12

4

o45

2 2

1 o

d 12 4 12.65

4tan 18.43

12

= + =

= =

T

dVd ( )

o

T

o

V T

o

V

18.43 4563.43

d dsin

12.65sin 63.43

d 11.31cm

8 2 cm

= +=

=

=

=

=

AC

o65 o353.281

1L

E1

0 2 10 2

G

F

A

B

C

D

20 ft 20 ft

o65 o351m

30

2436

rr

A

B

CD

0

( )4,4

2y

2y 4x=

1y

2x 4y=

dx

r 1.5

1.5r

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& 4th


( )

( )

2 o 2

2

102 2

10

Slicing medthod:

V=Ax thickness

1A 2y sin60 3y

2

V Adx 3y dx

3 10 x dx

V 2309.4

= =

= =

=

=

880

10100

5

( )

( )

4

3

1 2

1 2

4 2

2 1

0

x 64x

x x 64 0

x 0 x 4

y 0 y 4

xA y y dx 4x dx 5.33

4

=

=

= =

= =

= = =

67. Ans: 17.5 ft

2

1 1

222

2

1

2

1

1

x y

yx

y5030100

y 7.5

y 10 17.5 ft

=

=

=

+ =

68. Ans: 743.22

( )

( ) ( )

( ) ( )

oo

o

2

B

2 o 3

1 36036

2 5

3sin 36

a

a 5.1

1V A h 5 a sin 2 12

2

1V 5 5.1 sin 72 12 742.1 cm

2

= =

=

=

= =

= =

69. Ans: 358.15 gal

( )

( ) ( )

( )

( )

sector

2 2

2

2 o

3

A A A

1 1r sin 2 r

2 2

10.45 sin 141.06

2

10.45 218.94

2 180

A 0.451

265galV A xL 0.451 3 1.352m

1m

V 358.17 gal

= +

= +

=

+

=

= = =

=

70. Ans: 4066.67

( )( )

( )( )

( )

( ) ( )( )

1

2

1 2 1 2

A 80 8 640

A 100 10 1000

hV A A A A

3

5640 1000 640 1000

3

V 4066.67

= =

= =

= + +

= + +

=

71. Ans: 1 m

( )

( )( )

2

2

3 2

hV 3r h3

h5.236 3 2 h

3

0 h 6.283h 5.2363

Use calcu:

h 1m

=

=

= +

=

72. Ans: 2309.4

73. Ans: 0.66

2 2

2 2

2 2 2 2

16x 9y 32x 128 0

a 16 b 9

a 4 b 3

c a b 4 3 7

e c a 7 4 0.66

+ + =

= =

= =

= = =

= = =

74. Ans: 64x 5y 0+ =

2 2

To solve the equation of the diameter:

Differentiate the equation of the ellipse

and replace y' by the slope of the chords.

64x 25y 1600

128x 50yy' 0

1128x 50y 0

5

128x 10y 0

64x 5y 0

+ =

+ =

+ =

+ =

+ =

75. Ans: 2 27x 16y 112 0+ =

( )

2 2 2 2

2 2 2 2

2 2 2 2

2 22 2

The curve is an ellipse.

2a 8

a 4

The center is at origin and major

axis along the x-axis.

c 3

b a c 4 3 7

Standard equation:

x y x y1 1

a b 4 7

x y1 7x 16y 112 0

16 7

=

=

=

= = =

+ = + =

+ = + =

76. Ans: 4Reverse Engineering

77. Ans:9/2Use calculator.

78. Ans: y = x + 1

( ) ( ) ( )

( )

( ) ( )

( )

( )

3 2

2

3 2

2

1 1

y x 3x 4xy ' 3x 6x 4

y" 6x 6

At point of inflection, y" 0

0 6x 6

x 1

y 1 3 1 4 1 2

1,2 pointof inf lection

y ' 3 1 6 1 4 1

y y m x x

y 2 1 x 1

y x 1 Ans

= += +

=

=

=

=

= + =

= + =

=

=

= +

79. Ans: 2

( )

( )

( )

( ) ( ) ( ) ( ) ( ) ( ) ( )

4

2 2 4

2 2 4

2 2 3

2 2 3

xy x y x y

x y xy x y

dx y xy x y

dx

2xy x y ' y 2xyy ' 1 4y y '

At 1,1 ,

2 1 1 1 y ' 1 2 1 1 y ' 1 4 1 y '

y ' m 2

+ = +

+ = +

+ = +

+ + + = +

+ + + = +

= =

80. Ans: x y = 1Use reverse engineering. Assign a value ofx in the equation and solve for y. Example:

( ) ( ) ( ) ( )2 2 2 2

Example : x 2,

x y 1

2 y 1 y 1

So,the point at this curve is (2,1)

Test if the distance to the two points

are equal.

-1- 2 4 1 5 2 2 1

3 2 3 2 satisfy

=

=

= =

+ = +

=

81. Ans: (1, 0)

( )

2

2

Every point on the parabola is equidistant

to its directrix which is the y-axis and

the point which is the focus.

So, solve for the focus.

y 2x 1

y 2x 1

y 2 x 0.5

V :(0.5,0)

4a 2 a 0.5

=

=

=

= =

( )focus: 1,0

200

10

40

y

x

( )100,30

( )150,y

( )0,0

1y 10+

50

3 3

a

0.45

0.15 0.45

( )

o

0

0 o

o

0.15cos

0.45

70.53

360 2

360 2 70.53

218.94

=

=

=

=

=

h

2 2 2x y 10+ =

dxx

z

y

y

y

10

10

12

6

0.9

3

0.6

0

f

( )1, 0( )0.5,0

8/10/2019 Math R3 Soln36-97

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& 4th


82. Ans: x

( )

( ) ( ) ( )

( ) ( )

( ) ( ) ( ) ( )

( )

Since the divisor is a quadratic expression,

the remainder is in this format: Ax + B

P x Ax BQ x

x 1 x 1 x 1 x 1

P x Q x x 1 x 1 Ax B

Use remaider theorem:

Ifdivided by x 1,substitute x 1 to P x

tosolvethe remainder w

+= +

+ +

= + + +

=

( ) ( ) ( ) ( ) ( )

( )

( ) ( ) ( ) ( ) ( )

hich is 1:

P 1 Q 1 1 1 1 1 A 1 B

1 A B eq.1

If divided by x 1,substitute x 1 to P x

tosolvethe remainder which is -1:

P 1 Q 1 1 1 1 1 A 1 B

1 A B eq.2

Solving the equation 1 and 2.

A 1, B 0

Remainder :Ax B 1x 0 x

= + + +

= +

+ =

= + + +

= +

= =

+ = + =

83. Ans:1

e

Use calculator.

84. Ans:2

4

( )1

1

0

A ydx x tan x dx

A 0.2854

= =

=

85. Ans: 4

( ) ( )

2A r

dA dr2 r 2 4cm 0.5cm / s 4

dt dt

=

= = =

86. Ans: 2a

( )

2 2 2

2 2

2

2

The maximum rectangle that can be

inscribed in a circle is a square. So,

the maximum rectangle inscribed in asemi-circle is half-the square.

x x a

2x a

a ax

2 2

a2x 2 a 2

2

aArea a 2 a

2

+ =

=

= =

= =

= =

87. Ans: n2

( )

st

22

Use Reverse Engineering:

Using calculator:

d 1 1 1x

dx 1 2x 2 2

Substitute : n 1 1 derivative

1 1 1n 1 Since the same value, it

2 2 2

is the correct answer.

= =

=

= =

88. Ans: 8

( )

8 8 1

32 1

0 0

yA x x dy y dy 88

= = =

89. Ans:1

n 1

( )

( )

( )

( )( ) ( )

( )( )( )

[ ]

( )

n

n

e e

n 1

e

n 1 n 1

n 1n 1

n

e

Changelog to ln:

ln x dxdx

xx ln x

ln x

n 1

1ln lne

1 n

1 1 lne1 n ln

10 1

1 n

dx 1

n 1x ln x

+

+ +

+

=

= +

=

=

=

=

90. Ans: 256/105Cylindrical Coordinates:

( ) ( ) ( ) ( )

( )

( )

( )

( )

( )

( )

3 3

3 3

4 3 3

2 2s in24 3 3

0 0

2 2s in2

5 3 3

0 0

2s in22 63 3

0 0

6

3

z x y xy

r cos r sin r cos r sin

z r cos sin cos sin

dV z rdrd

V r cos sin cos sin rdrd

V r cos sin cos sin drd

rcos sin cos sin d

6

2sin2cos6

= +

= +

= +

=

= +

= +

= +

=

( )

2

3

0sin cos sin d

256V

105

+

=

91. Ans:2

( ) ( ) ( )

1 1 1

2 2 2

1 1 1

2 2 2

2 2 2

Ax By Cz Dd

A B C

x 2y 2z 6

1 2 2

0 2 0 2 0 6d 2

1 2 2

+ + +=

+ +

+ + =

+ +

+ + = =

+ +

92. Ans: 2

( ) ( )

( )

( )

2 2 2 2 2

2 2 2

2 2

2

2

Distance of any point to origin:

d x 0 y 0 d x y

dd x y

dx

0 2x 2yy '

xy ' Eq.1

y

d3x 4xy 3y 20dx

6x 4y 4xy' 6yy ' 0 Eq.2

Subst. Eq.1 to Eq. 2

x x6x 4y 4x 6y 0

y y

4x6x 4y 6x 0

y

4x4y

y

y x

Sub

= + = +

= +

= +

=

+ + =

+ + + =

+ + + =

+ =

=

=

( ) ( )

( ) ( )

2 2

22

2

2 22 2

st. y x to equation of the curve.

3x 4xy 3y 20

3x 4x x 3 x 20

10x 20

x 2 , y 2

d x y 2 2 2 ans

=

+ + =

+ + =

=

= =

= + = + =

93. Ans: 3 3 44x y 3y 7+ =

Use Reverse Engineering. If by longmethod, the D.E. is Bernoulli Equation.

94. Ans: 3 i+

Change log to ln.

( ) ( )3 ii

ln e ln 20.086e

ln20.086 lne

3 i

=

= +

= +

95. Ans: 62

( )6

6

6

6

6

5i 3 2

6

52 6

6

2 5

2 rectangular form

=

=

=

=

96. Ans: 52%

( )

( ) ( )

( )

( )( ) ( )

2

11

2

0.1z 2

2.1

50X 100 X

X 40

np 100 0.5 50

npq 100 0.5 0.5 5

P 39.5 x 50.5

x 39.5 50z 2.1

5

50.5 50z 0.1

5

1A e dz 0.52 or

2

P 100CX 0.5 0.5 0.52

P 52% correct answer.

=

= = =

= = =

= = =

= =

= =

= =

=

97. Ans: II and III only

Suppose the four tables are W, X, Y and Z.Since the question asks which must be true,we can quickly break statement I with acounterexample. Let tables W, X and Y have15 flowers each; then table Z has 16flowers, and no table has 13 or fewerflowers. We cannot break statement II andIII. So, II and III must be true.

All is well.

a x

2x

x

x 1=

( )1y x tan x=

dx

y 8x= 3y x=

dy

y 8=

2x1x

( )

dA rdrd

dV z rdrd

x r cos

y r sin

=

=

=

=

rd

dr

z

x y

r

dV

dA

02

math r3 soln36-97

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