Math Coaching1 2ndbooklet(final)

PROBLEM A Polar Equation r  a sin  represents a  CIRCLE TANGENT FROM X –AXIS PROBLEM A cube with vertex at the midpoint of the upper edge creates a pyramid inside the cube. What is the total surface area? 4

D

8

h1  8 2  8 2  11.3137

4

h2

h1

A

Slant height h1: 2 h1  8 2  8 2

V

C

B

Slant height h2: 2 h2  8 2  4 2 h2  8 2  4 2  8.9443

A1=AVCD 1 A1  811.3137  45.2548 2

A2=AVBC=AVAD 1 A2  88.9443  35.7772 2

A3=AVAB 1 A3  88  32 2 Total Surface Area  A1  2A 2  A 3  A 4

A4=AABCD=ABASE A4  8 2  64

 212.8092

PROBLEM A roof of a galvanized aluminium rest house. Sides are 12m with a height of 6m. (a) Find the volume of the galvanized aluminium roof. (b) Find the lateral area with a slant height of 10m. (a)

1 V  bh 3 1  12 2 6 3  288m 3

 

(b)

1 PB hS 2 1  4 x1210 2  240m 2

AL 

PROBLEM A spherical ball with a radius of 14cm was drop on a cone with a depth of 20cm. Find the height from the base of the cone to the point of tangency of the ball and the cone.

14cm

14cm

C

A 8 O

h 8

D

20cm

B h=?

8 AB  OD 20

h  OD  8 AB 2  20 2  14 2

OD 

AB  24.4131

h  OD  8 h  6.5539  8

820  6.5539 24.4131

h  14.5539cm

PROBLEM A trough which is triangular ends lie in parallel rectangle which is 50cm by 150cm. If the altitude is 40cm. What is the depth of the water if the volume is 12 liters?

150

40

50 X h

1 xh (150) 2 15  12000   h h150 24  V 

x 50  h 40 x

5 h 4

h  11.3137

PROBLEM A rectangular parallelepiped whose length is 4m and 5m and the altitude is 6m. What is the area of the cross section of the opposite edges of the parallelepiped?

4

6 x 5

A  xh

x  4 2  52 x  41

A  416  A  38.42m 2

PROBLEM A rectangular parallelepiped whose dimension is in the ratio 1:3:4. If the volume is 9000cu.m. What is the longest dimension? V  LWH

4 x  49.09

9000cu.m.  x3x 4 x 

 36.34m

9000  12 x3 x  9.09

PROBLEM A trapezoidal figure has an altitude of 40cm and a length of 72cm. The upper base is 25cm. and the lower base is 37cm. What is the volume of the figure?

72

25 40

37 1 V  25  37 4072 2 V  89280cm 3 PROBLEM When it rains 10cm. water fell. Determine the volume of the water fell on a level 10 hectare golf course.

A  10hectare 

10000sq.m 1hectare

 100cm  A  100000sq.m     1m 

2

V  100000000010 

1 1000cm 3

V  10000000mm3

A  1000000000cm 2 PROBLEM A crown hat has a base of 90cm2 and depth of 10cm. If the head covers 2/3, find the area for ventilation.

V  Bh

V  90cm 2 10cm  V  900cm 3 1  900cm 3  3 Area of Ventilatio n  300cm 3 Area of Ventilatio n 

PROBLEM If a smoke stick of a boat is 5m in height and it was sometime filled with 20cm3 of smoke, find the lateral area ignoring the thickness.

L  2rh

L  2 1.1285 L  35.437m 2 PROBLEM How much material should be used in making 2500 dice with edge of 3cm.

V  (3cm ) 3 V  27cm 3 M aterial used :





 27cm 3 2500   67500

PROBLEM If a log has an altitude of 8m, base diameter of 0.5m. Right diameter of 0.4m. if it rolls out to the ground covering 1 hectare of land. How many revolutions will it make?

Perimeter :

Pk  2r  2 0.2  Pk  0.4

solving for e :

Lateral area : L  Pk e 

L  0.4 10  L  4

e 0.5  8 0.4

1hectare  10000 sq.m

e  10

n  795.77

10000m 2  4n

PROBLEM A __________ of initial value or boundary value is a function y(x) which solves the differential equation and satisfies all the subsidiary conditions.  SOLUTION PROBLEM It is the set of all solution in the differential equation.  GENERAL SOLUTION PROBLEM What is the order of the differential equation?  THE HIGHEST ORDER OF THE DERIVATIVE FOUND IN THE EQUATION

PROBLEM Evaluate the integral of

?

Solution:

ANSWER: PROBLEM Find the length of the arc of

on [1,4].

Solution:

By Calculator: L = 7.63 units

PROBLEM Find the volume of the solid formed by revolving the area bounded by the x-axis. Solution:

By Calculator: V = 53.62 cu. units

, x =0, y = 0, about

PROBLEM Find the volume of solid of revolution formed by revolving the area bounded by about the line y = 4. Solution:

By Calculator: V = 41.88 cu. units

PROBLEM Solve the differential equation

with initial condition y(0) = 4.

Solution:

y(0) = 4 means that when x = 0, then y = 4 4=C ANSWER:

PROBLEM What kind of graph will you have when plotting r = a cos3θ?  3 leafed rose

PROBLEM What kind of graph will you have when plotting r = a cos2θ?  4 leafed rose

PROBLEM

, x = 0, y = 2,

A salad bowl in a shape of a hemisphere has a diameter of 40 cm. If the salad occupies a height of 15 cm, determine the volume of the salad and percentage volume of the salad with respect to the volume of the bowl.

Given: D = 40 cm h = 15 cm Required: (a)VS (b)%V Solution: using Pythagorean theorem, 52+r2=202 r2 = 375cm2 (a) VS = VS = 10,602.875 cm3 (b) Vbowl = Vbowl = 16,755.16 cm3 %V = % %V = 63.28 % PROBLEM A cone with a diameter of 6 cm is inscribed in a sphere with a radius of 8 cm. Compute for the volume and surface area of the cone.

Given: R = 8 cm d = 6 cm Required: (a) Vcone (b) Acone Solution: using Pythagorean theorem, 32+h2=82 h2 = 55cm2 (a) Vcone =

= 994.08 cm3

(b) Acone = πR(2h+r) = 448.18 cm2 PROBLEM A buoy is made up of a spherical segment and a cone. The angle of depreciation of the cone is 30° and the slant side of the cone is 2 cm long. Determine the volume and surface area.

Given: θ= 30° R = 2 cm Required: (a) V (b) A Solution: h = 2 – 2cos30° = 0.268 cm sin30 = r = 2sin30 = 1 cm (a) V =

= 2.245 cm3

(b) A =

+r) = 9.65 cm2

PROBLEM Three circles are tangent to each other inside and a big circle is tangent to them. The radius of the 3 circles is 10 cm each. Find the area of the bigger circle

cos 30 = b=11.25 R=10+11.25 A= πR2 A=1418.63

PROBLEM

Ans: cot

PROBLEM A cylindrical smokestack in a ship has a height of 5. In an instant it fills with a 25m 3 smoke. What is its radius? V= R=1.26

PROBLEM

V=AL V= V=150,000cm3 V=150,000cm3 x V=150 L

PROBLEM Filled with water: height is 25 cm

= X= 31.25 V=AL V= V= 58,593.75 cm3 x V= 58.59 L

PROBLEM Find the volume bounded by

, x = 0, and y = 1 rotated along the x-axis.

V= 2π R=y

V= 2π

L=

 V= 47.88

PROBLEM Find the volume bounded by

, x=1, x=0, and y=0 rotated along the y-axis.

V= 2π R=x L= V= 2π  V= 7.85

PROBLEM Find the volume of the region bounded by

, y=2x rotated along the x-axis.

V= 2π R=y L =X right – X left L= V= 2π  V= 13.40 = 64 PROBLEM Find the volume of the solid region bounded by

and y=0 along x-axis.

V= π R= V= π  V= π/30

PROBLEM A force 5 pounds compresses a spring 2 inches from its original length 14 inches. Find the work done if it compresses a total of 6 inches. W= F = kx 5 = k (2) K = 5/2 W=  W = 45 in-lb

PROBLEM At 12:00 noon, ship B is 100 miles east of ship A. Ship B sails west at 10 mi/hr and ship A sails south at 20 mi/hr. When is the two ships closest to each other?

-2000 + 200t + 800t = 0 t = 2 hrs time = 12 noon + 2 hrs = 2:00 pm

PROBLEM At 12:00 noon, ship B is 100 miles east of ship A. Ship B sails west at 10 mi/hr and ship A sails south at 20 mi/hr. What is the closest distance between the ships?

-2000 + 200t + 800t = 0 t = 2 hrs z= z= z = 89.44 = 40

miles

PROBLEM Find the volume of the solid formed by revolving the eq. f(x)=4-x^2, y=1,x=0 about y=1.

y=4-x^2 y-1=(4-x^2)-1 to find the limits, use the eq. and subs.y=1 to the eq. and find x. you will get  3 therefore limits are from x=0 and x=+sq.root of 3 3

V    (3  x 2 )dx 0

(hanggang jan na lang po ung natandaan ko, ndi ko na po sure ung kasunod) Ans. 26.1 --- yan po sagot sa booklet PROBLEM A right circular cone tank has a dimensions of 4ft @ the top and is 12 ft high. Find the work done to pump the water 4ft above the top of the tank. (Senxa na, mali ung sagot na lumabas dun sa computation ko kaya ndi ko na nilagay)

Ans. 43680*pi ft-lb.  sagot sa booklet PROBLEM Find the work done to stretch a 2.5cm bar, w/ a force of F= 4, to an additional of 2cm. F=kx 4=k(2.5) k=1.6

4.5

W   1.6 xdx 2.5

W  11.2

PROBLEM Find the area bounded by f(x)=sine(x) and g(x)=cosine(x) and the value of x are 0<= X <= 5pi/4. Soln. f(x)=sine(x) g(x)=cos(x) solving for the limit (sin(x)) / (cos(x)) =1 tan(x) = pi/4 A= A= 3.242640687 PROBLEM Find the area bounded by y-8x-x2-2x and x-axis Soln. Solving for the limit: 0=-x2+2x+8 (x+4)(x-2) X=-4 x=2 A= A= 60

PROBLEM Find the Surface area revolve about x=y3 0<= y <= 2 Soln S= S= 122.3229834 =

Ans. tanβ

h

r

2

Soln. 2 = 81.44888889 ˚ θ = 270 – (81.44888889 ˚) θ = 188.55111111˚