A SIMPLE BUT VERY USEFUL CAL TECH TO DETERMINE DISTANCE BETWEEN TWO POINTS . MY VERY FIRST CAL

TECH INTRODUCED: YEAR 2009FIND THE DISTANCE BETWEEN A( 3, 4) and B ( 4 , -6).Solution:MODE 2 

INPUT | A - B |A? 3 + 4i =B? 4 – 6i = 

Result: √101

PAST ECE BOARD EXAMFind log ( 3 + 4i) base ( 5 – 3i )Solution: RADIAN MODE 

log ( 3 + 4i ) base ( 5 – 3i) = ln ( 3 + 4j) / ln ( 5 – 3j )Store 3 + 4i to A 

Store 5 – 3i to B 

ENTER: ( ln |A| + arg(A) i ) / ( ln |B| + arg(B)

i ) 

Result: 0.687 + 0.7365i

Find the positive square root of 4 – 6i 

Solution:Store 4 – 6i to A 

ENTER: √ |A|∟| Arg( A) / 2| = 

RESULT: 2.367 - 1.267iCHECK: Ans^2 =Result: 4- 6i

WSD CAL TECH DESIGN: PAST CE BOARD EXAMDesign a reinforced beam with a balanced stress reinforcement that will resist a bending moment of 140 knm. Assume d = 1.5bfc = 12 Mpa fs = 160 Mpa and n = 8.

Solution: For a balance design , fc and fs/n must occur simultaneously.

ENTER: MODE 3 2INPUT:X Y0 121 -160/8Then get the x intercept: 0 SHIFT 1 5 4 =Result: 0.375The NA is at 0.375d from the top.ENTER: SHIFT 1 5 1 = ( Result: A = 12 , store to A)ENTER: SHIFT 1 5 2 = (Result: B = -32, store to B )MODE 1 Get the Moment of the compressive area with respect to the steel.∫ (A + Bx)( 1 –x ) dx limit from 0 to 0.375Result: 63/32 Then 63/32 bd^2 = M bd^2 = (140 x 106)/ (63/32) = 71.111 x 106if d = 1.5b b(1.5b)^2 = 71.111 x 106 b = 316.16 or 316 mmand d = 158 mm

The dimension is 316 m x 158 mmFor Area of steel:Area of Compression = Asfs½ (0.375d)(b) fc = As (160)b = 316 and d = 474 As = 2106 mm2

A CAL TECH LECTURE IN MECHANICS OF MATERIALS FROM TOLENTINO AND ASSOCIATESCAL TECH FOR DETERMINING PRINCIPAL STRESSES.SEE FIGURE BELOW Write as:-80 -25-25 50

Note: sigma x = -80 sigma y = 50 (main diagonal)Shearing stress -25 ( other diagonal)(right face of shearing stress = downward (-) )MODE 5 3 INPUT [ 1 -trace determinant ] (eigenvalues )[ 1 -( -80 + 50) -80(50) – 25^2 ] Ans. 54.64 Mpa -84.64 Mpa

Volume of a prismatoid using CAL TECHA prismatoid is a polyhedron where all vertices lie in two parallel planes.The cross sectional area of a prismatoid with some reference ( the easiest is the base ) is a quadratic function of its distance from the reference. (Taken from my books on CALCULATOR TECHNIQUES )Example:Given a frustum of a cone with base radius 5 m and upper radius 4 m with height 6 m.

1. Find the volume of the frustum of a cone.2. Find the area 1 m above the base.3. Find the depth where the volume of the frustum is equal above and below it. Solution:MODE 3 3 INPUT:X Y

0 pi(5)^2 6 pi(4)^2 3 pi ( 4.5)^2 ( Average of the 2 radii )ENTER: ACSHIFT 1 5 1 = A = 78.5398 Store to ASHIFT 1 5 2 = B = -5.236 Store to BSHIFT 1 5 3 = C = 0.087266 Store to C

To get the area 1 m above the base: 1 SHIFT 1 5 6 = ( 1 y hut ) ( 75.391 m2)To get the volume:MODE: 1 ∫ A + BX + CX^2 dx from 0 to 6. Result: 122pi.Half of this volume is 61piTo get the depth where the volume is equal above and below it:Volume is Ax + Bx^2/2 + Cx^3/3 (Integrate A + Bx + Cx^2 from 0 to x )This is the volume at any depth.MODE 5 4Input:[ C/3 B/2 A - 61pi ] = Result: x = 2.671

130. A set has 5 items and it has a range of 7. The set is composed of the following : { 1, 2, m, 5, m2 } with m > 0. Find the average number in the set. PAST CE EXAMSolution:Range = highest – lowest = m^2-1=7m=√8To get the average:MODE 3 1Input:

12√858ENTER: AC SHIFT 1 4 2 = 3.766

Binomial Expansion CAL TECH from my book AMAZING CALCULATOR TECHNIQUES Expand ( 2x - 3y)^7

179. A boat makes 25 mph in still water. It is headed N 45deg E. Find the direction of the course of the boat. The speed of the current is 7.5 mph PAST CE EXAM 

Sketch Below

Direction of the boat:MODE 2 Arg ( 25∟45 + 7.5∟0 ) SHIFT 2 1 is the Argument Result: 35.0730 ( Angle with the x axis ) Course Bearing N 90- 35.073 E = N 54.930 E

160. In a certain triangle ABC, A = 95deg, B = 50deg and C = 35deg. Which of the following expression correctly defines the lengths of the sides of a triangle?1. AB < BC < CA PAST CE EXAM2. AC < BC < AB3. AB < AC < BC4. BC < AC < AB See Figure belowAssume AB = 1 MODE 5 1 Input cos 95 cos 50 1 sin 95 -sin 50 0AC = 1.34 and BC = 1.74 1 < 1.34 < 1.74

AB < AC < BC Answer.

111. One of the diameters of a frustum of a sphere is 9.8 cm while the other is 10.4 cm. If the thickness of the frustum of a sphere is 4.2 cm, find the following:a. volume of the frustum b. radius of the sphere c. area of the zone formed 

a = 10.4 b = 9.8 and h = 4.2CAL TECH:Input:X=(A^2-B^2-Y^2)/2y:M=√(X^2+A^2 ):πY/6 (Y^2+3A^2+3B^2 ):2πYMSee Figure belowIn this CAL TECH ,X = distance of base with radius a from the center, A = radius of circle with base a , B = radius of the circle with base b, Y is the height πY/6 (Y^2+3A^2+3B^2 ) is the Volume 

and 2πM is the curved area.

CALC A? 10.4 /2 B? 9.8/2 Y? 4.2 Result:X = -487/280 M = 5.4831 ( Radius )πY/6 (Y^2+3A^2+3B^2 )=375.587 Volume2πYM=144.69 Curved Area

84. A man moves from A to B at an average speed of 4 kph. Had he moved at the rate of 3 2/3 kph, he would have taken 3 more hours to reach his destination. Find the distance between A and B. Solution:See Figure 

Distance = rate x time y = 4x (1)y = 11/3 ( x + 3) (2) Arrange:-4x + y = 0-11/3 x + y = 11 Use MODE 5 1x = 33 kph y = 132 km

33. A rubber ball is dropped from a height of 27 m. Each time that it hits the ground it bounces to a height of 2/3 of that from which it fell. Find the total distance traveled by the ball until it comes to rest.Solution:See figure below

Total distance =27+2(18+12+⋯..)27+ 18/(1-2/3)=81 m.Note: For an infinite geometric progression with |r| < 1 , S∞=a1/ ( 1- r)

32. An arch 18 m high has the form of a parabola with vertical axis. The length of a horizontal beam placed across the arch 8 m from the top is 64 m. Find the width of the arch at the bottom.PAST CE EXAMSolution:See figure belowModel the parabola using MODE 3 3.Input:X Y0 6432 10-32 10 To find the width of the arc at the bottom:ENTER: 0 SHIFT 1 5 5 = ( then x 2 ) =

Result: 69.674 m

95. A hemispherical tank having a top diameter of 40 ft is filled with oil having a density of 52.4 lb/ft3. Find the work done in pumping all the water to the top of the tank.Solution:

Use Tolentino’s Prismatoid CAL TECHSee Figure below

MODE 3 3Input Depth Area X Y0 020 π(20)240 0SHIFT 1 5 1 = A = 0

SHIFT 1 5 2 = B = 125.664SHIFT 1 5 3 = C = - 3.142 Work to pump all the water to the top =52.4 (0∫20 (A+Bx+Cx^2 )(20-x)dx = 6,584,778.2 ft lbs

101. Given the curves y = sin x and y = cos x a. Find all possible intersections.b. Find the area bounded by the curves from x = π/4 to 5π/4. Solution:a. y = sin x , y = cos x sin x = cos x tan x = 1 period of tan x = π and tan x is positive in the 1st and 3rd quadrant.then x = π/4 + k(π) and x = 5π/4 + k(π) k = 0, ±1,±2…See Figure below:

b. A=∫(π/4) to (5π/4) (sin x - cos x ) dx= 2.83

117. Two measurements for angle of elevations were made on the top of an inaccessible cliff, one from point A and another from point B. Point B is 425 nearer to the cliff than A but 25.6 m lower in elevation. If the angles of elevation from A and B are 18.650 and 36.20 respectively, find the elevation of the top of the cliff if A has an elevation of 625.4 m. PAST CE BOARDSolution:See Figure below Equation of A: y = x tan 18.65 Equation of B: y=-25.6+(x-425) tan 36.2 Get the intersection:Then: x tan 18.650 = -25.6+(x-425) tan 36.2 SHIFT CALC x = 853.625 mAnd y = x tan 18.650 = 288.106 Elevation of the top of the cliff is 288.106 + 625.4 = 913.506 m