# Mathematics

September 6, 2017 | Author: Johneil Asi | Category: Logarithm, Rational Number, Polynomial, Equations, Zero Of A Function

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Laws of Exponents: 1. a m a n = a m +n

m

2. 3.

a = a m-n n a

(a )

m n

Generally, an equation is said to be of quadratic form if it has the form ax2n + bxn + c = 0, where n is an integer or a fraction; such as x4 – 5x2 + 6 = 0 and y-3 + y-3/2 + 6 = 0

= a mn

4.

( ab ) n

5.

an a   = n b b 

= a nb n

X=

n

The expression b2 – 4ac is called the discriminant 1. When b2 – 4ac > 0, the roots are real and unequal. 2. When b2 – 4ac = 0, the roots are real and equal (or quadratic equation is a perfect trinomial) 3. When b2 – 4ac < 0, the roots are imaginary and unequal (complex conjugates)

Laws of Radicals: 1. n a n = a 2.

n

ab = n a n b

3.

n

a na = b nb

4.

m n

The roots:

a = mn a

X1 =

Laws of Logarithms: 1. log b MN = logb M + logb N 2. 3.

M = logbM - log bN N logbMN =NlogbM

logb

Important Properties: 1: a 0 = 1 provided a ≠ 0 2: a

-n

=

1 an

m

( a)

m

3:

a n = n am =

4:

a logab = b

5:

a m = a n implies that m = n log M =N implies that M = b N b

6:

n

or

7:

log M =logbN b

8:

log aN =

9:

log b =1 b log 1 =0 b

10:

− b ± b 2 − 4ac 2a

e lnb = b

implies that M = N

logbN logb a provided b > 0, b ≠ 1 provided b > 0, b ≠ 1

Sum of the roots,

− b + b 2 − 4ac 2a

X2 =

− b − b 2 − 4ac 2a

x1 + x2 = - b/a

Product of the roots, x1 ∙ x2 = c/a THE BINOMIAL THEOREM EXPANSION BINOMIAL (x + y)0 = 1 (x ≠ -y) (x + y)1 = (x + y)2 = (x + y)3 = (x + y)4 = (x + y)5 = (x + y)6 =

1 x+y x2 + 2xy + y2 3 x + 3x2y + 3xy2 + y3 4 x + 4x3y + 6x2y2 +4xy3 + y4 5 x + 5x4y + 10x3y2 +10x2y3 + 5xy4 + y5 6 x + 6x5y + 15x4y2 +20x3y3 + 15x2y4 + 6xy5 + y6

PASCAL’S TRIANGLE 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1 1 6 15 20 15 1

This array of numbers is called the Pascal’s Triangle. Any lower row is formed by adding any two adjacent numbers of the upper row and place 1 at both ends so as to form a triangle. Pascal’s Triangle is used to easily recall the numerical coefficients of the expansion of the powers of a binomial. But for large powers of a binomial, Pascal’s Triangle becomes inconveniently to use. For such, use Binomial Theorem. The rth term of (x + y)n =

n(n-1)(n-2)…(n-r+2) (r – 1)!

xn-r+1yr-1

LOWEST COMMON MULTIPLE (LCM) The lowest common multiple (LCM) of several natural numbers is the smallest natural number of which each of the given numbers is a factor. It mat be found by taking the product of all the different prime factors of the numbers, each taken the greatest number of times that it occurs in any of those numbers. Example: Find the lowest common multiple of 24, 10, 18, and 25. Solution: 24 = 2x2x2x3, 10 = 2x5, 18 = 2x3x3, 25 = 5x5 LCM = 2x2x2x3x3x5x5 = 1800 HIGHEST COMMON FACTOR (HCF) The highest common factor (HCF) of several natural numbers is the largest natural number which is a factor of all the given numbers. It may be found by taking the product of all the different prime factors common to the given numbers, each taken the smallest number of times that it occurs in any of those numbers. If the given numbers have no prime factors in common, the HCF is defined to be 1, in this case the numbers are said to be relatively prime. Example: Find the highest common factor of 24, 30, 18 and 150. Solution: 24 = 2x2x2x3, 30 = 2x3x5, 18 = 2x3x3, 150 = 2x3x5x5 HCF = 2x3 = 6 PROGRESSION Arithmetic Progression (A. P.) - a sequence of terms in which each term after the first is obtained by adding a fixed number to the preceding term. - a sequence of terms in which any two consecutive terms has a common difference. That is, the sequence a1, a2, a3 are in arithmetic progression if and only if: a2 – a1 = a3 – a2 Let: a1 = first term of an A. P. an = nth term of an A. P. d = common difference n = number of terms Sn = the sum of n terms Then, an = a1 + (n – 1)d Sn = n/2 (a1 + an) Sn = n/2 [ 2a1 + (n – 1)d]

Arithmetic Mean The arithmetic mean between two numbers is the number which when placed between the two numbers, forms with them an arithmetic progression. In general, for n terms, arithmetic mean (AM) = a1 + a2 + a3 + … + an n Geometric Progression (G.P.) - a sequence of terms in which each term after the first is found by multiplying the preceding term by a fixed number called common ratio. - The sequence a1, a2, a3 are in G.P. if and only if: a2/a1 = a3/a2 = r The nth term, an an = a1rn-1 Sum of the first n terms in G.P. Sn = a1 1-rn 1-r where a1 = first term r = common ratio n = number of terms Infinite Geometric Progression The sum of terms in geometric progression can be found if the common ratio | r | 0 c. b2 – 4ac < 0 2 b. b – 4ac = 0 d. b2 – 4ac < 0

25.

The sum of the integers between 288 and 887 that are exactly divisible by 15 is a. 23,700 b. 22,815 c. 21,800 d. 24,150

26. 27. 31. 32.

The sum of the prime numbers between 1 and 15 a. 42 b. 41 c. 39 What is the sum to infinity of the sequence 1 + 1/3 + 1/9 + … a. 2/5 b. 5/6 c. 2/3 The term free of y in the expansion of is a. 46 b. 84

39.

If f(x) = x + 2 and g(y) = y + 2, then f[g(2)] equals x–2 a. 6 b. 5 c. 4

the

In the quadratic equation Ax2 + Bx + C = 0, the product of the root is: a. C/A b. –B/A c. –C/A d. B/A

40. Two students were solving a problem that would reduce it to a quadratic equation. The first student committed an error in the constant term and found the roots to be 5 and 7 while the second student made an error in the first degree term and gave the roots as 2 and 16. if you were to check their solutions, the right equation is: a. x2 + 12x + 35 = 0 b. x2 + 18x + 32 = 0 2 c. x + 7x – 14 = 0 d. x2 – 12x + 32 = 0 41. Determine the value of k so that the equation x2 + (k-5)x + k – 2 = 0 is a perfect trinomial square. Ans. 3 or 11

d. 38 d. 3/2 d. 49

d. 3

33. If x3 + 3x2 + (K + 5)x + 2 – K is divided by x + 1 and the remainder is 3, then the value of K is a. -2 b. -4 c. -3 d. -5 34.

37. How many terms in the progression 3, 5, 7, 9, … must be taken in order that sum is 2,600? a. 53 b. 52 c. 51 d. 50 38. The other form of logaN = b is a. N = ab b. N = ba c. N = a/b d. N = ab

42. The expression x4 + ax3 + 5x2 + bx + 6 when divided by (x – 2) leaves the remainder 16, and when divided by (x – 1) leaves the remainder 10. Find the values of a and b. Ans. a = - 11/3, b = 5/3 43.

c. 47

If 3x = 9y and 27y = 81z, then is equal to a. 3/7 b. 3/5

The value of K which will make 4x2 – 4kx + 5k a perfect square trinomial is a. 6 b. 5 c. 4 d.3

Given the equation x4 + x2 + 1 = 0. Which of the following is not a root? a. 1 /120° b. 1 /135° c. 1 /240° d. 1 /300°

44. The area of a square field exceeds another square by 56 square meters. The perimeter of the larger field exceeds one half of the smaller by 26 meters. What are the sides of each field? Ans. larger field, 9m or 25/3m; smaller field, 5m or 11/3m 45. The sum of the areas of two unequal square lots is 5,200 square meters. If the lots were adjacent to each other, they would require 320 meters of fence to enclose the combined area formed by them. Find the dimensions of each lot. Ans. 60m and 40m or 68m and 24m BINOMIAL EXPANSION

1. 1.

2. 3.

Solve the following equations: A) Find the value of x: (a + b)x = (a2 + 2ab + b2)x-1 B) Find the roots of the equation 4x4 + 1 = 0

2. Maria is 36 years old, Maria was twice as old as Anna was when Maria was as old as Anna is now. How old is Anna now? Ans. 24 years old

Without expanding, find the term involving x4 of (3x2 – 2x-1)8. Ans. 90,720x4 3. The sum of the ages of the father and his son is 99. If the age of the son is added to the inverted age of the father, the sum is 72. If the inverted age of the son is subtracted from the age of the father, the difference is 22. What are their ages? Ans. 74 & 25

A) Expand to 4 terms (x2/3 – ½)x13 B) Find the 9th term in the expression of (x2 + ½)13 C) Write the first four terms and the last term of the expansion of (3/x – x/3)65 D) Find the term independent of y in the expansion of (y 2 – y-1)9. Ans. 84

4. Maria is 24 years old now. Maria was twice as old as Ana when Maria Anna is now. How old is Anna now? Ans. 16 years old 5.

4.

Which of the following has no middle term? a. (x + y)3 b. (a – b)4

c. (u + v)6

5.

Find the middle term of (x2 – 2y)10 Ans. -8,064x10y5

6.

If the middle term in the expansion of (x + 2y) n is kx4y4, find k and n. Ans. n = 8, k = 1,120

7.

A father is twice older than his son and the sum of their ages is 48. How old is each? a. 8, 40 b. 12, 36 c. 16, 32 d. none of these

was as old as

A father is twice as old as his son and the sum of their ages is 48. How old is each? a. 8, 40 b. 12, 36 c. 16, 32 d. nota

d. (x – y)8 6. Pedro is as old as Juan was when Juan is twice as old as Pedro was. When Pedro will be as old as Juan is now, the difference between their ages is 6 years. Find the age of each now. Ans. Juan, 24 years old and Pedro, 18 years old 7. I am three times as old as you were when I was as old as you are now. When you got to be my age together our ages will be 84. How old are we now? a. 24 & 36 b. 24 & 8 c. 16 & 8 d. 18 & 27

If the rth term of (x2 – 2y3)n is Cx8y12, find the value of C. Ans. 1120

8.

Without expanding, find the 10th term of the expansion of (S – 2t2)14

9.

In the expansion of (x2 + 1/x)12 find: a) the 6th term b) the middle term c) the term involving x6 d) the term free of x

10.

Find the sixth term in the expansion of (x/2 + y) 9 Ans. 63/8x4y5

11.

Find the term containing x26 in the expansion of (x-2 + x3)12 Ans. 66x26

12.

The term containing x9 in the expansion of (x3 + 1/x)15

13.

Find the coefficient of the expansion of (x2 + y)10 containing x10y5 a. 149 b. 252 c. 105

AGE PROBLEMS

8. The sum of the ages of two boys is four times the sum of the ages of a certain number of girls. Four years ago, the sum of the ages of the girls was one eleventh of the sum of the ages of the boys and eight years hence, the sum of the ages of the girls will be one half that of the boys. How many girls are there? Ans. 4 girls 9. In a family, there are 8 children, two of them are twins. The youngest is 3 years old and the eldest is 21 years old. Their ages are in arithmetic progression. There are three children younger than the twins. How old are the twins? Ans. 12 years 10. The sum of the ages of two men equals 99. If the inverted age of the elder is added to the age of the younger, the sum is 108. However, if the age of the younger is inverted and subtracted from the age of the older, the difference is 44. Find the age of the older man. a. 67 b. 32 c. 53 d. 46 INTEGER AND DIGIT PROBLEMS d. 10,818

1. Separate 132 into 2 parts such that the larger divided by the smaller the quotient is 6 and the remainder is 13. What are the parts? Ans. 17 and 115

2. A number of two digits divided by the sum of the digits the quotient is 7 and the remainder is 6. If the digits of the number are interchanged, the resulting number exceeds three times the sum of the digits by 5. What is the number? Ans. 83 3. Six times the middle of a three digit number is the sum of the other two. If the number is divided by the sum of the digits, the answer is 51 and the remainder is 11. If the digits are reversed, the number becomes smaller by 198. Find the number. Ans. 725 4. Find the number such that their sum multiplied by the sum of their squares is 65, and their difference multiplied by the difference of their squares is 5. Ans. 2 and 3 5.

Three numbers are in the ratio 2:5:8. If their sum is 60, find the numbers. Ans. 8, 20, 32

2. How much silver and how much copper must be added to 20kg of an alloy containing 10% silver and 25% copper to obtain an alloy containing 36% silver and 38% copper? a. 14kg, 12kg b. 16kg,14kg c. 12kg,10kg d. 16kg, 18kg 3. A tank full of alcohol is emptied of one third of its content and then filled up with water and mixed. If this is done six times, what fraction of the volume (original) of alcohol remains? Ans.64/729 4. How much tin and how much iron must be added to 50 kilograms of an alloy containing 10 percent tin and 25 percent iron to obtain an alloy containing 25 percent tin and 50 percent iron? Ans. 27.5 kg(tin), 52.5 kg (iron)

6. The sum of the digits of a three-digit number is twelve. The sum of the squares of the hundreds’ digit and the tens’ digit is equal to the square of the units’ digits. If the hundreds’ digit is increased by two, the digits will be reversed. Find the number. Ans. 345

5. How many liters of water must be added to 45 liters of solution which is 90% alcohol in order to make the resulting solution 80% alcohol? Ans. 5.63L

7. April 1978. The square of a number increased by 16 is the same as 10 times number. Find the number. Ans. 8, 2

the

6. A 40-gram solution of acid and water is 20% acid by weight. How much pure acid must be added to this solution to make it 30% acid? Ans. 5.71 grams

8. The sum of the digits of a three-digit number is 12. The middle digit is equal to the sum of the other two digits and the number shall be increased by 198 if its digits are reversed. Find the number. Ans. 264

7. How much water must be evaporated form 80 liters of 12% solution of salt in order to obtain a 20% solution of salt? Ans. 32 L 39. How many liters of water must be added to 100 liters of 85% sulfuric acid solution to produce 60% sulfuric acid solution? Ans. 41.67 L

9. Find three consecutive odd integers such that twice the sum of the first and the second integers plus four times the third is equal to 60. Ans. 5, 7, 9

8. A certain solution should contain 8% alcohol. If it was mistakenly mixed to contain 6% alcohol, how much must be drawn from a 5-liter tank and replaced by 10 percent alcohol solution to provide the proper concentration? Ans. 2.5 L

10. The sum of the digits of a three-digit number is 12. The sum of the squares of the hundreds digit and the tens digit is equal to the square of the units digit. If the hundreds digit is increased by 2 and the units digit is decreases by 2, the digits of the original number will be reversed. Find the number. Ans. 345 11. The excess of the sum of the fifth and the seventh parts over the difference of the half and third parts of number is 259. What is the number? Ans. 1470 12. The sum of the digits of the three-digit number is 6. The middle digit is equal to the sum of the two other digits and the number shall be increased by 99 if the digits are reversed. Find the number. Ans. 132

9. A certain amount of 80% sugar solution added to another amount of 40% sugar solution yields a solution that contains 14 kg of sugar. Had the amount been reversed, the solution would have contained 16 kg sugar. How much of the 80% solution was there? Ans. 10 kg 10. Ten liters of 25% salt solution and 15 liter of 35% salt solution are poured into a drum originally containing 30 liters of 10 % salt solution. What is the percent concentration of salt in the mixture? a.19.55% b. 22.15% c. 27.05% d. 25.72% RATE AND MOTION PROBLEMS

MIXTURE PROBLEMS 1. The tank of a car contains 50 liters of alcogas 25% of which is pure alcohol. How much of the mixture must be drawn off which when replaced by pure alcohol will yield a 5050% alcogas? a. 16 2/3 b. 15 1/3 c. 14 d. 20

1. A motorist is traveling from town A to town B at 60 kph and returns from town B to town A at 30 kph. His average velocity for the roundtrip is A. 45 kph B. 40 kph C. 35 kph D. NOTA

2. At the recent Olympic games in Montreal, Canada, a team which participated in 1600 meters relay event had the following individual speed. First runner, 24 kph, second runner, 20 kph, third runner, 22 kph and fourth runner 23 kph. What was the team’s speed. Ans. 22.149 kph 3. A troop of soldiers marched 15 km, going to the concentration camp after they were forced to surrender, at the same time that the victorious general who is supervising the “march” rode from the rear of the troop to the front and back at once to the rear. If the distance covered by the victorious general is 25 km. and both the troop and general traveled at uniform rate, how long is the troop? Ans. 8 km. 4. Two cyclists are practicing on a circular tract of circumference 276 meter. Starting at the same instant and from the same place, when they run in opposite directions they pass each other every 6 seconds and when they run the same direction the faster passes the slower at every 23 seconds. Determine their rates. Ans. F= 29 m/s, S= 17m/s 5. Two cars A and B start at the same point and at the same time and travel in opposite directions, car B traveling 20 km/hr slower than A. If they are 420 kilometers apart after 3 hours, find the rate of each. Ans. 60 kph, 80 kph 6. Two cars A and B are to race around a 1,500-meter circular track. If they will start at the same point and travel opposite directions, they will meet for the first time in 3 minutes. But if they will travel in the same direction, with the same starting point, car A will reach the starting point with car B trailing behind by 500 meters. What should be the rates of each? Ans. 300 m/min, 200 m/min 7. A one kilometer long caravan of men is walking at a constant rate. A man from the rear ends walk towards the head and back to the rear at the instant when the caravan has covered a distance of one kilometer. Find the total distance traveled by the man. Ans. 2.414 km

per hour. The airplane can fly 280 kilometers per hour in still air. If the package carrier takes 3 2/3 hours in going from A to C and 3 1/6 hours for the return trip, what is the total distance of travel covered by the man? D = 605 km. t1 = 1.5 hr., t 2 = 5/3 hr. 12. A motorcycle messenger left the rear of a motorized troop 8 kilometers long and rode to the front of the troop, returning at once to the rear. How far did he ride, if the troop traveled 15 kilometers during this time and each traveled at a uniform rate? Ans. 25kms. 13. An army officer made the first part of the trip on a plane which flew at the rate of 210 kilometers per hour. At the landing field, he was met by a jeep which took him the rest of the way to his destination at a rate of 40 kilometers per hour. The trip required 3 hours and 15 minutes. On his return trip, the jeep traveled at the rate of 50 kilometers per hour and the plane which he took flew at the rate of 200 kilometers per hour. The return journey required the same amount of time, but this included a minute which he spent waiting for the plane to take off. Find the total distance that he flew and the total distance that he traveled by jeep. Ans. 532km (by plane) and 28 2/3km (by jeep) 14. A BMW car drives from A toward C at 30 miles/hr. Another car starting from B at the same time, drives towards A at 20 mi/hr. If AB = 20 miles, find when the cars will be nearest together. Ans. 24 min. 15. Two boats started their voyages is in a straight line towards each other. One has an average navigational speed of 30 km/h and the other one has an average of 20kph. Assuming that they can not avoid a collision, how long will it take before the collision occurs? How far would each boat have traveled before the collision? Ans. 4 hrs, 120 km, 80 km 16. A man traveling 40 km finds that by traveling one more km per hour, he would the journey in 2 hours less time. How many kilometer per hour did he actually travel? a. 4 b. 8 c. 18 d. 6

PROGRESSION 8. A boy started one hour and twenty minutes earlier than a man. If the man ran at 6 kph faster than the boy and overtook the boy in 40 minutes, find the rate of each. Ans. 3 kph for the boy and 9 kph for the man 9. A man walked 24 km in time T. During the first part of this time, he walked at 6 kph and the last part at 4 kph. Had he reversed his rates, he would have walked two km more. Find the time. Ans. 5 hrs. 10. A traffic check counted 390 cars passing a certain spot on one day and 430 cars at the same spot on the second day. On the first day, there were three times as many cars going east and half as many going west on the second day. What was the total number of west bound cars for the two days? Ans. 280 east, 540 west 11. A man is sent to deliver an important package ant travels by car 75 kilometers per hour from point A to B and then by airplane to point C against a wind blowing 40 kilometers

1. Two numbers differ by 40 and their arithmetic mean exceeds their positive geometric mean by 2. The numbers are a. 45, 85 b. 64, 104 c. 81, 121 d.100, 140 2. A sets out to walk at the rate of four km per hour. After he had been walking for 2-3/4 hours, B sets out to overtake and went 4-1/2 km the first hour, 4-3/4 km the second hr., 5 km the third hr and so on gaining 1 quarter of a km. every hour. In how many hours would B overtake A? Ans. 8 hours 3. A besieged fortress is held by 5700 men who have provisions for 66 days. If the garrison loses 20 men each day, for how many days can the provisions hold out? Ans. 76 days

4. The sum of three numbers in arithmetic progressions is 60. If the numbers are increased by 2, 1, and 28, respectively, the new numbers will be in geometric progression. Find the arithmetic progression. 5. Three numbers are in arithmetic progression. Their sum is 15, and the sum of squares is 83. Find the numbers.

7. A rubber ball is dropped from a height of 27 meters. Each time that it hits the ground it bounces to a height 2/3 of that from which it fell. Find the distance that it travels up to the time that it hits the ground for the 5 th time. The total distance traveled by the ball until it comes to rest. 8. A man wishes to buy a piece of land worth 150,000 pesos. If it were possible for him to save one centavo on the first day, two centavos on the second day, 4 centavos on the third day and so on, in how many days would he save to be able to buy the land? Ans. about 24 days 9. The sum of the two numbers is 20 and their positive geometric mean is one greater than one half of their arithmetic mean. Find their difference. 10. A man cuts a piece of paper 0.03 mm thick into three equal parts. Then he cuts each of these parts into three equal parts again and the process is repeated 10 times after which he piles together the pieces of paper. How thick is the pile? 11. A rich man called his seven sons. He had with him a number of pebbles, each pebble representing a gold bar. To his first son, he gave half the pebbles that he had and one pebble more. To his second son, he gave half the remaining pebbles and one pebble more. He did the same to each to his five other sons and then found out that he had one pebble left. How many pebbles were there initially? Ans. 382 12. A man receives a salary of P36,000 per annum for the first year and a 10% raise every year for ten years. What is his salary during the fifth year? Ans. P52,707.60 13. A car running at 25 kilometers per hour can cover a certain distance in 8 hours. By how many kilometers per hour must its rate be increased in order to cover the same distance in three hours less? Ans. 15km/hr Find the harmonic mean 7, 1, 5, 2, 6 and 3 a. 2.36 b. 2.46

c. 2.56

d.2.66

15.

The 8th term of an AP is 3 while its 84th term is 273. Find the 35th term.

16.

Find the sum to infinity of 1/3, 1/27,1/243… Ans. 3/8

In the series 1.01, 1.0, .099, .098… Find the 80th term.

18.

Find the sum of 3+0.4+0.05+0.004+0.0005+… Ans. 38/11

their

6. A 20-liter container is filled with pure acid. Five liters are drawn off and replaced with water; then 5 liters of the mixture drawn off and replaced with water, and so on until 5 drawings and 5 replacements have been made. Find the amount of acid in the final mixture.

14.

17.

19. An arithmetic progression starts with 1, has 9 terms, and the middle term is 21. Determine the sum of the first 9 terms. 20. A pendulum swings 24 inches for the first time. It is swinging 11/12 of its previous swing. What would be the total distance traveled when the pendulum stopped? a. 246 b.264 c.288 d. 312 21. A small line truck hauls poles from substation stockyard to pole sites along a proposed distribution line. The truck can handle only one pole at a time. The first pole site is 150 meters from the substation and the poles are to be 50 meters apart. Determine the total distance traveled by the line truck, back and forth, after returning from delivering the 30th pole. a. 35.0km b. 30.0km c. 37.5km d. 40.0km 22. Two positive numbers may be inserted between 3 and 9 such that the first three are in geometric progression, while the last three are in arithmetic progression. What is the sum of these two positive integers? a.1.25 b. 12.25 c. 11.25 d.6.25 23. A man piles 150 logs in layers so that the top layer contains 3 logs and each lower layer has one more log than the layer above. How many logs are at the bottom? Ans. 17 logs 24. A body falls 16.1 meters during the first second, 48.3 meters during the second, 80.5 meters during the third second and so on. How long will it take the body to reach the ground if it was released at an altitude of 15,000 meters? Ans. 30.5 seconds 25. The 18th and the 52nd terms of an arithmetic progression are 3 and 173, respectively. The 25th term is a. 38 b. 35 c. 28 d. 25 25. Find the number of terms of a geometric progression in which the first term last term is 384 and the sum of the terms is 720. Ans. 4 terms

is 48, the

26. The sum of three numbers in A.P. is 27. If the first number is increased by 2, the second by 7, and the third by 20, the resulting numbers will be in G.P. Find the original numbers. a. 3, 9, 15 b. 4, 9, 14 c. 5, 9, 13 d. 6, 9, 12 WORK AND DISCHARGE PROBLEMS

1. If 4 men can plow 12 hectares in 8 hours, how many men are needed to plow 24 hectares in 24 hours? Ans. 6 men 2. A garden can be cultivated by 8 boys in 5 days. The same job can be done 5 men in 6 days. How long will it take to finish the job if A) the 8 boys and 5 men will work together? B) only 6 boys ands 3 men will work together? C) two days after the 5 men were working the 8 boys arrived to help? 3. A, B and C can do a piece of work in 10 days, A and B can do it in 12 days, A and C in 20 days. How many days would it take each to do the work alone? Ans. 30, 20 ,60 4. A boat’s crew rowing at half their usual rate can negotiate 2km. down a river and back in one hour and 40 minutes. At their usual rate in still water, they would have gone over the same course in 40 min. Find their rate of rowing in still water. Ans. 32/5 km/hr 5. Two pipes running simultaneously can fill a tank in 3 hours and 20 minutes. If both pipes run for 2 hours and the first is then shut off, it requires 2 hours more for the second to fill the tank. How long does it take each pipe to fill it alone? 6. A and B can do a job in 12 days. A and C can do the same job in 18 days while B and C can do it in 24 days. How will it take A, B and C to do the job together? 7. A man can finished a certain job in three-fourths the time that the boy can; the boy can finish the same job in two-thirds the time that a girl can; and the man and the girl working only jointly can finish the job in 4 hours. How long will it take to finish the job if they all work together? Ans. 8/3 hr.

12. A steel company has three blast furnaces of varying sizes. If furnaces A, B, C are used full time, 800 metric tons of steel are produced per day. If A and B are used half time and C full time, 545 metric tons are produced. If A is not used, B is used full time, and C half dime, 410 metric tons are produced. How many metric tons per day does each furnace produce? 13. Three observation planes A, B, and C, working together, can map the region in 4 hours. Planes A and B can map the region in 6 hours, planes B and C can map it in 6 hours and 40 minutes. How long would it take each of the planes working alone to map this region? 14. A pump discharging 9 gpm requires 36 hours to fill a tank. If the pump is replaced by one that will discharge 16 gpm, how long will it take to fill the tank? a. 64 hr b. 16 hr. c. 20.25 hr d. 40.5 hr 15. Two pipes running simultaneously can fill a swimming pool in 6 hours. If both pipes run for 3 hours and the first pipe is then shut off, it requires 4 hours more for the second to fill the pool. How long does it take each pipe running separately to fill the pool? Ans. 8 & 24 16. A man and a boy can dig a trench in 20 days. It would take the boy 9 days longer dig it alone than it would take the man. How long would it take the boy to dig alone? a. 45 days b. 16 days c.25 days d. 4 days

to

17. A job can be done in as many days as there are men in the group. If the number of men is reduced by 3, the job will be delayed by 4 days. How many men are there originally in the group? a. 6 b. 12 c. 20 d. 30

10. A swimming pool holds 54 cubic meters of water. It can be drained at a rate of one cubic meter per minute faster than it can be filled. If it takes 9 minutes longer to fill it than to drain it, find the drainage rate. Ans. 3 cu.m/min

VENN DIAGRAM 1. A certain part can be defective because it has one or more out of three possible defects; insufficient tensile strength a burr or diameter outside of tolerance limits. In a lot of 500 pcs: 19 have a tensile strength defect 17 have a burr 11 have an unacceptable diameter 12 have tensile strength and burr defects 7 have tensile strength and diameter defects 5 have burr and diameter defects 2 have all three defects a. how many have four defects b. how many pcs have only a burr defect c. how many pcs have exactly two defects. Ans. 475, 2, 18

11. One input pipe can fill a tank alone in 8 hrs. another input pipe can fill it alone in 6 hours and a drain pipe can empty the full tank in 10 hours. If the tank is empty and all the pipes are wide open, how long will it take to fill the tank? Ans.5.22 hours

2. During the election, the total number of votes recorded in a certain municipality was 12,400 had 2/5 of the supporters of LABAN candidate stampede away from the pools and ½ of the supporters of GAD candidate behaved likewise, the LABAN candidates majority would

8. The intake pipe to a reservoir is controlled by a valve which automatically closes when the reservoir is full and opens again when four-fifths of the water had been drained off. The intake pipe can fill the reservoir in 4 hours and the outlet pipe can drain it in 10 hours. If the outlet pipe remains open, how much time elapses between the two instants that the reservoir is full? Ans. 13.3 hr. 9. Two brothers washed the family car in 24 minutes. Previously, when each had washed the car alone, the younger boy took 20 min. longer to do the job than the older boy. How long did it take the older boy to wash the car alone? Ans. 40 minutes

have been reduced by 100. How many votes did the LABAN & GAD candidates actually received? Ans. 7,000, 5,400 3. The President just recently appointed 25 Generals of the Phil. Army of these 14 have already served in the war of Korea, 12 in the war of Vietnam and 10 in the war of Japan. Therefore 4 who have served both in Korea and Japan, 6 have served both in Vietnam and Korea and 3 have served in Japan, Korea, and Vietnam. Ans. 2 generals 4.

A survey of 500 T.V. viewers proceed the following result: 285 watch football games 195 watch hockey games 115 watch basketball games 45 watch football and basketball 70 watch football and hockey 50 watch hockey and basketball 50 do not watch any of the 3 games, How many watch the basketball games only? a.50 b.40 c.30

8. A student left his home to attend a party one morning at past 6 o’clock and returned at past 3 o’clock. He noticed the hands of the wall clock have exchanged position. What exact time did he arrive? a. 3:26.07 b. 3:34.62 c. 3:31.47 d. 3:32.19 9. How many times in one complete day will the hour and the minute hands with each other? a. 24 b. 23 c. 22 d. 25

coincide

RATIO, PROPORTION AND VARIATION 1. The kilowatt that can be transmitted safely by a shaft varies directly as the number of revolution it makes per minute and the cue of its diameter. If a shaft 3 centimeters in diameter making 200 revolutions per minute can safely transmit 60 kilowatt, what kilowatt can be safely transmitted by a 2-centimeter shaft making 300 revolutions per minute? d.60

2. The time required for an elevator to lift a weight varies directly with the weight and distance through which it is lifted and inversely as the power of the motor. If it takes 30 seconds for a 10 HP motor to lift 100 lbs through a height of 50 ft, what size of motor is required to lift 800 lbs in 40 sec through the height of 40 ft? Ans. 48 HP

CLOCK PROBLEMS 1.

At what time between 7 and 8 o’clock are the hands of the clock are A) at right angles B) straight line C) coinciding

2. The time is 3:00 o’clock and the hands of the clock are at right angles to each other. What is the nearest time of the clock such that the hands of it will be at right angles again? Ans. 32.72 minutes 3. How long will it be from the time the hour hand and the minute hand of a clock are together until they will be together again? Ans. 1 hr. and 5.45 min 4. At what time between 4 and 5 o’clock do the hands of the clock coincide? Ans. 4:21.82 o’clock 5. It is exactly 3 o’clock. In how many seconds will the angle formed by the hour hand and the minute hand be twice the angle formed by the hour and the second hand? Ans. 22.4 seconds 6. It is now between 9 and 10 o’clock. In 4 minutes, the hour hand will be exactly opposite the position occupied by the minute hand 3 minutes ago. What is the time now? Ans. 9:20 7. How many minutes after 2:00 o’clock will the hands of the clock extend in the opposite directions for the first time? a. 40.636 b. 41.636 c. 43.636 d. 42.636

3. Eight men can excavate 15m3 of drainage open canal in 7 hrs. Three men can backfill 10m3 in 4 hrs. How long will it take 10 men to excavate and back fill 20 m 3 in the project? Ans. 9.87 hrs. 4. A man is sent to deliver an important package and travels by car 75 kilometers per hour from point A to B and then by airplane to point C against a wind blowing 40 kilometers per hour in still air. If the package carrier takes 3 2/3 hours in going from A to C and 3 1/6 hours for the return trip, what is the total distance of ravel covered by the man? 5. A sphere 30 cm in diameter is divided into two segments. One of which is two times as high as the other. Find the volume of the bigger segment. 6. Two flywheels are connected by a belt. The radius of the flywheels are 30 in and 50 in. The small flywheel has a speed of 350 rpm. Determine the velocity of the belt in ft/sec. What would be the angular velocity of the larger flywheel? 7. A cylindrical tin can has its height equal to the diameter of its base. Another cylindrical tin can with the same capacity has its height equal to twice the diameter of its base. Find the ratio of the amount of tin required for making the two cans with covers. Ans. 0.9524 8. The diameters of two spheres are in the ratio 2:3 and the sum of their 1,260 cubic meters. Find the volume of the larger sphere. Ans. 972 cu. M

volumes

is

9. If the square root of x varies directly as y and inversely as the square of z and if x = 16 when y = 24 and z = 2, find z when x = 9 and y =2. Ans. 2/3 10.

If a:b = 2:3, b:c= 4:5, what is a:b:c? a. 2:3/4:5 b. 8:12:16

c. 8:12:16

d. 6:9:12

ANALYTIC GEOMETRY Distance Between Two Points P1 (x1,y1) and P2 (x2,y2) y P2(x2,y2) d

o

x

d = √ (x2 – x1)2 + (y2 – y1)2 Area of Polygon (Non-overlapping) of n-sides Given Vertices Given vertices (x1, y1), (x2, y2), ……… (xn, yn) oriented counterclockwise x2

y2

Angle Between Two Concurrent Lines Let α and β be the inclinations of lines L1 and L2 respectively and let θ be the angle between the two lines y L1 θ = β - α ;m1 = tan α, m2 = tan β L2 tan θ = tan (β - α) θ tan θ = tan β - tan α__ 1 + tan β tan α tan θ = m2 – m1 α β 1 + m1m2

Division of Line Segment Let P(x, y) be a point on the line joining P 1(x1, y1) and P2(x2, y2) and located in such a way that segment P1P is a given fraction k of P1P2, that is P1P = kP1P2. y P2(x2,y2) P(x,y)

2. Find the locus of points P(x, y) such that the distance from P to (3, 0) is twice its distance to (1, 0). Ans. 3x2 – 3y2 – 2x – 5 = 0 3. Find the length of the segment joining the two midpoints of the sides of the triangle if the length of the third side opposite to it is 30 cm. Ans. 15 cm.

5. Two vertices of a triangle are (0, -8) and (6, 0). If the medians intersect at (9, -3), find the third vertex of the triangle. Ans. (-3, -1)

y3 …………y1

+ + + A = ½ [ (x1y2 + x2y3 + x3y4 ……. + xny1) – (y1x2 + y2x3 + y3x4 ……. + ynx1)]

x = x1 + k (x2 – x1) y = y1 + k (y2 – y1)

P1(x1,y1) o

y0 = ½ (y1 + y0)

4. A line from P(1, 4) to Q(4, -1) is extended to a point R so that PR = 4PQ. Find the coordinate of R. Ans. R(13, -16)

x3 …………x1

A= y1

;

SUPPLEMENTARY PROBLEMS 1. A point P(x, 3) is equidistant from points A(1, 5) and B(-1, 2). Find x. Ans. ¾

P1(x1,y1)

x1

x0 = ½ (x1 + x2)

x

If k = ½, then formula above becomes a midpoint formula

6. The area of a triangle with vertices (6, 2), (x, 4) and (0, -4) is 26. Find x.Ans. – 2/3 and 50/3 7. Find the length of the median from A of a triangle ABC given vertices A(1, 6), B(-1, 3) and C(3, -3). Ans. 6 8. If the midpoint of a segment is (5, 2) and one endpoint is (7, -3), what are the coordinates of the other end? Ans. (3, 7) 9. Given vertices of a triangle ABC : A(1, 5),B(-1, 1) and C(6, 3). Find the intersection of the median. Ans.(2, 3) 10. Find the inclination of the line 2x + 5y = 10. Ans. 158.2° Locus point.

– the curve traced by an arbitrary point as it moves in a plane is called locus of a

– the locus of an equation is a curve containing only those points whose coordinates satisfy the equation.

y 4. Two- Intercept Form x +y =1

EQUATION OF A STRAIGHT LINE

P2(0, b)

Line – is a locus of points which has constant slope.

a

Theorems :

where a = x intercept

• •

b

P(x, y) b = y intercept

Every straight line can be represented by a first-degree equation. The locus of an equation of the first degree is always a straight line.

P1(a, 0)

x

General Equation of a Line Ax + By + C = 0

; A, B, C are constants

5. Normal Equation of a Straight Line

; A and B, not zero at the same time

y

Standard Equation of a Line 1. Two Point Form Y

P2(x2,y2)

P(x,y)

By similarity of triangles

P1(x1,y1)

x

x

(1)

y – y1 = y2 – y1 (x – x1)

Given ρ = normal intercept

x2 – x1

N(ρcosθ,ρsinθ) ρ

= segment from the origin y

θ

perpendicular to the required line θ = normal angle

2. Point-Slope Form In (1) replacing y2 – y1 by m,

P(x, y)

= inclination of the normal intercept

x2 – x1

From the point slope form :

θ

y – y1 = m (x – x1)

y – y1 = m (x – x1)

x

y - ρsin θ = (-1/ tan θ ) (x–ρcos θ) Simplifying, xcosθ + ysinθ = ρ

0° ≤ θ ≤ 180°

where m = slope

where x1 = ρcos θ, y1 = ρsin θ mL = -1 / tan θ

m = tan θ

3. Slope-Intercept Form y = mx + b

ρcosθ

y

Reduction to Normal Form : Given the line Ax + By + C = 0

(0, b)

The normal form is :

P(x,y)

A

b = y intercept

x

x

+

B

y +

C

=0

ρsinθ

±√ A2 + B2

±√ A2 + B2

±√ A2 + B2

Note : The sign of the radicand must be chosen such that the last term will become negative since ρ > 0.

y

A. Equation of the x – axis: Equation of a horizontal line :

y=0 y=b

B. Equation of the Y-axis : Equation of a vertical line :

x=0 x=a

where b is a constant

Distance Between Parallel Lines L1 : Ax + By + C1 = 0

L2

L2 : Ax + By + C2 = 0

Find the equations of the line/s satisfying the given conditions.

The distance between the two

1. Passing through (1, -2) and perpendicular to the line through (2, -1) and (-3, 2) Ans. 5x – 3y – 11 = 0

lines is given by the formula

2. With x intercept of 5 and passing through (3, 4) Ans. 2x + y – 10 = 0

d = C2 – C1_

3. Passing through (-3, 4) and with equal intercepts Ans. x – y + 7 = 0 and x + y – 1 = 0 Making an angle of 45° with the x-axis and passing through (2, 3) Ans. x – y – 1 = 0

5. With slope -12/5 crosses the first quadrant and forms with the axes a triangle with perimeter of 15. Ans. 5x + 12y – 3 = 0

Here, A = 3,

B = -4,

Using the formula

P0(x0, y0) ↔ (3, -1)

C = -2

d = Ax0 + By0 + C 2

=

3(3) + (-4)(-1) – 3 2

+√ 3 + 42

d = 2 units (the point (3, -1) and the origin are on the opposite side of the line)

A = 8, B = 15, C2 = 18, C1 = 1

Find the equations of the lines parallel to the line x + 2y – 5 = 0 and passing at a distance 2 from the origin Ans. x + 2y + 2√ 5 = 0 and x + 2y - 2√ 5 = 0

11. Given vertices of a triangle ABC, A(2, 0); B(3, -2) and C(7, 5)

1. Find the distance from point (3, -1) to the line 3x – 4y – 3 = 0 Solution :

2. Find the distance between parallel lines 8x + 15y + 18 = 0 and 8x + 15y + 1 = 0. Solution :

8. Find the value of parameter k so that the line 3x – 5ky + 5 = 0 a) will pass through (0, 1) b) will be parallel to x + 2y = 5 c) will be perpendicular to 4x + 3y = 2 d) has the y-intercept equal to 3 Ans. a) 1 b) –6/5 c) 4/5 d) 1/3

10. Find the equation of the perpendicular bisector of the segment joining (2, 5) and Ans. x – y + 1 = 0

Sample Problems:

2

7. Passing through the midpoint of the segment joining the points (1, 3) and (5, 1) and parallel to the line 2x – y + 5 = 0 Ans. 2x – 3y – 5

d

√ A2 + B 2

√A +B

6. Passing through (7, -4) and at a distance of 1 unit from the point (2, 1) Ans. 4x + 3y – 16 = 0 ; 3x + 4y – 5 = 0

9.

L1

Let the parallel lines be given by the equations :

where a is a constant

SUPPLEMENTARY PROBLEMS:

4.

c) (4, 1)

12. Find the normal intercept and the normal angle of line 5x+12y–39 = 0 Ans. ρ =3, θ = 67.38°

Special Cases of a Straight Line Α

a) find the equation of the median from A b) find the equation of the altitude from B c) the intersection of medians from B to C Ans. a) x – 2y – 2 = 0 b) x + y – 1 = 0

(4, 3).

D = C2 – C1 =

18 – 1

√ A2 + B 2

√ 82 + 152

= 1 unit

Distance from a Point to a Line The directed distance from a point P(x0,y0) to a line Ax + By + C = 0 is given by the formula: d = Ax0 + By0 + C ± √ A2 + B2

where the sign of the radical is chosen to be the opposite that of C. Remarks: 1. If d > 0, the origin and P lie on opposite sides of the given line. 2. If d < 0, the origin and P lie on the same side of the line. Notes: Regardless of the location of the point P0(x0, y0), the distance being always positive the formula can be expressed using the absolute value as: d =Ax0 + By0 + C √ A2 + B2

ASYMPTOTE - a straight line which the curve f(x, y) = 0 approaches indefinitely near as its tracing point approaches to infinity. • To find the vertical asymptote, solve the equation for y in terms of x and set the linear factors of the denominator equal to zero. • To find the horizontal asymptote, solve the equation for x in terms of y and set the linear factors of the denominator equal to zero. CIRCLE Circle is the locus of a point which moves so that it is always equidistant from a fixed point. Note: fixed point is called the center Fixed distance is called the radius

Line Through the Intersection of Two Lines Let Ax + By + C = 0 and Dx + Ey + F = 0 be two intersecting lines, where A, B, C, D, E and F are constants and A = B ≠ 0, E = F ≠ 0. The equation of the family of lines passing through the intersection of the two given lines is given by, (Ax + By + C) + k (Dx + Ey + F) = 0

Equation of a Circle In normal form Consider a circle of radius r with center at C(u, k) Let P(x, y) be a point in the circle y

where k is an arbitrary constant. r INTERCEPT OF A CURVE • x-intercept – directed distance from the origin to the point where the curve crosses the xaxis To find the x intercept of a curve, set y = 0, then solve for x. • y-intercept – the directed distance from the origin to the point where the curve crosses the y-axis to find the y- intercept of a curve, set x = 0, then solve for y. SYMMETRY • If the equation of a curve does not change upon replacement of y by –y, then the locus is symmetric with respect to the x-axis. f(x, -y) = f(x,y) =0 • If an equation of a curve does not change upon replacement of x by –x, then the locus is symmetric with respect to the y-axis f(-x,y) = f(x,y) = 0 • If an equation of a curve does not change upon replacement of x by –x and y by –y, then the locus is symmetric with respect to the origin. f(-x, -y) = f(x, y) = 0

C(h,k)

y–k

P(x,y) By Pythagorean Theorem (x – h)2 + (y – k)2 = r2 → standard form x–h

Center at the origin C(0, 0)

x 2 + y 2 = r2 x 0

General Form Expanding the form (x – h)2 + (y – k)2 = r2 becomes x2 + y2 – 2xh – 2ky + h2 + k2 – r2 = 0 This is of the form: x2 + y2 + Dx + Ey + F = 0 → general form where D, E, F are constants not all zero at a time. Note: By equation of coefficients: -2h = D ; h = -½ D → abscissa of center -2k = E ; k = -½ E → ordinate of center h2 + k2 – r2 = F ; r = √(h2 + k2 – F)

Radical Axis of Two Circles Consider the two non-concentric circles x2 + y2 +D1x + E1y + F1 = 0 x2 + y2 +D2x + E2y + F2 = 0 The equation: x2 + y2 +D1x + E1y + F1 + k (x2 + y2 +D2x + E2y + F2) = 0 represent a circle for any value of k except for k = -1

D. All tangents drawn to two circles from a point on their radical axis have equal lengths. y

Radical Axis P T1 & T2 are points of tangency

T1 if k = -1, the equation of the family of circles above becomes: (D1 – D2) x + (E1 – E2) y + (F1 – F2) = 0 This represents a straight line called the RADICAL AXIS of two circles.

T2 PT1 = PT2 x

Properties of the Radical Axis A. If two circles intersect at two distinct points, their radical axis is the common chord of the circles.

y

2. Find the area of the circle whose equation x2 + y2 = 6x – 8y (ECE Board Problem – Mar. 1981) Ans: 25π sq. units

Common Chord

x 0 Condition for Orthogonality 2 2 The two non-concentric circles : x + y + D1x + E1y + F1 = 0 x2 + y2 + D2x + E2y + F2 = 0, meet at right angles (orthogonal) if : D1D2 + E1E2 = 2(F1 + F2) B. If two circles are tangent, their radical axis is the common tangent to the circles at their point of tangency. Radical Axis

y

Supplementary Problems 1. Find the center and radius of the circle whose equation is x 2 + y2 – 4x –6y –12 = 0 (ECE Board Problem – Oct 1981) Ans: C(2, 3) r = 5

3. Find the equation of the circle whose center is at (3, -5) and whose radius is 4 units. Ans: (x – 3)2 + (y + 5)2 = 16 For Problems 4 – 9, determine the equation of the circle given the following conditions 4. Passes through the point (2, 3), (6, 1) and (4, -3) Ans: x2 + y2 – 10y = 0 5. Center on the y – axis, and passes through the origin and point (4, 2). Ans: x2 + y2 – 10y =0 6. Passes through the points of intersection of the circles x 2 + y2 = 5, x2 + y2–x + y = 4, and through the point (2, -3) Ans: x2 + y2 –2x + 2y –3 = 0 7. Center on the line x – 2y –9 = 0 and passes through the points (7, -2) and ( 5, 0) Ans: x2 + y2 – 10x + 4y +25 = 0

x

0

C. The radical axis of two circles is perpendicular to their line of centers.

y

9. Given the endpoints of the diameter (5, 2) (-1, 2) Ans: x2 + y2 – 4x – 4y – 1 = 0

Radical Axis AB – line of centers A

0

8. Circumscribe the triangle determine by the lines x – u – 8 = -y and y = -1. Ans: x2 + y2 –8x + 2y + 8 = 0

B

x

10. Find the equation of the line tangent to the circle x 2 + y2 – 8x – 8y + 7 = 0 at the point (1, 0) Ans: 3x + 4y – 3 = 0

PARABOLA The locus of a point that moves in a plane such that its distance from a fixed point equals its distance from a fixed line. Notes: • Fixed point is called focus • Fixed line is called directrix • Axis – the line passing through the focus and perpendicular to the directrix • Vertex – The midpoint of the segment of the axis from the focus to the directrix. • Latus rectum – a segment passing through the focus and perpendicular to the axis of the parabola. • Focal distance – distance from vertex to focus = a Standard Equations of Parabola A. Vertex at V(h,k), Vertical Axis (x-h)2 = 4a(y-k) if a is positive (+a) ----- concave upward if a is negative(-a) ----- concave downward Notes: 1. Equation of axis : x=h 2. Focus : F(h,k + a) 3. End of Latus Rectum L(h-2a, k+a) R(h+2a, k+a) 4. Equation of Directrix y = k-a

0 a>0

x

C. Vertex at the Origin, Vertical Axis x2 = 4ay if a is positive (+a) ----- concave upward if a is negative (-a) ----- concave downward Notes: y 1. Axis : the y axis axis 2. Focus: F(0,a) 3. Latus Rectum: /4a/ F Ends: L(-2a,a) L(-2a,a) R(2a,a) R(2a, a) 4. Equation of directrix y = -a V(0,0) directrix D. Vertex at the Origin, Horizontal Axis y2 = 4ax if a is positive (+a) ----- concave to the right if a is negative (-a) ----- concave tot he left

y axis L

F

R

Notes:

V(h,k) directrix 0

x

B. Vertex at V(h,k), Horizontal Axis (y-k)2 = 4a(y-k) if a is positive (+a) ----- concave to the right if a is negative (-a) ----- concave to the left

1. Axis : the x-axis 2. Focus: f(a,0) 3. Latus Rectum = 4a  Ends: L(a,2a) R(a,-2a) 4. Equation of Directrix x = -a

y directrix L V

F x axis R

Remarks: 1. The vertex and focus always lie on the axis of the parabola. 2. Focus is always located on the concave side of the parabola. Notes: 1. Equation of the axis: y=k 2. Focus: F(h+a, k) 3. Ends of Latus Rectum: L(h+a, k+2a) R(h+a, k-2a) 4. Equation of Directrix x = h-a

y

directrix L V F (h,k) R

axis

General Equations of Parabola 1. Vertical Axis • Ax2 + Dx + Ey + F = 0, E or A must not be zero 2. Horizontal Axis • Cy2 + Dx + Ey + F = 0, D or C must not be zero

Supplementary Problems 1. Find the vertex, focus, and end points of the Latus Rectum of each of the following parabolas a. 4y2 – x + 2y = 0 Ans: V(-1/4, -1/4), F(-3/16, -1/4) EL(-3/16, -1/4 ± 1/8) b.

2y2 – 5x + 3y - 7 = 0 Ans: V(-13/8, -3/4 ), F(-1, -3/4 ), EL(-1, -3/4 ± 5/4 )

2. Find the equation of the parabola determined by the given conditions a. focus at (-11/4, 1) and the endpoint of latus rectum is (-11/4, 5/2) Ans: y2 + 3x – 2y + 7 = 0 b. vertex at (1, -1) and focus at (1, -3/4) Ans: x2 – 2x – y = 0 c.

vertex at (0, 3) directrix x = -1 Ans: y2 – 4x – 6y + 9 = 0

d. axis vertical, vertex (-1, -1) and passing through (2, 2) Ans: x2 + 2x – 3y – 2 = 0 3. A chord passing through the focus of the parabola y2 = 16x has one end at the point (1, 4). Where is the other end of the chord?Ans: (4, 8) 4. Find the equation of the line tangent to the parabola x 2 + 2x + 3y – 1 = 0 Ans: 2x + 3y – 1 = 0 5. Find the equation of the circle that passes through the vertex and the endpoints of latus rectum of the parabola y2 = 8x. Ans: x2 + y2 – 10x = 0 6. Find the equation of parabola whose axis is horizontal, vertex is on the y – axis and which through (2, 4) and (8, -2) Ans: y2 = 20y –18 +100 = 0, y2 = 4y + 2x + 4 =0 7. An arch in the form of parabolic curve, with a vertical axis is 60 m, across the bottom. The highest point is 16 m above the horizontal base. What is the length of a beam placed horizontally across the arch 3m below the top? Ans: 26 m 8. Assume that water issuing from the end of a horizontal pipe, 25 ft. above the ground describe a parabolic curve, the vertex of the parabola being at the end of the pipe, the flow of water has curve outward 10 ft. beyond a vertical line through the end pipe, how far beyond this vertical line will the water strikes the ground? Ans: 17. 68 ft. ELLIPSE Ellipse is the locus of a point P(x, y) in a plane which moves such that the sum of its distances from two fixed points is constant.

   

Major axis – the segment cut by the ellipse on the line containing the foci - a segment joining the two vertices of an ellipse of length equal to I2aI Diameters– the chords of an ellipse that pass through the center Vertices – the endpoints of the diameter through the foci - the endpoints of the major axis Latus Rectum – the segment cut by the ellipse passing through the foci and perpendicular to the major axis Eccentricity – measure the degree of flatness of an ellipse

Standard Equation of Ellipse Center at C(h, k), Horizontal Major Axis y (x – h)2 (y – k)2 ------------ + ----------- = 1 a2 b2 Notes: 1. Major axis : y = k V1 2. Minor axis ; x = h 3. Vertices : V1(h-a, k) V2(h+a, k) 4. Foci : F1(h-c, k) F2(h+c, k) 0 Center C(h, k), Vertical Major Axis

The two fixed points are called foci.

C(h,k)

F2

V2

b

c

c

a

a

y

V1

F1

a

(x – h)2 (y – k)2 –––––––– + –––––––– = 1 a2 b2 Notes: 1. Major axis : x = h 2. Minor axis ; y = k 3. Vertices : V1(h, k+a) V2(h, k-a) 4. Foci : F1(h, k+c) F2(h, k-c)

(y)2

b

b c F2 V2

Center at the Origin, Horizontal Major Axis (x)2

c

C(h,k)

0

Notes: 

b F1

x

a

------- + ------ = 1 a2 b2 Notes: 5. Major axis : x = axis 6. Minor axis ; y = axis 5. Vertices : V1(-a, 0) V2(a, 0) 7. Foci : F1(-c, 0) F2(c, 0)

y

General Equation of an Ellipse Ax2 + Cy2 + Dx + Ey + F = 0 where A ≠ C but of the same sign

V1

F1 C(0,0)

F2

Supplementary Problems: 1. In the ellipse below determine the following: a) Center b) Vertex c)Foci d) Major Axis e) Major Axis f) Latus Rectum g) Eccentricity

V2

c

A. 25x2 + 16y2 – 50x + 32y – 1559 = 0 B. 144x2 + 169y2 +864x – 23,760 = 0 Ans: A. a) C(1, -1); b) V1(1, 9), V2(1, -11); c) F1(1, 5), F2(1, -7) d) 20 e) 16 f) 12.8 g) 0.6

c

a

a

B. a) C(-3, 0); b) V1(-16, 0), V2(10, 0); c) F1(-8, 0), F2(2, 0) d) 26 e) 24 f) 288/13 g) 5/13 Center at the Origin, Vertical Major Axis

(x)2 ------a2

(y)2 + ------ = 1 b2

Notes: 8. Major axis : y = axis 9. Minor axis ; x = axis 10. Vertices : V1(0, a) V2(0. -a) 11. Foci : F1(0, c) F2(0, -c)

a

F1

c b

General Remarks 1. Vertices and foci lie on the major axis 2. IaI is the distance from the center to the vertex 3. IcI is the distance from the center to the foci ( focal distance) 4. The ellipse is symmetrical to the major, minor axes and the center. Important Relations 1. a > b, a > c 2. a2 = b2 + c2 3. e = eccentricity = c/a < 1 4. Latus Rectum, LR = 2b2/a

2.

V1

b C(0,0) F2 V2

c

a

In each of the following find the equation of ellipse satisfying the given conditions A. center at (0, 0), focus at (±√3, 0), and b = 1 Ans: x2 + 4y2 = 4 B. center at (1, 0) focus at (1, √3) e= √3/2 Ans: 4y2 + y2 = 8x C. focus at (0, -1), (-4, -1), a = √6 Ans: x2 + 3y2 + 4x + 6y +1 = 0 D. center at (-1/2, 2) a = 5/2 b = 2 2 2 Ans: 16x + 25y + 16x + 4 = 100y E. center at (0, 0), vertex (0, 4) e = ½ Ans: 4x2 + 3y2 – 48 = 0

3. A satellite orbits around the earth in an ellipse orbit of eccentrically of 0.80 and semi – major axis of length 20,000 km. If the center of the earth is at one focus, find the maximum altitude (apogee) of the satellite. Ans: 36, 000 km 4.

Find the equation of the locus of a point which moves so that the sum of its distance from (-2, 2) and (1, 2) is 5. Ans: 16x2 + 25y2 + 16x – 100y +4 = 0

5.

Find the eccentricity of an ellipse whose major axis is thrice a long as its minor axis Ans: 2√ 2/3

6.

What is the quadrilateral formed by joining the foci of an ellipse to the endpoints of the minor axis? Ans: rhombus

7.

Find the distance of the point (3, 4) to the foci of the ellipse whose equation 4x 2 + 9y2 = 36 Ans: 3 ±√ 5

8.

9.

An arch in the form of a semi – ellipse has a span at 45 m and its greatest height is 12m. There are two vertical supports equidistant from each other and the ends of the arc. Find the height of the support. Ans: 8 √ 2 m.

1. Transverse axis: y = k 2. Conjugate axis: x = h transverse 3. Vertices: V1 (h – a, k) V2 (h + a, k) 4. Foci: F1 (h – c, k) F2 (h + c, k) 5. Asymptotes: y – k = ± b/a ( x – h)

Determine the locus of a point P(x, y) so that the product of the slopes joining P(x, y) to (3, -2) and (-2, 1) is –6. Ans: 6x2 + y2 – 6x +y – 20 = 0

10. What is the area of the ellipse whose equation is 25x2 + 16y2 = 400. Ans: 20π sq. units HYPERBOLA Hyperbola is the locus of point P(x, y) in a plane which moves such that the difference of its distances from two fixed points is a positive constant.

0

x

Center at C(h, k), Vertical Transverse Axis Asymptote

Notes:       

  

(y – k)2 _ a2

The two fixed points are called foci Transverse axis – a line segment joining the two vertices of hyperbola - the length of the transverse axis is I 2a I Conjugate Axis – the perpendicular bisector of the transverse axis. - the length of the conjugate axis is I2b I Center – point of intersection of transverse and conjugate axis Central Rectangle – the rectangle whose area is (2a) (2b) and whose diagonals are asymptotes of hyperbola Vertices – the endpoints of the transverse axis Asymptotes of the hyperbola – two intersecting lines containing the diagonal of the central rectangle –To find the equation of the asymptote, set the right side of the equation of hyperbola in standard form to zero then solve for y. Central Circle – the circle of radius c with center at the center of the hyperbola circumscribing the central rectangle Equilateral Hyperbola – hyperbola whose transverse axis equals its conjugate axis Conjugate Hyperbolas – hyperbolas whose transverse axis of one is the conjugate axis of the other

1. 2. 3. 4. 5.

(x – h)2 a2

_ (y – k)2 = 1 b2

= 1

Transverse axis

F1

Notes: axis Transverse axis: x = h Conjugate axis: y = k Vertices: V1 (h, k + a) V2 ( h, k – a) Foci: F1 (h, k + a) F2 ( h, k – c) Asymptotes: y – k = ± a/b( x – h)

x2 a2

y Asymptote

(x – h)2 b2

Conjugate

V1

C

V2 F2

Center at the Origin, Horizontal Transverse Axis

Standard Equations of Hyperbola Center at C(h,k), Horizontal Transverse Axis

Notes:

axis conjugate axis

F1

_

y2 b2

= 1

Notes: 1. Transverse axis: x–axis F1 y–axis V2 2. Conjugate axis: V1 3. Vertices: V1 (-a, 0) C V2 (a, 0) 4. Foci: F1 (-c, 0) F2 (c, 0) C Asymptotes: y = ± b/a x 5. V V F 1

2

y

F2

2

Center at the origin , Vertical Transverse Axis

x

y2 a2

_

x2 = 1 b2

y c. F1

Notes:

V1 C

1. Transverse axis: y – axis 2. Conjugate axis: x – axis 3. Vertices: V1 (0, a) V2 ( 0, -a) 4. Foci: F1 (0, c) F2 (0, -c) 5. Asymptotes: y = ± a/b x

V2 F2

c.

General Relations 1. c > a, c > b ( a = b or a < b or a > b) If a = b, then the hyperbola is called equilateral hyperbola 2. c2 = a2 + b2 3. Length of Latus Rectum = 2b2 a 4. Eccentricity, e = c/a > 1 where A and C are of opposite signs

Supplementary Problems 1. Find the center, vertices, foci and asymptotes, transverse axis, conjugate axis, latus rectum and eccentricity of hyperbola below. a. 9x2–16y2+18x+64y=91 b. 16x2–4y2–62x+24y+92=0 c. 25x2–16y2=400 Ans. a. C(–1,2); V(–1± 2, 2); F(–1± 5/2, 2); asymptote (3x+4y–5=0, 3x–4y+11=0); TA = 4; CA = 3;LR = 9/4; e = 5/4 b.

2. Find the equation of the hyperbola satisfying the conditions given in each case a. Center(3,–1); vertex(1,–1); focus(0,–1) Ans. 5x2–4y2–30x–8y+21=0 b. vertices at(0,4) and (4,4); foci at (5,4) and (–1,4) Ans. 5x2–4y2–20x+32y–64=0

General Remarks 1. Vertices and foci are on the transverse axis 2. IaI is the distance from the center to the vertex 3. IcI is the distance from the center to the focus. 4. The hyperbola is symmetrical to the transverse and conjugate axis and to the center.

General Equation of Hyperbola Ax2 + Cy2 + Dx + Ey + F = 0

C(0,0); V1(–4,0),V2(4,0); F(–√ 41 ,0), F2(√ 41 ,0); Asymptote ( y = ± 5/4 x ); TA = 8; CA = 10;LR = 25/2; e = √41/4

C(2,3); V1(2,1),V2(2,7); F(2,3+√ 20 ), F2(2, 3–√ 20 ); asymptote(2x–y–1=0, 2x+y–7=0); TA = 8; CA = 4;LR = 2; e = √5/2

Center at (1,1), vertex(1,3), eccentricity=2 Ans. x2–3y2–2x +10 =0

d. Directrices: y=±4; asymptotes: y=±3/2x Ans. 9y2–324x2–208 = 0 e. Asymptotes: 3y=±4x ; foci (±6,0) Ans. 400x2–225y2–5184=0 f.

Foci(0,0), (0,10); asymptote: x+y=5 Ans. 2x2–2y2+20y–25=0

g. Asymptotes: x+y=1 and x–y=1 and passing through (–3,4) and (5,6) Ans. x2–y2–2x–2=0 h. Axes along the coordinates axes, passing through (–2,5) Ans. 4y2–5x2=19 i. Vertices at (0, ±4) passing through (–2,5) Ans. 4y2–9x2 = 64 3. Find the eccentricity of a hyperbola whose transverse axis and conjugate axis are equal in length Ans. e= √ 2 THE CONIC SECTIONS (a summary) A conic section is the locus of a point which moves such that its distance from a fixed point called focus is in constant ratio and called eccentricity to its distance from a fixed straight line called directrix. A. General Form of a Quadratic Equation in x and y Ax2 + Cy2 + Dx + Ey + F = 0 1. Ellipse : A ≠ C, same signs 2. Circle : A = C, same signs 3. Hyperbola : A and C have opposite signs B. Eccentricity e = c/a

1. Circle : e = 0 2. Parabola :e=1 3. Ellipse : e < 1 4. Hyperbola :e>1 Note:  The circle, parabola, ellipse and hyperbola are called conic sections (or conics) because any one of them can be obtained geometrically by cutting a cone with a plane.

Circle

   

Ellipse

Parabola

Hyperbola

If the cutting plane is perpendicular to the axis of the cone, the section is a circle. If the cutting plane is making an angle ( other than 90 o) with the axis of the cone, the section is an ellipse. If the cutting plane is parallel to one of the elements of a cone, the section is a parabola If the cutting plane is parallel (but not coincident) to the axis of the cone, the section is a hyperbola. In each of the cases, the cutting plane should not pass through the vertex of the cone, otherwise the section hat will be formed is a degenerate conic.

Degenerate Conic ( one point, one line, two lines) is a conic formed if the cutting plane is passing through the vertex along one of its elements Principal Axis of a Conic – is the line through the focus and perpendicular to the directrix Diameter of a Conic – the locus of the midpoints of a system of parallel chords. Direction: Encircle the letter corresponding to the correct answer.

1. The sum of the digits of a two-digit number is 11. If the digits are reversed, the resulting number is seven more than twice the original number. What is the original number? a. 38 b. 53 c. 83 d. 44 2. A metal washer 1-inch in diameter is pierced by ½-inch hole. What is the volume of the washer if it is 1/8 inch thick. a. 0.074 b. 0.047 c. 0.028 d. 0.082 3. If a regular polygon has 27 diagonals, then it is a, a. nonagon b. pentagon c. hexagon d. heptagon 4. Find the probability of getting exactly 12 out of 30 questions on a true or false question. a. 0.12 b. 0.08 c. 0.15 d. 0.04 5. Find the area bounded by the curve y = 9 − x 2 and the x-axis a. 18 units2 b. 25 units2 c. 36 units2 d. 30 units2 6. It is a measured of relationship between two variables. a. Function b. Relation c. Correlation d. Equation 7. A central angle of 45 degrees subtends an arc of 12 cm. what is the radius of the circle? a. 15.28 cm b. 12.82 cm c. 12.58 cm d. 15.82 cm 8. Two posts, one 8 m and the other 12 m high are 15 cm apart. If the posts are supported by a cable running from the top of the first post to a stake on the ground and then back to the top of the second post, find the distance from the lower post to the stake to use minimum? a. 6 m b. 8 m c. 9 m d. 4 m 9. A regular octagon is inscribed in a circle of radius 10. Find the area of the octagon. a. 228.2 b. 288.2 c. 238.2 d. 282.2 10. The volume of the two spheres is in the ratio 27:343 and the sum of their radii is 10. Find the radius of the smaller sphere. a. 5 b. 4 c. 3 d. 6 11. Find the approximate change in the volume of a cube of side “x” inches caused by increasing its side by 1%. a. 0.30x2 in3 b. 0.02 in3 c. 0.1x3 in3 d. 0.03x3 in3 12. The time required for the examinees to solve the same problem differ by two minutes. Together they can solve 32 problems in one hour. How long will it take for the slower problem solver to solve a problem? a. 5 minutes b. 2 minutes c. 3 minutes d.4 minutes 13. If a = b, then b = a. This illustrates which axiom in Algebra? a. Transitive axiom b. Replacement axiom c. Reflexive axiom d. Symmetric axiom 14. Find the distance of the directrix form the center of an ellipse if its major axis is 10 and its minor axis is 8. a. 8.5 b. 8.1 c. 8.3 d. 8.7 15. A regular hexagonal pyramid has a slant height of 4 cm and the length of each side of the base is 6 cm. Find the lateral area. a. 62 cm2 b. 52 cm2 c. 72 cm2 d. 82 16. At the surface of the earth 9 = 9.806 m/s 2. Assuming the earth to be a sphere of radius 6.371 x 106 m, compute the mass of the earth. a. 5.12 x 1024 kg b. . 5.97 x 1023 kg 24 c. 5.97 x 10 kg d. . 5.62 x 1024 kg 17. The perimeter of an isosceles right triangle is 6.6824. Its area is

a. 4 b. 2 c. 1 d. ½ 18. Determine the vertical pressure due to a column of water 85-m high. a. 8.33 x 105 N/m2 b. 8.33 x 104 N/m2 c. 8.33 x 106 N/m2 d. 8.33 x 103 N/m2 19. a 40-gm rifle with a speed of 300 m/s strikes into a ballistic pendulum of mass 5 kg suspended from a cord 1 m long. Compute the vertical height through which the pendulum rises. a. 28.87 cm b. 29.42 cm c. 29.88 cm d. 28.45 cm 20. What is the area of the largest rectangle that can be inscribed in a semi-circle of radius 10? a. 2 50units 2 b. 100 units2 c. 1000 units2 d. 50units 2 21. The amount of heat needed to change solid to liquid is a. condensation b. cold fusion c. latent heat of fusion d. solid fusion 22. Mr J. Reyes borrowed money from the bank. He received from the bank P1,842 and promise to repay P 2,000 at the end of 10 months. Determine the simple interest. a.19.45% b. 15.70% c. 16.10% d. 10.29% 23. Find the length of the vector (2,4,4) a. 7.00 b. 6.00 c. 8.50 d. 5.18 24. According to this law, “The force between two charges varies directly as the magnitude of each charge and inversely as the square of the distance between them.” a. Law of Universal Gravitation b. Coulomb’s Law c. Newton’s Law d. Inverse Square Law 25. A loan of P 5,000 is made for a period of 15 months, at a simple interest rate of 15 %, what future amount is due at the end of the loan period. a. P 5,637.50 b. P5,937.50 c. P 5,900.90 d. P5,842.54 26. If y = x ln x, find a.

−1 x

d2y . dx 2 −1 b. x2

c.

1 x

d.

1 x2

27. The integral of any quotient whose numerator is the differential of the denominator is the ____________. a. cologarithm b. product c. logarithm d. derivative 28. Find the nominal rate, which if converted quarterly could be used instead of 12% compounded semi-annually. a. 14.02% b. 21.34% c. 11.29% d. 11.83% 29. Evaluate the expression (1 + i2 )10 where I is an imaginary number. a. 1 b. 0 c. 10 d. -1 30. A VOM has a selling price of P 400. If its selling price is expected to decline at a rate of 10% per annum due to obsolence, what will be its selling price after 5 years? a. P 213.10 b. 249.50 c. 200.00 d. 236.20 31. If tan 4 A = cot 6 A, then what is the value of angle A. a. 90 b. 120 c. 100 d. 140

32. Find the sum of the rots of 5 x 2 − 10 x + 2 = 0. a. -1/2 b. -2 c. 2 d. ½ 33. In a box there are 25 coins consisting of quarters, nickels and dimes with a total amount of \$ 2.75. If the nickels were dimes, the dimes were quarters and the quarters were nickels, the total amount would be \$ 3.75. How many quarters arethere? a. 12 b. 16 c. 10 d. 5 34. A point moves so that its distance from the point (2,-1) is equal to its distance from the axis. The equation of the locus is. a. x 2 − 4 x + 2 y + 5 = 0 b. x 2 − 4 x − 2 y + 5 = 0 c. x 2 + 4 x + 2 y + 5 = 0 d. x 2 + 4 x − 2 y − 5 = 0 35. You loan from a loan firm an amount of P 100,000 with a rate of simple interest of 20% but the interest was deducted from the loan at the time the money was borrowed. If at the end of one year you have to pay the full amount of P 100,000 what is the actual rate of interest. a. 25.0% b. 27.5% c. 30.0% d. 18.8% 36. It is a polyhedron of which two faces are equal polygons in parallel planes and the other faces are parallelograms. a. Tetrahedron b. Prism c. Frustum d. Prismatoid 37. A railroad is to be laid –off in a circular path. What should be the radius if the track is to change direction by 30 0 at a distance of 157.08 m? a.150 m b. 200 m c. 250 m d. 300 m 38. A 200-gram apple is thrown from the edge of a tall building with an initial speed of 20 m/s. What is the change in kenetic energy of the apple if it strikes the ground 50 m/s? a. 130 Joules b. 210 Joules c. 100Joules d. 82 Joules 39. A machine costs P 8, 000 and an estimated life of 10 years with a salvage value of P 500. What is its book value after 8 years using straight-line method? a. P 2,000 b. P 4,000 c. P 3, 000 d. P 2, 500 40. The distance between the points AB defined by A( cos A,−sin A) and B ( sin A, cos A) is equal to a. cos A b. 1 c. 2 d. 2 tan A 41. What nominal rate, compounded semi-annually, yields the same amount as 16% compounded quarterly? a. 16.64% b. 16.16 c. 16.32 d. 16.00% 42. If (2log x to the base 4) – (log9 to the base 4) = 2, find x. a. 10 b. 13 c. 12 d. 11 x + 5 y − 2 z = 9 ,3x − 2 y + z = 3 and x + y + z = 2 43. The point of intersection of the planes is at a. (1,2,1) b. ( 2,1,-1) c. (1,-1,2) d. ( - 1,-1,2) 44. To compute the value of n factorial, in symbolic form (n!); where n is large number, we use a formula called a. Richardson-Duchman Formula b. Diophantine Formula c. Stirling’s Approximation d. Matheson’s Formula 45. A boat can travel 8 miles per hour in still water. What is it velocity with respect to the shore if it heads 350 East of North? a. 6,743 b. 8,963 c. 5,400 d. 4,588

46. What is the distance in cm between two vertices of a cube which are farthest from each other, if an edge measures 8 cm? a. 13.86 b. 16.93 c. 12.32 d. 14.33

1 3

47. If arctan  =

π , then the value of x is 4

a. 1/3 b. 1/2 c. 1/4 d. 1/5 48. The energy stored in a stretched elastic material such as a spring is a. mechanical energy b. elastic potential energy c. internal energy d. kinetic energy 49. Given the points (3, 7) and (-4, -7). Solve for the distance between them. a. 15.65 b. 17.65 c. 16.65 d. 14.65 50. What rate of interest compounded annually is the same as the rate of interest of 8% compounded quarterly? a. 8.48% b. 8.42% c. 8.24% d. 8.86% ANSWERS 1A 2A 3A 4B 5C 6C 7A 8 A 9 D 10 C 11D 12D 13D 14C 15C 16C 17B 18A 19D 20B 21C 22D 23B 24B 25B 26C 27C 28D 29D 30D 31A 32C 33D 34A 35A 36B 37D 38B 39A 40C 41C 42C 43B 44C 45D 46A 47B 48B 49A 50C Differential Equation - an equation involving differential coefficient or differentials. Consider the equation : (d2y / dx ) + P (x) =Q (x ) * When an equation involves one or more derivatives with respect to a particular variable, that variable is called independent variable. For the given equation, the independent variable is x. * If a derivative of a variable occurs, that variable is a dependent variable. For the given equation; dependent variable is x. Types of Differential Equation: 1. Ordinary Differential Equation – an equation which all differential coefficient have reference to a single independent variable. 2. Partial Differential Equation – an equation which there are two or more independent variables and partial differential coefficients with respect to any of them. Order of Differential Equation: - The order of the highest derivative appearing in the equation.

Degree of Differential Equation: - The power to which the highest - order derivative is raised , if the equation is written as a polynomial in the unknown function . Linearity of a Differential Equation - A differential equation is linear if it has the form: Pn(x) dny + Pn-1 (x) dn-1y +……+ P1(x) dy + Po(x)y = Q(x) dxn dxn-1 dx where: y – the unknown function x – the independent variable Q(x) , Pn(x), Pn-1(x)…P0(x) - presumed known and depend only on the

variable x

* Differential equations that cannot be reduced or put in this form are nonlinear. Sample Problems: Determine the order, degree, linearity, unknown function, and independent variable of the differential equation. 1. y’’’ – 2xy ‘ = ln x +2 2. 5x (d2y / dx2 ) 9x3 (dy / dx) tanx (y) = 0 3. (d2b / dp2 ) + p (db / dp)2 = 0 Answers: 1. third order , first degree, linear (P 3(x) =1 , P1(x) = -2x , P2(x ) = P0 (x) =0 , Q(x) = lnx +2, the unknown function is y, and the independent variable is x. 2. second order , first degree, linear, unknown function is y, and the independent variable is x. 3. second order , second degree , nonlinear ( one of the derivatives is raised to a power other than the first) unknown function is b and its derivatives, and the independent variable is p. Supplementary Problems: Determine a)order b) degree c) linearity d) unknown function and e)independent variable for the following differential equations. 1. (y’’’)3 - 5x(y’)2 = e-x + 1 2. 5y d2z / dy

+ 3y2 (dz/dy) - (siny) z = 0

3. (d4x / dy4)5 + 7(dx / dy)10 + x7 –x5 = y 4. ty’’ + t2y’ - (sin t ) √ y

= t2 – t + 1

Families of Curves: Families of curves may be represented by equation involving parameters. If the constants of this equation is treated as an arbitrary constant, the result is called differential equation of the family represented by equation.

2. Parabolas with foci at the origin and axis Ox. Ans. y( y’’)2 + 2xy’ –y =0 3. Circles tangent to the x-axis. Ans. [ 1+ (y’ )2 ]3 = [ yy ’’ +1 + (y’)2 ]2

Sample Problems : 1. Obtain the differential equation of the family of:

4. All tangents to the parabola y2 =2x Ans. 2x (y’ )2 –2yy’ + 1 = 0

a). straight lines with slope and y- intercept equal. Solution: y = mx +b

Solution of Differential Equations

- the slope intercept form

A function y = f(x) is a solution of differential equation if it is identically satisfied when y and its derivatives are replaced throughout by f(x) and its corresponding derivatives. A differential equation of order n will, in general, possesses a solution involving n arbitrary constants. This solution is called general solution. It will be necessary to assign specific values to these arbitrary constants in order to meet the prescribed initial conditions.

In these case, m = b y = mx+m

- the constant to be eliminated

differentiating dy/dx = m

dy = m

substituting to the original equation

y = (dy / dx) (x+1)

b). circles with center on the x-axis y x

First Order Equations 1. First Order : Variables Separable A first order differential equation can be solved by integration if it is possible to collect all y terms with dy and all x terms with dx. That is, if it is possible to write it in the form: f(y) dy + g(x) dx = 0 , Then the general solution is ∫ f(y) dy + ∫ g(x) dx = C where C is an arbitrary constant. Sample Problems: 1. Find the general solution of the following differential equation a. y’ =(x+1) / y b. ds/dt = s2 + 2s + 2

Solution: C( h,0) (x-h)2 + y2 = r2 2(x-h) + 2yy’ = 0 (x-h) + yy ‘ = 0 x+yy’ = h 1+yy’’+ (y ‘)2 = 0

the arbitrary constants are h and r, therefore we must differentiate the equation two times. first differentiation

2. Obtain a particular solution which satisfies the given initial condition. dy/dx = 3x3 / √ y , y = 4 when x = 1 Solution: 1. a). The equation maybe written in the form

dy/dx = (x+1)/y ydy = (x+1) dx

Supplementary Problems: For each of the following, obtain the equation of the family of plane curves. 1. Straight lines whose distance is a from the origin. Ans. (xy’ –y ) 2 = a2 ( 1+ (y’)2)

The solution is ∫ydy = ∫ (x+1) dx ½ y2 = ½ x 2 + x + C y2 – x2 – 2x = C

Ans.

b. Separate the variables to obtain ds = dt s2 + 2s + 2 which is the same as ds = dt (s + 1)2 + 1 The solution is ds ∫ = ∫ dt (s + 1)2 + 1 C + t = arctan (s+1) s = tan (C+t) – 1

2. The variables are separable, which can be written in the form; √ y dy = 3x3 dx

b. dy/dx = sec y tan x

x=0, y=0 Ans. sin y + ln cos x = 0

2. First Order : Homogeneous Equation Polynomials in which all terms are of the same degree such as x 2 + y2 and x2 sin (y/x) are called homogeneous polynomial. Equations with homogeneous coefficient. Consider M (x,y) dx + N (x,y) dy = 0  1 If the coefficients M and N are of the same degree in x,y , they are called homogeneous functions. Let y = Vx, and dy = Vdx + xdV, substitute this to equation 1 to make the variables separable. Sample Problem: 1. Solve y’ = (y + x )/ x Solution: Let y = Vx;

dy = Vdx + xdV

integrating both sides

dy/dx = (y + x)/ x = (Vdx + xdV ) / dx = (Vx + x) / x

∫ √ y dy = ∫ 3x dx

or simply

3

2

x (dV/dx) = 1

/3 y3/2 = ¾ x4 + C  this is the general solution

if y = 4, x = 1

V = ln x + C

C = 2/3 (4)3/2 – ¾ (1)4

V = ln C x *

C = 55/12

But v = y/x

2

/3 y

3/2

4

Therefore, y = x ln C x

55

= ¾ x + /12

* The ln of a constant is still a constant.

or simply 8 y3/2 – 9 x4 = 55

Supplementary Problems: 1. Obtain the general solution of the following: a. xy’ + y = 0 Ans. xy + C = 0 b. x dx = y dy = 0

Ans. x2 + y2 = C

c. x3 dx + (y + 1)2 dy = 0

Ans. (y + 1)3 + ¾ x3 = C

d. ds/dt = t2 / s2 + 6s + 9

Ans. (s + 3)3 – t3 = C

e. y dy + (y2 + 1) dx

Ans. ln (1 + y2) + 2x

2. Obtain the particular solution satisfying the given initial conditions. a. dy/dx + 2y = 3 x=0, y=1 Ans. ln (3 – 2y) + 3x = 0

Supplementary Problems: Find the general solution: 1. y’ = 3xy / 2.

y 2 – x2

Ans. (y2 + 2x2)3 = Cy2 Ans. 3y2 - x2 = C√ x

y’ = (x2 + y2 ) / 4xy

3. y’ = y / ( x + √ xy)

Ans. – 2 √ x/y + ln y = C

3. First Order : Linear Equation A differential equation of first order, which is also linear, can be written in the form: dy/dx + P(x)y =Q (x) 1

To obtain the general solution we must find a function ∅ = ∅ (x) such that if the equation is multiplied by this function derivative of the product ∅y. This function is called the integrating factor ∅. ∅ = e ∫Pdx And the solution is y e∫Pdx = ∫ Q e∫Pdx dx + C  2 Sample Problem: 1. Obtain the general solution of the differential equation. x dy + y dx = sin x dx Solution: The differential equation must be first reduced in the form of equation 1, hence,

Integrating factor φ = e∫Pr (x) dx Where: P(r)(x) = (1-n) P(x) Q(r)(x) = (1-n) Q(x) z = y1-n General Solution: zφ = ∫ φ Qr (x)dx + c Sample Problem: Solve dy / dx – 2y / x = 4x3 y3 Solution: dy / dx + y (-2/x) = y3 (4x3) n=3 z = y1-3 z = y –2

dy/dx + y/x = (sin x) / x

P(x) = -2 / x

P(x) = 1/x ; Q(x) = (sin x) / x

Pr (x) = (1-n) Px

Solving for the integrating factor,

Pr (x) = (1-3) Px

∅=e

Pr (x) = -2 (-2/x) = 4/x

∅ = eln x

Q (x) = 4x3

∅=x

Qr (x) = (1-n) Q(x)

∫dx/x

Qr (x) = (1-3) (4x3)

using the formula,

Qr (x) = -8x3

yx = ∫ sinx /x (x) dx + C

φ = e ∫Pr (x) dx = e ∫ (4/x) dx + e 4 lnx = x4

simplifying, xy + cos x = C

y-2x4 = -8∫ x7 dx + c

Ans. y = c e

2. e-2ydx + 2 (xe2y – y) dy

(-x^4)/4

Ans. x e2y = y2 + C

3. dT/dθ = cos θ + y cot θ 4. y’ = x – 2y cot 2x

General solution: zφ = ∫ φ Qr (x)dx + c (y-2)(x4) = ∫ x4 (-8x3) dx + c

Supplementary Problems: Obtain the general solution: 1. dy/dx + x3y = 0

Reduced polynomial in x

Ans. T = c sin θ - cos θ

Ans. 4y sin 2x = c + sin 2x – 2x cos 2x

5. y’ = x3 2xy; when x = 1, y = 1 3. Bernoulli’s Equation Standard Form: dx/dy + y P(x) = yn Q(x) provided, n ≠ 0, 1

Ans. 2y = x2 – 1 + 2 e(1-x²)

y-2x4 = - x8 + c x4 = (c –x8) y2 4. First Order: Exact equations Any equation that can be written in the form M(x,y)dx + N(x,y)dy = 0 And we have the property ∂M

∂N ------- = ------- is said to be an exact equation.

∂y

∂x

k < 0 – exponential decay

The technique is to find a function f(x,y) such that f’(y) – the common term of f(x,y) and N(x,y) g’(y) – the common term of f(x,y) and M(x,y) Sample Problem: Determine whether the equation 2xy dx + (1+x2) dy = 0 is exact. If so, obtain the general solution. Solution: Given in the equation that N = 1+ x2

M = 2xy,

∂M / ∂y = ∂N / ∂x = 2x

∴ The equation is exact

F = ∫ 2xy dx + f(y)

Solution: Given: Qo = 50 mg Q(2) = (50 mg) (1-0.1) Q(2) = 45 mg Solving for k, 45 = 50 e2k k = -0.053 ∴ amount of material at any time t

F = x2y + f(y)

Q = 50 e -0.053 t

∂F/ ∂y = x + f’(y) = x + 1 2

Sample Problem: 1. A certain radioactive material is known to decay at a rate proportional to the amount present. If initially there is 50 mg of the material present and after 2 hrs. it is observed that the material has lost 10% of its original mass. Determine mass of the material after 4 hours?

2

In the problem t = 4 hrs.

f’(y) = 1

Q = 50 e -0.053 t

f(y) = y + c

Q = 40.5 mg.

x2y + y + c = 0

Supplementary Problems: 1. (2x ey + ex) dx + (x2 +1) ey dy = 0

Ans. ex + ey(x2 +1) = c

2. (2xy + y2) dx + (x2 +2xy - y) dy = 0

Ans. 2x2y +2xy2 – y2 = c

3. (x + sin y) dx + (x cos y – 2y) dy = 0 4. dy + (y- sin x) / x dx = 0

Ans. ½ x2 + x sin y – y2 = c

Ans. xy + cos x = c

2. Newton’s Law of Cooling The time rate of change of the temperature of a body is proportional to the temperature difference between body and its surrounding medium. T = temperature of the body at any time t To = initial temperature Tm = temperature of the surrounding medium dT / dt = time rate of change of the temp. of the body

Elementary Applications: 1. Law of Exponential Change (Growth /Decay) The rate of which the amount of a substance changes, is proportional to the amount present or remaining at any instant. Q = amount of substance at any time t dQ / dt = rate of change of the amount Q dQ / dt ∝ Q dQ / dt = kQ Q = Qo ekt

k > 0 – exponential growth

dT / dt = - k (T – Tm) (dT / dt) + kT = kTm , k is always positive, dT / dt = negative for cooling T = Ce-kt + Tm

or

T = (To – Tm) e-kt + Tm

Sample Problem: A body at a temperature off 36oC is placed outdoors where the temperature is 60oC. If after 5 mm, the temperature of the body is 40 oC. How long will it take to reach a temperature of 45oC? Solution: Given: Tm = 60oC

To = 36oC

into the tank at Ri = (gal/min. , liter/min.) while simultaneously, the well-stirred solution leaves the tank at the rate R o ( gal/min.)

o

T @ 5 mm = 40 C

Ci , Ri

Solve for k: 40 = (36-60) e

–k(5)

Mixture

k = 0.03646 o

Solving for t at T = 45 C 45oC = (36-60) e –0.03646 t + 60 t = 12.9 min.

3. Simple Chemical Conversion In certain reactions in which a substance A is being converted into another substance the time of change of the amount x of unconverted substance is proportional to x. Let x = xo the uncovered substance at to = 0, then the amount x at any time t > 0, is given by the differential equation as dx/dt = - kx  ← The proportionality is chosen to be –k because x is decreasing as time increases, from ← x = ce-kt but x = xo at t = 0, hence xo= ce-k(1) c = xo X = xoe-kt Sample Problem: Suppose that a chemical reaction proceed such that the time rate of change of the unconverted substance is proportional to the amount of it. If half of substance A has been converted at the end of 10 sec. Find when 9/10 of the substance will have been converted. Solution: Given: x = ½ xo

when 9/10 is converted only 1/10 is unconverted

Solution: At t = 0, Vo = 0 , ci = 1, Ri = 4,Ro = 2 The volume of brine in the tank at any time t equals to Vo + Ri t - Ro t = 10 + 4t – 2t = 10+2t overflow will occur if the volume of the brine in the tank equals to the volume of tank now. T = ½ (50-10) = 20 min.

integrating factor = e ∫ (dt / 5+t) = e ln (5+t) = 5+t Q (5+t) = 4 ∫ (5+t) dt

∴ x = 1/10 x

Q (5+t) = 4 [ 5t +1/2 t2] + c

/10 xo = xo e –0.069 t

Q = 20t +12 t2 + c / 5+t

At t = 0 Q = a = 0 0 = [20(0) + 2(0)2 + c] / (5+t)

4. Dilution / Flow Problem: The tank initially hold Vo (gal., liter,etc.) of solution that contains S o (lb.,kg.,N) of a substance. Another solution containing a substance at C I = lb/gal.; N / L is poured

Ri > Ro , then Vo will overflow

Sample Problem: A 50 gal. tank contains 10 gal. of fresh water. At t = 0, a brine solution containing 1 lb. Of salt per gallon is poured into the tank at the rate of 4 gal/min. , while the well-stirred mixture leaves the tank at the rate of 2 gal/min,a.) find the amount of time required for overflow to occur, b.) Find the amount of salt in the tank at the amount of overflow.

dQ/dt + 2Q/10+2t = 4

k = 0.069

t = 33 sec.

Outgoing

In general, dS/dt Vo=,SSoo / [Vo + (Ri , Ro) t] = Ci Ri RCi =o ,RRo o, Vo = c Ro < RI , then Vo↓ Let S = amount of substance present in the tank dS/dt = rate of change of substance S

b.) dQ/dt + Q/10+2t = 4

½ xo = xo e k(-10)

1

ds/dt = Ci Ri - Co Ro Co = S / [Vo + (Ri , Ro) t]

+ 60

c = 0; now Q = 20t + 2 t2 / 5+t Req’d: Q =? At t = 20 min.

Q = [20(20) + 2(20)2 / (5+20) Q = 48 lb.

5. Orthogonal Trajectories Given a family of curves given by f(x,y,c) = 0 The curves that intersects a family f(x,y,c) = 0 at right angles, whenever they do intersect is given by the family of another curve g(x,y,k) and are called orthogonal trajectories. This two curves are said to be orthogonal to each other, because each point of intersection, the slopes of the curves are negative reciprocals to each other.

5. If a population of a country doubles in 50 years, in how many years will it treble under the assumption that the rate of increase is proportional to the no. of inhabitants? Ans. 79 6. A cylindrical tank contains 40 gal. of a salt solution containing 2 lb. of salt per gallon. A salt solution of concentration 3 lb/gal flows into the tank at 4 gal/min. How much salt is in the tank at any time t if the well-stirred solution flows at 4 gal/min? Ans. 120 - 40e –t/10 7. An inductance of 1 henry and a resistance of 2 ohms are connected in series with an emf of 100e-t volts. If the current is initially zero, what is the maximum current attained? Ans. 25A 8. Find the orthogonal trajectories of x2=cy. Ans. 2y2+x2=c DIFFERENTIAL CALCULUS

Let Mdx + Ndy = 0 where M and n are f(x,y) dy/dx = - M/N Now the DE of the orthogonal trajectories is dy/dx = N/M Sample Problem: Find the orthogonal trajectories of all circles whose center is at the origin.

LIMIT OF FUNCTIONS Definitions • The limit of a function of x or f(x) as x approaches a is L; written as lim f(x) = L x→a

• Solution: x2 + y2 = c2 Differentiating implicitly, 2xdx + 2ydy = 0 Simplifying dy/dx = - x/y = - M/N ∴ N/M = dy/dx  the D. E. of trajectories ∫ y/dy = ∫ x/dx then integrating y = kx

lim f(x) = L if and only if, for any chosen positive number ∈, x→a however small, there exists a positive number δ such that, whenever 0 < x-a < δ , then ƒ(x) – L  < ∈

Theorems on Limits 1. Limit of a constant c lim c = c x→a

2. Limit of a variable x lim x = a x→a

3. Limit of sum of two functions lim [ ƒ(x) + g(x) ] = lim ƒ(x) + lim g(x) x→a

Supplementary Problems: 1. A body at an unknown temperature is placed in a room which is held at a constant temperature of 30oC and after 10 min., the temperature of the body is 0 oC and after 20 min., the temperature of the body is 15oC. Find the expression for the temperature at any time t. Ans. T = - 60 e –0.069t + 30 2. At exactly 10:30 pm, a body at a temperature of 50 oF is placed In an oven whose temperature is kept at 150oF, If at 10:40, the temperature of the body is 75 oF, at what time will it reach 100oF. Ans.10:54 p.m. 3. Find the required for a radioactive substance to disintegrate half of its original mass if three quarters of it are present after 8 hrs. Ans. 19.3 hrs. 4. After 2 days, log of a radioactive material is present. Three days later, 5g. is present. How much of the chemical was present initially, assuming the rate of disintegration is proportional to the amount present. Ans. 15.87 g.

x→a

x→a

4. Limit of product of two functions lim [ ƒ(x) g(x) ] = lim ƒ(x) [ lim g(x) ] x→a

x→a

5. Limit of a quotient lim ƒ(x) = lim x→a x→a g(x) lim g(x)

ƒ(x)

x→a

6. Limit of a radical lim n √ ƒ(x) = n√ lim ƒ(x) x→a

x→a

x→a

Limit to Infinity or Zero Given a constant c and variable x 1. lim cx = + ∞ for positive c x→∞ = - ∞ for negative c 2. lim c/x = 0 x→∞

3.

/c = + ∞ for positive c = - ∞ for negative c 4. lim c/x = + ∞ for positive c x→0 = - ∞ for negative c lim

x

x→∞

4.

L’Hospitals Rule If the functions f(x) and g(x) are continuous in an interval containing x = a, and if their derivatives exist and g’(x) ≠ 0 in this interval (except possibly at x = a), then when f(a) = 0 and g(a) = 0 lim f(x) = lim f ’(x) x→a x→a g(x) g ’(x) Provided that the limit on the right side exists. Evaluation of Limits Let lim N(x) = L x→a D(x) where N (x) and D (x) are polynomials in x a = any real number L = limit Methods of Evaluation 1. direct substitution ( obvious limit ) 2. rationalizing N (x) or D (x) 3. expanding N(x) or D (x) 4. combining terms in N (x) or D (x) 5. factoring N (x) or D (x) 6. applying L’Hospital’s Rule Limits to Infinity of a Fraction Let lim

N(x)

N(x) D(x)

= L x→∞ D(x) 1. If the degree of the numerator N (x) is less than the degree of denominator D (x), then L = 0. 2. If the degree of N (x) equals the degree of D (x) then L = coefficient of N (x) with highest degree coefficient of D (x) with highest degree 3. if the degree of N (x) > D (x), then L = ∞

INTERMEDIATE FORMS A. 0/0 or ∞/∞ ( L’Hospital’s Rule is applicable) B. ∞ - ∞ , 0. ∞ ( L’Hospital’s Rule is not directly applicable) C. 00, ∞0, 1∞ Note: If the evaluated function turned out to be in the form of those in B and C, change the form of the given function to obtain an evaluated function in the form of that in A. •

The Indeterminate Form 0. ∞ If f(x) approaches zero and g(x) approaches infinity as x approaches a (or x → ± ∞) , the product ƒ(a). g(a) is undefined and will be of the form 0. ∞. If the limit f(x) . g(x) exists as x → a (or as x → ± ∞), it may be found by writing the product as a fraction lim ƒ(x) g(x) = lim ƒ(x) = lim g(x) x→a x→a 1 x→a 1 /g(x) /f(x) Then apply L’Hospital’s Rule The Indeterminate Form ∞ - ∞ If ƒ(x) and g(x) both increase without bound as x → a (or x → ± ∞), the difference ƒ(a) – g(a) is undefined and will be of the form ∞ - ∞ If the limit of ƒ(x) – g(x) exists as x → a (or x → ± ∞), then by algebraic means. lim [ ƒ(x) – g(x)] = lim x→a

x→a

1

/g(x) - 1/f(x)__ 1___

g(x) f(x) •

The form 0 , ∞ , 1 If the limit of ƒ(x)g(x) exists as x → a (or as x → ± ∞), then by logarithm, lim ƒ(x)g(x) =y can be evaluated 0

0

x→a

Let

L = lim ƒ(x)g(x) x→a

Taking the logarithm of both sides ln L = ln lim ƒ(x)g(x) = lim ƒ(x)g(x) lim g(x) ln ƒ(x) = k x→a

x→a lnL

In L = k , then e = e

k

Supplementary Problems Evaluate the limits given below 1. lim sin 5x x→0 x 2. lim 1 – cos θ θ →∞ 2θ 3. lim (1 + 1/x)3x

x→a

or L = ek

Ans: 5 Ans: 0

x→∞

Slope of tangent line =- 1 dy/dx =- 1 f’(xo)

Ans: e3 1

1

4. lim /3 – /x x→3 x–3 5. lim sinθ θ →0 θ

Ans: 1/9 Ans: 1

ANGLE BETWEEN TWO CURVES (Angle of Intersection) – is defined as the angle between their tangents at their point of intersection. To determine the angle of intersection of two curves, f(x) and g(x) 1. Solve the equations simultaneously to find the points of intersection. 2. Find the slopes m1 = f’ (xo) and m2 = g’ (xo)

6. lim (1/x – 1/3sinx) x→0

Ans: 0

7. 7. lim (x+1) ln x x→0

Ans: 1

8. 8. lim x csc 5x x→0

Ans: 1/5

9. ECE Board exam 1987 Evaluate lim x3 – 2x + 9 x→∞ 2x3 – 8

Then the acute angle θ of intersection is given by

Ans: ½

10. ECE Board Exam 1987 Evaluate lim 2x4 – 2x3 + 9x2 – x + 7 x→∞ x3 – 8 Interpretation of Derivative 1. Derivative as Slope

m1 – m 2 Tan θ = --------------1 + m1m2

Ans: ∞ 2. Derivative as Rate of Change dy /dx = lim ∆y / ∆x x→0

y y=ƒ(x) Tangent Line

Po(xo,yo) θ 0

x

Definitions 

The tangent to the curve with equation y = f(x) at Po(xo, yo) is the line through Po(xo, yo) with slope f’(xo) Slope of Tangent Line = tan θ = dy /dx = f’(xo) The normal to the curve with equation y = f(x) at Po(xo, yo) is the line through Po which is perpendicular to the tangent line at Po Slope of Normal Line =

1

The value of the derivative of the function is the instantaneous rate of change of the function with respect to the independent variable. Rectilinear Motion – motion in a straight line Assumption: 1. Motion will always be assumed to take place along straight line, although the object in motion may go either direction. 2. The body in motion is idealized to be a point or a particle. Let S = displacement of a particle at any time t • Speed (v) – of a particle moving along a curve is the absolute value of the time rate of change of the displacement (or distance), measured along the curve, of the point with reference to some fixed point on the curve. V = ds/ dt  Acceleration (a) – of a moving point is the time rate of change of the velocity of the point a = dv/ dt = d2s/dt2 Differentiation Formulas Let u, v be any functions of x n = any integer C= any constant

I. Basic Formulas 1. 2. 3. 4. 5. 6.

7.

8.

d (c) = 0 dx d (x) = 1 dx d (x n ) = nx n-1 dx d du (cu) = c dx dx d du dv (u + v) = + dx dx dx d udv vdu (uv) = + dx dx dx

du dv v − u d dx dx (u / v) = dx v2 du −k d dx (k / u) = 2 dx u

d n (u ) = nu n −1 dx d 1 ( u) = 10. dx 2 u 9.

du dx du dx

d - n du 11. (1/ u n ) = n+1 dx u dx 12.

dy dy du (The Chain Rule) = • dx dx dx

13.

dy 1 = dx dx / dy

II. Differentiation of Trigonometric Function 1. 2. 3.

d du (sinu) = cosu dx dx d du (cosu) = -sinu dx dx d 2 du (tanu) = sec u dx dx

4. 5. 6.

d du (cotu) = −csc 2 u dx dx d du (secu) = sec u tan u dx dx d du (cscu) = −cscucotu dx dx

III. Differentiation of Inverse Trigonometric Function 1. 2. 3. 4. 5. 6.

d 1 du (Arcsinu) = 2 dx 1 − u dx d - 1 du (Arcosu) = dx 1 − u 2 dx

d 1 du (Arctanu) = dx 1 + u 2 dx d − 1 du (Arcotu) = dx 1 + u 2 dx

d 1 du (Arcsecu) = 2 dx u u − 1 dx d -1 du (Arcscu) = dx u u 2 − 1 dx

IV. Differentiation of Logarithmic Functions

d 1 du (lnu) = dx u dx d 1 du 2. (logu) = (log e) dx u dx 1.

3.

log b e du d (log b u) = dx u dx

V. Differentiation of Exponential Functions

d u du (e ) = e u dx dx d u du 2. (a ) = a u (ln a ) dx dx d v du dv 3. (u ) = u v + lnu dx dx dx 1.

VI. Differentiation of Hyperbolic Function

d du (sinhu) = coshu dx dx d du 2. (coshu) = sinhu dx dx d du 3. (tanhu) = sec 2 u dx dx d du 4. (cothu) = −csch 2 u dx dx d du 5. (sechu) = −sechutanhu dx dx d du 6. (cschu) = −cschucothu dx dx 1.

VII. Differentiation of Inverse Hyperbolic Function

d 1 du (sinh −1u) = 2 dx u + 1 dx d 1 du (cosh −1u) = 2. 2 dx u − 1 dx 1.

d 1 (tanh −1u) = dx 1 − u2 d -1 4. (coth −1u) = dx 1 − u2 3.

du dx du dx

5.

d -1 du (sech −1u) = dx u 1 − u 2 dx

6.

d -1 du (csch −1u) = 2 dx u 1 + u dx

Supplementary Problems: 1. Find the slope of the curve y = x + 2x-2 at point (1, 2).Ans: -3 2. Find the point on the curve y = x2 – 6x + 3 where the tangent is horizontal. Ans: P(3, -6) 3. At what point on the curve y = x4 + 1 is the normal line parallel to 2x + y = 5?Ans: P(1/2, 17/16) 4. Find the point on the curve y = 7x – 3x2 + 2 where the inclination of the tangent is 450.Ans: P(1, 6) 5. Find the point where the normal to y = x + x1/2 + 1 at (4, 7) crosses the y- axis. Ans: P(0,10 1/5)

6. The tangent to y = x3 – 6x2 + 8x at (3, -3) intersects the curve at another point. Determine this point. Ans: P(0, 0) 7. For the curve y = x2 + x, at what point does the normal line at (0, 0) intersect the tangent line at (1, 2)? Ans: P(3/10, -1/10) 8. Find the angle of intersection between the curves y = x 2 + 2 and y = x + x-1 + 1. Ans: θ = 63.40 9. A boy 3 ft. tall walks away from alight which is on top of post 7.5 ft. high. Find the rate of change of the length of his shadow with respect to his distance from the lamp post. Ans: 2/3 ft/ft 10. Find the rate of change the area of an equilateral triangle with respect to the side of this triangle when the latter is 2 ft. Ans: √3 ft2/ft. 11. If the particle moves according to the law S = t 2 – t3 + 3, find the velocity when acceleration is zero.Ans: -1/3 12. If the motion of the body is described by S = 3t 5 – 30t2 + 5, when will the acceleration be zero? Ans: 1 13. Two particles have position at time t given by the equation S 1=t3 + 6t2 – 7t + 1 and S2 = 2t3 – 3t2 – t + Find their position when they have the same acceleration. Ans: S1 = 61; S2 = 25 14. Find y’ and y”, given x3y + xy3 = 2 and x = 1. Ans: y’ = -1;y” = 0 15. Find the equation of the tangent and normal to x2 + 3xy + y2 = 5 at (1, 1). Ans: Tangent: x + y – 2 = 0, Normal: x = y Other Applications of Derivatives Increasing and Decreasing Functions:  A function y = f(x) is said to be increasing if its value increases as y increases  A function y = f(x) is said to be decreasing if y decreases as x increases Given a function f(x) differentiable in the interval a ≤ x ≤ b 1. If f’(x) > 0, then f(x) is increasing 2. If f’(x) < 0, then f(x) is decreasing 3. If f’(x) = 0, then f(x) is stationary Concavity, Critical Points, Inflection Points  Concave Upward The graph of a function is said to be concave upward if the function is decreasing then increasing.  Concave Downward The graph of a function is concave downward if the function is increasing then decreasing.  Maximum Point A point where the function from increasing to decreasing and the function is said to have a relative minimum value.  Critical Point The point at which y’ = 0 and value of x at this point critical value.  Inflection Point A point at which the curve changes its direction of concavity. First Derivative Test

Substitute in the expression for the first derivative a value slightly less than and then a value slightly greater than the critical value under consideration. 1. If f’(x) changes from positive to negative as x increases through the critical value, then the critical is a maximum point. 2. If f’(x) changes from negative to positive as x increases through the critical value, then the critical is a minimum point. 3. If f’ (x) does not changes sign, the critical is neither a maximum or a minimum point. It is the point of inflection with horizontal tangent. maximum point y’= 0 Inflection Point

y = f(x)

y’=0 minimum point

concave downward

x=a concave upward

Second Derivative Test 1. The function y = f(x) has a maximum value at x = a if f’(a) = 0 and f”(a) < 0. The curve is concave downward at that point. 2. The function y = f(x) has a minimum value at x = a if f’(a) = 0 and f”(a) > 0. The curve is concave upward at that point. 3. If at f’(a) = 0, f”(a), then his test fails. Use the first derivative test. Third Derivative Test A function y = f(x) has an inflection point at x = a if f”(a) = 0 and f”’(a) ≠ 0. Supplementary Problems 1. Find the maximum, minimum and inflection point of the curve y=-x3 + 2x2 – x +2 Ans:Max.(1, 2); Min. (1/3, 50/27); Inflection (2/3, 52/27) 2. Find the value of k such that the curve y = x3 – 3kx2 + 5x – 10 Ans: k = 1 3. Determine the equation of the parabola y = ax2 + bx + c passing through (2, 1) and be tangent to the line y = 2x + 4 at point (1, 6) Ans: y = -7x2 + 16x –3 4. What curve of the form y = ax3 + bx2 + cx + d will have critical points at (0, 4) and (2, 0). Ans: y = x3 – 3x2 + 4

5. Determine a, b, c and d so that the curve y = ax 3 + bx2 + cx + d will have horizontal tangents at the points (1, 2) and (2, 3) 6. Find the equation of the line normal to the curve y = 3x 5 – 10x3 + 15x + 3 at its point of inflection. Ans: x + 15y – 45 = 0 7. Find a such that the curve y = 2x 3 – 3ax2 + 12x – 1 will have are of its critical points where x = 2 Ans: a = 3 8. Find the all values of x where the curve y = x2 – 2x + 5 is increasing. Ans: x > 1 MAXIMA AND MINIMA APPLICATIONS Step in Solving Problems Involving Maxima and Minima 1. Identify the quantity to be maximized or minimized 2. Use the information in the problem to eliminate all quantities so as to have a function variables. Determine the possible domain of this function. 3. Differentiate this function with respect to the variable whose maximum/minimum value is to be determined 4. Equate the derivative of step 4 to zero and solve for the unknown. Supplementary Problem 1. ECE Board Exam 1973 An open rectangular tank with square base is to have a volume of 10 cu. m and the material for the bottom is to cost per square meter and that for the sides 6 cents per square meter. Find the height of the tank if the coil of making the tank is to be a minimum. Ans: 2.5 cm. 2. ECE Board Exam Feb. 1973 In problem no. 1 above, find the most economical dimension for the tank. Ans: 2m x 2m x 2.5m 3. ECE Board Exam 1981 Divide 60 into two parts so that the product of one part and the square of the other is a maximum. Ans: 20, 40 4. ECE Board Exam April 1988 Find the altitude of the largest circular cylinder that can be inscribed in a circular cone of radius r and height h. Ans: 1/3 h 5. ECE Board Exam 1989 Find the greatest volume of a right circular cylinder that can be inscribe in a sphere of radius r. Ans: r = 2.418 r3 6. ECE Board Exam Nov. 1995 Find the radius of a right circular cylinder that can be inscribe in a cone of a radius R and height H. Ans: r = 2/3 R 7. ECE Board Exam Feb. 1978 The sum of the two numbers is 36. What are these numbers if their product is to be the largest possible. Ans: 18 and 18 8. ECE Board Exam Feb. 1978 A square sheet of galvanized iron 100cm x 100cm will be used in making an open top container by cutting a small square from each corners and bending up the sides.

9.

10. 11.

12.

Determine how large the square should be cut from each corner in order to obtain the largest possible volume. Ans: 16.67cm x 16.67 ECE Board Exam, Mar. 1989 A rectangular field containing a given area is to be fenced off along a straight river. If no fencing is needed along the river, what should be the dimension of the field so that least amount of fencing materials will be used? Ans: L = 2W. ECE Board Exam, Nov. 1989 Find the minimum volume of a right circular cylinder that can be inscribe in a sphere having a radius equal to r. EE Board Exam , April 1981 A telephone company agrees to put up a new exchange for 100 subscribers or less at a uniform change of P40 each. To encourage more subscriber the company agrees to deduct 20 centavos from their uniform rate for each subscriber in excess of 100. If the cost to serve each subscriber is P 14, what number of subscriber would give the telephone company the maximum net income. Ans: 115 Find the equation of the tangent line to the curve y = x3 – 3x2 + 5x = 2 that has the least slope Ans: 2x – y + 3 = 0

2.

3.

4.

5. 13. Find the area of the largest rectangle with sides parallel to the coordinate axes which can be inscribe in the bounded by x2 = 28 – 4 and x2 = y – 4. Ans: 64 sq. units 14. An isosceles trapezoid is 6 cm long on each side. How long must be the longest side if the area is maximum. Ans: 12 cm. 15. Find the dimension of the right circular cone of minimum volume which can be circumscribed about a sphere of radius 8 cm. Ans: radius = 8√2 cm ; height = 32 cm. 16. Find the dimension of the cylinder of maximum lateral area which can be inscribe in a sphere of radius 6√2 cm. Ans: radius = 6 cm. ; height = 12 cm. 17. Find the ratio of the volume of the right circular cylinder of maximum volume to that of the circumscribing cone. Ans: Vcyl/Vcone = 4/9 18. Find the equation of the line passing through the point (3, 4) which cuts from the first quadrant of a triangle of minimum area. Ans: 4x + 3y – 24 = 0 19. Find the dimension of the right circular cone of maximum volume which can be inscribe in a sphere of radius 12 cm. Ans radius=8√2 cm ; height = 16cm. 20. Find the area of the largest rectangle that can be inscribe in an ellipse 9x2 + 4y2 = 36. Ans: 12π square units. RELATED RATES If a quantity x is a function of time t, the time rate of x given expressed as dx/dt. When two or more time varying quantities are related by an equation, the relation between their rates of change may be obtained by differentiating both members of the equation with respect to time t. Supplementary Problems: 1. ECE Board Exam, Sept. 1986

6.

7. 8.

9.

10.

11.

A baseball diamond is a square, 27 m on each side. The instant a runner is halfway from home to first base, he is giving towards first base at 9 m/s. How fast is his distance from the second base changing at this instant? Ans: -4.025 m/s ECE Board Exam , Sept. 1983 A boat is being towed to a pier. The pier is 20 ft. above the boat. The remaining length of the rope to be pulled is 25 ft. It is being puled at 6 ft. per second. How fast does the boat approaches the pier? Ans: 10 ft/s ECE Board Exam, March 1981 Given a conical funnel of radius 5 cm and height 15 cm. The volume is decreasing at the rate of 15 cu. cm/s. Find the rate of change in height when the water is 5 cm from the top. Ans: 0.43 cm/s ECE Board Exam, April 1998 / ECE Board Exam, Oct. 1985 Sand is pouring from a hole at the rate of 25 cu, ft. per second and is forming a conical pile on the ground. If the conical formation has an altitude always ¼ of the diameter of the base, how fast is the altitude increasing when the conical pile is 5 ft. high? Ans: 1/12.75 ft/s increasing ECE Board Exam, Feb. 1978 A helicopter is rising vertically from the ground at a constant rate of 4.5 m/s. When it is 75 m off the ground, jeep passed beneath the helicopter travelling in a straight line at a constant speed of 80 kph. Determine how fast the distance between them changing after 1 second? Ans: 10.32 m/s (increasing) ECE Board Exam, Feb. 1977 Two boats starts at the same point. One sail due east starting 10 A.M. at a constant rate of 20 kph. The other sail due south starting 11 A.M. at a constant rate of 9 kph. How fast are they separating at noon? Ans: 21.49 kph. ECE Board Exam, August 1976 A dive bomber loss altitude at a rate of 400 mph. How fast is the visible surface of the earth decreasing when the bomber is 1 mile high? Ans: 2792 A man lifts a bag of sand to a scaffold 30 m above his head by means of a rope which passes over a pulley on the scaffold. The rope is 60 m long. If he keeps his end of the rope horizontal and walks away from beneath the pulley at 4 m/s, how fast is the bag rising when he id 22.5 m away?Ans: 2.4 m/s Water is passing through a conical filter 24 cm deep and 16 cm across the top into a cylindrical container of radius 6 cm. At what rate is the level of water in the cylinder rising if when the depth of the water in the filter is 12 cm its level is falling at the rate of 1 cm/min? Ans: 4/9 cm/min A particle starts at the origin and travel up the line y = √3 x at a rate of 5 cm/sec. Two seconds later, another particle starts at the origin and travels up the line y = x at the rate of 10 cm/s. At what rate are they separating 2 seconds after the last particle started? Ans: 0.37 ft/s A particle travels along a parabola y = 5x2 + x + 3. At what point do its abscissa and ordinate change at the same rate? Ans: P(0, 3)

12. At a certain instant the semi major axis and semi minor axis of an ellipse are 12 and 8 respectively and the semi major axis is increasing ½ unit each minute. At what rate is the semi major axis decreasing if the area remains constant? Ans: 1/3 unit/min. 13. A clock hands are 1 and 8/5 inches long respectively. At what rate are the ends approaching each other when the time is 2’o clock? Ans:0.095 in/min. 14. An elevated train on a track 30 ft above the ground crosses a street at the rate of 20 ft. per sec. At the instant that the car approaching at the rate of 30 ft/s and the car are separating 1 sec later/ Ans: 2.67 ft/sec Differential Approximation Approximation of Error If y = f(x), then dy = f’(x) dx dx = change or error in x dy = change or error in y dx/x = relative error in x dy/y = relative error in y dx/x (100) = percentage error in x dy/y (100) = percentage error in y Supplementary Problems: 1. What is the maximum allowable error in the edge of a cube to be used to contain 10 cubic meters if the error in the volume is not to exceed 0.015 cubic meter? Ans: 0.00108

C(h,k), Center of Curvature Normal Circle of Curvature

θ 0

x

Definition: - The curvature K of a curve y = f(x), at any point P(x, y) on it , is the rate of change indirection (that is, the angle of inclination - θ of the tangent line per unit arc length S. K = dθ/ds = lim ∆θ/∆S ∆s→0

y” K = -------------------[1 + (y’)2]3/2

or

- x” K = ---------------------[1 + (x’) 2]3/2

Notes:   2. The semi major axis and semi minor axis of an elliptical plate is measure to be 8 cm and 6 cm respectively. If there is an approximate error of 0.01 cm and .02 cm in measuring error in computing for the area. Ans: 0.628 3. The altitude of a certain circular cone is the same as the radius if the base and is measured as 12 cm with a possible error f 0.04 cm. Find approximately the percentage error in the calculated value of the volume. Ans: 1% 4. Find the allowable percentage error in the radius of a circle if the area is to be correct to within 5%. Ans: 2.5% 5. What is the percentage error made in the computed surface area of a sphere if the error made in measuring the radius is 3%.Ans: 6%. CURVATURE y

tangent

If K > 0, the point P is on the arc that is concave upward If K < 0, the point P is on the arc hat is concave downward

The curvature K is given by: g’ h” - g” h’ K = -----------------------[ (g’)2 + (h)2]3/2 In Polar form, r = f(θ) Where r’ = dr/dθ , r” = d2r / dθ2 r2 + 2(r’)2 – rr” K = ------------------------[r2 + (r’)2]3/2

y = ƒ(x) Radius of Curvature - the radius curvature R for a point P on the curve is the reciprocal of its curvature at that point P(x,y)

R

R = 1/K Notes:  If R > 0, the curve is concave upward  If R r2

r1 r2

b

s

Supplementary Problems 1. Bisectors of the 3 angles of a triangle meet at a common point called the __________ a. orthocenter b. centroid c. incenter d. circumcenter 2. The perpendicular bisector of the sides of a triangle pass through a common point called the __________ a. orthocenter b. centroid c. incenter d. circumcenter 3. Which of the following is not a property of a circle? a) through 3 points not in the straight line one circle and only 1 can be drawn b) a tangent to a circle is perpendicular to the radius at the point of tangency and conversely. c) an inscribed angle is measured by ½ of the intercepted arc. d) the arc of 2 circles subtended by equal central angle are equal. 4. Which of the following is not a property of a triangle? a) the sum of the 3 angles of the triangle is equal to two right triangles. b) the sum of the 2 side of the triangle is less than the 3 rd side c) if the 2 sides of the triangle are unequal, the angles opposite are unequal. d) the altitude of a triangle meets in a point 5. The radius of the circle inscribed in a polygon is called as a) internal radius b) radius of gyration c) apothem d) hydraulic radius 6. A polygon with 12 sides is called as a) bidecagon b) dodecagon c) nonagon d) pentedecagon 7. A polyhedron having bases 2 polygons in parallel plane and for lateral faces triangles or trapezoid with 1 side lying on 1 base and the opposite vertex or side lying on the other base of the polyhedron is a) pyramid b) cone c) prismatoid d) rectangular parallelepiped 8. An angle greater than a straight angle but less than 2 straight angles is called as a) complement b) supplement d) complex d) reflex 9. A part of a circle is often called as a) sector b) cord b) arc d) segment 10. An angle whose vertex is appoint on the circle and whose sides are cords is known as a) interior angle b) vertical angle

c) acute angle d) inscribed angle 11.Two angles whose sum is 360 degrees are said to be a) supplementary b) complimentary b) elementary d) explementary 12. All circles having the same center but with unequal radii are called as a. eccentric circles b. concentric circles c. inner circles d. Pythagorean circles 13. A circle is _________ outside the triangle if it is tangent to one side and the other two sides prolonged. a. inscribed b. escribed c. circumscribed d. tangent 14. A triangle having three sides of unequal length is known as a. equilateral triangle b. scalene triangle c. isosceles triangle d. equiangular triangle 15. In a proportion of four quantities, the first and the fourth terms are referred to as the a.means b. extremes c. denominators d. axiom 16. A statement the truth of which is admitted without proof is a. theorem b. corollary c. postulate d. axiom 17. The part of the theorem which is assumed to be true is the a. corollary b. hypothesis c. postulate d. conclusion 18. In Geometry,the construction or drawing of lines and figures, the possibility of which is admitted without proof is called the: a. corollary b. theorem c. postulate d. hypothesis 19. A statement the truth of which follows with little or no proof from the theorem is a. corollary b. axiom c. postulate d. conclusion 20. A polygon is ______ when no side, when extended, will pass through the interior of the polygon. a. convex b. equilateral c. isoperimetric d. regular 21. A circle is said to be _______ to a polygon having the same perimeter with that of the circle a. congruent b.isoperimetric c.proportional d. similar 22. The intersection of the sphere and the plane through the center is the a. great circle b. small circle c. poles d. polar distance

23. Points that lie on the same plane are said to be a. collinear b. coplanar c. dihedral d. parallel

36. Find the area of the folded triangle shown below.

24. What kind of a quadrilateral is always formed by connecting the midpoints of the consecutive sides of a quadrilateral? Ans. parallelogram 25. The angles of a pentagon are in the ratio 3: 3: 3: 4: 5. Find the largest angle. Ans. 150deg 26. The legs of a triangle are in the ratio 3: 3 and its area is 108 sq. cm. Find the length of the legs. Ans. 12 cm., 18 cm 27. Find the side of a regular octagon inscribed in a circle of radius 19 cm. 28. The upper and lower bases of a trapezoid are 6cm and 12cm respectively, and the altitude is 4cm. The non-parallel sides of the trapezoid are produced until they meet at P. Find the altitude of the triangle whose vertex is P and whose base is the lower base of the trapezoid. Ans. 8cm 29. A secant and a tangent are drawn to circle from the same point. If the internal segment is 1cm longer than the external segment, and if the tangent is 6cm long, find the length of the secant. Ans.√145 cm. 30. Two parallel chords of a circle are 16cm and 30 cm long respectively, and the distance between them is 23cm long. Find the distance each is from the center, and radius of the circle. 15cm and 8cm., r = 17 cm. 31. The diagonals of a rhombus have the ratio of 3:4. The area of the rhombus is 96 sq.cm. Find a side of the rhombus. Ans. 10cm 32. The areas of two similar triangles are to each other as 9:25. The perimeter of the first is 36 cm. What is the perimeter of the second triangle? Ans. 60cm. 33. The side of a rhombus is 13cm, and one of its diagonals is 24cm. Finds its area. 120 sq.cm. 34. In a circle whose radius is 10cm, a chord bisects the radius at right angles. Find the area of the smaller of the two segments into which the chord divides the circle. 25/3(2π - 3√3)sq cm. 35. From a point P, exterior to circle C, the tangents are drawn to the circle. The distance of P from the center of C is 4cm. The length of each tangent is 2cm. Find the diameter of the circle. Find the area bounded by the tangents and the arc between them. Ans. d = 4√3 cm, A = (4√3 - 2π) sq. cm.

a) 15 b) 20 c) 25 d) 40 37. The area of a triangle whose angles are A = 69.159˚, B = 34.246˚, C = 84.595˚ is 680.60 m2. The longest side is a) 15.387 b) 52.431 c) 37.853 d) 64.974 38. Find the area of a decagon that can be inscribed in a circle with radius 10 cm. a)11.387459 b)10.066446 c)37.853527 d)64.974136 39. Find the area of a regular 5-pointed star that can be inscribed in a circle with radius 10 cm. a) 112.257 b) 56.129 c) 114.535 d) 124.431 40. Find the radius of the largest circle that can be inscribed in the triangle with sides a = 8cm, b = 15cm, and c= 17cm Aa) 3 b) 1.8 c) 30 d) 8.5 41. Find the radius of the smallest circle that circumscribe the triangle in the previous problem. a) 8.5 b) 10.2 c) 9.51 d) 7.75 42. Chords AB = 12 cm and BC = 8cm of a circle forms 120˚. Find the radius of the circle. a) 10 b) 12.164 c) 14.742 d) 24.634 43. The radius of the smallest circle that can be circumscribe a right triangle of sides a, b, and c. a. a + b + c b. a + b – c 2 2 C. a – b + c d. c 2 2 44. The radius of the largest circle that an be inscribed in a right triangle of sides a, b and c. a. a + b + c b. a + b – c 2 2 c. a – b + c d. c 2 2 Find CD.

45. Find the area of the cyclic quadrilateral.

AP = 50cm, BP = 28cm, DP = 56cm, Ө = 30degrees a) 456 b) 627 d) 364 d)525 46. The internal angle of a polygon is 150˚ greater than its external angle, how many side has the polygon? a) 9 b) 6 c) 7 d) 8 47. The 30˚ angle of a right triangle is bisected. Find the ratio of which opposite side is divided. a) 1:2 b) 1:3 c) √2:2 d) 1:4 48. A certain angle has a supplement 5 times the compliment. Find the angle. a. 67.5˚ b. 58.5˚ c. 30˚ d. 27˚ 49. Find the supplement of an angle whose compliment is 62 degrees a. 30˚ b. 28˚ c.152˚ d. 118˚ 50. The sum of the interior angles of a polygon is 540 degrees. Find the number of the sides. a. 5 b. 6 c. 8 d. 11 51. Six lines are situated in the plane so that no two are parallel and no three are congruent. How many points of intersection are there? A. 13 b. 14 c. 15 d. 16 52. The area of a square field exceeds another square by 56 square meters. The perimeter of the larger field exceeds ½ of the smaller by 26 m. What are the sides of each field? Ans. larger field, 9m or 25/3m; smaller field, 5m or 11/3m 53 . The sum of the areas of two unequal square lots is 5,200 square meters. If the lots were adjacent to each other, they would require 320 meters of fence to enclose the combined area formed by them. Find the dimensions of each lot. Ans.60m and 40m, or 68m and 24m.

54. Six lines are situated in the plane so that no two are parallel and no three are congruent. How many points of intersection are there? a. 13 b. 14 c. 15 d. 16 55. A triangle is inscribed in ac ircle such that one side of the triangle is a diameter of the circle. If one of the acute angles of the triangle has a measure of 60˚ and the opposite side of that angle has a length 12, then the nearest value of the radius of the circle is. a. 6.93 b. 1.93 c. 9.6 d. 5.8 56. The hypotenuse of a right triangle is 34 cm. Find the length of the two legs if one leg is 14 cm longer than the other. a. 18 and 32 cm b. 15 and 29 cm c. 17 and 32 cm d. 16 and 30 cm 57. The sum of the interior angles of a polygon is 540 degrees. Find the number of the sides. a. 5 b. 6 c. 8 d. 11 58. A circle of radius 6 has half its area removed by cutting off a border of uniform width. Find the width of the border. a. 22 b. 135 c. 375 d. 176 59. A circle is inscribed in a 3, 4, 5 right triangle. How long is the line segment joining the point of tangency of the “3-side” and the “5-side”? a. 1.28 b. 1.35 c. 1.46 d. 1.79 60. Two figures having equal perimeter are said to a. congruent b. isoperimetric c. equal d. similar 61. A right triangle whose length and side may be expressed as ratio of integral units a. isosceles triangle b. scalene c. pedal triangle d. primitive triangle 62. The perpendicular bisector of the sides of a triangle intersect at the point known as the a. orthocenter b. cicumcenter c. centroid d. incenter 63. The bisectors of the 3 angles of a triangle meet at a common called the a. circumcenter b. centroid c. orthocenter d. incenter 64. An isosceles trapezoid ABCD, AB=5, BC=AD = 3 and CD = 8. Find the length of the diagonals. a. 7.5 b. 8 c. 7 d. 6 65. The diagonals of rhombus have length 4 and 6 inches. Find the area of the region inside the rhombus but outside the circle that is inscribed in a rhombus. a. 3.3 b. 3.4 c. 2.9 d. 2.85 66. Two equilateral triangles, each with 12-cm sides, overlap each other to form a 6-point “Star of David”. Determine the overlapping area, in sq-cm. a. 36.64 b. 41.57 c. 28.87 d. 49.88 67. Let D be the set of vertices of a regular dodecagon (12 sided plane polygon). How many triangles may be constructed having D as the vertices?

a. 220

b. 120

c. 240

d. 180 Rectangular Parallelepiped – polyhedron whose six faces are all rectangles.

68. A group of children playing with marble place 50 pieces of the marble inside the cylindrical container with water filled to a height of 20 cm. If the diameter of aech marble is 1.5 cm and that of the cylindrical container 6 cm, what would be the new height of water inside the cylindrical container after the marbles were placed inside? a. 23.125 b. 24.125 c. 26.125 d. 25.125

Prism – polyhedron of which two faces are equal polygon in parallel planes and the other faces are equal parallelogram. Pyramid – a polyhedron of which one faces is a regular polygon and other faces are triangles which have a common vertex.

69. A horizontal cylindrical tank with hemispherical ends is to be filled with water to a height of 762mm. If the total inside length of the cylinder is 3,600mm, find the volume of water in cubic meters that will have to filled into the tank up to the required height. If one edge of a cube and the volume of the cube respectively in sq.cm. and cu.cm. a. 864 and 1728 b. 684 and 1728 c. 864 and 1728 d. 864 and 1729

Regular Prism – prism whose lateral edges are perpendicular to its bases.

70. A pyramid with square base has an altitude of 25 cm. If the edge of the base is 15cm, calculate for the volume of the pyramid. a. 1885 b. 1875 c. 1785 d. 1958

Section – polyhedron is the plane figure formed by a plane passing through the solid.

71. If a right circular cone has a base radius of 35 cm and an altitude of 45cm, solve for the total surface area in square cms. a.10116 b. 10117 c. 11117 d. 12117

Frustum of Pyramid – section of the pyramid between the base and a section parallel to the base.

Regular Pyramid – pyramid whose base is a regular polygon and whose altitude passes through the center of base. Slant height – altitude of lateral faces.

Convex Polygon – a polygon whose each interior angle of which is less than 18 0.

Cone – a solid bounded by a conic surface and a plane intersecting all the elements. 72. Two circles of radius 4 and 6.93 are placed in a plane so that their circumference intersect at right angles. Find the area of the interlapping region. a. 14.81 b. 13.92 c. 14.18 d. 15.1

Dihedral Angle – the divergence of two intersecting planes.

73. One side of a regular octagon is 2. Find the area of the region inside the octagon. a. 3.91 b. 13.92 c. 14.18 d. 15.1

Sphere – a solid bounded by a surface all points of which are equidistant from a point called center. Great circle – the intersection and a plane passing through the center.

SOLID GEOMETRY

Small circle – the intersection of a sphere and a plane not passing through the center.

Definition of Terms: Polyhedron – a solid bounded by planes. Regular Polyhedron – polyhedron whose faces are congruent regular polygons and whose polyhedral angles are regular polygons and whose polyhedral angles are equal.

Quadrant – one-fourth of a great circle.

There are only five regular polyhedron: Tetrahedron – 4 faces Hexahedron – 6 faces Octahedron – 8 faces Dodecahedron – 12 faces Icosahedron – 20 faces Cube – a polyhedron whose six faces are all squares

Zone – portion of sphere bounded by a spherical polygon and the plane of its sides. Lune – portion of a sphere lying between two semi – circles of great circle. Spherical pyramid – portion of sphere bounded by a spherical polygon and the plane of its sides. Spherical Sector – portion of sphere generated by the revolution of circular sector about any diameter of the circle of which the sector is apart. Spherical segment - portion of sphere included between two parallel lines.

Spherical wedge – portion bounded by a lune and the planes of two great circles. Focus – a solid formed by revolving a circle about a line not intersecting it.

Prismoidal Formula: V =

1 h( b + B + 4M) 6

Formulas in Solid Mensuration 1. Cube with edge s: Volume : V = s 3 Surface Area : A = 6s 2

6. Sphere of radius r or diameter D: Volume : V =

4 3 1 πr or V = πD 3 3 6

Surface Area : S = 4πr 2 or S = πD 2 7. Right circular cylinder with radius r and altitude h:

2. Rectangular parallelepiped with edges a, b, c and diagonal D: Volume : V = abc Surface Area : A = 2( ab + ac + bc ) Diagonal : D = a 2 + b 2 + c 2 3. Volume of a prism with base B and altitude h:

V = Bh

Volume : V = πr 2 h Lateral Area

: S = 2πrh

8. Right circular cone with radius r and altitude h: Volume : V = Lateral Area

1 2 πr h 3

: S = πrl

(l = slant height)

4. Volume of a pyramid with base B and altitude h:

V=

1 Bh 3

5. Volume of a prismatoid with bases b and B, midsection M and altitude h:

9. Frustum of a right circular cone with base radii r and R and altitude h: slant height l: Volume : V = Lateral Area

(

1 πh r 2 + R 2 + rR 3

)

: S = πl( r + R )

10. Frustum of a pyramid with bases b and B and altitude h:

Volume : V =

(

1 πh b + B + bB 3

)

Volume : V = where

1 (p + P )l 2

Lateral Area

: V=

where

l = slant height

1 2 ZR or V = πR 2 h 6 3

V = volume of spherical sector

Z = area of the zone which forms

p = perimeter of base b

the base of the sector

R = radius of the sphere

P = perimeter of base B

h = altitude of the sphere

11. Area Z of a zone with altitude h on a sphere of radius R: 15. Ellipsiod V = 4/3 π abc

Z = 2πRh

b a

For oblate spheroid V = 4/3 π ab2

12. Spherical segment of one base and altitude h on a sphere of radius R: Volume : V = Total Area

1 2 πh ( 3R − h) 3

c

For prolate Spheriod V = 4/3 π a2 b

16Paraboloid

: T = πh( 4R − h)

h

V = ½ π a2 h a

13. Spherical segment of two bases with radii a and b and altitude h on a sphere 17. Ungula V = 2/3 r2h S = 2 rh

of radius R: Volume : V = Total Area

(

1 πh 3a 2 + 3b 2 + h 2 6

(

: T = π 2Rh + a + b 2

) 2

h

r

)

14. Spherical Cone = a spherical sector having only one conical surface Supplementary Problems:

1.

The slant height of a regular hexagonal pyramid is 9 cm. and a side of the base is 6 cm. Find the volume of the inscribe cone. Ans. 27 √6π cu. cm.

2.

A rectangle, 4 cm wide and 9 cm long is revolved about its longer side. Find the total surface area and the volume of the cylinder generated. Ans. T = 104π sq. cm. ; V = 144π cu. cm

3. A solid wooden cone is cut into two parts, the cut being parallel to the base and halfway between the base and the vertex. Find the ratio of the weights of the two parts. Ans. 7 : 1 4.

Find the area of a zone of one base, if the base is a circle of radius 6 cm and is 8 cm from the center of the sphere. Ans 40 π sq. cm.

5. Three spheres of radii a, 2a, and 3a are melted and formed into a new sphere. Find the surface of this sphere. Ans. 24 3√6 πa2 6.

A sphere whose radius is 10 cm is cut into three parts by parallel planes 8 cm and 6 cm. from the center respectively, and on the opposite sides of the center. Find the total surface of the part in the middle. Ans. 380 π sq. cm.

7. In a frustum of a regular pyramid the bases are squares with sides 6 cm and 12 cm respectively. If the lateral area of the frustum is just half of its total area, what is the volume? Ans: 420 m3 8.

The volume of two similar polyhedron are 64 and 125 cu cm. respectively. If the total surface of the first is 112 square centimeters, find the total surface area of thesecond.Ans.380πsq.cm

9. Find the total surface and the volume of a tetrahedron whose edges are equal to a. Ans: T = √3 a2 ,V = (√2/12) a3 10. A sphere of radius 5 cm and a right circular cone of bass radius 5 cm and height 10 cm stand on a plane. Find the position of a plane that cuts the two solids in equal circular sections. Ans: d = 2 cm and h = 10 cm 11. A cylindrical tin can has its height equal to the diameter of its base. Another cylindrical tin can with the same capacity has its height equal to twice the diameter of its base. Find the ratio of the amount of tin required for making the two cans with covers. Ans. 0.9524 12. Find the volume of the frustum of a regular square pyramid whose base edges are 4 cm and 10 cm and whose slant height is 5 cm. Ans. 208 cu cm

13. Find the volume of a prism having an altitude of 13 cm and a rectangular base of 8 cm long and 4 cm wide. Ans.416 cm3 14. The base of a right prism is a rhombus whose sides are each 10 cm long and whose shorter diagonal is 12 cm. Find its volume if the altitude is 8 cm. Ans. 768 cu cm 15. The diameter of two spheres are in the ratio 2:3 and the sum of their volumes is 1,260 cu m. Find the volume of the larger sphere. Ans. 972 cu m If the volume of a cube is 625 m3, find the length of the diagonal. Ans. 14.8 m 16. Find the volume of a regular triangular pyramid whose slant height is 17 cm and whose altitude is 15 cm.Ans. 960 3 cu cm 17. Find the capacity in liters of a pail in the form of a frustum of a circular cone if the radii of the bases are 10 and 15 cm and the depth of the pail is 36 cm. Ans. 17.9 L 18. The space occupied by the water in a reservoir is the frustum of a right circular cone. Each axial section of this frustum has an area of 28 sq m and the diameter of the upper and lower bases are in the ratio 4:3. If the reservoir contains 148 Π/3 cu m, find the depth of the water in the reservoir. Ans. 4m 19. The lateral surface area of a right circular cylinder is 330 Π sq cm. If its altitude is 20 cm, find the diameter of its base. Ans. 16.5 cm 20. A reservoir is in the form of the frustum of an inverted square pyramid with upper base 24 m, lower base edge 18 m and altitude 9 m. How many hours will it require for an inlet pipe to fill the reservoir if water flows in at the rate of 5,000 liters per minute? Ans. 13.32 hr 21. A horizontal cylindrical tank with diameter of 0.60 m and 3.66 m long is filled with water to a depth of 0.46 m. Find the number of liters of water in the tank. Ans. 851 L 22. A sphere is inscribed in aright circular cone with altitude 15 cm. If the slant height of the cone is equal to the diameter of its base, find the surface area of the sphere. Ans. 100 sq cm 23. Use the prismoidal formula to find the volume of the common part of two cylinders, each with a radius of 6 cm, which intersect at right angles. Ans. 1,152 cu cm 24. If the radius of a sphere is inscribed by 6 cm, its volume is multiplied by 27. Find the radius of the sphere. Ans.3 cm

1. Find the slope of x2y = 8 at point (2, 2) a. –2 b. 2 c. 8 2. If y = x lnx, find y” a. 1/x b. ln x c. 1/lnx 3. Evaluate the limit (x - 4)/(x2 – x – 12) as x approaches 4

d. 4 d. x

a. 1/7 4. Evaluate the limit

b. 0

c. infinity

lim

d. indeterminate

x – sin x x3 a. 1/3 b. ¼ c. 1/5 d. 1/6 5. Find the maximum point on the curve y = x3 – 3x2 – 9x + 5 a. (1, 15) b. (3, -22) c. (-1, 10) d. (-3, 21) 6. Determine the velocity of a body which moves according to the law S = 2t 3 – t2 + 4 where S is displacement in ft. and t is time in sec at t = 1sec. a. 2ft b. 6ft. c. 4ft. d. 10ft. 7. Find the equation of the tangent to the curve y = x 4 – x2 + 2 at x = -1. a. 2x + y = 0 b. 2x – y = 1 c. 2x – y = 0 d. y + x = 1 8. Determine the altitude of the largest circular cylinder that can be inscribed in a right circular cone of radius 6 inches and of height 15 inches. a. 12 inches b. 8 inches c. 5 inches d. 10 inches 9. Given a cone of radius R and of altitude H, what percent is the volume of the largest cylinder which can be inscribed in the cone to the volume of the cone? a. 49% b. 44% c. 50% d. 60% 10. Find the most economical proportion for a box with an open top and a square base. a. b = h b. b = 4h c. b = 3h d. b = 2h 11. The 5m picture hung on a wall so that its bottom edge is 4m above an observer’s eye. How far should the observer stand from the wall so that the angle subtended by the picture at the eye is a maximum. a. 4.9 b. 5.5 c. 6.7 d. 7.2 12. What is the largest area of a rectangle that can be inscribed within an ellipse whose major diameter is 10ft and whose minor diameter is 6ft. a.32.50 sq. ft. b. 28.00 sq. ft c. 30.00 sq. ft d. 34.00 sq.ft 13. The hands of the tower clock are 4 ½ ft and 6 ft long respectively. How fast are the ends approaching at 4 o’clock in ft per minute? a. – 0.246 b. – 0.203 c. –0.264 d. –0.256 14. A man on the wharf (pier) is pulling a rope tied to a raft at a time rate of 0.60 m/sec if the hands of the man pulling the rope is 3.66m above the level of the water, how fast is the raft approaching the wharf when there are 6.10m of rope out? a. –0.75 m/sec b. –0.55 m/sec c. –0.45 m/sec d. –0.65 m/sec 15. A balloon is rising vertically over a point A on the ground at the rate 15 ft/sec. A point B on the ground level is with the same horizontal plane as A and 30 ft from it, when the balloon is 40 ft from A, at what rate is its distance from B changing? a. 12 ft/sec b. 15 ft/sec c. 18 ft/sec d. 21 ft/sec 16. What is the percentage error made in the computed surface area of a sphere if the error made in measuring the radius is 3% a. 3% b. 4% c. 5% d. 6% 17. What is the allowable error in measuring the edge of the cube that is intended to hold 8 cu.m. if the error of the computed volume is not to exceed 0.03 cu.m.? a. 0.002 b. 0.003 c. 0.0025 d. 0.001 18. Find the radius of curvature of the curve y = x3/3 at x = 1. a. 2 b. 2 /3 c. 3/ 2 d. 2 /2 x

19. 20.

0

21. 22.

23.

24.

Find the area bounded by the curves y = 4x – x2, y = 0, x = 1, x = 3. a. 23/3 b. 22/3 c. 21/4 25/3 Find the volume generated by revolving the region bounded by y = x 2 and y = x about the y= axis. a. π/8 b. π/10 c. π/6 d. π/12 If the first derivative of a function is constant, then the function is said to be: a. constant b. linear c. sinusoid d. exponential Three sides of the trapezoid are each 8 cm long. How long is the 4 th side when the area of the trapezoid has the greatest value? a. 16 cm b. 12 cm c. 15 cm d. 10 cm Water is running out from a canonical funnel at a rate of 2 cu. in. per second. If the radius of the top of the funnel is 4 inches and the altitude is 8 inches, find the rate at which the water level is dropping when it is 2 inches from the top. a. 2/9π in/sec b. -2/9π in/sec c. -3/2π in/sec d. 5/9π in/sec Find the area bounded by the parabola x2 = 8y and its latus rectum. a. 10.67 sq. units b. 32 sq. units c. 48 sq. units d. 16.67 sq. units

25. What theorem is used to solve for centroid?

a. Pappus

b. Varignon’s

c. Castiglliano’s d. Pascal’s

26. Find the area (in sq. units) bounded by the parabola x 2 – 2y = 0 and x2 + 2y = 8.

a. 11.7 b. 4.7 c. 9.7 d. 10.7 27. Find the equation of the curve at every point of which the target line has a slope of 2x. a. x =-y2 + C b. y = -x2 + C c. y = x2 + C d. x = y2 + C 28. The rate of change of a function of y with respect to x equals 2 – y, and y = 8 when x = 0. Find y when x = ln 2 b. 5 b. 2 c. –5 d. –2 29. Solve the differential equation (x2 + y2)dx + 2xydy = 0 c. 3xy2 + x3 = C b. 2xy2 = C c. 3xy + 2 = C d. 3x2 + 2y = C 30. Solve the differential equation y” – 5y’ + 6y = 0 d. Y = Ae2x + Be3x b. Y = Ae-2x + Be-3x 5x 3x c. Y = Ae + Be d. Y = Ae2x + Be5x 31. Find the equation f the family of orthogonal trajectories of the system of parabolas y 2 = 2x + C. e. y = ce-x b. y = ce2x c. y = cex d. y = ce-2x 32. The rate of change of a certain substance is proportional to the amount of substance is 10 grams at the start and 5 grams at the end of 2 minutes, find the amount of substance remaining at the end of 6 minutes. f. 1.25 grams b. 2.67 grams c. 3.46 grams d. 2.98 grams 33. The rate of population growth of a country is proportional to the number of inhabitants. If the population of a country now is 40 million and is expected to double in 25 years, in how many years will the population be 3 times the present? g. 39.6 yrs b. 39.5 yrs c. 37.9 yrs d. 36.9 yrs

34. Water at 100°C is transferred to a room which is at constant temperature of 60°C. After 3 minutes the water temperature is 90°C, find the water temperature after 6 minutes. h. 82.5°C b. 85.2°C c. 80°C d. 75°C 35. A body at 90°C cools in 10 mins to 70°C in a room temperature of 25°C. When will its temperature be 40°C? i. 39.8mins b. 38.8mins c. 36.8mins d. 34.7mins 36. Evaluate ln (3 + j4) j. 1.77 + j0.843 b. 1.61 + j0.927 c. 1.95 + j0.112 d. 1.46 + j0.102 37. Find the value of (1 + i)12 where i is an imaginary number. k. –64 b. 64 c. 4 d. 4i 38. Simplify the expression i1997 + i1999 l. 0 b. –1 c. 1 + i d. 1- i 39. Express e0.32 + j0.56 in rectangular form m. 1.167 + j0.732 b. 1.193 + j1.163 c. 1.452 – j0.315 d. 1.684 – j1.462 40. Evaluate cos (0.492+j0.942) n. –1.032 + j0.541 b. 1.302-j0.504 c. 3.12+j1.54 d. 1.48+j0.01 41. Evaluate the value of log (-5) o. 5+jπ log e b. 5π+j log e c. log 5+jπ log e d. log 5π+j log e 42. Find the Laplace transform of t3 e4t p. 6/(s+4)4 b. 6/(s-4)4+ c. 6/(s-2)2 d. 6(s+4)4 43. Find the Laplace transform of (1 – e-at) q. 1/[s(s+a)] b. 1/(s2 + a2) c. 1/[s(s-a)] d. 1/(s+a)2 44. In complex algebra, we use a diagram to represent a complex plane commonly called: a. De Moivre’s diagram b. Argand diagram c. Venn Diagram d. Funicular diagram 45. When the corresponding elements of two rows of a determinant are proportional, then the value of the determinant is: a. unknown b. one c. zero d. multiplied by the ratio 46. A sequence of number where the succeeding term is greater than the preceeding term. a. isometric series b. divergent series c. convergent series d. dissonant series 47. In any square matrix, when the two elements of any two rows are exactly the same, the determinant is: a. unity b. positive integer c. zero d. negative integer 48. Which of the following cannot be an operation of matrices? a. Subtraction b. multiplication c. addition d. division 49. Convergent series is a sequence of decreasing numbers or when the succeeding term is ___________ than the preceding term. a. ten times more b. equal c. greater d. lesser 50. Find the length of the vector (2, 4, 4). a. 8.75 b. 7.00 c. 6.00 d. 5.18 51. Find the value of (1 + i)5, where i is an imaginary number.

a. 1 – i

b. 1 + i

c. -4(1 + i)

d. 4(1 + i)

52. Find the equation of the family of orthogonal trajectories of the system of the

parabolas y2 = 2x + C. a. y = Ce-x b. y = Ce2x c. y = Cex d. y = Ce2x 53. What is the quotient when 4 + 8i is divided by i3. a. 8-4i b. 8+4i c. -8+4i d. -8-4i 54. Evaluate the expression (1+i2)10, where i is an imaginary number. a. -1 b. 10 c. 0 d. 15 55. Solve for x in (x+yi) (2+4i) = 14 + 8i. a. 3 b. 4 c. 14 d. 8

TRIGONOMETRY Trigonometry – the branch of mathematics that deals with the solution of triangles. Angle – the space between two line meeting at a point called vertex. Kinds of Angles: 1.

Acute angle – an angle which measures between 0° to 90°

2.

Right angle – an angle measuring exactly 90°

3.

Obtuse angle – an angle which measures between 90° to 180°

4.

Straight angle – an angle measuring exactly 180°

5.

Reflex angle – an angle greater than 180° but less than 360°

Two General Classes of Triangles 1. Right Triangle – a triangle with a right angle 2. Oblique Triangle – a triangle without a right angle Oblique Triangles can be further classified as: 1. Acute Triangle – a triangle whose all angles are acute. 2. Obtuse Triangle – a triangle with one obtuse angle.

The Pythagorean Theorem -

4. CscA=1/SinA

states that the sum of the squares of the legs is equal to the square of the 5. Versed SinA=1-CosA

hypotenuse. Pythagorean Triple – three positive integers satisfying the Pythagorean principle. Example: 3, 4, and 5; 5, 12, and 13; 20, 21 and 29; 8, 15, 17; 7, 24, 25; etc.

6. Coversed SinA=1-SinA 7. ExsecantA=SecA-1

Supplementary Problems: 1. A storm broke a tree 50 ft high so that its top touched the ground 30 ft from the foot of the tree. What is the height of the part standing? Ans. 16ft.

III. Pythagorean Identities 1. Cos2A+Sin2A=1

2. 3. How far from the center of a circle is its chord 8 inches long if its radius is 5 inches. Ans.

2. 1+Tan2A=Sec2A

3” 4. For what positive value of x will the following lengths be sides of a right triangle 2x + 1, 5x

3. Cot2A+1=Csc2A

– 1, 8x – 3? (The last being the longest) Ans. X = 1 IV. Sum and Difference of Angles Identities 1. Sin(A+B)=SinACosB+CosASinB

I. Circular Functions 1.SinA=y/r

4. CotA=x/y P(x,y)

2. Sin(A-B)=SinACosB-CosASinB

2.CosA=x/r

5.SecA=r/x

3. Cos(A+B)=CosACosB-SinASinB

3.TanA=y/x

6.CscA=r/y

4. Cos(A-B)=CosACosB+SinASinB 5.

Tan( A +B) =

TanA + TanB 1 − TanATanB

6.

Tan( A − B ) =

TanA − TanB 1 + TanATanB

II. The Relation Among Functions 1. TanA=SinA/CosA 2. CotA=CosA/SinA=1/TanA

V. Double Angle Identities 3. SecA=1/CosA

1. Sin2A = 2SinACosA

2. Cos2A = Cos2A-Sin2A 2

3.

2. 2CosASinB=Sin(A+B)-Sin(A-B)

= 2cos A-1

3. 2CosACosB=Cos(A+B)+Cos(A-B)

= 1-2Sin2A

4. 2SinASinB=Cos(A-B)-Cos(A+B)

Tan2 A =

2 TanA 1 − Tan2 A

VI. Complementary Angle Identities SinA = Cos(90-A) CosA = Sin (90-A) TanA = Cot(90-A) CotA = Tan(90-A)

IX. Sum and Difference of Sine and Cosine Identities 1. SinA+SinB=2Sin ½ (A+B)Cos ½ (A-B) 2. SinA-SinB=2Cos ½ (A+B)Sin ½ (A-B) 3. CosA+CosB=2Cos ½ (A+B)Cos ½ (A-B) 4. CosA-CosB=2Sin ½ (A+B)Sin ½ (A-B) X. Sine Law 1.

SecA = Csc(90-A) CscA = Sec(90-A) VII. Half Angle Identities 1.

1 − cosA Sin A / 2 = 2

a b c = = sina sinb sinc

2. A+B+C = 180 XI. Cosine Law

2

1 + cosA 2

2.

Cos 2 A / 2 =

3.

1 − cosA Tan 2 A / 2 = 1 + cosA

VIII. Product of Sine and Cosine Indentities 1. 2SinACosB=Sin(A+B)+Sin(A-B)

1. a2 = b2+c2-2bcCosA 2. b2 = a2+c2-acCosB 3. c2 = a2+b2-2abCosC 4. A+B+C = 180 XII. Tangent Law

π/b 1.

a − b Tan1/ 2(a − b) = a + b Tan1/ 2(a + b)

– for tangent and cotangent – distance of one complete wave of the function

3) Phase Shift  /b c

– property true for any functions –

2.

a − c Tan1/ 2(a − c) = a + c Tan1/ 2(a + c)

distance the graph is shifted to the right or to the left from its

standard position – if c < 0, the graph is shifted to the right – if c > 0, the graph is shifted to the left

3.

b − c Tan1/ 2(b − c) = b + c Tan1/ 2(b + c)

4) Vertical translation d

– true for all functions – defined as the distance the graph is shifted upward or downward

4. A+B+C = 180

– if d > 0, the graph is shifted upward – if d < 0, the graph is shifted downward

5. Properties of Graph of Circular Functions

Sample Problem: Determine the maximum value and the period of the function f(x) = –3 sin( π/4 x + 5) - 7

Consider the function

Solution: a = 3

y = a sin (bx + c) + d

d = –7 The graph has an amplitude of 3, but it is shifted downward by 7 units. Therefore the

a

maximum value of the function is 3 – 7 = –4. Period = 2π/(π/4) ;

d

c

/b

Period = 8

Supplementary Problems 1. If tan y = 1/3 and tan m = ½, y and m being acute, find y + m.

/b

2. 1) amplitude a – highest value of the function in standard form, a property is true for sine, cosine functions only 2) Period 2π/b

since b = π/4

– for sine, cosine, secant and cosecant functions.

If tan A = ¾ and Sin B = 12/13, A being greater than 180° and B is obtuse , find a) Sin (A + B) b) Cos (A + B) c) Tan (A + B)

3. Find the value of Cos (A + B) if tan A = ¾ and csc B = 13/15, both angles are acute. 4.

If

tan x + tan y

Ans: P2 + Q2 = A2 + B2

= √3 and csc y = √2, x and y are acute, find x. 19.

1 – tan x tan y\

(ECE Board Exam April 6, 1991) Simplify cos A + cosB

5. Tan x = 3, x is acute. Find

+

Sin A - sin B

a) sin 2x

sin A

+ sin B

Ans: 0

cos A - cos B

b) tan (x + 45°) 20.

6. If tan (x + y) = 2 and tan x = 1, find tan y.

(ECE Board Exam April 6, 1991) Simplify

7. Given the acute angles A and B, sin A = 3/5 and sin B = 12/13, find the value of

2sin B cos B – cos B

Ans: 2 sin B – 1

1 – sin B + sin2 B –cos2 B

sin (A + B) + cos (A + B). 21.

8. Find the value of tan (A + 2B) if cot A = tan B = 2.

(ECE Board Exam April 6, 1991) Simplify

9. Find sin 2A if tan A = -5/12 and cos A is negative.

1 – sin B

sin2 150 + sin2 750

Ans:1

10. Evaluate sin x(cot x/2 + tan x/2). 11. Given A, an angle between 0° and 360° such that sin A = - 4/5 and whose tangent is

22.

Simplify

positive, construct A and find the value of sin 2A – cot ½ A. 12. Find the value of cos 2x – sin (90° + x) if tan x = -3/4 and sin x is positive. Find all values of x less than 360° satisfying the equations below. 13. 5sin 2x – 25cos x = 10 – 4sin x 14. cot 2x – tan x = 1

1 – cos x ––––––––

17.

0

0

Cos 0 + cos 1 + cos 2 + . . . + cos 90 23.

Ans: 1 0

(ECE Board Exam April 16, 1991) Simplify: sin20o + sin21o + sin22o + . . . sin290o

Ans: 45.5

observer to an object on the higher level than the observer.

sin x +

––––––––

Object

Ans: 2cscx

Line of sight

1 – cos x

ECE Board Exam April 3, 1993 Solved for θ in the equation: sin 2θ = cos θ

18.

0

Angle of elevation – is the angle made with the horizontal by the line of sight from an

(ECE Board Exam Nov. 4, 1995)

sin x

sin 00 + sin 10 + sin 20 + . . . + sin 900

Angles of Elevation and Depression

15. sin 2x + sin x = 0 16.

(ECE Board Exam April 6, 1991)

Angle of elevation Ans: 300 and 1500

ECE Board Exam April 3, 199 Given the relations P = Asin θ + Bcos θ and Q = Acos θ - Bsin θ, derive another equation showing the relationship between P, Q, A and B not involving any of the trigonometric functions of angle θ

Observer

Horizontal Line

Angle of depression – the angle made with the horizontal by the line of sight from an

angle of elevation of the top of the lighthouse is 24º. Find the height of the lighthouse.

observer to an object of lower level that the observer. Horizontal Line

Ans. 95.181ft 6. If in right triangle eight times the product of the legs equals the square of the hypotenuse,

Observer

find the angles of the triangle. Ans. [ArcTan (4 + 2√15)], [ArcTan ( 2/4 + 2√15)], 90º

Angle of depression

7. Two ladders, one of which is twice as long as the other, rest on the floor and reach the Line of sight

same vertical height on the wall. The shorter ladder makes an angle of 60º with the floor. What angle does the longer ladder make with the floor? Ans. 25.66º Object

8.

From a point midway between two objects, one being three times as tall as the other, the

Supplementary Problems:

angle of elevation of the taller is twice the angle elevation of the shorter. Find this angle

1. Two buildings 450 ft. and 600 ft. in height are opposite each other. From the roof of the

of elevation. Ans. . 30° , 60°

higher building, the angle of depression of the edge of the roof of the lower building is 38º40’. How wide is the street? Ans. 187.45ft

9.

A man standing at a certain point on a level filed determines the angle of elevation of the top of a tower 50 feet high, he then finds that by going 90 feet nearer the tower, the angle

2. A tower and a monument stand on a level ground. The angles of depression of the top of the monument viewed from the top of the tower are 13º and 31º, respectively; the height of the tower is 145 m. Find the height of the monument. Ans. 89.3 3. A flagpole 25m high stands on the top of a tower, which is 10.5 m high. At what distance from the base of the tower will the flagpole subtend an angle of 3º20’. Ans. 34.59 m 4. The angle of depression of a barge from the top of a lighthouse is 16º. After the barge has

traveled 100 m toward the lighthouse, the angle of depression of the barge is 20º.

of elevation is increased by 45º. At the first observation how far was he from the tower, and what was the angle of elevation. Ans. 108.443ft., 27.753° 10. A garage is 12 feet high and fixed on its top is a flagpole 15 feet high. On the opposite side of the street from the garage at a given point, the garage and the flagpole subtend equal angles. How wide is the street? Ans. 36ft. 11. The angles of a triangle are in the ratio 3:4:5. Express the ratio of the sine of the smallest angle. Ans. √2/2

Find the height of the lighthouse. Ans. 135.147 12. Two parallel chords of a circle of radius 8 inches are on the same side of the center and 5 5. From a boat directly south of a lighthouse, the angle of elevation of the top of the lighthouse is 32º. From a second boat lying directly east of the first and 150 ft. from it, the

inches apart. One subtends a central angle twice as large as the other. Find the length of the shorter chord.

13. A man standing at a certain point on a level field determines the angles of elevation of the

Two General Forms of Oblique Triangle

top of a tower 50 feet high. He then finds that by going 2/3 of the distance to the base of the tower, toward it, the angle of elevation has been doubled. At the first observation, how far was he from the tower, and what was the angle of elevation?

A triangle has three angles and three sides. If we are given the measures of three out of this six principal parts, at least one of which is a side, we can possibly find the measures of the other three parts.

14. A and C, the bases of two towers AB and CD standing on a horizontal plane, are 120 feet apart. The angle of elevation of D as observed from A is double the angle of elevation of B as observed from C. From a point midway between A and C the angles of elevation of

Case I. Given two angles and one side. Use Law of Sines:

B

B and D are complementary. Find the height of the towers. Ans. AB = 40 ft., CD = 90 ft.

a

=

b

sin A sin B

=

c

c

a

sin C

15. (ECE Board Exam Nov. 4, 1995) The angle which the line of sight to the object makes with the horizontal is above the

A

b

C

eye of the observer. Ans. Angle of Elevation Case II. Given two sides and an included angle. 16. (ECE Board Exam Nov. 4, 1995)

Use Law of Cosines:

The angle which the line of sight to object makes with the horizontal is below the eye of a2 = b2 + c2 – 2bc cos A

an observer. Ans. Angle of Depressio

b2 = a2 + c2 – 2ac cos B c2 = a2 + b2 – 2ab cos C

Solutions of Oblique Triangles

C

C

Case III. Given two sides and the angle opposite one of them. This is known as the ambiguous case, you can use sine law and examine the possibility

b

a

b

of no solution, one solution or two solutions.

a Case IV. Given three sides. A

c

B

A

c

B Use cosine law, to solve for each angle.

A = cos-1

b2 + c 2 – a 2

= 45.178 ft.

2bc 2. The sides of a triangle are 3, 5 and 7. Find the largest angle. Solution : B = cos

-1

2

2

2

c +a –b

Using Cosine Law a2 = b2 + c2 – 2bcCosA

2ac

here, a = longest side = 7 ; b = 3; c = 5 C = cos

-1

2

2

a +b –c

A = largest angle → unknown

2

2ab

Then, b2 + c2 – a2 = Cos A 2bc Cos A = 32 + 52 – 72 = -0.5

Sample Problem. 1.

2(3)(5)

Two angles of a triangle are 52°15′ and 59°30′, and the shortest side is 38.46 ft. long.

A = Cos-1(-0.5)

Find the length of the largest side.

A = 120°

Solution : Using Sine Law (given 2 angles and any side) Note : • The shortest side is opposite of the smallest angle 38.46 ft. is opposite of 52°15′.

Supplementary Problems 1. In an oblique triangle ABC, it is known that tan A = ¾, cos B = 5/13 and AB = 10. Find

• The longest unknown side must be opposite of the largest angle.

a) sin C

Largest angle is the third angle :

b) side AC

180 – (52°15′ + 59°30′) = 68.25°

c) side BC

Thus : a =

2.

b__

A chord, AC, of a certain circle equals 13″, and the angle B of the inscribed triangle ABC is 49.35°. Find the radius of the circle.

Sin A Sin B a = longest side, A = 68.25 b = shortest side = 38.46 ft. B = 52°15′ Now,

An observer in a balloon 1 mile high observes the angle of depression of an object on the ground to be 35°40′. After ascending vertically at uniform rate for 10 minutes, he finds the

a = b SinA SinB

angle of depression of the same object is to be 55°20′. Find the rate of ascent of the o

= (38.46)(Sin 68.25 ) Sin52°15’

3.

balloon in miles per hour.

4.

From the top of a lighthouse 85 ft. high standing on a rock, the angle of depression of a ship was 7°38′ and from the bottom of the lighthouse the angle of depression was 3°.

12. The area of a triangle is 50 sq. in. and the lengths of two of its sides are 20 in. and 10 in. Find the included angle.

What was the height of the rock? 5.

A tower 51.63′ high makes an angle of 113°12′ with the inclined plane on which it stands.

13. Two automobiles leave the same town at the same time. One goes north at the rate of 30

The angle subtended by the tower at the same point down the plane from its base is

miles per hour, the other goes in the direction 47° E of N at the rate of 20 miles per hour.

23°27′. How far is this point from the base?

How far apart are they at the end of 5 hours? 14. The acute angle between the diagonals of a parallelogram is 45°. The diagonals are 8 and 12 inches. Find the lengths of the sides of the parallelogram.

6.

A house is situated on a hillside which is inclined 29°39′ to the horizontal plane. A ladder 32.75′ long just reaches the bottom of a window 25′ from the ground. What is the distance of the foot of the ladder from the house, measured down the slope?

7.

A man standing at a point due to west of a tower 150′ high found the angle of elevation of the top of the tower to be 68°. He then walked to a second point southwest of the first and

15. The sides of a triangle are 9, 12, and 14 and the largest angle is bisected, find the length of the bisector of this angle and the length of the median to the shortest side. Inverse Trigonometric Functions 1. Inverse Sine Function y = Arcsin x

found the angle of elevation of the top of the tower to be 39°. Determine how far he

if and only if x = sin y y

y

walked. π

1 8.

/2

A circle is inscribed in a ∆ ABC whose sides are a = 25″, b = 33″ and c = 38″. How long are tangents to the circle from vertex A? vertex B? vertex C? –1

9. A circle is inscribed in a triangle whose sides are 5, 12 and 13. Find the radius of the circle.

x - π/ 2

π

/2

x 1

10. The lower base of an isosceles trapezoid is 70.23″. Each of the nonparallel sides is 35.18″. each lower base angle is 81°30′. How long is each diagonal?

11. If the diameter of a circle is 32.68″, find the angle at the center determined by a chord

-π/2

-1 y = sin x

12.9″ long. 2. Inverse Cosine Function

y = Arcsin x

y = Arccos x

if and only if x = cos y

y = tan x

y

y = Arctan x

y where, ∞ ≤ x ≤ ∞, -π/2 < y < π/2 → principal value π

1

Inverse Identities π

π

/2

π

x

–1

–1

1

y = sin x

/2

x y = Arccos x

where, –1 ≤ x ≤ 1, 0 ≤ y ≤ π → principal value

1. Arcsin (sin θ) = θ

for -π/2 ≤ θ ≤ π/2

2. sin (Arcsin x) = x

for –1 ≤ x ≤ 1

3. Arccos (cos θ) = θ

for 0 ≤ θ ≤ π

4. cos (Arccos x) = x

for –1 ≤ x ≤ 1

5. Arctan (tan θ) = θ

for -π/2 ≤ x ≤ π/2

6. tan (Arctan x) = x

for all values of x

Supplementary problems 3. Inverse Tangent Function Y = Arctan x

Solve the following trigonometric equations:

if and only if x = tan y

1. 3sinA = 2cos2A , 00

y

≤ A ≤ 3600

ans: 300, 1500

1 2. sin2x + 2sin2 2 x = 1 , 0 < x < 1800

y π

1

3 .Tan -12x +Tan-13x =

/2

Π 4

4. Tan-1[2sin(Cos-1x)] = 600

ans:300,900, 1500

ans: 1 / 6 ans: 1 / 2

0 x

0 -1

-1

x

5. Arctanx + Arctan2x = Arctan3 ans: -1, 1 / 2

1 6. Arcsin(1 – x) + Arccosx = 900 ans: 1 / 2

π

- /2 Solve the following problems 1. If sinA = 0.80, cosA > 0, find cotA. Ans. 0.75

2. If sin θ =

13. From the top of the tower 33 m high, the angles of depression of the top

12 and cos θ < 0, find tan θ . Ans. -12/5 13

of another tower standing on the same horizontal plane are found to be 28.93 and 53.680

3. If A – B = 450 and tanA = ¾ , find tanB. Ans. -1/7 4. If cot θ = -24 / 7,

and bottom 0

respectively. Find the distance between the tops of the towers. Ans. 27.7m

θ in Quadrant II, find sec θ . Ans. -25/24

14. A surveyor at a certain distance measured the angle of elevation of a cliff. He then walked 20 m on a level ground toward the cliff. The angle of elevation from this second station was then the complement of the former angle. The surveyor again walked 5 m

5. If cosx = 1 / 3, find cos4x. Ans. 17/81 6. If cos θ = -1 / 3 and 00 <

θ < 3600, find tan

nearer to the cliff in the same line and found the angle of elevation from the third station

1 θ . Ans. Sqrt(2) 2

to be double the first angle. How high is the cliff? Ans. 22.91m

If sinA = 3 / 5 and A is in Quadrant II, find cos2A. ans. 7/25 15. The minute hand of a clock is 23 cm long while the hour hand is 15 cm long. The plane of rotation of the minute hand is 5 cm above the plane of rotation of the hour hand. Find the

If tanA = 3 / 4 , tanB = -15 / 8, A in Q III, B in Q IV, find

distance between the tips of the hands of the clock at 2:30 p.m. Ans. 30.9cm

cos(A – B). Ans. 13/85 straight tracks

16. A point P within an isosceles right triangle is at a distance of 6, 5 and 4 cm from the

diverging at an angle of 68 degrees. If one train runs at 32 kph and the other at 46 kph,

vertices A, B and C respectively. Find the length of the hypotenuse of the right triangle if

how far apart are they at the end of 3 hours? Ans. 135.4km

the right angle is at C. Ans. 10.65cm

9. Two trains start at the same time from the same station and upon

10. The bearing of B from A is N20 0E; the bearing of C from B is S30 0E; and the bearing of A

17. The angle of elevation of the top of a pole at a point 30 m from the pole is three times the

from C is S400W. If AB = 10 m, find the area of the triangle formed by A, B and C. Ans.

angle of elevation of the top of the same pole at a point 150 m from the pole. Find the

13.94sq.m.

height of the pole. Ans. 56.7cm

11. An airplane flying an altitude of one km directly away from a stationary observer on the

18. A corner lot of land is 35 m on one street and 25 m on the other street, the angle between

ground, has an angle of elevation of 48 degrees at a certain instant and an angle of

two lines of the streets being 80 degrees. The other two lines of the lot are respectively

elevation of 20 degrees one minute later. Find the speed of the plane. Ans. 110.82kph

perpendicular to the lines of the streets. What is the worth of the lot at P1,000 per square meter?

12. From a point on a level ground, the angle of elevation of the top of a building is observed

ans: P725,475.00

to be twice the angle of elevation of a window one third of the way up the building. Find the angle of elevation of the top of the building. Ans. 60degrees

19. At noon, a ship A is sailing on a course eastward at the rate of 20 kph. At the same instant, another ship B, 100 km east of ship A is sailing on a course N30 0W at the rate of 10 kph. How far away from each other are the ships after one hour? Ans 75.5km

Note: Null set is a subset of all sets 20. A pole casts a shadow 15 cm long when the angle of elevation of the sun is 61 degrees. If the pole leans 15 degrees from the vertical toward the sun, what is the length of the

Unit Set – a set with only one element

pole? Ans. 54.23cm

Infinite Set – if it is impossible to list down all elements of a set

Finite Set – if it is possible to write all of its elements

21. Find the interior angles of a triangle whose sides are 21, 28 and 17. ans: 48.400,

94.340, 37.260

Cardinal Number

22. The sides of a triangle ABC are a = 50 cm, b = 64 cm and c = 20 cm. Find the length of

– denoted by n(A) → no. of elements of set A

the median drawn from B to AC. Ans. 20.64cm 23. Two stations B and C are situated on a horizontal plane 366 m apart. A balloon is directly above a point A in the same horizontal plane as B and C. At B, the angle of elevation of

Operations on Sets •

Union of Sets The union of sets A and B is another set denoted by A∪B whose elements

the balloon is 62 degrees and the angle at B subtended by AC is 53 degrees and at C,

are in A, or in B or both A and B.

the angle subtended by AB is 72 degrees. Find the height of the balloon. Ans. 799.19m PROBABILITY

– refers to the number of elements of a certain set

In symbol, A⋃B = {x x ∈ A or x ∈ B} •

Intersection of Sets

Introduction to Set Theory

The intersection of sets A and B is another set denoted by A ⋂B whose

Definitions

elements are common to both A and B.

those objects clearly described. (Denoted by capital letters) •

In symbol, A⋂B = { x x ∈ A and x ∈ B }

Set – a collection of objects called elements of any sort with restriction to •

The relative complement of the set B with respect to A is another set denoted

Element – any object that belong to a set (Denoted by small letters)

by A – B whose elements are all A but not in B.

∈ - symbol used to indicate that an element belong to a given set. •

In symbol, A – B = { x x ∈ A and x ∉ B }

Subset – If all elements of set A are in set B or if there are elements of set B not belonging to set A, then A is a subset of B denoted by A ⊂ B.

elements do not belong to A but in the universal set U In symbol A′ = { x x ∈ u and x ∉ A }

Equivalent Set – Two sets are equivalent if they have equal number of •

elements. A ⇔ B •

Product Set or Cartesian Product The product set denoted by A x B is a set of ordered pairs (x, y) such that x is

Size of Set 

Absolute Complement The absolute complement of set A denoted by A′ is another set whose

Equal Sets – Two sets are equal if they have exactly the same elements. A=B

Relative Complement

Null Set or Empty Set – a set without an element denoted by { } or φ

an element of set A and y is an element of set B. In symbol, A x B = { (x, y) x ∈ A and y ∈ B}

Note: A x B ≠ B x A •

n(B⋃S) = 593 total

Disjoint Set Two sets A and b are considered disjoint sets if they have no element in

Fundamental Counting Principle If an element E, can happen in n1 ways, and for each of these another event E 2 can

common.

happen in n2 ways, and for each of these events, another event E 3 can happen in n3 ways,

In symbol, A⋂B = { } or A⋂B = φ

and so on and so forth, then all in all the events can happen simultaneously in N ways equal to: Venn Diagram –

N = n1 n2 n3 …nk

Pictorial representation of set relations and operations Introduced by John Venn, an English Logician

PERMUTATION / COMBINATION Factorial Notation

Number of elements in a union of Sets: COUNTING FORMULA

- the factorial n, where n is any positive integer is denoted by n!

a.) n(A⋃B) = n(A) + n(B) – n(A⋂B)

Factorial n is defined as the product of positive consecutive integers from 1 to n

b.) n(A⋃B⋃C) = n(A) + n(B) + n(C) – n(A⋂B) – n(A⋂C) – n(B⋂C) + n(A⋂B⋂C)

inclusive That is,

Sample Problems: 1. In a survey, 458 men like basketball, 385 like softball, 250 like both. How many men were there? How many like softball only. In using Venn Diagram, always prioritize to indicate the number of elements in the intersections. u

or n! = n(n-1)(n-2) …3.2.1

n(B⋂S)=number

S

of

men

who

basketball and softball n(B⋃S) = total number of men in the survey

Now,

n(B⋃S) = n(B) + n(S) - n(B∩S)

like

- is any linear ordering of the elements of a set CASES OF PERMUTATION CASE 1: Permutation of n different things taken r at a time - defined as an arrangement of r out the n objects with attention given to the order of

n(B) – n(B⋂S) = 458 – 250 = 208 like basketball only n(S) - n(B⋂S) = 385 – 250 = 135 like softball only

By definition 0! = 1

- is arrangement of all or part of a set of objects in a definite order

n(S) = number of men who like softball

208 250 135

= n(n-1)(n-2)!

PERMUTATION

Let n(B) = numbers who like basketball

= 458 + 385 –250

n! = 1(2)(3) …(n-3)(n-2)(n-1)n Observe that n! = n(n-1)!

Solution:

B

no. of ways