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VARIATION BY EDSEL M. LLAVE VARIATION • DIRECT VARIATION – Definition of Direct Variation – Direct Variation as the nth Power • INVERSE VARIATION – Definition of Inverse Variation – Inverse Variation as the nth Power • JOINT VARIATION – Definition of Joint Variation • COMBINED VARIATION DIRECT VARIATION • The variable y varies directly as the variable x, or y is directly proportional to x, if and only if where k is a constant called constant of proportionality or the variation constant. DIRECT VARIATION • The statement "y varies directly as x," means that when x increases, y increases by the same factor. • In other words, y and x always have the same ratio: Example 1: If y varies directly as x, and x = 12 when y = 9, what is the equation that describes this direct variation? • Solution: Given: x = 12 y=9 Example 2: If y varies directly as x, and the constant of variation is , what is y when x = 9? • Solution: Given: x=9 Example 3: If y varies directly as x, and y = 15 when x = 10, then what is y when x = 6? • Solution: Given: Condition 1: y = 15 x = 10 Condition 2: y=? x=6 1. Find the variation constant. (condition 1) Example 3: If y varies directly as x, and y = 15 when x = 10, then what is y when x = 6? • Solution: Given: Condition 1: y = 15 x = 10 Condition 2: y=? x=6 2. Find the value of y. (Condition 2) Example 3: If y varies directly as x, and y = 15 when x = 10, then what is y when x = 6? • Solution 2: (Proportion) Given: Condition 1: y1 = 15 x2 = 10 Condition 2: y=? x=6 Example 3. The distance sound travels varies directly as the time it travels. If sound travels 1340 meters in 4 seconds, find the distance sound will travel in 5 seconds. • Solution: CONDITION Distance (d) Time (t) Given: 1 d1 = 1340m t1 = 4sec 2 d=? t = 5sec By proportion: Example 3. The distance sound travels varies directly as the time it travels. If sound travels 1340 meters in 4 seconds, find the distance sound will travel in 5 seconds. • Solution: CONDITION Distance (d) Time (t) Given: 1 d1 = 1340m t1 = 4sec 2 d=? t = 5sec Find the value of k. Find the value of d. Example 4. Hooke’s Law states that the distance a spring stretches varies directly as the weight on the spring. A weight of 80 pounds stretches a spring 6 inches. How far will a weight of 100 pounds stretch the spring? • Solution: CONDITION Distance (d) Weight (w) Given: 1 d1 = 6in w1 = 80lbs 2 d=? w = 100lbs Find the value of k. Find the value of d. Example 4. Hooke’s Law states that the distance a spring stretches varies directly as the weight on the spring. A weight of 80 pounds stretches a spring 6 inches. How far will a weight of 100 pounds stretch the spring? • Solution: CONDITION Distance (d) Weight (w) Given: 1 d1 = 6in w1 = 80lbs 2 d=? w = 100lbs By proportion: DIRECT VARIATION AS THE nth POWER • If y varies directly as the nth power of x, then where k is constant. Example 5. The distance s that an object falls from rest (neglecting air resistance) varies directly as the square of the time t that it has been falling. If an object falls 64 feet in 2 seconds, how far will it fall in 10 seconds? • Solution: CONDITION Distance (s) Time (t) Given: 1 s1 = 64ft t1 = 2sec Formula: 2 s=? t = 10sec Example 5. The distance s that an object falls from rest (neglecting air resistance) varies directly as the square of the time t that it has been falling. If an object falls 64 feet in 2 seconds, how far will it fall in 10 seconds? • Solution: CONDITION Distance (s) Time (t) Given: 1 s1 = 64ft t1 = 2sec 2 s=? t = 10sec By proportion: Problem 6. The range of a projectile is directly proportional to the square of its velocity. If a motorcyclist can make a jump of 140 feet by coming off a ramp at 60 mph, find the distance the motorcyclist could expect to jump if the speed coming off the ramp were increased to 65 mph. • Solution: CONDITION Distance (s) Velocity (v) Given: 1 s1 = 140ft v1 = 60 mph 2 s=? v = 65 mph Formula: Problem 6. The range of a projectile is directly proportional to the square of its velocity. If a motorcyclist can make a jump of 140 feet by coming off a ramp at 60 mph, find the distance the motorcyclist could expect to jump if the speed coming off the ramp were increased to 65 mph. • Solution: CONDITION Distance (s) Velocity (v) Given: 1 s1 = 140ft v1 = 60 mph 2 s=? v = 65 mph By proportion: INVERSE VARIATION • The variable y varies inversely as the variable x, or y is inversely proportional to x, if and only if where k is the variation constant INVERSE VARIATION • The statement "y varies inversely as x” means that when x increases, y decreases by the same factor. • In other words, the expression xy is constant: where k is constant of variation. Example 1: If y varies inversely as x, and y = 6 when , write an equation describing this inverse variation. • Solution : • Find the value of k. Write an equation. or Example 2: If y varies inversely as x, and the constant of variation is , what is y when x = 10? • Solution: Example 3: If y varies inversely as x, and y = 10 when x = 6, then what is y when x = 15? • Solution: CONDITION y x Given: 1 y1 = 10 x1 = 6 2 y=? x = 15 Example 4. Boyle’s Law states that the Volume V of a sample of gas (at constant temperature) varies inversely as the pressure P. The volume of a gas in a J-shaped tube is 75 milliliters when the pressure is 1.5 atmospheres. Find the volume of the gas when the pressure is increased to 2.5 atmospheres. • Solution: CONDITION Volume(V) Pressure(P) Given: 1 V1 = 75ml P1 = 1.5atm 2 V=? P = 2.5atm Find the value of k. Example 4. Boyle’s Law states that the Volume V of a sample of gas (at constant temperature) varies inversely as the pressure P. The volume of a gas in a J-shaped tube is 75 milliliters when the pressure is 1.5 atmospheres. Find the volume of the gas when the pressure is increased to 2.5 atmospheres. • Solution: CONDITION Volume(V) Pressure(P) Given: 1 V1 = 75ml P1 = 1.5atm 2 V=? P = 2.5atm INVERSE VARIATION AS nth POWER • If y varies inversely as the nth power of x, then where k is a constant Example 5. The illumination a source of light provides is inversely proportional to the square of the distance from the source. If the illumination at a distance of 10 feet from the source is 50 footcandles, what is the illumination at a distance of 15 feet from the source? • Solution: CONDITION Illumination(ℓ) Distance(d) Given: 1 ℓ1 = 50fc d1 = 10ft 2 ℓ=? d = 15ft Find the value of k. Example 5. The illumination a source of light provides is inversely proportional to the square of the distance from the source. If the illumination at a distance of 10 feet from the source is 50 footcandles, what is the illumination at a distance of 15 feet from the source? • Solution: CONDITION Illumination(ℓ) Distance(d) Given: 1 ℓ1 = 50fc d1 = 10ft 2 ℓ=? d = 15ft JOINT VARIATION • The variable z varies jointly as the variables x and y if and only if where k is constant. • If one variable, z, varies as the product of other variables, x and y, it is called joint variation. Example 1. The cost of insulating the ceiling of a house varies jointly with the thickness of the insulation and the area of the ceiling. It costs Php8,750 to insulate a 2100-square-foot ceiling with insulation 4 inches thick. Find the cost of insulating a 2400-square-foot ceiling with insulation that is 6 inches thick. • Solution: CONDITION Cost (C) Area (A) Thickness (T) Given: 1 C1 = Php8,750 A1 = 2100ft2 T1 = 4in 2 C=? A = 2400ft2 T = 6in Formula: Example 1. The cost of insulating the ceiling of a house varies jointly with the thickness of the insulation and the area of the ceiling. It costs Php8,750 to insulate a 2100-square-foot ceiling with insulation 4 inches thick. Find the cost of insulating a 2400-square-foot ceiling with insulation that is 6 inches thick. • Solution: CONDITION Cost (C) Area (A) Thickness (T) Given: 1 C1 = Php8,750 A1 = 2100ft2 T1 = 4in 2 C=? A = 2400ft2 T = 6in Formula: Example 2. The load L that a horizontal beam can safely support varies jointly as the width w and the square of the depth d. If a beam with width 2 inches and depth 6 inches safely supports up to 200 pounds, how many pounds can a beam of the same length that has width 4 inches and depth 4 inches be expected to support? • Solution: CONDITION Load (L) Width (w) Depth (d) Given: 1 L1 = 200lbs w1 = 2in d1 = 6in 2 L=? w = 4in d = 4in Formula: Example 2. The load L that a horizontal beam can safely support varies jointly as the width w and the square of the depth d. If a beam with width 2 inches and depth 6 inches safely supports up to 200 pounds, how many pounds can a beam of the same length that has width 4 inches and depth 4 inches be expected to support? • Solution: CONDITION Load (L) Width (w) Depth (d) 1 L1 = 200lbs w1 = 2in d1 = 6in Given: 2 L=? w = 4in d = 4in COMBINED VARIATIONS • Combined variations involve more than one type of variation. • It involves the direct and inverse variations. Problem 1. The weight that a horizontal beam with a rectangular cross section can safely support varies jointly as the width and square of the depth of the cross section and inversely as the length of the beam. If a 4-inch by 4-inch beam 10 feet long safely supports a load of 256 pounds, what load L can be safely supported by a beam made of the same material and with a width w of 4 inches, a depth d of 6 inches, and a length l of 16 feet? • Solution: CONDITION Load (L) Width (w) Depth(d) Length (ℓ) Given: 1 L1 = 256lbs w1 = 4in d1 = 4in l1 = 10ft 2 L=? w = 4in d = 6in l = 16ft Formula: Problem 1. The weight that a horizontal beam with a rectangular cross section can safely support varies jointly as the width and square of the depth of the cross section and inversely as the length of the beam. If a 4-inch by 4-inch beam 10 feet long safely supports a load of 256 pounds, what load L can be safely supported by a beam made of the same material and with a width w of 4 inches, a depth d of 6 inches, and a length l of 16 feet? • Solution: CONDITION Load (L) Width (w) Depth(d) Length (ℓ) 1 L1 = 256lbs w1 = 4in d1 = 4in l1 = 10ft Given: 2 L=? w = 4in d = 6in l = 16ft Formula: Example 2. The Ideal Gas Law states that the volume V of a gas varies jointly as the number of moles of gas n and the absolute temperature T and inversely as the pressure P. What happens to V when n is tripled and P is reduced by a factor of one-half? • Solution: CONDITION Volume No. of Temperature Pressure Given: (V) Moles (n) (T) (P) 1 V1 = V n1 = n T1 = T P1 = P 2 V2 = ? n2 = 3n T2 = T P2 = 0.5P Formula: Example 2. The Ideal Gas Law states that the volume V of a gas varies jointly as the number of moles of gas n and the absolute temperature T and inversely as the pressure P. What happens to V when n is tripled and P is reduced by a factor of one-half? • Solution: CONDITION Volume No. of Temperature Pressure (V) Moles (n) (T) (P) Given: 1 V1 = V n1 = n T1 = T P1 = P 2 V2 = ? n2 = 3n T2 = T P2 = 0.5P Formula: Thank you