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§ 6.8 Modeling Using Variation Direct Variation Direct Variation If a situation is described by an equation in the form y = kx where k is a constant, we say that y varies directly as x. The number k is called the constant of variation. Solving Variation Problems 1) Write an equation that describes the given English statement. 2) Substitute the given pair of values into the equation in step 1 and find the value of k. 3) Substitute the value of k into the equation in step 1. 4) Use the equation from step 3 to answer the problem’s question. Blitzer, Intermediate Algebra, 4e – Slide #121 Direct Variation EXAMPLE Use the 4-step procedure for solving the variation problem. y varies directly as x. y = 45 when x = 5. Find y when x = 13. SOLUTION 1) Write an equation. We know that y varies directly as x is expressed as y = kx. 2) Use the given values to find k. y = kx Given equation. Blitzer, Intermediate Algebra, 4e – Slide #122 Direct Variation CONTINUED 45 = k(5) Replace y with 45 and x with 5. Divide both sides by 9. 9=k 3) Substitute the value of k into the equation. y = 9x Replace k from the equation from step 1 with 9. 4) Answer the problem’s question. y = 9(13) y = 117 Replace x from the equation from step 3 with 13. Multiply. Blitzer, Intermediate Algebra, 4e – Slide #123 Direct Variation EXAMPLE An object’s weight on the moon, M, varies directly as its weight on Earth, E. Neil Armstrong, the first person to step on the moon on July 20, 1969, weighed 360 pounds on Earth (with all of his equipment on) and 60 pounds on the moon. What is the moon weight of a person who weighs 186 pounds on Earth? SOLUTION 1) Write an equation. We know that y varies directly as x is expressed as y = kx. By changing letters, we can write an equation that describes the following English statement: Moon weight, M, varies directly as Earth weight, E. Blitzer, Intermediate Algebra, 4e – Slide #124 Direct Variation CONTINUED M = kE. 2) Use the given values to find k. We are told that an object with an earth weight of 360 pounds has a moon weight of 60 pounds. Substitute 360 for E and 60 for M in the direct variation equation. Then solve for k. M = kE 60 = k360 1/6 = k Direct variation equation. Replace M with 60 and E with 360. Divide both sides by 360. 3) Substitute the value of k into the equation. Blitzer, Intermediate Algebra, 4e – Slide #125 Direct Variation CONTINUED M = kE M = (1/6)E Direct variation equation. Replace k with 1/6. 4) Answer the problem’s question. How much does an object with a 186 pound earth weight weigh on the moon? Substitute 186 for E in the preceding equation and then solve for M. M = (1/6)E M = (1/6)(186) M = 31 Equation from step 3. Replace E with 186. Multiply. An object that weighs 186 pounds on the earth weighs 31 pounds on the moon. Blitzer, Intermediate Algebra, 4e – Slide #126 Direct Variation With Powers Direct Variations With Powers y varies directly as the nth power of x if there exists some nonzero constant k such that y kx . n Blitzer, Intermediate Algebra, 4e – Slide #127 Direct Variation EXAMPLE On a dry asphalt road, a car’s stopping distance varies directly as the square of its speed. A car traveling at 45 miles per hour can stop in 67.5 feet. What is the stopping distance for a car traveling at 60 miles per hour? SOLUTION 1) Write an equation. We know that y varies directly as the square of x is expressed as y kx . 2 By changing letters, we can write an equation that describes the following English statement: Distance, d, varies directly as the square of the speed, s. Blitzer, Intermediate Algebra, 4e – Slide #128 Direct Variation CONTINUED d ks 2 2) Use the given values to find k. We are told that a car traveling at 45 miles per hour can stop in 67.5 feet. Substitute 45 for s and 67.5 for d. Then solve for k. d ks 2 67 . 5 k 45 This is the equation from step 1. 2 Replace d with 67.5 and s with 45. 67 . 5 2025 k Simplify. 0 . 033 k Divide both sides by 2025. 3) Substitute the value of k into the equation. 2 d ks This is the equation from step 1. 2 d 0 . 033 s Replace k with 0.033. Blitzer, Intermediate Algebra, 4e – Slide #129 Direct Variation CONTINUED 4) Answer the problem’s question. What is the stopping distance for a car traveling at 60 miles per hour? Substitute 60 for s in the preceding equation. d 0 . 033 s 2 d 0 . 033 60 This is the equation from step 3. 2 d 0 . 033 3600 d 120 Replace s with 60. Simplify. Simplify. The stopping distance for a car traveling at 60 miles per hour is 120 feet. Blitzer, Intermediate Algebra, 4e – Slide #130 Inverse Variation Inverse Variation If a situation is described by an equation in the form y k x where k is a constant, we say that y varies inversely as x. The number k is called the constant of variation. Blitzer, Intermediate Algebra, 4e – Slide #131 Inverse Variation EXAMPLE The water temperature of the Pacific Ocean varies inversely as the water’s depth. At a depth of 1000 meters, the water temperature is 4 . 4 Celsius. What is the water temperature at a depth of 5000 meters? SOLUTION 1) Write an equation. We know that y varies inversely as x is expressed as y k . x By changing letters, we can write an equation that describes the following English statement: Temperature, T, varies inversely as the water’s depth, D. Blitzer, Intermediate Algebra, 4e – Slide #132 Inverse Variation CONTINUED T k D 2) Use the given values to find k. The temperature of water in the Pacific Ocean at a depth of 1000 meters is 4 . 4 Celsius. Substitute 1000 for D and 4.4 for T in the inverse variation equation. Then solve for k. T k This is the equation from step 1. D k 4 .4 1000 4400 k Replace T with 4.4 and D with 1000. Multiply both sides by 1000. Blitzer, Intermediate Algebra, 4e – Slide #133 Inverse Variation CONTINUED 3) Substitute the value of k into the equation. T T k D 4400 D This is the equation from step 1. Replace k with 4400. 4) Answer the problem’s question. We need to find the temperature of the water at a depth of 5000 meters. Substitute 5000 for D. T 4400 D This is the equation from step 3. Blitzer, Intermediate Algebra, 4e – Slide #134 Inverse Variation CONTINUED T 4400 5000 T 0 . 88 Replace D with 5000. Simplify. The temperature of the Pacific Ocean at a depth of 5000 meters is 4 . 4 Celsius. Blitzer, Intermediate Algebra, 4e – Slide #135 Direct & Inverse (Combined) Variation EXAMPLE One’s intelligence quotient, or IQ, varies directly as a person’s mental age and inversely as that person’s chronological age. A person with a mental age of 25 and a chronological age of 20 has an IQ of 125. What is the chronological age of a person with a mental age of 40 and an IQ of 80? SOLUTION 1) Write an equation. We know that y varies directly as x and inversely as z is expressed as y kx . z By changing letters, we can write an equation that describes the following English statement: IQ, I, varies directly as the mental age, M, and inversely as chronological age, C. Blitzer, Intermediate Algebra, 4e – Slide #136 Direct & Inverse (Combined) Variation CONTINUED I kM C 2) Use the given values to find k. A person with a mental age of 25 and a chronological age of 20 has an IQ of 125. Substitute 125 for I, 25 for M, and 20 for C. I kM C 125 k 25 20 2500 25 k 100 k This is the equation from step 1. Replace I with 125, M with 25, and C with 20. Multiply both sides by 20. Divide both sides by 25. Blitzer, Intermediate Algebra, 4e – Slide #137 Direct & Inverse (Combined) Variation CONTINUED 3) Substitute the value of k into the equation. I I kM C 100 M This is the equation from step 1. Replace k with 100. C 4) Answer the problem’s question. We need to determine the chronological age of a person who has a mental age of 40 and an IQ of 80. Substitute 40 for M and 80 for I. I 80 100 M C 100 40 This is the equation from step 3. Replace M with 40 and I with 80. C Blitzer, Intermediate Algebra, 4e – Slide #138 Direct & Inverse (Combined) Variation CONTINUED 80 4000 Multiply. C 80 C 4000 Multiply both sides by C. C 50 Divide both sides by 80. When a person has a mental age of 40 and an IQ of 80, his/her chronological age is 50 years old. Blitzer, Intermediate Algebra, 4e – Slide #139 Joint Variation Joint Variation If a situation is described by an equation in the form y kxz where k is a constant, we say that y varies jointly as x and z. The number k is called the constant of variation. Blitzer, Intermediate Algebra, 4e – Slide #140 Joint Variation EXAMPLE Kinetic energy varies jointly as the mass and the square of the velocity. A mass of 8 grams and velocity of 3 centimeters per second has a kinetic energy of 36 ergs. Find the kinetic energy for a mass of 4 grams and velocity of 6 centimeters per second. SOLUTION E kMV 36 k 8 3 36 72 k Translate “Kinetic energy, E, varies jointly as the mass, M, and the square of the velocity, V.” 2 2 If M = 8, V = 3, then E = 36. Simplify. Blitzer, Intermediate Algebra, 4e – Slide #141 Joint Variation CONTINUED 0 .5 k E 0 . 5 MV Divide both sides by 72. Substitute 0.5 for k in the model for kinetic energy. 2 E 0 . 5 4 6 2 Find E when M = 4 and V = 6. E 72 Simplify. The kinetic energy is 72 ergs. Blitzer, Intermediate Algebra, 4e – Slide #142