1. § 6.8 Modeling Using Variation

2. Blitzer, Intermediate Algebra, 5e – Slide #2 Section 6.8 Variation Certain situations occur so frequently in applied situations that they are given special names. Variation formulas show how one quantity changes in relation to other quantities. y varies directly as x y varies jointly as x and z y varies inversely as x varying directly as a power of x Quantities can vary directly, inversely, or jointly.

3. Blitzer, Intermediate Algebra, 5e – Slide #3 Section 6.8 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.

4. Blitzer, Intermediate Algebra, 5e – Slide #4 Section 6.8 Direct VariationEXAMPLE 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. ( For example: Purchase 5 pounds of steak for $45.) SOLUTION 1) Write an equation. We know that y varies directly as x is expressed as 2) Use the given values to find k. y = kx. y = kxGiven equation.

5. Blitzer, Intermediate Algebra, 5e – Slide #5 Section 6.8 Direct Variation 3) Substitute the value of k into the equation. For example: $9 a pound. 45 = k(5) Replace y with 45 and x with 5. CONTINUED 9 = k Divide both sides by 9. y = 9x Replace k from the equation from step 1 with 9. 4) Answer the problem’s question. For example: How much will it cost for 13 pounds? y = 9(13) Replace x from the equation from step 3 with 13. y = 117 Multiply. Will cost $117.

6. Blitzer, Intermediate Algebra, 5e – Slide #6 Section 6.8 Direct VariationEXAMPLE 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.

7. Blitzer, Intermediate Algebra, 5e – Slide #7 Section 6.8 Direct Variation 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. CONTINUED M = kEDirect variation equation. 60 = k360Replace M with 60 and E with 360. 1/6 = kDivide both sides by 360. 3) Substitute the value of k into the equation.

8. Blitzer, Intermediate Algebra, 5e – Slide #8 Section 6.8 Direct VariationCONTINUED 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 = kEDirect variation equation. M = (1/6)EReplace k with 1/6. M = (1/6)EEquation from step 3. M = (1/6)(186)Replace E with 186. M = 31 An object that weighs 186 pounds on the earth weighs 31 pounds on the moon. Multiply.

9. Blitzer, Intermediate Algebra, 5e – Slide #9 Section 6.8 Direct Variation Check Point 1 on page 465 The number of gallons of water, W used when taking a shower varies directly as the time, t, in minutes, in the shower. A shower lasting 5 minutes uses 30 gallons of water. How much water is used in a shower lasting 11 minutes? SOLUTION 1) Write an equation. We know that y varies directly as x is expressed as y = kx. W = kt.

10. Blitzer, Intermediate Algebra, 5e – Slide #10 Section 6.8 Direct Variation 2) Use the given values to find k. A shower lasting 5 minutes uses 30 gallons of water. Use the given values to find k. CONTINUED W = ktDirect variation equation. 30 = k5Replace W with 30 and t with 5. 6 = kDivide both sides by 5. 3) Substitute the value of k into the equation. Replace k with 6. W = 6t

11. Blitzer, Intermediate Algebra, 5e – Slide #11 Section 6.8 Direct VariationCONTINUED 4) Answer the problem’s question. Find W when t is 11 minutes. W = 6tEquation from step 3. W = (6)(11)Replace t with 11. W = 66 An 11 minute shower will use 66 gallons of water. Multiply.

12. Blitzer, Intermediate Algebra, 5e – Slide #12 Section 6.8 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 Next, we consider y varying directly as a power of x:

13. Blitzer, Intermediate Algebra, 5e – Slide #13 Section 6.8 Direct VariationEXAMPLE 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 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.

14. Blitzer, Intermediate Algebra, 5e – Slide #14 Section 6.8 Direct Variation 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. CONTINUED This is the equation from step 1. Replace d with 67.5 and s with 45. Simplify. Divide both sides by 2025. 3) Substitute the value of k into the equation. This is the equation from step 1. Replace k with 0.033.

15. Blitzer, Intermediate Algebra, 5e – Slide #15 Section 6.8 Direct Variation 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. CONTINUED This is the equation from step 3. Replace s with 60. Simplify. The stopping distance for a car traveling at 60 miles per hour is 120 feet.

16. Blitzer, Intermediate Algebra, 5e – Slide #16 Section 6.8 Direct Variation Check Point 2 on page 467 The distance required to stop a car varies directly as the square of its speed. If it requires 200 feet to stop a car traveling 60 miles per hour, how many feet are required to stop a car traveling 100 miles per hour? SOLUTION 1) Write an equation. By changing letters, we can write an equation that describes the following English statement: Stopping distance, s, varies directly as the square of the speed of the car, v. 2) Use the given values to find k.

17. Blitzer, Intermediate Algebra, 5e – Slide #17 Section 6.8 Direct VariationCONTINUED 3) Substitute the value of k into the equation. This is the equation from step 1. 4) Answer the problem’s question: find s when v=100. Replace v with 100. Simplify. About 556 feet are required to stop a car traveling 100 miles per hour.

18. Blitzer, Intermediate Algebra, 5e – Slide #18 Section 6.8 Inverse Variation If a situation is described by an equation in the form where k is a constant, we say that y varies inversely as x. The number k is called the constant of variation. And now we consider inverse variation…

19. Blitzer, Intermediate Algebra, 5e – Slide #19 Section 6.8 Inverse VariationEXAMPLE The water temperature of the Pacific Ocean varies inversely as the water’s depth. At a depth of 1000 meters, the water temperature is 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 By changing letters, we can write an equation that describes the following English statement: Temperature, T, varies inversely as the water’s depth, D.

20. Blitzer, Intermediate Algebra, 5e – Slide #20 Section 6.8 Inverse Variation 2) Use the given values to find k. The temperature of water in the Pacific Ocean at a depth of 1000 meters is Celsius. Substitute 1000 for D and 4.4 for T in the inverse variation equation. Then solve for k. This is the equation from step 1. CONTINUED Replace T with 4.4 and D with 1000. Multiply both sides by 1000.

21. Blitzer, Intermediate Algebra, 5e – Slide #21 Section 6.8 Inverse Variation 3) Substitute the value of k into the equation. This is the equation from step 1. CONTINUED 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. This is the equation from step 3.

22. Inverse VariationCONTINUED Replace D with 5000. Simplify. The temperature of the Pacific Ocean at a depth of 5000 meters is Celsius. Blitzer, Intermediate Algebra, 5e – Slide #22 Section 6.8

23. Blitzer, Intermediate Algebra, 5e – Slide #23 Section 6.8 Inverse Variation Check Point 3 on page 468 The length of a violin string varies inversely as the frequency of its vibrations. A violin string 8 inches long vibrates at a frequency of 640 cycles per second. What is the frequency of a 10-inch string? SOLUTION 1) Write an equation. We know that y varies inversely as x is expressed as By changing letters, we can write an equation that describes the following English statement: Frequency, f. varies inversely as the water’s depth, D. 2) Use the given values to find k.

24. Blitzer, Intermediate Algebra, 5e – Slide #24 Section 6.8 Inverse VariationCONTINUED 3) Substitute the value of k into the equation. This is the equation from step 1. 4) Answer the problem’s question: find f when l=10. Replace v with 100. Simplify. A string length of 10 inches will vibrate at 512 cycles per second.

25. Combined Variation Blitzer, Intermediate Algebra, 5e – Slide #25 Section 6.8 In combined variation, direct and inverse variation occur at the same time. For example, as the advertising budget, A, of a company increases, its monthly Sales, S, also increase. Then S increases directly as A and S = kA But by contrast, as the price of the company’s product P increases, the company’s monthly sales decrease. Therefore, monthly sales S vary inversely as the price P of the product. That is, We can combine these two variation equations into one equation: This is an example of combined variation.

26. Blitzer, Intermediate Algebra, 5e – Slide #26 Section 6.8 Direct & Inverse (Combined) VariationEXAMPLE 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 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.

27. Blitzer, Intermediate Algebra, 5e – Slide #27 Section 6.8 Direct & Inverse (Combined) Variation 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. This is the equation from step 1. CONTINUED Replace I with 125, M with 25, and C with 20. Multiply both sides by 20. Divide both sides by 25.

28. Blitzer, Intermediate Algebra, 5e – Slide #28 Section 6.8 Direct & Inverse (Combined) Variation 3) Substitute the value of k into the equation. This is the equation from step 1. CONTINUED Replace k with 100. 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. This is the equation from step 3. Replace M with 40 and I with 80.

29. Blitzer, Intermediate Algebra, 5e – Slide #29 Section 6.8 Direct & Inverse (Combined) VariationCONTINUED When a person has a mental age of 40 and an IQ of 80, his/her chronological age is 50 years old. Multiply. Multiply both sides by C. Divide both sides by 80.

30. Blitzer, Intermediate Algebra, 5e – Slide #30 Section 6.8 Direct & Inverse (Combined) Variation Check Point 4 on page 470 The number of minutes needed to solve an exercise set of variation problems varies directly as the number of problems and inversely as the number of people working to solve the problems. It takes 4 people 32 minutes to solve 16 problems. How many minutes will it tak 8 people to solve 24 problems? SOLUTION 1) Write an equation. y varies directly as x and inversely as z m=the number of minutes needed to solve an exercise set (32) p =the number of people working on the problems (4) x= the number if problems in the exercise set (16) 2) Use the given values to find k.

31. Blitzer, Intermediate Algebra, 5e – Slide #31 Section 6.8 Direct & Inverse (Combined) Variation 3) Substitute the value of k into the equation. This is the equation from step 1. CONTINUED Replace k with 8 4) Answer the problem’s question. Find m when p=8 and x = 24 Replace M with 40 and I with 80. It will take 24 minutes for 8 people to solve 24 problems.

32. Blitzer, Intermediate Algebra, 5e – Slide #32 Section 6.8 Joint Variation If a situation is described by an equation in the form where k is a constant, we say that y varies jointly as x and z. The number k is called the constant of variation. Finally, we have joint variation.

33. Blitzer, Intermediate Algebra, 5e – Slide #33 Section 6.8 Joint VariationEXAMPLE 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 Translate “Kinetic energy, E, varies jointly as the mass, M, and the square of the velocity, V.” If M = 8, V = 3, then E = 36. Simplify.

34. Blitzer, Intermediate Algebra, 5e – Slide #34 Section 6.8 Joint Variation Substitute 0.5 for k in the model for kinetic energy. CONTINUED Find E when M = 4 and V = 6. Simplify. The kinetic energy is 72 ergs. Divide both sides by 72.

35. Blitzer, Intermediate Algebra, 5e – Slide #35 Section 6.8 Joint Variation Check Point on page 471 The volume of a cone, V, varies jointly as its height, h, ad the square of its radius r. A cone with a radius measuring 6 feet and a height measuring 10 feet has a volume of 120pi cubic feet. Find the volume of a cone having a radius of 12 feet and a height of 2 feet. SOLUTION

36. DONE