1. Copyright © Cengage Learning. All rights reserved. Graphs; Equations of Lines; Functions; Variation 3

2. Copyright © Cengage Learning. All rights reserved. Section 3.7 Variation

3. 3 Objectives Solve a direct variation application. Solve an inverse variation application. 1.Solve a joint variation application. 2.Solve a combined variation application. 1 1 2 2 3 3 4 4

4. 4 Solve a direct variation application 1.

5. 5 Solve a direct variation application The most basic type of variation is called direct variation. Direct Variation The words y varies directly with x mean that y = kx for some constant k. The constant k is called the constant of variation.

6. 6 Solve a direct variation application The more force that is applied to a spring, the more it will stretch. Scientists call this fact Hooke’s law: The distance a spring will stretch varies directly with the force applied. If d represents distance and f represents force, this relationship can be expressed by the equation (1)d = kf where k is the constant of variation.

7. 7 Solve a direct variation application If a spring stretches 5 inches when a weight of 2 pounds is attached, we can find the constant of variation by substituting 5 for d and 2 for f in Equation 1 and solving for k. d = k f 5 = k (2)

8. 8 Solve a direct variation application To find the distance that the spring will stretch when a weight of 6 pounds is attached, we substitute for k and 6 for f in Equation 1 and solve for d. The spring will stretch 15 inches when a weight of 6 pounds is attached.

9. 9 Example At a constant speed, the distance traveled varies directly with time. If a bus driver can drive 105 miles in 3 hours, how far would he drive in 5 hours? Solution: We let d represent the distance traveled and let t represent time. We then translate the words distance varies directly with time into the equation (2)d = kt

10. 10 Example – Solution To find the constant of variation, k, we substitute 105 for d and 3 for t in Equation 2 and solve for k. d = k t 105 = k (3) 35 = k We can now substitute 35 for k in Equation 2 to obtain Equation 3. (3)d = 35t Divide both sides by 3. cont’d

11. 11 Example – Solution To find the distance traveled in 5 hours, we substitute 5 for t in Equation 3. d = 35(5) d = 175 In 5 hours, the bus driver would travel 175 miles. cont’d

12. 12 Solve an inverse variation application 2.

13. 13 Solve an inverse variation application A second common type of variation is called inverse variation. Inverse Variation The words y varies inversely with x mean that for some constant k. The constant k is the constant of variation.

14. 14 Solve an inverse variation application Comment Since k is a constant, the equation defines a special function called a rational function. In this section, k will always be a positive number. Under constant temperature, the volume occupied by a gas varies inversely with its pressure. If V represents volume and p represents pressure, this relationship is expressed by the equation (4)

15. 15 Example A gas occupies a volume of 15 cubic inches when placed under 4 pounds per square inch of pressure. How much pressure is needed to compress the gas into a volume of 10 cubic inches? Solution: To find the constant of variation, we substitute 15 for V and 4 for p in Equation 4 and solve for k.

16. 16 Example – Solution To find the pressure needed to compress the gas into a volume of 10 cubic inches, we substitute 60 for k and 10 for V in Equation 4 and solve for p. Multiply both sides by 4. cont’d

17. 17 Example – Solution 10p = 60 p = 6 It will take 6 pounds per square inch of pressure to compress the gas into a volume of 10 cubic inches. Multiply both sides by p. Divide both sides by 10. cont’d

18. 18 Solve a joint variation application 3.

19. 19 Solve a joint variation application A third type of variation is called joint variation. Joint Variation The words y varies jointly with x and z mean that y = kxz for some constant k. The constant k is the constant of variation.

20. 20 Solve a joint variation application The area A of a rectangle depends on its length l and its width w by the formula A = lw We could say that the area of the rectangle varies jointly with its length and its width. In this example, the constant of variation is k = 1.

21. 21 Example The area of a triangle varies jointly with the length of its base and its height. If a triangle with an area of 63 square inches has a base of 18 inches and a height of 7 inches, find the area of a triangle with a base of 12 inches and a height of 10 inches.

22. 22 Example – Solution We let A represent the area of the triangle, b represent the length of the base, and h represent the height. We translate the words area varies jointly with the length of the base and the height into the formula (5)A = kbh We are given that A = 63 when b = 18 and h = 7. To find k, we substitute these values in Equation 5 and solve for k. A = kbh

23. 23 Example – Solution 63 = k(18)(7) 63 = k(126) Thus,, and the formula for finding the area is (6) Divide both sides by 126. Simplify. cont’d

24. 24 Example – Solution To find the area of a triangle with a base of 12 inches and a height of 10 inches, we substitute 12 for b and 10 for h in Equation 6. The area is 60 square inches. cont’d

25. 25 Solve a combined variation application 4.

26. 26 Solve a combined variation application The final type of variation, called combined variation, involves a combination of direct and inverse variation.

27. 27 Example The pressure of a fixed amount of gas varies directly with its temperature and inversely with its volume. A sample of gas at a pressure of 1 atmosphere occupies a volume of 3 cubic meters when its temperature is 273 Kelvin (about 0  Celsius). Find the pressure after the gas is heated to 364 K and compressed to 1 cubic meter.

28. 28 Example – Solution We let P represent the pressure of the gas, T represent its temperature, and V represent its volume. The words the pressure varies directly with temperature and inversely with volume translate into the equation (7) To find k, we substitute 1 for P, 273 for T, and 3 for V in Equation 7.

29. 29 Example – Solution Sincethe formula is cont’d

30. 30 Example – Solution To find the pressure under the new conditions, we substitute 364 for T and 1 for V in the previous equation and solve for P. The pressure of the heated and compressed gas is 4 atmospheres. cont’d