1. 3.6 - 1 10 TH EDITION LIAL HORNSBY SCHNEIDER COLLEGE ALGEBRA

2. 3.6 - 2 3.6 Variation Direct Variation Inverse Variation Combined and Joint Variation

3. 3.6 - 3 Direct Variation When one quantity is a constant multiple of another quantity, the two quantities are said to vary directly. For example, if you work for an hourly wage of $6, then [pay] = 6 [hours worked]. Doubling the hours doubles the pay. Tripling the hours triples the pay, and so on. This is stated more precisely as follows.

4. 3.6 - 4 Direct Variation y varies directly as x, or y is directly proportional to x, if there exists a nonzero real number k, called the constant of variation, such that

5. 3.6 - 5 Solving Variation Problems Step 1 Write the general relationship among the variables as an equation. Use the constant k. Step 2 Substitute given values of the variables and find the value of k. Step 3 Substitute this value of k into the equation from Step 1, obtaining a specific formula. Step 4 Substitute the remaining values and solve for the required unknown.

6. 3.6 - 6 Example 1 SOLVING A DIRECT VARIATION PROBLEM Solution The area of a rectangle varies directly as its length. If the area is 50 m 2 when the length is 10 m, find the area when the length is 25 m. Step 1 Since the area varies directly as the length, where A represents the area of the rectangle, L is the length, and k is a nonzero constant.

7. 3.6 - 7 Example 1 SOLVING A DIRECT VARIATION PROBLEM Solution The area of a rectangle varies directly as its length. If the area is 50 m 2 when the length is 10 m, find the area when the length is 25 m. Step 2 Since A = 50 when L = 10, the equation A = kL becomes

8. 3.6 - 8 Example 1 SOLVING A DIRECT VARIATION PROBLEM Solution The area of a rectangle varies directly as its length. If the area is 50 m 2 when the length is 10 m, find the area when the length is 25 m. Step 3 Using this value of k, we can express the relationship between the area and the length as Direct variation equation.

9. 3.6 - 9 Example 1 SOLVING A DIRECT VARIATION PROBLEM Solution The area of a rectangle varies directly as its length. If the area is 50 m 2 when the length is 10 m, find the area when the length is 25 m. Step 4 To find the area when the length is 25, we replace L with 25. The area of the rectangle is 125 m 2 when the length is 25 m.

10. 3.6 - 10 Direct Variation as nth Power Let n be a positive real number. Then y varies directly as the nth power of x, or y is directly proportional to the nth power of x, if there exists a nonzero real number k such that

11. 3.6 - 11 Inverse Variation as nth Power Let n be a positive real number. Then y varies inversely as the nth power of x, or y is inversely proportional to the nth power of x, if there exists a nonzero real number k such that If n = 1, then and y varies inversely as x.

12. 3.6 - 12 In a certain manufacturing process, the cost of producing a single item varies inversely as the square of the number of items produced. If 100 items are produced, each costs $2. Find the cost per item if 400 items are produced. Example 2 SOLVING AN INVERSE VARIATION PROBLEM

13. 3.6 - 13 Example 2 SOLVING AN INVERSE VARIATION PROBLEM Solution Step 1 Let x represent the number of items produced and y represent the cost per item. Then for some nonzero constant k, y varies inversely as the square of x.

14. 3.6 - 14 Example 2 SOLVING AN INVERSE VARIATION PROBLEM Solution Step 2 Substitute, y = 2 when x = 100. Solve for k.

15. 3.6 - 15 Example 2 SOLVING AN INVERSE VARIATION PROBLEM Solution Step 3 The relationship between x and y is Step 4 When 400 items are produced, the cost per item is

16. 3.6 - 16 Joint Variation Let m and n be real numbers. Then y varies jointly as the nth power of x and the mth power of z if there exists a nonzero real number k such that

17. 3.6 - 17 Caution Note that and in the expression “y varies jointly as x and z” translates as the product y = kxz. The word “and” does not indicate addition here.

18. 3.6 - 18 Example 3 SOLVING A JOINT VARIATION PROBLEM The area of a triangle varies jointly as the lengths of the base and the height. A triangle with base 10 ft and height 4 ft has area 20 ft 2. Find the area of a triangle with base 3 ft and height 8 ft.

19. 3.6 - 19 Example 3 SOLVING A JOINT VARIATION PROBLEM Solution Step 1 Let A represent the area, b the base, and h the height of the triangle. Then for some number k, A varies jointly as b and h.

20. 3.6 - 20 Example 3 SOLVING A JOINT VARIATION PROBLEM Solution Step 2 Since A is 20 when b is 10 and h is 4,

21. 3.6 - 21 Example 3 SOLVING A JOINT VARIATION PROBLEM Solution Step 3 The relationship among the variables is the familiar formula for the area of a triangle,

22. 3.6 - 22 Example 3 SOLVING A JOINT VARIATION PROBLEM Solution Step 4 When b = 3 ft and h = 8 ft,

23. 3.6 - 23 The number of vibrations per second (the pitch) of a steel guitar string varies directly as the square root of the tension and inversely as the length of the string. If the number of vibrations per second is 5 when the tension is 225 kg and the length is.60 m, find the number of vibrations per second when the tension is 196 kg and the length is.65 m. Example 4 SOLVING A COMBINED VARIATION PROBLEM

24. 3.6 - 24 Example 4 SOLVING A COMBINED VARIATION PROBLEM Solution Let n represent the number of vibrations per second, T represent the tension, and L represent the length of the string. Then, from the information in the problem, write the variation equation. (Step 1) n varies directly as the square root of T and inversely as L.

25. 3.6 - 25 Example 4 SOLVING A COMBINED VARIATION PROBLEM Solution Substitute the given values for n, T, and L to find k. (Step 2) Let n = 5, T = 225, L =.60. Multiply by.60. Divide by 15.

26. 3.6 - 26 Example 4 SOLVING A COMBINED VARIATION PROBLEM Solution Substitute for k to find the relationship among the variables (Step 3).

27. 3.6 - 27 Example 4 SOLVING A COMBINED VARIATION PROBLEM Solution Now use the second set of values for T and L to find n. (Step 4) Let T = 196, L =.65. The number of vibrations per second is approximately 4.3.