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COLLEGE ALGEBRA

COLLEGE ALGEBRA. LIAL HORNSBY SCHNEIDER. 3.6. Variation. Direct Variation Inverse Variation Combined and Joint Variation. Direct Variation.

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COLLEGE ALGEBRA

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  1. COLLEGE ALGEBRA LIAL HORNSBY SCHNEIDER

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

  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. Direct Variation y varies directly asx, oryis directly proportional to x, if there exists a nonzero real number k, called the constant of variation, such that

  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. SOLVING A DIRECT VARIATION PROBLEM Example 1 The area of a rectangle varies directly as its length. If the area is 50 m2 when the length is 10 m, find the area when the length is 25 m. Solution 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. SOLVING A DIRECT VARIATION PROBLEM Example 1 The area of a rectangle varies directly as its length. If the area is 50 m2 when the length is 10 m, find the area when the length is 25 m. Solution Step 2 Since A = 50 when L = 10, the equation A = kL becomes

  8. SOLVING A DIRECT VARIATION PROBLEM Example 1 The area of a rectangle varies directly as its length. If the area is 50 m2 when the length is 10 m, find the area when the length is 25 m. Solution Step 3 Using this value of k, we can express the relationship between the area and the length as Direct variation equation.

  9. SOLVING A DIRECT VARIATION PROBLEM Example 1 The area of a rectangle varies directly as its length. If the area is 50 m2 when the length is 10 m, find the area when the length is 25 m. Solution Step 4 To find the area when the length is 25, we replace L with 25. The area of the rectangle is 125 m2 when the length is 25 m.

  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 isdirectly proportional to the nth power of x,if there exists a nonzero real number k such that

  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. SOLVING AN INVERSE VARIATION PROBLEM Example 2 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.

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

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

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

  17. CautionNote 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. SOLVING A JOINT VARIATION PROBLEM Example 3 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 ft2. Find the area of a triangle with base 3 ft and height 8 ft.

  19. SOLVING A JOINT VARIATION PROBLEM Example 3 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. SOLVING A JOINT VARIATION PROBLEM Example 3 Solution Step 2 Since A is 20 when b is10 and h is 4,

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

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

  23. SOLVING A COMBINED VARIATION PROBLEM Example 4 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.

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

  27. SOLVING A COMBINED VARIATION PROBLEM Example 4 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.

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