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College Algebra K /DC Monday, 10 March 2014

College Algebra K /DC Monday, 10 March 2014. OBJECTIVE TSW apply (1) direct variation and (2) inverse variation to solve applications. ASSIGNMENTS DUE WEDNESDAY Sec. 3.5: p. 373 (37-46 all, 61-89 odd) WS Sec. 3.5 Sec. 3.6: p. 384 (11-20 all)

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College Algebra K /DC Monday, 10 March 2014

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  1. College Algebra K/DCMonday, 10 March 2014 • OBJECTIVETSW apply (1) direct variation and (2) inverse variation to solve applications. • ASSIGNMENTS DUE WEDNESDAY • Sec. 3.5: p. 373 (37-46 all, 61-89 odd) • WS Sec. 3.5 • Sec. 3.6: p. 384 (11-20 all) • TEST: Sec. 3.4 – 3.6 has been moved up to Wednesday, 12 March 2014. • ASSIGNMENT DUE FRIDAY • Sec. 3.6: pp. 384-385 (21-27 odd, 28-31 all, 33, 35-38 all)

  2. Variation 3.6 Direct Variation ▪ Inverse Variation ▪CombinedVariation ▪ Joint Variation

  3. Direct Variation • yvaries 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 • y = kx. • Steps to Solve Variation Problems • 1) Write the general relationship (use the constant k). • 2) Substitute the given values to find k. • 3) Substitute this value for k into the equation. • 4) Find the required unknowns and answer the question that is asked.

  4. Direct Variation • Ex: If y varies directly as x, and y = 35 when x = 9, find y when x = 7. • y = kx • 35 = k(9) • k = 35/9 • y = (35/9)(7) • y = 245/9

  5. Direct Variation • At a given average speed, the distance traveled by a vehicle varies directly as the time. If a vehicle travels 156 miles in 3 hours, find the distance it will travel in 5 hours at the same average speed. Step 1: Since the distance varies directly as the time, d = kt. Step 2: Substitute d = 156 and t = 3 to find k.

  6. Direct Variation Step 3: The relationship between distance and time is d = 52t. Step 4: Solve the equation ford with t = 5. The vehicle will travel 260 miles in 5 hours. Be sure to properly label all units !!!

  7. Direct Variation • 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. Step 1: Since the area varies directly as the length, A = kL. Step 2: Substitute A = 50 and L = 10 to find k.

  8. Direct Variation Step 3: The relationship between Area and Length is A = 5L. Step 4: Solve the equation forA with L = 25. The area will be 125 m2 when the length is 25 m.

  9. Inverse Variation Problem If n = 1, then and yvaries inversely as x. • Let n be a positive real number. Then yvaries 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

  10. Inverse Variation Problem Step 1: Let x represent the number of items produced and y represent the cost per item. • 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 $1.50. Find the cost per item if 250 items are produced.

  11. Inverse Variation Problem Step 3: The relationship between x and y is Step 2: Substitute y = 1.50 and x = 100 to find k. Step 4: Solve the equation fory with x = 250. The cost per item will be $0.24.

  12. Combined Direct and Inverse Variation • If y varies directly as x and inversely as p and q, and y = 4 when x = −3, p = 2, and q = 5, find y when x = 2, p = 4, and q = 6.

  13. Combined Direct and Inverse Variation • If y varies directly as x and inversely as p and q, and y = 4 when x = −3, p = 2, and q = 5, find y when x = 2, p = 4, and q = 6.

  14. Joint Variation • Let m and n be real numbers. Then yvaries jointly as the nth power of x and the mth power of z if there exists a nonzero real number k such that

  15. Joint Variation • Ex: If y varies jointly as the square of x and z, and y = 24 when x = 3 and z = 4, find y when x = 5 and z = 7. y = kx2z 24 = k(3)2(4)  k = 2/3 y = 2/3x2z y = 2/3(5)2(7)  y = 350/3

  16. Inverse Variation • Ex: If y varies inversely as the cube of x, and y = 6 when x = 4, find y when x = 2.  k = 384  y = 48

  17. Assignment • Sec. 3.6: p. 384 (11-20 all) • Due on Wednesday, 12 March 2014. Sec. 3.6: pp. 384-385 (21-27 odd, 28-31 all, 33, 35-38 all) • Due on Friday, 14 March 2014.

  18. Assignment: Sec. 3.6: p. 384 (11-20 all) • Solve each variation problem. • 11) If y varies directly as x, and y = 20 when x = 4, find y when x = −6. • 12) If y varies directly as x, and y = 9 when x = 30, find y when x = 40. • 13)If m varies jointly as x and y, and m = 10 when x = 2 and y = 14, find m when x = 11 and y = 8. • 14)If m varies jointly as z and p, and m = 10 when z = 2 and p = 7.5, find m when z = 5 and p = 7. • 15) If y varies inversely as x, and y = 10 when x = 3, find y when x = 20. • 16) If y varies inversely as x, and y = 20 when x = ¼, find y when x = 15.

  19. Assignment: Sec. 3.6: p. 384 (11-20 all) • Solve each variation problem. • 17)Suppose r varies directly as the square of m, and inversely as s. If r = 12 when m = 6 and s = 4, find r when m = 6 and s = 20. • 18)Suppose p varies directly as the square of z, and inversely as r. If p = 32/5 when z = 4 and r = 10, find p when z = 3 and r = 36. • 19)Let a be directly proportional to m and n2, and inversely proportional to y3. If a = 9 when m = 4, n = 9, and y = 3, find a when m = 6, n = 2, and y = 5. • 20)If y varies directly as x, and inversely as m2 and r2, and y = 5/3 when x = 1, m = 2, and r = 3, find y when x = 3, m = 1, and r = 8.

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