Variations

1. 13 Variations Case Study 13.1Direct Variations 13.2 Inverse Variations 13.3 Joint Variations 13.4Partial Variations Chapter Summary

2. The cost of making 200 badges is $1000. Thus, the average cost of making a badge is $5. Will the average cost per badge be lowered if we make 500 badges? Case Study Andy and Betty are responsible for making the school badges. After comparing the cost of different companies, they find that the cost of making school badges consists of two parts, one is fixed for design while the other part depends on the number of badges made. So, the cost per badge can be lowered if more school badges are made. In this case, the cost of making school badges and the quantity to be made demonstrate a partial variation.

3. 13.1 Direct Variations Variation describes a relation between two variables. There are different kinds of variations. In this section, we will discuss direct variations first. Suppose Mrs. Chan goes to a store and buys some flour for baking cakes. The following table shows the total payment required for buying different amounts of flour. From the above table, we observe that the ratio of the total payment to the amount of flour bought is a constant, that is,

4. 13.1 Direct Variations The relation can be expressed as P 8x, where $P is the total payment and x kg is the amount of flour. The relation between the total payment and the amount of flour can also be represented by a graph. From the graph, we note that the total payment $P increases as the amount of flour x kg increases.

5. 13.1 Direct Variations This kind of relation is an example of direct variation. Direct Variation Both statements ‘yvaries directly as x’ and ‘y is directly proportional to x’ mean that ykx for a non-zero constant k. Symbolically, we write yx. Notes: 1. k is called the variation constant. Since ykx, if y1 is the corresponding value of x1, then . 2. ‘’ means ‘varies directly as’. In general, if y is directly proportional to x, that is, y kx, then the graph of y against x is a straight line that passes through the origin with slope k.

6. 13.1 Direct Variations Example 13.1T Suppose y is directly proportional to x 3 and y 32 when x 5. (a) Find an equation connecting x and y. (b) Find the values of (i) y when x 4; (ii) x when y 48. Solution: (a) Since yx 3, we have yk(x  3), where k 0. (b) (i) When x 4, y 4(4  3) Substituting y 32 and x  5 into the equation, we have  28 (ii) When y 48, 32 k(5  3) 48  4(x 3) 32  8k 12 x 3 k 4 x9  y 4(x  3)

7. Since for any pairs of the corresponding values of x any y, we can find x without finding the variation constant k. 13.1 Direct Variations Example 13.2T Suppose 2y 1 x and y 14 when x 4. Find the value of x when y  5. Solution: Since 2y 1 x, for any points of(x1, y1) and (x2, y2),we have: Substituting x1 4, y1 14 and y2 5 into the equation,we have

8. 13.1 Direct Variations Example 13.3T The height h cm of an average person varies directly with their foot length f cm. If the height of a person is 168 cm, then his/her foot length is 24 cm. What is the height of David if his foot length is 26.5 cm? Solution: Since hf , we haveh  kf, where k 0. Substituting h 168 and f  24 into the equation, we have 168  24k k 7  h  7f When f 26.5, h  7(26.5)  185.5  The height of David is 185.5 cm.

9. 13.1 Direct Variations Example 13.4T A ball is dropped onto the ground. Let v m/s be the speed of the ball after it falls a distance of s m. Suppose . If s increases by 44%, find the percentage change in v. Solution: Since , we havev  , where k 0. Letv1ands1be the original speed and the original distance respectively. Percentage change in v New distance s2  (1  44%)s1  1.44s1 New speed v2  v increases by 20%.

10. 13.2Inverse Variations A florist is going to construct a rectangular greenhouse with area 84 m2. He can build the greenhouse with materials of different lengths and widths. The following table shows several pairs of length and width for the greenhouse: From the above table, we observe that w increases when l decreases and the product of l and w is a constant, that is, lw 84 or .

11. Inverse Variation Both statements ‘yvaries inversely as x’ and ‘y is inversely proportional to x’ mean that xykor for a non-zero constant k. Symbolically, we write . 13.2Inverse Variations The relation between the length and the width can also be represented graphically. The relation between the length and the width is called an inverse variation.

12. 0.5 0.25 0.17 0.13 0.1 Remark: We observe that the graph of y against is a straight line. In fact, from the relation of , we have . This means y varies directly as . 13.2Inverse Variations Suppose .

13. 13.2Inverse Variations Example 13.5T Suppose y varies inversely as the square of x 1 and y 8 when x 3. Find the values of (a) y when x 10; (b) x when y 2. Solution: Substituting y 8 and x  3 into the equation, we have (a) When x 10,

14. 13.2Inverse Variations Example 13.5T Suppose y varies inversely as the square of x 1 and y 8 when x 3. Find the values of (a) y when x 10; (b) x when y 2. Alternative Solution: Solution: (b) When y 2, Let x0 be the desired value of the variable when y 2. x 1 4 x 4  1 or 4  1  2(x0 1)2 8(3  1)2 x 3 or 5 (x0 1)2 16 x0 3 or 5  When y 2, x  3 or 5.

15. 13.2Inverse Variations Example 13.6T Suppose . Find the percentage change in y when x increases by 15%. (Give the answer correct to 3 significant figures.) Solution: Letxandybe the new values of x and y respectively. Thenx  (1  15%)x Percentage change 1.15x (cor. to 3 sig. fig.) y decreases by 13.0% when x increases by 15%.

16. 13.2Inverse Variations Example 13.7T At a fixed temperature, the pressure P of any gas with a fixed mass varies inversely as its volume V. A gas is compressed to 80% of its original volume without a temperature change. Find the percentage change in the pressure of the gas. Solution: Percentage change in the pressure of the gas LetVandPbe the new volume and the new pressure respectively. Then V  0.8V.  The pressure of the gas increases by 25%.

17. 13.3Joint Variations Consider the volume V of a cylinder with base radius r and height h. The volume of the cylinder can be calculated by the formula Vpr2h. In this formula, p is a constant. 1. If h is kept constant, then V varies directly as r2; 2. If r is kept constant, then V varies directly as h; 3. If neither r nor h is kept constant, then V varies directly as r2h. We say that the volume V varies jointly as the height h and the square of the base radius r. Such a relation is called a joint variation. The variation can be represented by Vr2h or V kr2h, where k is the variation constant. Joint Variation If one variable z varies jointly as two (say x and y) or more other variables (either directly or inversely), it is called a joint variation.

18. 13.3Joint Variations Example 13.8T Suppose P varies jointly as u2 and . When u 3 and v  16, P 54. (a) Express P in terms of u and v. (b) Find the value of P when u 6 and v 64. Solution: Substituting u 3, v 16 and P  54 into the equation, we have (b) When u 6 and v 64,

19. 13.3Joint Variations Example 13.9T Suppose z varies inversely as x2 and directly as . When x 4 and y  25, z . (a) Express z in terms of x and y. (b) Find the value of z when x 6 and y 64. Solution: Substituting x 4, y  25 and into the above equation, (b) When x 6 and y  64,

20. 13.3Joint Variations Example 13.10T The mass M kg that a wooden bar of a certain thickness can support varies directly as its width W cm and inversely as its length L m. If a bar of width 3 cm and length 15 m can support a mass of 300 kg, what mass can be supported by a bar of width 4 cm and length 30 m? Solution: Let , where k 0. Substituting W 3, L  15 and M  300 into the equation, When W 4 and L  30,  The bar can support a mass of 200 kg.

21. 13.3Joint Variations Example 13.11T The current I flowing through a resistor varies directly as the voltage V across it and inversely as its resistance R. (a) Write down an equation connecting I, V and R. (b) Find the percentage change in I if V increases by 10% and R decreases by 10%. (Give the answer correct to 1 decimal place.) Solution: (a) The equation is , where k 0. (b) Let I,V and R be the new current,voltageandresistancerespectively. Percentage change in I (cor. to 1 d. p.)

22. 13.4Partial Variations In many practical situations, a variable is the sum of two or more parts; each part may be either fixed (a constant) or may vary as other variables. For example, an association organizes a seminar. The organizer has to book a hall for running this function which involves a fixed expense. Moreover, the organizer needs to prepare refreshments for the participants. If the rent of the hall is $5000 and the cost of refreshments for one participant is $30, then the total expense $E can be expressed as E 5000  30N, where N is the number of participants. We see that E is partly constant, and it partly varies directly as N. Such a relation is called partial variation.

23. 13.4Partial Variations Consider another example of partial variation. The solid in the figure consists of two parts, one part is a sphere with radius r while the other part is a cube with sides s. The total volume V of the solid is given by: In this case, V is the sum of two parts such that one part varies directly as the cube of r and the other part varies directly as the cube of s.

24. 13.4Partial Variations Example 13.12T Suppose x is partly constant and partly varies directly as . When y 16, x 116; when y  64, x 132. (a) Find an equation connecting x and y. (b) Find the value of x when y 25. Solution: (a) Since x is partly constant and partly varies directly as , we have , where c and k are non-zero constants. Substituting y 16 and x  116 into the equation, (2)  (1): 16  4k k 4 Substituting k 4 into (1), Substituting y 64 and x  132 into the equation, (b) When y 25,

25. 13.4Partial Variations Example 13.13T Suppose Q partly varies directly as the square of x and partly varies inversely as x. Q 26 when x 1 or 3. (a) Express Q in terms of x. (b) Find the value of Q when x 6. Solution: (a) Since Q partly varies directly as x2 andpartly varies inversely as x, we have , where k1 and k2 are non-zero constants. (2)  (1): 26k1 52 Substituting x 1 and Q  26 into the equation, k1 2 Substituting k1 2 into (1), Substituting x 3 and Q  26 into the equation,

26. 13.4Partial Variations Example 13.13T Suppose Q partly varies directly as the square of x and partly varies inversely as x. Q 26 when x 1 or 3. (a) Express Q in terms of x. (b) Find the value of Q when x 6. (c)Find the possible values of x when Q 20. (Give the answers in surd form if necessary.) Solution: (b) When x 6, Let P(x) x3 10x  12. P(2)  23 10(2)  12  0 (c) When Q 20,  x  2 is a factor of P(x). By long division,

27. 13.4Partial Variations Example 13.14T The total cost $C of printing a magazine is partly constant and partly varies directly as the number of copies N printed. If 500 copies are printed, the cost of printing per copy is $15.5. If 1200 copies are printed, the cost of printing per copy is $8.5. (a) Express C in terms of N. (b) What is the cost of printing per copy if 1000 copies are printed? Solution: (a) Since C is partly constant and partly varies directly asN, we have Ck1k2N, where k1 and k2 are non-zero constants. Substituting N 500 and C  15.5  500 into the equation, Substituting N 1200 and C  8.5  1200 into the equation,

28. 13.4Partial Variations Example 13.14T The total cost $C of printing a magazine is partly constant and partly varies directly as the number of copies N printed. If 500 copies are printed, the cost of printing per copy is $15.5. If 1200 copies are printed, the cost of printing per copy is $8.5. (a) Express C in terms of N. (b) What is the cost of printing per copy if 1000 copies are printed? Solution: (b) When N 1000, (2)  (1): 2450  700k2  The total cost is $9500. k2 3.5 Substituting k2 3.5 into (1), Cost of printing per copy 7750 k1 500(3.5) k1 6000  C 6000  3.5N

29. 13.4Partial Variations Example 13.15T The cost $C of making a wooden cube is partly constant and partly varies directly as the surface area of the cube. When the side s cm of the cube is 4 cm, the cost is $18.4; when the side of the cube is 7 cm, the cost is $31.6. (a) Express C in terms of s. Solution: (a) Since C is partly constant and partly varies directly asthe surface area, we have Ck1k2s2, where k1 and k2 are non-zero constants. Substituting s 4 and C  18.4 into the equation, (2)  (1): 33k2 13.2 k2 0.4 Substituting k2 0.4 into (1), Substituting s 7 and C  31.6 into the equation, 18.4 k1 16(0.4) k1 12

30. 13.4Partial Variations Example 13.15T The cost $C of making a wooden cube is partly constant and partly varies directly as the surface area of the cube. When the side s cm of the cube is 4 cm, the cost is $18.4; when the side of the cube is 7 cm, the cost is $31.6. (a) Express C in terms of s. (b) In order to make a profit percentage of 25%, what is the selling price of a cube with sides of 6 cm? Solution: (b) For a cube with sides of 6 cm, cost Selling price  $[26.4  (1  25%)]  $33

31. Chapter Summary 13.1 Direct Variations Variation describes a relation between two changing quantitiesand we can use an equation to express the relation. Both statements ‘y varies directly as x’ and ‘y is directly proportional to x’ mean that ykx, where k is a non-zero constant. Symbolically, we write yµx.

32. Chapter Summary 13.2 Inverse Variations Both statements ‘y varies inversely as x’ and ‘y is inversely proportional to x’ mean that xy k or , where k is a non-zero constant. Symbolically, we write .

33. Chapter Summary 13.3Joint Variations If one variable z varies jointly as two (say x and y) or more other variables (either directly or inversely), it is a joint variation. Symbolically, if z varies jointly as x and y, we write zµxy.

34. Chapter Summary 13.4Partial Variations 1. If z is partly constant and partly varies directly as x, then zckx, where c is a constant and k is the variation constant. 2. If z partly varies directly as x and partly varies directly as y, then zk1xk2y, where k1 and k2 are variation constants.