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1313 13.1Direct Variations 13.2Inverse Variations 13.3Joint Variations Chapter Summary Case Study Variations 13.4Partial Variations.

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Presentation on theme: "1313 13.1Direct Variations 13.2Inverse Variations 13.3Joint Variations Chapter Summary Case Study Variations 13.4Partial Variations."— Presentation transcript:

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

2 P. 2 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. Case Study 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. 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?

3 P. 3 Variation describes a relation between two variables. There are different kinds of variations. In this section, we will discuss direct variations first Direct Variations 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. Total payment ($P) Amount of flour (x kg)13579 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 P. 4 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 Direct Variations From the graph, we note that the total payment $P increases as the amount of flour x kg increases.

5 P Direct Variations 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. Direct Variation Both statements ‘y varies 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. This kind of relation is an example of direct variation. Notes: 1.k is called the variation constant. Since y  kx, if y 1 is the corresponding value of x 1, then. 2.‘  ’ means ‘varies directly as’.

6 P. 6 Example 13.1T Solution: 13.1 Direct Variations 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. (a)Since y  x  3, we have y  k(x  3), where k  0. Substituting y  32 and x  5 into the equation, we have 32  k(5  3) 32  8k k  4  y  4(x  3) (b)(i)When x  4, y  4(4  3)  28 (ii)When y  48, 48  4(x  3) 12  x  3 x  9

7 P. 7 Example 13.2T 13.1 Direct Variations 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 (x 1, y 1 ) and (x 2, y 2 ), we have: Since for any pairs of the corresponding values of x any y, we can find x without finding the variation constant k. Substituting x 1  4, y 1  14 and y 2  5 into the equation, we have

8 P. 8 Example 13.3T 13.1 Direct Variations  h  7f  The height of David is cm. 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 have h  kf, where k  0. Substituting h  168 and f  24 into the equation, we have k  7 When f  26.5,h  7(26.5)   24k

9 P. 9 Example 13.4T 13.1 Direct Variations Percentage change in v  v increases by 20%. 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 have v , where k  0. Let v 1 and s 1 be the original speed and the original distance respectively. New distance s 2  (1  44%)s 1  1.44s 1 New speed v 2

10 P Inverse Variations A florist is going to construct a rectangular greenhouse with area 84 m 2. 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: Width (w) Length (l) 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 P Inverse 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. Width (w) Length (l) Inverse Variation Both statements ‘y varies inversely as x’ and ‘y is inversely proportional to x’ mean that xy  k or for a non-zero constant k. Symbolically, we write.

12 P Inverse Variations Suppose. x y 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 P. 13 Example 13.5T Solution: 13.2 Inverse Variations 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. Substituting y  8 and x  3 into the equation, we have (a)When x  10,

14 P. 14 Example 13.5T 13.2 Inverse Variations Solution: 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. (b)When y  2, x  1   4 x   4  1 or 4  1 x   3 or 5 Alternative Solution: Let x 0 be the desired value of the variable when y  2.  2(x 0  1) 2  8(3  1) 2 (x 0  1) 2  16  x 0   3 or 5  When y  2, x   3 or 5.

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

16 P. 16 Example 13.7T 13.2 Inverse Variations Percentage change in the pressure of the gas  The pressure of the gas increases by 25%. 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: Let V and P be the new volume and the new pressure respectively. Then V  0.8V.

17 P Joint 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   r 2 h. In this formula,  is a constant. 1.If h is kept constant, then V varies directly as r 2 ; 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 r 2 h. 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  r 2 h or V  kr 2 h, 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 P. 18 Example 13.8T Solution: 13.3 Joint Variations Suppose P varies jointly as u 2 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. Substituting u  3, v  16 and P  54 into the equation, we have (b)When u  6 and v  64,

19 P. 19 Example 13.9T Solution: 13.3 Joint Variations Substituting x  4, y  25 and into the above equation, (b) When x  6 and y  64, Suppose z varies inversely as x 2 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.

20 P. 20 Example 13.10T 13.3 Joint Variations When W  4 and L  30,  The bar can support a mass of 200 kg. 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,

21 P Joint Variations Percentage change in I (cor. to 1 d. p.) 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, voltage and resistance respectively.

22 P Partial 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 P Partial 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 P. 24 Example 13.12T, where c and k are non-zero constants. Substituting y  16 and x  116 into the equation, Substituting y  64 and x  132 into the equation, (2)  (1): Substituting k  4 into (1), 13.4 Partial Variations 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 16  4k k  4 (b)When y  25,

25 P. 25 Example 13.13T Solution: 13.4 Partial Variations 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. Substituting x  1 and Q  26 into the equation, Substituting x  3 and Q  26 into the equation, (2)  (1): Substituting k 1  2 into (1), (a)Since Q partly varies directly as x 2 and partly varies inversely as x, we have, where k 1 and k 2 are non-zero constants. 26k 1  52 k 1  2

26 P. 26 Example 13.13T 13.4 Partial Variations  x  2 is a factor of P(x). 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, (c)When Q  20, Let P(x)  x 3  10x  12. P(2)  2 3  10(2)  12  0 By long division,

27 P. 27 Example 13.14T 13.4 Partial Variations Substituting N  500 and C  15.5  500 into the equation, 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 as N, we have C  k 1  k 2 N, where k 1 and k 2 are non-zero constants. Substituting N  1200 and C  8.5  1200 into the equation,

28 P. 28 Example 13.14T 13.4 Partial Variations (b)When N  1000,  The total cost is $9500. Cost of printing per copy 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: (2)  (1):2450  700k 2 k 2  3.5 Substituting k 2  3.5 into (1), 7750  k 1  500(3.5) k 1  6000  C  6000  3.5N

29 P. 29 Example 13.15T 13.4 Partial Variations 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 as the surface area, we have C  k 1  k 2 s 2, where k 1 and k 2 are non-zero constants. Substituting s  4 and C  18.4 into the equation, Substituting s  7 and C  31.6 into the equation, (2)  (1):33k 2  13.2 k 2  0.4 Substituting k 2  0.4 into (1), 18.4  k 1  16(0.4) k 1  12

30 P. 30 Example 13.15T cost Selling price  $[26.4  (1  25%)] 13.4 Partial Variations Solution: (b)For a cube with sides of 6 cm,  $33 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?

31 P Direct Variations Variation describes a relation between two changing quantities and we can use an equation to express the relation. Chapter Summary 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 y  x.

32 P Inverse Variations Chapter Summary 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 P Joint Variations Chapter Summary 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 z  xy.

34 P Partial Variations Chapter Summary 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  k 1 x  k 2 y, where k 1 and k 2 are variation constants.

35 Follow-up Direct Variations Solution: Suppose y is directly proportional to for x  0 and y  6 when x  4. (a)Find an equation connecting x and y. (b)Find the value of y when x  36. (a)Since y , we have y , where k  0. Substituting y  6 and x  4 into the equation, we have (b)When x  36,  18  3(6)

36 Follow-up Direct Variations Suppose y  2 is directly proportional to x and y  8 when x  4. Find the value of y when x  14. Solution: Since y  2  x, for any points of (x 1, y 1 ) and (x 2, y 2 ), we have: Since for any pairs of the corresponding values of x any y, we can find y without finding the variation constant k. Substituting x 1  4, y 1  8 and x 2  14 into the equation, we have

37 Follow-up Direct Variations  y  0.17x  The weight of the object on the Moon is 25.5 units. An object that weighs 110 units on Earth weighs 18.7 units on the Moon. How much would an object weigh on the Moon if its weight on Earth is 150 units? Solution: Let x units be the weight on Earth and y units be the weight on the Moon. Substituting y  18.7 and x  110 into the equation, we have k  0.17 When x  150,y  0.17(150)   110k Since y  x, we have y  kx, where k  0.

38 Follow-up Direct Variations Percentage change in C  C decreases by 36%. The cost $C of making a square carpet varies directly as the square of the side x cm of the carpet. If x decreases by 20%, find the percentage change in C. Solution: Since C  x 2, we have C  kx 2, where k  0. Let C 1 and x 1 be the original cost and the original side length respectively. New side length x 2  (1  20%)x 1  0.8x 1 New cost C 2  kx 2 2 Then C 1  kx 1 2.  k(0.8x 1 ) 2  0.64kx 1 2  0.64C 1

39 Follow-up Inverse Variations Suppose y  2 varies inversely as x and y  8 when x  8. Find the value of y when x  20. Solution: Substituting y  8 and x  8 into the above equation, we have When x  20,

40 Follow-up Inverse Variations Solution: Let x and y be the new values of x and y respectively. (cor. to 3 sig. fig.) Suppose y is inversely proportional to 2x. If x decreases by 10%, find the percentage change in y. (Give the answer correct to 3 significant figures.) Then x  (1  10%)x  0.9x Percentage change  y increases by 11.1% when x decreases by 10%.

41 Follow-up Inverse Variations The time taken T hours for Jane to run from her home to the park and back home again varies inversely as her speed V km/h. She takes 0.8 hour for her trip if she is running at a speed of 10 km/h. (a)Write down an equation connecting T and V. Solution: Substituting T  0.8 and V  10 into the above equation, we have

42 Follow-up Inverse Variations Solution: (b)Let T and V be the new time and the the new speed respectively. Percentage change in the time taken  The time taken for the trip increases by 11.1%. Then V  0.9V. (cor. to 3 sig. fig.) The time taken T hours for Jane to run from her home to the park and back home again varies inversely as her speed V km/h. She takes 0.8 hour for her trip if she is running at a speed of 10 km/h. (a)Write down an equation connecting T and V. (b)If her speed decreases by 10%, find the percentage change in the time taken. (Give the answer correct to 3 significant figures.)

43 Follow-up Joint Variations Solution: Suppose z varies jointly as and y. When x  9 and y  3, z  18. (a)Express z in terms of x and y. (b)Find the value of z when x  16 and y  6. Substituting x  9, y  3 and z  18 into the equation, we have (b)When x  16 and y  6,

44 Follow-up Joint Variations Solution: Substituting x  3, y  12 and into the above equation, (b) When x  2 and y  6, Suppose z varies directly as x 2 and inversely as y. When x  3 and y  12, z . (a)Express z in terms of x and y. (b)Find the value of z when x  2 and y  6.

45 Follow-up Joint Variations When r  2 and t  5,  The amount of interest is $5000. The simple interest $I earned on a fixed amount of money varies jointly as the interest rate r% p.a. and the time t years of deposit. If the interest rate is 5% p.a. and the time of deposit is four years, then the interest is $ Find the amount of interest if the interest rate is 2% p.a. and the time of deposit is five years. Solution: Let I  kr%t, where k  0. Substituting r  5, t  4 and I  into the equation,  k(5%)(4) k  I  (2%)(5)  5000

46 Follow-up Joint Variations  BMI decreases by 13.2%. Percentage change in BMI (cor. to 3 sig. fig.) Body Mass Index (BMI) is used to measure the proportion of a person’s height and weight. BMI varies directly as a person’s weight w and inversely as the square of a person’s height h. If w increases by 25% and h increases by 20%, find the percentage change in BMI. (Give the answer correct to 3 significant figures.) Solution: Let B be the BMI of a person. Then, where k  0. Let w, h and B be the new weight, height and BMI respectively.

47 Follow-up Partial Variations, where c and k are non-zero constants. Suppose z is partly constant and partly varies inversely as y. When y  12, z  ; when y  27, z . (a)Express z in terms of y. (b)Find the value of z when y  42. Solution: (a)Since z is partly constant and partly varies directly as y, we have

48 Follow-up (b)When y  42, 13.4 Partial Variations Suppose z is partly constant and partly varies inversely as y. When y  12, z  ; when y  27, z . (a)Express z in terms of y. (b)Find the value of z when y  42. Solution: (2)  2  (1): 15  30c

49 Follow-up Partial Variations Solution: Suppose P partly varies directly as x and partly varies inversely as x. When x  2, P  16; when x  3, P  19. (a)Express P in terms of x. (b)Find the value of P when x  6. When x  2 and P  16, When x  3 and P  19, (2)  (1): Substituting k 1  5 into (1), (a)Since P partly varies directly as x and partly varies inversely as x, we have, where k 1 and k 2 are non-zero constants. 5k 1  25 k 1  5

50 Follow-up Partial Variations Solution: Suppose P partly varies directly as x and partly varies inversely as x. When x  2, P  16; when x  3, P  19. (a)Express P in terms of x. (b)Find the value of P when x  6. (c)Find the values of x when P  23. (b)When x  6,(c)When P  23,

51 Follow-up Partial Variations The price $P of a globe is partly constant and partly varies as the square of the radius r m of the globe. When r  0.16, P  178; when r  0.2, P  250. (a)Express P in terms of r. (b)Find the price of a globe with radius 30 cm. Solution: Substituting P  178 and r  0.16 into the equation, (a)Since P is partly constant and partly varies as the square of r, we have P  c  kr 2, where c and k are non-zero constants. Substituting P  250 and r  0.2 into the equation,

52 Follow-up Partial Variations The price $P of a globe is partly constant and partly varies as the square of the radius r m of the globe. When r  0.16, P  178; when r  0.2, P  250. (a)Express P in terms of r. (b)Find the price of a globe with radius 30 cm. Solution: (b)When r  0.3,  The price of a globe with radius 30 cm is $500. (2)  (1):72  k k  5000 Substituting k  5000 into (2), 250  c  0.04(5000) c  50  P  50  5000r 2

53 Follow-up Partial Variations The average cost $C for a manufacturer to make a tennis racket is partly constant and partly varies inversely as the number of rackets n produced per day. If the manufacturer produces 1000 rackets per day, the cost per racket is $42; if he manufacturer produces 2000 rackets per day, the cost per racket is $27. (a)Express C in terms of n. Solution: (a)Since C is partly constant and partly varies inversely as n, we have, where k 1 and k 2 are non-zero constants. Substituting n  1000 and C  42 into the equation, Substituting n  2000 and C  27 into the equation,

54 Follow-up Partial Variations The average cost $C for a manufacturer to make a tennis racket is partly constant and partly varies inversely as the number of rackets n produced per day. If the manufacturer produces 1000 rackets per day, the cost per racket is $42; if he manufacturer produces 2000 rackets per day, the cost per racket is $27. (a)Express C in terms of n. Solution: (2)  (1):  1000k 1 k 1  12 Substituting k 1  12 into (1),  1000(12)  k 2 k 2 

55 Follow-up Partial Variations (cor. to the nearest dollar) The average cost $C for a manufacturer to make a tennis racket is partly constant and partly varies inversely as the number of rackets n produced per day. If the manufacturer produces 1000 rackets per day, the cost per racket is $42; if he manufacturer produces 2000 rackets per day, the cost per racket is $27. (a)Express C in terms of n. (b)Find the total cost of producing 1500 rackets a day. (c)If the manufacturer wants to make a profit of $ by making and selling 1500 rackets per day, find the selling price of each racket. (Give the answer correct to the nearest dollar.) Solution: (b)When n  1500, Total cost  $(32  1500)  $ (c)Total income  $(  )  $ Selling price of each racket


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