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FUNCTIONS AND THEIR CURVES 269 r = a sinq ao a Figure 31.12 scale factor ‘a’. Graphs of y D x C 1and y D 3x C 1 are shown in Fig. 31.13(a) and graphs of y D sin  and y D 2sin are shown in Fig. 31.13(b). 0 p 2 p 3p 2 2p 1 y 2 q y = 2 sin q y = sin q (b) (a) y 8 6 4 2 012 x y = 3( x + 1) y = x + 1 Figure 31.13 (ii) y = f .x/ Y a The graph of y D fx is translated by ‘a’ units parallel to the y-axis to obtain y D fx C a. For example, if fx D x, y D fx C 3 becomes y D x C 3, as shown in Fig. 31.14(a). Similarly, if f D cos Â, then y D f C 2becomesy D cos  C 2, as shown in Fig. 31.14(b). Also, if fx D x 2 , then y D fx C 3 becomes y D x 2 C 3, as shown in Fig. 31.14(c). 0 p 2 p 3p 2 2pq 1 3 y = cos q + 2 y = cos q (b) y 8 6 4 2 0 (c) −2 −112 x y = x 2 + 3 y = x 2 Figure 31.14 (iii) y = f .x Y a/ The graph of y D f x is translated by ‘a’ units parallel to the x-axis to obtain y D fx C a.If‘a’ > 0 it moves y D fx in the negative direction on the x-axis (i.e. to the left), and if ‘a’ < 0 it moves y D fx in the positive direction on the x-axis (i.e. to the right). For example, if fx D sin x, y D f x 3 becomes y D sin x 3 as shown in Fig. 31.15(a) and y D sin x C 4 is shown in Fig. 31.15(b). 270 ENGINEERING MATHEMATICS p 2 p 3p 2 2p x 0 −1 1 y p 3 y = sin x − p 3 y = sin x (a) p 3 ( ) p 2 p 3p 2 2p x 0 −1 1 y p 4 y = sin x + p 4 y = sin x p 4 (b) ( ) Figure 31.15 y 6 4 2 012−1−2 y = ( x + 2) 2 y = ( x − 1) 2 y = x 2 Figure 31.16 Similarly graphs of y D x 2 , y D x 1 2 and y D x C 2 2 are shown in Fig. 31.16. (iv) y = f .ax/ For each point (x 1 , y 1 ) on the graph of y D fx, there exists a point x 1 a ,y 1 on the graph of y D fax. Thus the graph of y D fax can be obtained by stretching y D fx parallel to the x-axis by a scale factor 1 a . For example, if fx D x 1 2 ,anda D 1 2 , then fax D x 2 1 2 . Both of these curves are shown in Fig. 31.17(a). Similarly, y D cos x and y D cos 2x are shown in Fig. 31.17(b). y = ( x − 1) 2 y 4 2 0−22 (a) 46 x y = 2 x () − 1 2 y 1.0 0 −1.0 y = cos xy = cos 2 x p 2 3p 2 p 2p x (b) Figure 31.17 y x 1 (a) −1 y = e x e y = − e x e y 8 4 (b) 0−112 x −2 −4 −8 y = x 2 + 2 y = −( x 2 + 2) Figure 31.18 (v) y = −f .x/ The graph of y Dfx is obtained by reflecting y D fx in the x-axis. For exam- ple, graphs of y D e x and y De x are shown in Fig. 31.18(a), and graphs of y D x 2 C2and y Dx 2 C 2 are shown in Fig. 31.18(b). (vi) y = f .−x / The graph of y D fx is obtained by reflecting y D fx in the y-axis. For exam- ple, graphs of y D x 3 and y D x 3 D x 3 are shown in Fig. 31.19(a) and graphs FUNCTIONS AND THEIR CURVES 271 y 20 10 (a) 0−223 x −1 −10 −20 y = x 3 y = (− x ) 3 y x −10 (b) 1 y = ln x y = −ln x Figure 31.19 of y D ln x and y Dln x are shown in Fig. 31.19(b). Problem 1. Sketch the following graphs, showing relevant points: (a) y D x 4 2 (b) y D x 3 8 (a) In Fig. 31.20 a graph of y D x 2 is shown by the broken line. The graph of y D x 4 2 is of the form y D fx Ca.Sincea D4, then y D x 4 2 is translated 4 units to the right of y D x 2 , parallel to the x-axis. (See section (iii) above). Figure 31.20 (b) In Fig. 31.21 a graph of y D x 3 is shown by the broken line. The graph of y D x 3 8isof the form y D f x C a.Sincea D8, then y D x 3 8 is translated 8 units down from y D x 3 , parallel to the y-axis. (See section (ii) above) Figure 31.21 Problem 2. Sketch the following graphs, showing relevant points: (a) y D 5 x C2 3 (b) y D 1 C3sin2x (a) Figure 31.22(a) shows a graph of y D x 3 . Figure 31.22(b) shows a graph of y D x C2 3 (see fx C a, section (iii) above). Fig- ure 31.22(c) shows a graph of y Dx C 2 3 (see fx, section (v) above). Fig- ure 31.22(d) shows the graph of y D 5 x C 2 3 (see fx C a, section (ii) above). (b) Figure 31.23(a) shows a graph of y D sin x. Figure 33.23(b) shows a graph of y D sin 2x (see fax, section (iv) above) 2−2 −10 −20 10 20 y = x 3 x y 0 (a) Figure 31.22 272 ENGINEERING MATHEMATICS 2−2 −10 −20 10 20 −4 x y (b) y = ( x +2) 3 0 2−2 −10 −20 10 20 −4 x y (c) y = −( x + 2) 3 0 2−2 −10 −20 10 20 −4 x y (d) 0 y = 5 − ( x + 2) 3 Figure 31.22 (continued) Figure 31.23(c) shows a graph of y D 3sin2x (see a fx, section (i) above). Figure 31.23(d) shows a graph of y D 1 C3sin2x (see fx Ca, section (ii) above). y 0 (a) 1 −1 p p 3p 2 x 2 y = sin x y y = sin 2 x 1 0 (b) −1 p 2p x 3p 2 p 2 y 3 2 1 0 (c) −1 −2 −3 p 2p x 3p 2 p 2 y = 3 sin 2 x y x 4 3 2 1 0 −1 −2 y = 1 + 3 sin 2 x p (d) 2pp 2 3p 2 Figure 31.23 Now try the following exercise Exercise 114 Further problems on simple transformations with curve sketching Sketch the following graphs, showing relevant points: (Answers on page 277, Fig. 31.33) 1. y D 3x 52.y D3x C4 3. y D x 2 C 34.y D x 3 2 5. y D x 4 2 C 26.y D x x 2 7. y D x 3 C 28.y D 1 C2cos3x FUNCTIONS AND THEIR CURVES 273 9. y D 3 2sin x C 4 10. y D 2lnx 31.3 Periodic functions A function fx is said to be periodic if fx C T D fx for all values of x,whereT is some positive number. T is the interval between two successive repetitions and is called the period of the function fx. For example, y D sin x is periodic in x with period 2 since sin x D sinx C 2 D sinx C 4, and so on. Similarly, y D cos x is a periodic function with period 2 since cos x D cosx C 2 D cosx C 4, and so on. In general, if y D sin ωt or y D cos ωt then the period of the waveform is 2/ω. The function shown in Fig. 31.24 is also periodic of period 2 and is defined by: fx D 1, when Ä x Ä 0 1, when 0 Ä x Ä f ( x ) 0 1 −1 − p−2pp2p x Figure 31.24 31.4 Continuous and discontinuous functions If a graph of a function has no sudden jumps or breaks it is called a continuous function,examples being the graphs of sine and cosine functions. How- ever, other graphs make finite jumps at a point or points in the interval. The square wave shown in Fig. 31.24 has finite discontinuities as x D ,2, 3, and so on, and is therefore a discontinuous func- tion. y D tan x is another example of a discontinuous function. 31.5 Even and odd functions Even functions A function y D fx is said to be even if fx D fx for all values of x. Graphs of even functions are always symmetrical about the y-axis (i.e. is a mirror image). Two examples of even functions are y D x 2 and y D cos x as shown in Fig. 31.25. −3 −2 −10 1 2 3 x 2 4 6 8 y y = x 2 (a) 0−pp/2 p x y y = cos x (b) −p/2 Figure 31.25 Odd functions A function y D fx is said to be odd if fx Dfx for all values of x. Graphs of odd functions are always symmetrical about the origin. Two examples of odd functions are y D x 3 and y D sin x as shown in Fig. 31.26. Many functions are neither even nor odd, two such examples being shown in Fig. 31.27. Problem 3. Sketch the following functions and state whether they are even or odd functions: (a) y D tan x (b) fx D 2, when 0 Ä x Ä 2 2, when 2 Ä x Ä 3 2 , 2, when 3 2 Ä x Ä 2 and is periodic of period 2 274 ENGINEERING MATHEMATICS −3 03 x 27 −27 y y = x 3 (a) −3p/2 −p −p/2 3π/20 p/2 p 2π x y = sin x y 1 −1 (b) Figure 31.26 (a) 0−1 −123 x y 20 10 y = e x (b) 0 x y Figure 31.27 (a) A graph of y D tan x is shown in Fig- ure 31.28(a) and is symmetrical about the origin and is thus an odd function (i.e. tanx Dtan x). (b) A graph of fx is shown in Fig. 31.28(b) and is symmetrical about the fx axis hence the function is an even one, (fx D fx). Problem 4. Sketch the following graphs and state whether the functions are even, odd or neither even nor odd: (a) y D ln x (b) fx D x in the range to and is periodic of period 2 −p 0 x y y = tan x −2p −p 0 p 2π x f ( x ) 2 −2 p 2p (a) (b) Figure 31.28 12 3 4 x y = ln x y 1.0 0.5 −0.5 −2p −p 0 p 2p x −p p y y = x (a) (b) 0 Figure 31.29 (a) A graph of y D ln x is shown in Fig. 31.29(a) and the curve is neither symmetrical about the y-axis nor symmetrical about the origin and is thus neither even nor odd. (b) A graph of y D x in the range to is shown in Fig. 31.29(b) and is symmetrical about the origin and is thus an odd function. FUNCTIONS AND THEIR CURVES 275 Now try the following exercise Exercise 115 Further problems on even and odd functions In Problems 1 and 2 determine whether the given functions are even, odd or neither even nor odd. 1. (a) x 4 (b) tan 3x (c) 2e 3t (d) sin 2 x (a) even (b) odd (c) neither (d) even 2. (a) 5t 3 (b) e x C e x (c) cos   (d) e x (a) odd (b) even (c) odd (d) neither 3. State whether the following functions which are periodic of period 2 are even or odd: (a) f D Â, when Ä Â Ä 0 Â, when 0 Ä Â Ä (b) fx D x, when 2 Ä x Ä 2 0, when 2 Ä x Ä 3 2 [(a) even (b) odd] 31.6 Inverse functions If y is a function of x, the graph of y against x can be used to find x when any value of y is given. Thus the graph also expresses that x is a function of y. Two such functions are called inverse functions. In general, given a function y D fx, its inverse may be obtained by inter-changing the roles of x and y and then transposing for y. The inverse function is denoted by y D f 1 x. For example, if y D 2xC1, the inverse is obtained by (i) transposing for x,i.e.x D y 1 2 D y 2 1 2 and (ii) interchanging x and y, giving the inverse as y D x 2 1 2 Thus if fx D 2x C 1, then f 1 x D x 2 1 2 A graph of fx D 2x C 1 and its inverse f 1 x D x 2 1 2 is shown in Fig. 31.30 and f 1 x is seen to be a reflection of fx in the line y D x. Figure 31.30 Similarly, if y D x 2 , the inverse is obtained by (i) transposing for x,i.e.x Dš p y and (ii) interchanging x and y, giving the inverse y Dš p x Hence the inverse has two values for every value of x. Thus fx D x 2 does not have a single inverse. In such a case the domain of the original function may be restricted to y D x 2 for x>0. Thus the inverse is then y DC p x.Agraphoffx D x 2 and its inverse f 1 x D p x for x>0 is shown in Fig. 31.31 and, again, f 1 x is seen to be a reflection of fx in the line y D x. Figure 31.31 It is noted from the latter example, that not all functions have a single inverse. An inverse, however, can be determined if the range is restricted. 276 ENGINEERING MATHEMATICS Problem 5. Determine the inverse for each of the following functions: (a) fx D x 1(b)fx D x 2 4 x > 0 (c) fx D x 2 C 1 (a) If y D fx,theny D x 1 Transposing for x gives x D y C 1 Interchanging x and y gives y D x C1 Hence if fx D x 1, then f −1 .x/ = x Y 1 (b) If y D fx,theny D x 2 4 x > 0 Transposing for x gives x D p y C4 Interchanging x and y gives y D p x C 4 Hence if fx D x 2 4 x > 0 then f −1 .x/ = p x Y 4ifx > −4 (c) If y D fx,theny D x 2 C 1 Transposing for x gives x D p y 1 Interchanging x and y gives y D p x 1, which has two values. Hence there is no single inverse of f .x/ = x 2 Y 1, since the domain of f(x) is not restricted. Inverse trigonometric functions If y D sin x,thenx is the angle whose sine is y. Inverse trigonometrical functions are denoted either by prefixing the function with ‘arc’ or by using 1 . Hence transposing y D sin x for x gives x D arcsin y or sin 1 y. Interchanging x and y gives the inverse y D arcsin x or sin 1 x. Similarly, y D arccos x, y D arctan x, y D arcsec x, y D arccosec x and y D arccot x are all inverse trigonometric functions. The angle is always expressed in radians. Inverse trigonometric functions are periodic so it is necessary to specify the smallest or principal value of the angle. For y D arcsin x,arctanx, arccosec x and arccot x, the principal value is in the range 2 <y< 2 .Fory D arccos x and arcsec x the principal value is in the range 0 <y<. Graphs of the six inverse trigonometric functions are shown in Fig. 31.32. Problem 6. Determine the principal values of (a) arcsin 0.5 (b) arctan1 (c) arccos p 3 2 (d) arccosec p 2) y 3p/2 p/2 p 0 −p −3p/2 +1 x −1 (b) y = arccos x −p/2 y p 0 −p 3p/2 p/2 −p/2 −3p/2 y = arcsec x x +1−1 (d) y y = arccot x −p 0 p x (f) y p/2 0 −p/2 y = arctan x (c) x y p 0 −p 3p/2 p/2 −p/2 −3p/2 y = arccosec x x +1 (e) −1 −p/2 p/2 y 3p/2 p/2 p −p/2 −p −3p/2 +1 x B y = arcsin x A −1 (a) 0 Figure 31.32 Using a calculator, (a) arcsin 0.5 Á sin 1 0.5 D 30 ° D p 6 rad or 0.5236 rad (b) arctan1 Á tan 1 1 D45 ° D − p 4 rad or −0.7854 rad (c) arccos p 3 2 Á cos 1 p 3 2 D 150 ° D 5 p 6 rad or 2 .6180 rad (d) arccosec p 2 D arcsin 1 p 2 Á sin 1 1 p 2 D 45 ° D 4 rad or 0 .7854 rad Problem 7. Evaluate (in radians), correct to 3 decimal places: arcsin 0.30 Carccos 0.65 arcsin 0.30 D 17.4576 ° D 0.3047 rad arccos 0.65 D 49.4584 ° D 0.8632 rad Hence arcsin 0.30 Carccos 0.65 D 0.3047 C 0.8632 D 1 .168, correct to 3 decimal places FUNCTIONS AND THEIR CURVES 277 Now try the following exercise Exercise 116 Further problems on inverse functions Determine the inverse of the functions given in Problems 1 to 4. 1. fx D x C 1[f 1 x D x 1] 2. fx D 5x 1 f 1 x D 1 5 x C 1 3. fx D x 3 C 1[f 1 x D 3 p x 1] 4. fx D 1 x C 2 f 1 x D 1 x 2 Determine the principal value of the inverse functions in problems 5 to 11. 5. arcsin1 2 or 1.5708 rad 6. arccos 0.5 3 or 1.0472 rad 7. arctan 1 4 or 0.7854 rad 8. arccot 2 [0.4636 rad] 9. arccosec 2.5 [0.4115 rad] 10. arcsec 1.5 [0.8411 rad] 11. arcsin 1 p 2 4 or 0.7854 rad 12. Evaluate x, correct to 3 decimal places: x D arcsin 1 3 C arccos 4 5 arctan 8 9 [0.257] 13. Evaluate y, correct to 4 significant figures: y D 3arcsec p 2 4arccosec p 2 C 5 arccot 2 [1.533] Answers to Exercise 114 Figure 31.33 Graphical solutions to Exercise 114, page 272. 278 ENGINEERING MATHEMATICS Figure 31.33 (continued) [...]... [36 20° ]4 D 34 6 4 ð 20° D 81 6 80 ° by de Moivre’s theorem D 1 5 7 746 144 .46 ° (Note, by considering the Argand diagram, 7 C j5 must represent an angle in the second quadrant and not in the fourth quadrant) Applying de Moivre’s theorem: p 7 C j5 4 D [ 746 144 .46 ° ]4 p D 744 6 4 ð 144 .46 ° D 547 66 577. 84 Problem 1 (a) [26 35° ]5 (a) Determine, in polar form: (b) 2 C j3 D 547 66 217. 84 or 6 547 66 217°... (b) 1 C j2 6 [(a) 7.5 946 75° (b) 1256 20.62° ] 3 04 ENGINEERING MATHEMATICS 2 Determine in polar and Cartesian forms (a) [36 41 ° ]4 (b) 2 j 5 (a) 81 6 1 64 , 77 .86 C j22.33 47 .17° , 38 (b) 55.906 6 C j5 3 [47 6 .46 119 .42 ° , 5 6 2 34 C j415] j8 5 [45 5306 12. 78 , 44 40 0 C j10 070] 3 2 C j7 52 C 122 D p D 131/2 6 16 6 38. 27 ð 106 6 176.15° , 106 38. 18 C j2.570 136 42 7. 38 1 2 ð 67. 38 and D 3.616 33.69°... f l VC (a) In Problems 6 to 8, evaluate in polar form 299 (b) V Figure 34. 8 6 (a) 36 20° ð 156 45 ° 21° (b) 2 .46 65° ð 4. 46 [(a) 45 6 65° 7 (a) 6 .46 27° ł 26 (b) 10.566 44 ° ] 15° 40 ° (b) 56 30° ð 46 80 ° ł 106 [(a) 3.26 42 ° C 36 6 8 6 120° C 5.26 58 (b) 2 (b) 26 150° ] 8 (a) 46 [(a) 6. 986 6 26. 78 1.66 40 ° (b) 7.1906 85 .77° ] Similarly, for the R-C circuit shown in Figure 34. 8( b), VC lags I by 90° (i.e... Since r 6  D r cos  C jr sin Â, 547 66 217. 84 D 547 6 cos 217. 84 C j 547 6 sin 217. 84 D 326 175° (b) 2 2C 3 2 C j3 D p 26 tan 3 2 136 123.69° , since 2 C j3 lies in the second quadrant p D [ 136 123.69° ]6 p D 136 6 6 ð 123.69° , by De Moivre’s theorem D i.e D 2 C j3 6 D 21976 742 . 14 D 21976 382 . 14 since 742 . 14 Á 742 . 14 360° D 382 . 14 43 25 j3359 −7 Y j 5 /4 = 43 25 − j 3359 in rectangular form Now... 56 45 ° 46 120° D 1.732 C j1.000 C 3.536 j3.536 2.000 C j3 .46 4 Problem 12 D 7.2 68 Determine, in polar form: (a) 86 25° ð 46 60° (b) 36 16° ð 56 44 ° ð 26 80 ° (a) 86 25° 46 60° D 8 4 (b) 6 D 7.2 682 C 6.0002 6 tan 166 75° (a) 26 15° Evaluate in polar form: 106 (b) 4 66 ð 126 3 1 2 6.000 7.2 68 Now try the following exercise Exercise 1 24 Problem 13 1 D 9 .42 56 −39. 54 or 9 .42 56 −39° 320 25° C60° D 326 85 °... 45 ° and F2 D 30 N at 125° act at a point Determine by drawing and by calculation (a) F1 Y F2 (b) F1 − F2 (a) 54. 0 N at 78. 16° (b) 45 . 64 N at 4. 66° 15 cos 290° D 6.99 27.67 1 The horizontal component of v2 − v1 − v3 D 40 cos 190° 1 Now try the following exercise D 14. 18 Thus v1 − v2 Y v3 = 28. 54 units at 14. 18 6.99 27.67 hence the required angle is 180 ° C 14. 18 D 1 94. 18 Thus v2 − v1 − v3 = 28. 54. .. component of force, V D 7 sin 0° C 4 sin 45 ° D 0 C 2 .82 8 D 2 .82 8 N The magnitude of the resultant of vector addition p D H2 C V2 D 9 .82 82 C 2 .82 82 p D 1 04. 59 D 10.23 N The direction of the resultant of vector addition D tan 1 V H D tan 1 2 .82 8 9 .82 8 D 16.05° 2 84 ENGINEERING MATHEMATICS Thus, the resultant of the two forces is a single vector of 10.23 N at 16.05° to the 7 N vector 4 Problem 5 Calculate the resultant... trigonometric ratios, x D 4 cos 30° D 3 .46 4 and y D 4 sin 30° D 2.000 Hence 46 30° = 3 .46 4 Y j 2.000 (b) 76 145 ° is shown in Fig 34. 7(b) and lies in the third quadrant Angle ˛ D 180 ° 145 ° D 35° 3 j4 is shown in Fig 34. 6 and lies in the third quadrant Hence x D 7 cos 35° D 5.7 34 Modulus, r D 5 and ˛ D 53.13° , as above and y D 7 sin 35° D 4. 015 2 98 ENGINEERING MATHEMATICS Imaginary axis 4 30° 0 166 75° 16... or 45 ° and joined to the end of y1 as shown in Fig 33 .8( b) yR is measured as 4. 6 units long and angle is measured as 27° or 0 .47 rad Alternatively, yR is the diagonal of the parallelogram formed as shown in Fig 33 .8( c) Hence, by drawing, yR = 4. 6 sin.!t Y 0 .47 / (b) 21 .49 D 4. 64 yR D D Hence D sin By calculation, 3 sin D 4. 64 from sin 135° 3 sin 135° D 0 .45 72 4. 64 1 0 .45 72 D 27.21° or 0 .47 5 rad yR = 4. 64. .. Alternatively 76 145 ° D 7 cos 145 ° C j7 sin D 2 cos 30° C j2 sin 30° D 1.732 Cj1.000 145 ° D −5.7 34 − j 4. 015 45 ° D 5 cos 45 ° C j sin 45 ° D 5 cos 56 45 ° C j5 sin 45 ° D 3.536 34. 7 Multiplication and division in polar form If Z1 D (i) (ii) r1 6 Â1 and Z2 D r2 6 46 120° D 4 cos 120° C j sin 120° D 4 cos 120° C j4 sin 120° Â2 then: Z1 Z2 D Â1 C Â2 and Z1 r1 D 6 Â1 Â2 Z2 r2 r1 r2 6 j3.536 2.000 C j3 .46 4 D Hence . 4 cos 45 ° D 7 C2 .82 8 D 9 .82 8 N Vertical component of force, V D 7sin0 ° C 4sin45 ° D 0 C2 .82 8 D 2 .82 8 N The magnitude of the resultant of vector addition D H 2 C V 2 D p 9 .82 8 2 C 2 .82 8 2 D p 1 04. 59. 4. 2 6.0 8 .4 9 .8 11 .4 y 15 .4 32.5 60.2 111 .8 150.1 200.9 (9) 6. Determine the law of the form y D ae kx which relates the following values: y 0.0306 0. 285 0. 84 1 x -4. 0 5.3 9 .8 y 5.21 173.2 1 181 x. 1.57 08 rad 6. arccos 0.5 3 or 1. 047 2 rad 7. arctan 1 4 or 0. 78 54 rad 8. arccot 2 [0 .46 36 rad] 9. arccosec 2.5 [0 .41 15 rad] 10. arcsec 1.5 [0. 84 1 1 rad] 11. arcsin 1 p 2 4 or