Trigonometric waveforms

Transcription

1 Trigonometric waveforms. Graphs of trigonometric functions By drawing up tables of values from to 6, graphs of y sin A, y cosa and y tana may be plotted. Values obtained with a calculator correct to decimal places which is more than sufficient for plotting graphs, using intervals, are shown below, with the respective graphs shown in Fig... a b c y = sin A A sin A A 7 6 sin A y = cos A A cos A A 7 6 cos A y = tan A A tan A A 7 6 tan A From Fig.. it is seen that: i ii iii Sine and cosine graphs oscillate between peak values of š The cosine curve is the same shape as the sine curve but displaced by 9. The sine and cosine curves are continuous and they repeat at intervals of 6 ; the tangent curve appears to be discontinuous and repeats at intervals of 8. a b c. y A... y. y.. y = cos A y = sin A A 6 9 Figure. y = tan A 8 7. Angles of any magnitude 6 A Figure. shows rectangular aes XX and YY intersecting at origin. As with graphical work, measurements made to the right and above are positive, while those to the left and downwards are negative. Let A be free to rotate about. By convention, when A moves anticlockwise angular measurement is considered positive, and vice versa. Let A be rotated anticlockwise so that is any angle in the first quadrant and let perpendicular AB be constructed to form the right-angled triangle AB in Fig... Since all three sides of the triangle are positive, the trigonometric ratios sine, cosine and tangent will all be positive in the first quadrant. Quadrant 9 Y Quadrant X A X Quadrant Quadrant Figure. 8 Figure. Y 7 9 Quadrant Quadrant A + + A + + q q + C q q E B + + A A Quadrant Quadrant Note: A is always positive since it is the radius of a circle. Let A be further rotated so that is any angle in the second quadrant and let AC be constructed to form the right-angled triangle AC. Then sin C C C cos C tan C Let A be further rotated so that is any angle in the third quadrant and let A be constructed to form the right-angled triangle A. Then sin C cos C tan C Let A be further rotated so that is any angle in the fourth quadrant and let AE be constructed to form the right-angled triangle AE. Then sin C cos C C C tan C 8 Sine Tangent Figure. 9 7 TRIGONOMETRIC WAVEFORMS 8 All positive Cosine 6 The above results are summarized in Fig... The letters underlined spell the word CAST when starting in the fourth quadrant and moving in an anticlockwise direction. In the first quadrant of Fig.. all of the curves have positive values; in the second only sine is positive; in the third only tangent is positive; in the fourth only cosine is positive eactly as summarized in Fig... A knowledge of angles of any magnitude is needed when finding, for eample, all the angles between and 6 whose sine is, say,.6. If.6 is entered into a calculator and then the inverse sine key pressed or sin key the answer 9. appears. However, there is a second angle between and 6 which the calculator does not give. Sine is also positive in the second quadrant either from CAST or from Fig..a. The other angle is shown in Fig.. as angle where Thus 9. and 6.97 are the angles between and 6 whose sine is.6 check that sin on your calculator. Be careful! Your calculator only gives you one of these answers. The second answer needs to be deduced from a knowledge of angles of any magnitude, as shown in the following worked problems. Problem. etermine all the angles between and 6 whosesineis.68 The angles whose sine is.68 occurs in the third and fourth quadrants since sine is negative in these quadrants see Fig..6.

2 8 ENGINEERING MATHEMATICS TRIGONOMETRIC WAVEFORMS 8 S T Figure. y..68. Figure.6 S 7 A q C y = sin A y.769 Figure.8 8 y = tan S T q Figure A q C 6 Problem. Solve the equation cos.8 for angles of between and 6 Now try the following eercise Eercise 8 Further problems on angles of any magnitude. etermine all of the angles between and 6 whosesineis: a.679 b.8 a.78 and 7. b 88. and.7. Solve the following equations for values of between and 6 : a cos.879 b cos.7 a 9.8 and.9 b.86 and 6.. Find the angles between to 6 whose tangent is: a.978 b.8 a. and. b. and 9. If all horizontal components such as OS are projected on to a graph of y against angle, then a cosine wave is produced. It is easier to visualize these projections by redrawing the circle with the radius arm OR initially in a vertical position as shown in Fig... From Figs.. and. it is seen that a cosine curve is of the same form as the sine curve but is displaced by 9 or / radians.. Sine and cosine curves Graphs of sine and cosine waveforms i A graph of y sin A is shown by the broken line in Fig.. and is obtained by drawing up a table of values as in Section.. A similar table may be produced for y sin A. A 6 9 A sin A A 8 A sin A T q Figure.7 7 q C 6 From Fig..7, sin Measured from, the two angles between and 6 whosesineis.68 are 8 C7.6,i.e.7.6 and 6 7.6, i.e..7 Note that a calculator only gives one answer, i.e Problem. etermine all the angles between and 6 whose tangent is.769 A tangent is positive in the first and third quadrants see Fig..8. From Fig..9, tan Measured from, the two angles between and 6 whose tangent is.769 are 6. and 8 C 6., i.e.. Cosine is positive in the first and fourth quadrants and thus negative in the second and third quadrants from Fig.. or from Fig..b. In Fig.., angle cos Figure. S T q q 9 7 A C 6 Measured from, the two angles whose cosine is.8 are i.e..8 and 8 C 76., i.e.6.. The production of a sine and cosine wave In Fig.., let OR be a vector unit long and free to rotate anticlockwise about O. In one revolution a circle is produced and is shown with sectors. Each radius arm has a vertical and a horizontal component. For eample, at, the vertical component is TS and the horizontal component is OS. From trigonometric ratios, sin TS TO TS, i.e. TS sin and cos OS TO OS, i.e. OS cos The vertical component TS may be projected across to T S, which is the corresponding value of on the graph of y against angle. If all such vertical components as TS are projected on to the graph, then a sine wave is produced as shown in Fig... A 7 6 A sin A A graph of y sin A is shown in Fig... ii A graph of y sin A is shown in Fig.. using the following table of values. iii A A sin A A 7 6 A 6 8 sin A A graph of y cos A is shown by the broken line in Fig.. and is obtained by drawing up a table of values. A similar table may be produced for y cos A with the result as shown. iv A graph of y cos A is shown in Fig..6 which may be produced by drawing up a table of values, similar to above.

3 86 ENGINEERING MATHEMATICS TRIGONOMETRIC WAVEFORMS y. T. T y = sin y. y = cos A y = cos A y. y = sin A 8 Angle R S S A A Figure Figure. y. y = cos A y = cos A. Figure.7 Problem. Sketch y sina from A to A radians Amplitude and period / rads or A Asketchofy sina is shown in Fig..8.. y y = sin A Figure. y. y = sin A y = sin A y. y = sin A y = sin A Figure.6 waveform is 6 /p or /p rad. Hence if y sin A then the period is 6/, i.e., and if y cos A then the period is 6/, i.e A A Figure. Periodic time and period i Each of the graphs shown in Figs.. to.6 will repeat themselves as angle A increases and are thus called periodic functions A Figure. ii iii y sin A and y cos A repeat themselves every 6 or radians; thus 6 is called the period of these waveforms. y sin A and y cos A repeat themselves every 8 or radians; thus 8 is the period of these waveforms. In general, if y sin pa or y cospa where p is a constant then the period of the Amplitude Amplitude is the name given to the maimum or peak value of a sine wave. Each of the graphs shown in Figs.. to.6 has an amplitude of C i.e. they oscillate between C and. However, if y sina, each of the values in the table is multiplied by and the maimum value, and thus amplitude, is. Similarly, if y cosa, the amplitude is and the period is 6 /, i.e. 8 Problem. Sketch y sin A between A and A 6 Amplitude and period 6 / Asketchofy sin A is shown in Fig..7. Figure.8 Problem 6. Sketch y cos from to 6 Amplitude and period 6 / 8. Asketchofycos is shown in Fig..9. Problem 7. cycle Amplitude ; period 6 Sketch y sin A over one 6 ð 6 Asketchofysin A is shown in Fig...

4 88 ENGINEERING MATHEMATICS TRIGONOMETRIC WAVEFORMS 89 y y = cos y y = sin A 6. y = sin A A y y = sin A y = sin A A p/w rads y y = cos wt y = cos wt p/ p/w p/w p/w p/w t. Figure.9 y y = sin A A Figure. Lagging and leading angles i ii iii iv A sine or cosine curve may not always start at. To show this a periodic function is represented by y sin A š or y cos A š where is a phase displacement compared with y sin A or y cos A. By drawing up a table of values, a graph of y sin A 6 may be plotted as shown in Fig... If y sin A is assumed to start at then y sin A 6 starts 6 later i.e. has a zero value 6 later. Thus y sin A 6 is said to lag y sin A by 6 By drawing up a table of values, a graph of y cos A C may be plotted as shown in Fig... If y cos A is assumed to start at then y cos A C starts earlier i.e. has a maimum value earlier. Thus y cos A C is said to lead y cos A by Generally, a graph of y sin A lags y sin A by angle, and a graph of y sin A C leads y sin A by angle. 6 Figure. y. y = cos A y = cosa A Figure. v A cosine curve is the same shape as a sine curve but starts 9 earlier, i.e. leads by 9. Hence cos A sin A C 9 Problem 8. Sketch y sin A C from A to A 6 Amplitude and period 6 / 6. sin A C leads sin A by i.e. starts earlier. A sketch of y sin A C is shown in Fig... Problem 9. Sketch y 7sin A / in the range A 6 Amplitude 7 and period / radians. Figure. In general, y = sin.pt a/ lags y = sin pt by a=p, hence 7 sin A / lags 7 sin A by / /, i.e. /6 rad or A sketch of y 7sin A / is shown in Fig... y 7 π/ A π/ π π/ π 7 π/6 Figure. y = 7sin A y = 7sin A π/ Problem. Sketch y cos ωt / over one cycle Amplitude and period /ω rad. cos ωt / lags cos ωt by /ω seconds. Asketchofy cos ωt / is shown in Fig... Now try the following eercise Eercise 8 Further problems on sine and cosine curves In Problems to 7 state the amplitude and period of the waveform and sketch the curve between and 6.. y cos A, Figure.. y sin,. y sint, 9. y cos. y 7 sin 8, 7 7, y 6sin t 6, 6 7. y cos C, 8. Sinusoidal form A sin.!t ± a/ In Fig..6, let OR represent a vector that is free to rotate anticlockwise about O at a velocity of ω rad/s. A rotating vector is called a phasor. After a time t seconds OR will have turned through an angle ωt radians shown as angle TOR in Fig..6. If ST is constructed perpendicular to OR, thensinωt ST/OT, i.e.st OT sin ωt. Figure.6 ω rads/s T y. ωt S R ωt π/ π π/ π ωt. y = sin ωt If all such vertical components are projected on to a graph of y against ωt, a sine wave results of amplitude OR as shown in Section..

5 9 ENGINEERING MATHEMATICS TRIGONOMETRIC WAVEFORMS 9 If phasor OR makes one revolution i.e. radians in T seconds, then the angular velocity, ω /T rad/s, from which, T = p=! seconds T is known as the periodic time. The number of complete cycles occurring per second is called the frequency, f i.e. Frequency f =! p Hz number of cycles second T ω Hz Hence angular velocity,! = pf rad/s Amplitude is the name given to the maimum or peak value of a sine wave, as eplained in Section.. The amplitude of the sine wave shown in Fig..6 has an amplitude of. A sine or cosine wave may not always start at. To show this a periodic function is represented by y sin ωt š or y cos ωt š, where is a phase displacement compared with y sin A or y cos A. A graph of y sin ωt lags y sin ωt by angle, and a graph of y sin ωt C leads y sin ωt by angle. The angle ωt is measured in radians i.e. ω rad s t s ωt radians hence angle should also be in radians. The relationship between degrees and radians is: 6 radians or 8 = p radians Hence rad 8 7. and, for eample, 7 7 ð.9 rad 8 Summarising, given a general sinusoidal function y = A sin.!t ± a/, then: i ii A amplitude ω angular velocity f rad/s iii periodic time T seconds ω ω iv frequency, f hertz v angle of lead or lag compared with y A sin ωt Problem. An alternating current is given by i sin t C.7 amperes. Find the amplitude, periodic time, frequency and phase angle in degrees and minutes i sin tc.7 A, henceamplitude = A. Angular velocity ω, hence periodic time, T ω. s or ms Frequency, f T Hz. Phase angle, a.7 rad.7 ð 8.7 or 8 leading i= sin.pt/ Problem. An oscillating mechanism has a maimum displacement of. m and a frequency of 6 Hz. At time t the displacement is 9 cm. Epress the displacement in the general form A sin ωt š Amplitude maimum displacement. m Angular velocity, ω f 6 rad/s Hence displacement.sin t C m When t, displacement 9 cm.9 m Hence,.9.sin C i.e. sin.9..6 Hence sin rad Thus, displacement =.sin.pt Y.68/ m a Problem. The instantaneous value of voltage in an a.c. circuit at any time t seconds is given by v sin t. volts. etermine the: a amplitude, periodic time, frequency and phase angle in degrees b value of the voltage when t c value of the voltage when t ms d time when the voltage first reaches V, and e time when the voltage is a maimum Sketch one cycle of the waveform Amplitude = V Angular velocity, ω Hence periodic time, T ω. s or ms Frequency f T Hz. Phase angle. rad. ð 8 b When t, c lagging v sin t v sin. sin 7. V When t = ms, then v sin. sin.98 sin 9 9. volts d When v volts, then sin t. sin t. Hence t. sin 6. or.688 rad e Voltage v t.688 C..698 Hence when v V, time, t ms When the voltage is a maimum, v V Hence sin t. sin t. t. sin 9 or.78 rad t.78 C..8 Hence time, t.8. ms Asketchofv sin t. volts is shown in Fig..7. Figure.7 tms 7.7. v = sin πt. v = sin πt Now try the following eercise Eercise 86 Further problems on the sinusoidal form A sin.!t ± a/ In Problems to find the amplitude, periodic time, frequency and phase angle stating whether it is leading or lagging sin ωt ofthe alternating quantities given.. i sin t C.9 ma,. s, Hz,.9 rad or 6 7 leading sin t. y 7 sin t. cm 7 cm,.7 s, 6.7 Hz,. rad or 6 lagging 7 sin t

6 9 ENGINEERING MATHEMATICS TRIGONOMETRIC WAVEFORMS 9. v sin t. V V,. s, Hz,. rad or 6 lagging sin t. A sinusoidal voltage has a maimum value of V and a frequency of Hz. At time t, the voltage is a zero, and b V. Epress the instantaneous voltage v in the form v A sin ωt š. a v sin t volts b v sin t C. volts. An alternating current has a periodic time of ms and a maimum value of A. When time t, current i amperes. Epress the current i in the form i A sin ωt š. i sin 8t 6 amperes 6. An oscillating mechanism has a maimum displacement of. m and a frequency of Hz. At time t the displacement is cm. Epress the displacement in the general form A sin ωt š.. sin t C.88 m 7. The current in an a.c. circuit at any time t seconds is given by: i sin t. amperes etermine a the amplitude, periodic time, frequency and phase angle in degrees b the value of current at t, c the value of current at t 8ms, d the time when the current is first a maimum, e the time when the current first reaches A. Sketch one cycle of the waveform showing relevant points. a A, ms, Hz, lagging b.9 A c.6 A d 6.7 ms e. ms.6 Waveform harmonics Let an instantaneous voltage v be represented by v V m sin ft volts. This is a waveform which varies sinusoidally with time t, has a frequency f, and a maimum value V m. Alternating voltages are usually assumed to have wave-shapes which are sinusoidal where only one frequency is present. If the waveform is not sinusoidal it is called a comple wave, and, whatever its shape, it may be split up mathematically into components called the fundamental and a number of harmonics. This process is called harmonic analysis. The fundamental or first harmonic is sinusoidal and has the supply frequency, f; the other harmonics are also sine waves having frequencies which are integer multiples of f. Thus, if the supply frequency is Hz, then the third harmonic frequency is Hz, the fifth Hz, and so on. A comple waveform comprising the sum of the fundamental and a third harmonic of about half the amplitude of the fundamental is shown in Fig..8a, both waveforms being initially in phase with each other. If further odd harmonic waveforms of the appropriate amplitudes are added, v v v Comple waveform Fundamental v Third harmonic t a Comple waveform Fundamental v Second harmonic t A B Third harmonic c e Figure.8 Comple waveform Fundamental Second harmonic t v Comple waveform Fundamental Third harmonic b Comple waveform Fundamental Second harmonic Second harmonic f d Comple waveform Fundamental t t t Third harmonic a good approimation to a square wave results. In Fig..8b, the third harmonic is shown having an initial phase displacement from the fundamental. The positive and negative half cycles of each of the comple waveforms shown in Figs..8a and b are identical in shape, and this is a feature of waveforms containing the fundamental and only odd harmonics. A comple waveform comprising the sum of the fundamental and a second harmonic of about half the amplitude of the fundamental is shown in Fig..8c, each waveform being initially in phase with each other. If further even harmonics of appropriate amplitudes are added a good approimation to a triangular wave results. In Fig..8c, the negative cycle, if reversed, appears as a mirror image of the positive cycle about point A. In Fig..8d the second harmonic is shown with an initial phase displacement from the fundamental and the positive and negative half cycles are dissimilar. A comple waveform comprising the sum of the fundamental, a second harmonic and a third harmonic is shown in Fig..8e, each waveform being initially in-phase. The negative half cycle, if reversed, appears as a mirror image of the positive cycle about point B. In Fig..8f, a comple waveform comprising the sum of the fundamental, a second harmonic and a third harmonic are shown with initial phase displacement. The positive and negative half cycles are seen to be dissimilar. The features mentioned relative to Figs..8a to f make it possible to recognise the harmonics present in a comple waveform.

7 Cartesian and polar co-ordinates a q q a CARTESIAN AN POLAR CO-ORINATES 9 Figure.. Introduction There are two ways in which the position of a point in a plane can be represented. These are a b by Cartesian co-ordinates, i.e., y, and by polar co-ordinates, i.e.r,, where r is a radius from a fied point and is an angle from a fied point.. Changing from Cartesian into polar co-ordinates In Fig.., if lengths and y are known, then the length of r can be obtained from Pythagoras theorem see Chapter since OPQ is a rightangled triangle. Hence r C y from which, r = Y y y O q r Figure. P Q y From trigonometric ratios see Chapter, from which tan y q = tan y r C y and tan y are the two formulae we need to change from Cartesian to polar coordinates. The angle, which may be epressed in degrees or radians, must always be measured from the positive -ais, i.e. measured from the line OQ in Fig... It is suggested that when changing from Cartesian to polar co-ordinates a diagram should always be sketched. Problem. Change the Cartesian co-ordinates, into polar co-ordinates. A diagram representing the point, is shown in Fig... q Figure. From Pythagoras theorem, r p C note that has no meaning in this contet. By trigonometric ratios, tan. or.97 rad note that.. ð /8 rad.97 rad. Hence, in Cartesian co-ordinates corresponds to,. or,.97 rad in polar co-ordinates. Problem. Epress in polar co-ordinates the position, A diagram representing the point using the Cartesian co-ordinates, is shown in Fig... From Pythagoras theorem, r p C By trigonometric ratios, tan 6.87 or.6 rad. Hence or.6.98 rad. Hence the position of point P in polar co-ordinate form is,. or,.98 rad. Problem. Epress, in polar co-ordinates. A sketch showing the position, is shown in Fig... r C and tan 67.8 or.76 rad. Hence 8 C or C.76.8 rad. a Figure. q Thus, in Cartesian co-ordinates corresponds to, 7.8 or,.8 rad in polar co-ordinates. Problem. Epress, in polar co-ordinates. A sketch showing the position, is shown in Fig... r C p 9.8 correct to decimal places tan 68. or.9 rad Figure. Hence or.9.9 rad Thus, in Cartesian co-ordinates corresponds to.8, 9.8 or.8,.9 rad in polar co-ordinates. Now try the following eercise Eercise 87 Further problems on changing from Cartesian into polar co-ordinates In Problems to 8, epress the given Cartesian co-ordinates as polar co-ordinates, correct to decimal places, in both degrees and in radians..,.8, 9. or.8,. rad. 6.8,. 6.6,.8 or 6.6,.6 rad.,.7, 6.7 or.7,. rad..,.7 6.,.8 or 6.,. rad. 7, 7.6,. or 7.6,. rad 6..,.6., 6. or.,. rad 7.,.8, 9. or.8,.7 rad ,..68, 7.7 or.68,.7 rad

8 96 ENGINEERING MATHEMATICS CARTESIAN AN POLAR CO-ORINATES 97. Changing from polar into Cartesian co-ordinates From the right-angled triangle OPQ in Fig..6. cos r and sin y r, from trigonometric ratios Hence = r cos q and y = r sin q q Problem 6. Epress 6, 7 incartesian co-ordinates. A sketch showing the position 6, 7 is shown in Fig..8. r cos 6 cos 7.88 which corresponds to length OA in Fig..8. y r sin 6 sin 7.9 which corresponds to length AB in Fig..8. q y r sin.sin.6.7 which corresponds to length AB in Fig..9. Thus.98,.7 in Cartesian co-ordinates corresponds to.,.6 rad in polar coordinates.. Use of R P and P R functions on calculators Another name for Cartesian co-ordinates is rectangular co-ordinates. Many scientific notation calculators possess R! P and P! R functions. The R is the first letter of the word rectangular and the P is the first letter of the word polar. Check the operation manual for your particular calculator to determine how to use these two functions. They make changing from Cartesian to polar co-ordinates, and vice-versa, so much quicker and easier. Now try the following eercise Eercise 88 Further problems on changing polar into Cartesian coordinates In Problems to 8, epress the given polar co-ordinates as Cartesian co-ordinates, correct todecimalplaces.., 7.9,.8..,. rad.97,.96. 7,,6,...6,. rad.88,...8, 9.,. 6., rad.6,.7 7..,.7, ,. rad.,. Figure.6 Figure.8 If length r and angle are known then r cos and y r sin are the two formulae we need to change from polar to Cartesian co-ordinates. Problem. Change, intocartesian co-ordinates. Thus 6, 7 in polar co-ordinates corresponds to.88,.9 in Cartesian co-ordinates. Note that when changing from polar to Cartesian co-ordinates it is not quite so essential to draw a sketch. Use of r cos and y r sin automatically produces the correct signs. A sketch showing the position, is shown in Fig..7. Problem 7. Epress.,.6 rad in Cartesian co-ordinates. Now r cos cos.9 and y r sin sin. A sketch showing the position.,.6 rad is shown in Fig..9. r cos.cos.6.98 which corresponds to length OA in Fig..9. q q Figure.7 Hence, in polar co-ordinates corresponds to.9,. in Cartesian co-ordinates. Figure.9

9 98 ENGINEERING MATHEMATICS Assignment 6 This assignment covers the material in Chapters to. The marks for each question are shown in brackets at the end of each question.. Fig. A6. shows a plan view of a kite design. Calculate the lengths of the dimensions shown as a and b.. In Fig. A6., evaluate a angle b angle. etermine the area of the plan view of a kite shown in Fig. A6... cm b q Figure A6. a. cm a 6. cm. If the angle of elevation of the top of a m perpendicular building from point A is measured as 7, determine the distance to the building. Calculate also the angle of elevation at a point B, m closer to the building than point A.. Evaluate, each correct to significant figures: a sin.78 b cos 6 c tan 8 6. Sketch the following curves labelling relevant points: a y cos C b y sin t Solve the following equations in the range to 6 a sin.6 b cot The current in an alternating current circuit at any time t seconds is given by: i sin t C.7 amperes. etermine a the amplitude, periodic time, frequency and phase angle with reference to sin t b the value of current when t c the value of current when t 6ms d the time when the current first reaches 8 A Sketch one cycle of the oscillation Change the following Cartesian coordinates into polar co-ordinates, correct to decimal places, in both degrees and in radians: a.,. b 7.6, Change the following polar co-ordinates into Cartesian co-ordinates, correct to decimal places: a 6., b, rad

10 Trigonometric identities and equations. Trigonometric identities A trigonometric identity is a relationship that is true for all values of the unknown variable. tan sin cos, cosec sin cos cot sin, sec cos and cot tan are eamples of trigonometric identities from Chapter. Applying Pythagoras theorem to the right-angled triangle shown in Fig.. gives: a C b c b c a q Figure. ividing each term of equation by c gives: a c C b c c c a b i.e. C c c cos C sin Hence cos q Y sin q = ividing each term of equation by a gives: a a C b a c a b c i.e. C a a Hence Y tan q = sec q ividing each term of equation by b gives: a b C b b c b i.e. a c C b b Hence cot q Y = cosec q Equations, and are three further eamples of trigonometric identities.. Worked problems on trigonometric identities Problem. Prove the identity sin cot sec sin With trigonometric identities it is necessary to start with the left-hand side LHS and attempt to make it equal to the right-hand side RHS or vice-versa. It is often useful to change all of the trigonometric ratios into sines and cosines where possible. Thus LHS sin cot sec cos sin sin cos sin by cancelling RHS Problem. Prove that: tan C sec sec C tan sec tan C sec LHS sec C tan sec cos sin cos C cos sin C cos cos Problem. sin C cos sin cos C cos cos sin C cos cos sin C cos C sin cos C sin by cancelling RHS Prove that: C cot C tan cot cos LHS C cot C C tan sin C sin cos sin C cos sin cos cot RHS sin Problem. Show that: cos sin sin cos cos C sin sin C cos sin cos C sin cos From equation, cos C sin, from which, cos sin Hence, LHS cos sin sin sin sin sin sin RHS Problem. Prove that: sin sec tan C sin LHS sin C sin TRIGONOMETRIC IENTITIES AN EQUATIONS 9 sin sin C sin sin sin sin Since cos C sin then sin cos sin LHS sin sin cos sin sin cos cos cos sec tan RHS Now try the following eercise Eercise 9 Further problems on trigonometric identities Prove the following trigonometric identities:. sin cot cos. p cos cosec. cos A cos A sin A. cos cos sin cos sin. C cot C cot cosec 6. sin sec C cosec cos tan C tan. Trigonometric equations Equations which contain trigonometric ratios are called trigonometric equations. There are usually an infinite number of solutions to such equations; however, solutions are often restricted to those between and 6. A knowledge of angles of any magnitude is essential in the solution of trigonometric equations and calculators cannot be relied upon to give all the solutions as shown in Chapter. Figure. shows a summary for angles of any magnitude.

11 ENGINEERING MATHEMATICS TRIGONOMETRIC IENTITIES AN EQUATIONS 8 Sine and cosecant positive Tangent and cotangent positive Figure. 9 7 All positive 6 Cosine and secant positive reduce equations to one of the above forms see Problems to.. Worked problems i on trigonometric equations Problem 6. Solve the trigonometric equation: sin C for values of from to 6 sin C, from which sin /.6 y. 8 9 y = tan.9.9 S.9 T 9 7 b 8 a A C 6...cott t 66.7 or 6.7. Worked problems ii on trigonometric equations Problem 9. Solve: cos A for values of A in the range <A<6 Equations of the type a sin A Y b sin A Y c = i When a =, b sin A C c, hence sin A c b and A = sin c b There are two values of A between and 6 that satisfy such an equation, provided c see Problems 6 to 8. b ii When b =, a sin A C c, hence sin A c a,sina c a and A = sin c a If either a or c is a negative number, then the value within the square root sign is positive. Since when a square root is taken there is a positive and negative answer there are four values of A between and 6 which satisfy such an equation, provided c see Problems 9 and. a iii When a, b and c are all non-zero: a sin A C b sin A C c is a quadratic equation in which the unknown is sin A. The solution of a quadratic equation is obtained either by factorising if possible or by using the quadratic formula: iv sin A = b ± p b ac a see Problems and. Often the trigonometric identities cos A C sin A, C tan A sec A and cot A C cosec A need to be used to y S T 9 a 9 7 b Figure. y = sin q a a A C q Hence sin.6. Sine is negative in the third and fourth quadrants see Fig... The acute angle sin shown as in Fig..b. Hence 8 C 6.87, i.e.6.87 or , i.e.. Problem 7. Solve:.tan.8 for 6.tan.8, from which tan.8.. Hence tan. Tangent is positive in the first and third quadrants see Fig... The acute angle tan..9. Hence, =.9 or 8 C.9.9 Figure. Problem 8. Solve: sec t for values of t between and 6 sect, from which sec t. Hence t sec. 8 S T 9 7 Figure. A C 6 Secant /cosine is positive in the first and fourth quadrants see Fig... The acute angle sec Hence t = 6.87 or Now try the following eercise Eercise 9 Further problems on trigonometric equations Solve the following equations for angles between and 6. 7sin.8 or.. cosec A C. A. or 6.9 cos A, from which cos A. Hence cos A p. š.77 and A cos š.77 Cosine is positive in quadrants one and four and negative in quadrants two and three. Thus in this case there are four solutions, one in each quadrant see Fig..6. The acute angle cos.77. Hence A =,, or Problem. Solve: cot y. for <y<6 cot y., from which, cot y..6 y Figure.6 y = cos A 8 6 A S T a 9 7 b A C 6

12 ENGINEERING MATHEMATICS TRIGONOMETRIC IENTITIES AN EQUATIONS Hence cot y p.6 š.6, and y cot š.6. There are four solutions, one in each quadrant. The acute angle cot.6.8. Hence y =.8, 8.9,.8 or 8.9 Now try the following eercise Eercise 9 Further problems on trigonometric equations Solve the following equations for angles between and 6. sin y.77, 9., y.77 or 9.. C cosec 8 9 or 7. cot., 7.68,. or Worked problems iii on trigonometric equations Problem. Solve the equation: 8sin C sin, for all values of between and 6 Factorising 8 sin C sin gives sin sin C Hence sin, from which, sin., or sin C, from which, sin. Instead of factorising, the quadratic formula can, of course, be used. sin..8 or 6., since sine is positive in the first and second quadrants, or sin. or, since sine is negative in the third and fourth quadrants. Hence q =.8, 6., or Problem. Solve: 6cos C cos 6 for values of from to 6 Factorising 6 cos C cos 6 gives cos cos C. Hence cos, from which, cos.6667, or cos C, from which, cos. The minimum value of a cosine is, hence the latter epression has no solution and is thus neglected. Hence q cos or.8 since cosine is positive in the first and fourth quadrants. Now try the following eercise Eercise 96 Further problems on trigonometric equations Solve the following equations for angles between and 6. sin A C sin A A 9.7, 6.,.8 or tan C tan.,.68,. or.68. cosec t cosect t.8, 6.,.8 or Worked problems iv on trigonometric equations Problem. Solve: cos t C sint for values of t from to 6 Since cos t C sin t, cos t sin t. Substituting for cos t incos t C sint gives sin t C sint sin t C sint sin t C sint C sin t sint Factorising gives sint C sin t. Hence sint C, from which, sin t., or sin t, from which, sin t. t sin..8 or 6., since sine is negative in the third and fourth quadrants, or t sin 9. Hence t = 9,.8 or 6. as shown in Fig..7. y... Figure.7 y = sin t t Problem. Solve: 8 sec A tana for values of A between and 6 C tan A sec A. Substituting for sec A in 8 sec A tana gives 8 C tan A tana i.e. 8 C 8 tan A tana 8 tan A tana Factorising gives 6tanA tana C Hence 6 tan A, from which, tan A. or tan A C, from which, 6 tan A.. Thus A tan. 6.7 or 6.7, since tangent is positive in the first and third quadrants, or A tan. 6.7 or.7, since tangent is negative in the second and fourth quadrants. Hence A = 6.7, 6.7, 6.7 or.7 Problem. Solve: cosec cot in the range <<6 cot C cosec. Substituting for cosec in cosec cot gives: cot C cot cot C cot cot cot Since the left-hand side does not factorise the quadratic formula is used. Thus, cot š š p 6 C š p 6 or.6 6 Hence cot.78 or.87, cot.78.7 or.7, since cotangent is positive in the first and third quadrants, or cot.87 =.8 or 9.8, since cotangent is negative in the second and fourth quadrants. Hence, q =.7,.8,.7 or 9.8 Now try the following eercise Eercise 97 Further problems on trigonometric equations Solve the following equations for angles between and 6. sin 6 cos 8.8, 8.8,. or.8. 6 sec tan.9 or 7.7. cot A 6cosecA C 6 A 9. sec t C tan t t 7.8 or.7..9cos a 7sina C a 7.8 or.7 6. cosec ˇ 8 7cotˇ ˇ 6.7, 6.,.7 or.

13 6 Compound angles 6. Compound angle formulae An electric current i may be epressed as i sin ωt. amperes. Similarly, the displacement of a bo from a fied point can be epressed as sin t C.67 metres. The angles ωt. and t C.67 are called compound angles because they are the sum or difference of two angles. The compound angle formulae for sines and cosines of the sum and difference of two angles A and B are: sin A C B sin A cos B C cos A sin B sin A B sin A cos B cos A sin B cos A C B cos A cos B sin A sin B cos A B cos A cos B C sin A sin B Note, sin ACB is not equal to sin A C sin B, and so on. The formulae stated above may be used to derive two further compound angle formulae: tan A C tan B tan A C B tan A tan B tan A tan B tan A B C tan A tan B The compound-angle formulae are true for all values of A and B, and by substituting values of A and B into the formulae they may be shown to be true. a Problem. Epand and simplify the following epressions: a sin C b cos 9 C ˇ c sin A B sin A C B sin C sin cos C cos sin from the formula for sin A C B cos C sin sin a b c cos 9 C ˇ cos 9 cos ˇ sin 9 sin ˇ cos ˇ sin ˇ sin b sin A B sin A C B sin A cos B cos A sin B sin A cos B C cos A sin B cosasin B Problem. Prove that: cos y C sin y C cos y cos y cos C sin y sin cos y C sin y cos y sin y C sin y cos C cos y sin sin y C cos y cos y Hence cos y C sin y C cos y C cos y Problem. Show that tan C tan tan C tan C tan tan tan from the formula for tan A C B C tan, tan tan C tan since tan tan tan tan C tan tan Hence, tan C tan C tan tan tan C tan tan C tan tan tan tan tan Problem. If sin P.8 and cos Q. evaluate, correct to decimal places: a sin P Q, bcos P C Q and c tan P C Q, using the compound angle formulae Since sin P.8 then P sin.8. Thus cos P cos..86 and tan P tan.. Since cos Q.,Q cos Thus sin Q sin and tan Q tan a b c sin P Q sin P cos Q cos P sin Q cos P C Q cos P cos Q sin P sin Q tan P C Q tan P C tan Q. C. tan P tan Q Problem. Solve the equation: sin cos for values of between and 9 COMPOUN ANGLES sin sin cos cos sin, from the formula for sin A B sin.997 cos..788 sin.68 cos Since sin cos then.788 sin.68 cos cos Rearranging gives:.788 sin cos C.68 cos 6.68 cos sin and cos i.e. tan.69, and tan or 9 7 Check: LHS sin 9.9 sin9.9. RHS cos cos 9.9. Now try the following eercise Eercise 98 Further problems on compound angle formulae. Reduce the following to the sine of one angle: a sin 7 cos C cos 7 sin b sin 7t cos t cos 7t sin t a sin 8 b sin t. Reduce the following to the cosine of one angle: a cos 7 cos sin 7 sin b cos cos C sin sin a cos cos 76 b cos. Show that: a sin C C sin C p cos b sin cos

14 6 ENGINEERING MATHEMATICS COMPOUN ANGLES 7. Prove that: a sin C sin p sin C cos cos 7 C b cos 6 tan. Given cosa. and sin B.7 evaluate a sin A B, bcos A B, c tan ACB, correct to decimal places. a.6 b.99 c.687 In Problems 6 and 7, solve the equations for values of between and 6 6. sin C 7cos 6.7 or.7 7. sin sin 67. or Conversion of a sin!t Y b cos!t into R sin.!t Y a i ii R sin ωt C represents a sine wave of maimum value R, periodic time /ω, frequency ω/ and leading R sin ωt by angle. See Chapter. R sin ωt C may be epanded using the compound-angle formula for sin ACB,where A ωt and B. Hence R sin ωt C Rsin ωt cos C cos ωt sin R sin ωt cos C R cos ωt sin If the values of a and b are known then the values of R and may be calculated. The relationship between constants a, b, R and are shown in Fig. 6.. Figure 6. a R a From Fig. 6., by Pythagoras theorem: R = a Y b and from trigonometric ratios: a = tan b a b Problem 6. Find an epression for sinωt C cosωt in the form R sin ωt C and sketch graphs of sin ωt, cosωt and R sin ωt C on the same aes Let sin ωt C cosωt R sin ωt C then sin ωt C cosωt Rsin ωt cos C cos ωt sin R cos sin ωt C R sin cos ωt Equating coefficients of sin ωt gives: R a Figure 6. From trigonometric ratios: tan. or.97 radians Hence, sin!t Y cos!t = sin.!t Y.97/ Asketchofsinωt, cosωt and sin ωt C.97 is shown in Fig rad y Figure rad y = cos wt y = sin wt y = sin wt +.97 p/ p p/ p wt rad Two periodic functions of the same frequency may be combined by Let.6sinωt 7.cosωt R sin ωt C then.6sinωt 7.cosωt Rsin ωt cos C cos ωt sin R cos sin ωt C R sin cos ωt Equating coefficients of sin ωt gives:.6 R cos, from which, cos.6 R Equating coefficients of cos ωt gives: 7. R sin, from which sin 7. R There is only one quadrant where cosine is positive and sine is negative, i.e. the fourth quadrant, as shown in Fig. 6.. By Pythagoras theorem: R.6 C By trigonometric ratios: 7. tan or.8 radians. Hence,.6sin!t 7.cos!t = 8.68 sin.!t.8/ Problem 8. Epress:.7sinωt.cosωt in the form R sin ωt C Let.7sinωt.cosωt R sin ωt C Rfsin ωt cos C cos ωt sin R cos sin ωt C R sin cos ωt iii iv R cos sin ωt C R sin cos ωt If a R cos and b R sin, wherea and b are constants, then R sin ωt C a sin ωt C b cos ωt, i.e. a sine and cosine function of the same frequency when added produce a sine wave of the same frequency which is further demonstrated in Chapter. Since a R cos, then cos a, and since R b R sin, thensin b R R cos, from which, cos R Equating coefficients of cos ωt gives: R sin, from which, sin R There is only one quadrant where both sin and cos are positive, and this is the first, as shown in Fig. 6.. From Fig. 6., by Pythagoras theorem: R C a plotting the functions graphically and combining ordinates at intervals, or b by resolution of phasors by drawing or calculation. Problem 6, together with Problems 7 and 8 following, demonstrate a third method of combining waveforms. Problem 7. Epress:.6sinωt 7.cosωt in the form R sin ωt C Figure 6..6 a R 7.

15 8 ENGINEERING MATHEMATICS COMPOUN ANGLES 9 Equating coefficients gives:.7 R cos, from which, cos.7 R and. R sin, from which, sin. R There is only one quadrant in which both cosine and sine are negative, i.e. the third quadrant, as shown in Fig. 6.. From Fig. 6., R.7 C..99. and tan Hence 8 C or. radians. Thus,.7sin!t.cos!t =.99 sin.!t Y./. An angle of 6.6 is the same as.7 or. radians. Hence.7sinωt.cosωt may be epressed also as.99 sin.!t./, which is preferred since it is the principal value i.e.. Problem 9. Epress: sin C cos in the form R sin C, and hence solve the equation sin C cos, for values of between and 6 Let 8 sin C cos R sin C a Rsin cos C cos sin R cos sin C R sin cos Figure 6..7 a q. R Equating coefficients gives: R cos, from which, cos R and R sin, from which, sin R Since both sin and cos are positive, R lies in the first quadrant, as shown in Fig From Fig. 6.6, R p C.8 and tan 9. Hence sin C cos.8 sin C 9. However sin C cos Thus.8 sin C 9., from which C 9. sin.8 i.e. C 9.. or 6.68 Hence or Since.7 is the same as.7 C 6, i.e..9, then the solutions are q = 77.6 or.9, which may be checked by substituting into the original equation. Let Problem. Solve the equation:.cosa.8sina 6. for A 6.cosA.8sinA R sin A C Rsin A cos C cos A sin R cos sin A C R sin cos A Figure 6.6 R a Equating coefficients gives:. R sin, from which, sin. R and.8 R cos, from which, cos.8 R There is only one quadrant in which both sine is positive and cosine is negative, i.e. the second, as shown in Fig Figure 6.7. R.8 q 9 7 a 6 From Fig. 6.7, R. C and. tan.8. Hence Thus.cosA.8sinA 6.77 sin A C Hence, sin A C , from which, 6.77 A C 8.88 sin or 6. Thus, A C or A C The solutions are thus A = 8.77 or 7.7, which may be checked in the original equation. Now try the following eercise Eercise 99 Further problems on the conversion of a sin!t Y b cos!t into R sin.!t Y a/ In Problems to, change the functions into the form R sin ωt š. sin ωt C 8cosωt 9. sin ωt C.. sin ωt cosωt sin ωt.6. 7sinωt C cosωt 8.6 sin ωt C.6. sinωt 6cosωt 6.78 sin ωt.. Solve the following equations for values of between and 6 : a sin C cos b sin 9cos 7 a 7. or 8.7 b 6.68 or Solve the following equations for <A<6 : a cos A C sina.8 b cos A sina a 7.7 or.6 b. or The third harmonic of a wave motion is given by. cos 6.9sin. Epress this in the form R sin š 8. sin C.8 8. The displacement metres of a mass from a fied point about which it is oscillating is given by.sinωt C.cosωt, where t is the time in seconds. Epress in the form R sin ωt C..sin ωt C Two voltages, v cosωt and v 8sinωt are inputs to an analogue circuit. etermine an epression for the output voltage if this is given by v Cv. 9. sin ωt C.8

16 ENGINEERING MATHEMATICS COMPOUN ANGLES 6. ouble angles i If, in the compound-angle formula for sin A C B, weletb A then sin A = sinacos A. Also, for eample, sin A sinacos A and sin 8A sinacos A, and so on. ii If, in the compound-angle formula for cos A C B, weletb A then cos A = cos A sin A Since cos A C sin A, then cos A sin A,andsin A cos A, and two further formula for cos A can be produced. Thus cos A cos A sin A sin A sin A i.e. cos A = sin A and cos A cos A sin A cos A cos A i.e. cos A = cos A Also, for eample, cos A cos A sin A or sin A or cos A andcos6a cos A sin A or sin A or cos A and so on. iii If, in the compound-angle formula for tan A C B, weletb A then tan A = tana tan A. Also, for eample, tan A tana tan A and tan A tan A tan and so on. A Problem. I sin is the third harmonic of a waveform. Epress the third harmonic in terms of the first harmonic sin, wheni When I, I sin sin sin C sin cos C cos sin, from the sin A C B formula sincos cos C sin sin, from the double angle epansions sin cos C sin sin i.e. sin sin C sin sin, since cos sin sin sin C sin sin sin q = sinq sin q Problem. LHS Prove that: cos sin cos sin tan sin sin cos sin sin cos sin tan RHS cos Problem. Prove that: cot C cosec cot LHS cot C cosec cos sin C cos C sin sin cos C sin cos sin cos sincos cos cot RHS sin Now try the following eercise Eercise Further problems on double angles. The power p in an electrical circuit is given by p v. etermine the power R in terms of V, R and cos t when V v V cos t. C cos t R. Prove the following identities: cos a cos tan C cos t b sin cot t t tan C tan c tan tan d cosec cos cot tan. If the third harmonic of a waveform is given by V cos, epress the third harmonic in terms of the first harmonic cos, when V. cos cos cos 6. Changing products of sines and cosines into sums or differences i ii iii iv sin A C B C sin A B sina cos B from the formulae in Section 6., i.e. sin A cos B = sin.a Y B/ Y sin.a B/ sin A C B sin A B cosasin B, i.e. cos A sin B = sin.a Y B/ sin.a B/ cos A C B C cos A B cosacos B, i.e. cos A cos B = cos.a Y B/ Y cos.a B/ cos A C B cos A B sinasin B, i.e. sin A sin B = cos.a Y B/ cos.a B/ Problem. Epress: sin cos as a sum or difference of sines and cosines From equation, sin cos sin C C sin.sin 7 Y sin / Problem. Epress: cos sin as a sum or difference of sines or cosines From equation, cos sin { } sin C sin sin 7q sin q Problem 6. Epress: cos t cos t as a sum or difference of sines or cosines From equation, { } costcos t cos t C t C cos t t.cos t Y cos t/ Thus, if the integral costcos t dt was required, then costcos t dt cos t C cos t dt sin t sin t Y Y c Problem 7. In an alternating current circuit, voltage v sinωt and current i sin ωt /6. Find an epression for the instantaneous power p at time t given that p vi, epressing the answer as a sum or difference of sines and cosines p vi sinωt sin ωt /6 sin ωt sin ωt /6. From equation, sin ωt sin ωt /6 fcos ωt C ωt /6 cosωt ωt /6 g fcos ωt /6 cos /6g i.e. instantaneous power, p = cos p=6 cos.!t p=6/ Now try the following eercise Eercise Further problems on changing products of sines and cosines into sums or differences In Problems to, epress as sums or differences:

17 ENGINEERING MATHEMATICS. sin 7t cos t sin 9t C sin t. cos 8 sin sin sin 6. sin 7t sin t cos t cos t. cos cos cos C cos. sin cos sin 6 C sin 6 6. etermine sintcos t dt cos t 7. Evaluate / cos cos cos t C c 8. Solve the equation: sin sin cos in the range to 8, 9 or 6. Changing sums or differences of sines and cosines into products From equation, C sin C sin sin cos Problem 9. product sinq cos q Epress: sin 7 sin as a From equation 6, 7 C 7 sin 7 sin cos sin Problem. product cos sin Epress: cos t cos t as a From equation 8, t C t cos t cos t sin sin sin 7 t sin t t t. sin C sin sin cos. sin 9 sin 7 cos8 sin. cos t C cos t cos t cos t. cos t cos t 8 sin t sin t. cos C cos 6. Show that:a COMPOUN ANGLES cos 7 sin sin cos C cos cos tan b fsin sin C g cos sin In the compound-angle formula let ACB X and A B Y Solving the simultaneous equations gives A X C Y and B X Y Thus sin A C B C sin A B sinacos B becomes X Y Y X Y sin X Y sin Y = sin cos Similarly, X Y Y X Y sin X sin Y = cos sin 6 X Y Y X Y cos X Y cos Y = cos cos 7 X Y Y X Y cos X cos Y = sin sin 8 Problem 8. product Epress: sin C sin as a sin 7 t sin t since sin t sin t Problem. Show that cos 6 C cos cot sin 6 C sin From equation 7, cos 6 C cos coscos From equation, sin 6 C sin sincos cos 6 C cos coscos Hence sin 6 C sin sincos cos cot sin Now try the following eercise Eercise Further problems on changing sums or differences of sines and cosines into products In Problems to, epress as products:

18 ENGINEERING MATHEMATICS Assignment 7 P This assignment covers the material in Chapters to 6. The marks for each question are shown in brackets at the end of each question. Q m P R. A triangular plot of land ABC is shown in Fig. A7.. Solve the triangle and determine its area. B. m 7 Figure A7.. m A. Figure A7. shows a roof truss PQR with rafter PQ m. Calculate the length of a the roof rise PP, b rafter PR, and c the roof span QR. Find also d the cross-sectional area of the roof truss. C Figure A7.. Prove the following identities: cos a tan cos b cos C sin 6. Solve the following trigonometric equations in the range 6 : a cos C b. cosec. c sin C sin. Solve the equation sin /6 8cos for values 8 6. Epress. cos t 7.sint in the form R sin t C. Hence solve the equation. cos t 7.sint. in the range t

19 Part 8 ifferential Calculus 76 ENGINEERING MATHEMATICS Now try the following eercise f B Introduction to differentiation. Introduction to calculus Calculus is a branch of mathematics involving or leading to calculations dealing with continuously varying functions. Calculus is a subject that falls into two parts: i ii differential calculus or differentiation and integral calculus or integration. ifferentiation is used in calculations involving velocity and acceleration, rates of change and maimum and minimum values of curves.. Functional notation In an equation such as y C, y is said to be a function of and may be written as y f. An equation written in the form f C istermedfunctional notation. The value of f when is denoted by f, and the value of f when is denoted by f and so on. Thus when f C, then and f C f C and so on. Problem. If f C find: f, f, f and f f f C f C f C 6 9 C 9 f C C C 9 f f 9 9 Problem. Given that f C 7 determine: i f ł f iii f C a f ii f C a iv f C a f a f C 7 i f C 7 f C 7 f ł f ii f C a C a C C a 7 9 C 6a C a C C a 7 C a C a C C a 7 Y a Y a iii f C 7 f C a f C a C a a Y a iv f C a f a a C a a Y a Eercise 7 Further problems on functional notation. If f 6 C findf, f, f, f and f.,,, 9, 6. If f C 7findf, f, f, f f.,,,. Given f C C prove that f 7 f. If f CC6findf, f Ca, f C a f f C a f and a 8, a a C 8, a a, a. The gradient of a curve a b If a tangent is drawn at a point P on a curve, then the gradient of this tangent is said to be the gradient of the curve at P. In Fig.., the gradient of the curve at P is equal to the gradient of the tangent PQ. f P Q Figure. For the curve shown in Fig.., let the points A and B have co-ordinates, y and, y, respectively. In functional notation, y f and y f as shown. The gradient of the chord AB BC B C AC E f f f A E C f Figure. f 8 6 A. Figure. c For the curve f shown in Fig..: i the gradient of chord AB ii iii iv v f f C the gradient of chord AC f f the gradient of chord A f. f. 9 B... if E is the point on the curve., f. then the gradient of chord AE f. f.... if F is the point on the curve., f. then the gradient of chord AF f =

20 f. f.... Thus as point B moves closer and closer to point A the gradient of the chord approaches nearer and nearer to the value. This is called the limiting value of the gradient of the chord AB and when B coincides with A the chord becomes the tangent to the curve. Now try the following eercise Eercise 8 A further problem on the gradient of a curve. Plot the curve f for values of from to C. Label the co-ordinates, f and,f as J and K, respectively. Join points J and K to form the chord JK. etermine the gradient of chord JK. By moving J nearer and nearer to K determine the gradient of the tangent of the curve at K. 6, 8. ifferentiation from first principles i In Fig.., A and B are two points very close together on a curve, υ delta and υy delta y representing small increments in the and y directions, respectively. y A,y f B+, y+ f + Figure. Gradient of chord However, Hence INTROUCTION TO IFFERENTIATION 77 υy As υ approaches zero, approaches υ a limiting value and the gradient of the chord approaches the gradient of the tangent at A. ii When determining the gradient of a tangent to a curve there are two notations used. The gradient of the curve at A in Fig.. can either be written as: { } υy f C υ f limit or limit υ! υ υ! υ In Leibniz notation, limit In functional notation, { } f. Y / f./ f./= limit iii is the same as f and is called the differential coefficient or the derivative. The process of finding the differential coefficient is called differentiation. Summarising, the differential coefficient, υy f limit υ! υ { } f C υ f limit υ! υ Problem. ifferentiate from first principles f and determine the value of the gradient of the curve at To differentiate from first principles means to find f by using the epression { } f C υ f f limit υ! υ f Substituting C υ for gives f C υ C υ C υ C υ, hence AB υy { υ f C υ C υ } limit υ! υ υy f C υ f { υy f C υ f υ C υ } limit limit f C υg υ υ υ! υ υ! 78 ENGINEERING MATHEMATICS As υ!, Cυ! C. Thus f./ =, i.e. the differential coefficient of is. At, the gradient of the curve, f Problem. of y By definition, Find the differential coefficient f { } f C υ f limit υ! υ The function being differentiated is y f. Substituting C υ for gives: f C υ C υ C υ. Hence f limit υ! limit υ! { } C υ υ { } υ limit υ fg υ! Since the term υ does not appear in the limiting value as υ! of is. Thus =, i.e. the differential coefficient of is. The equation y represents a straight line of gradient see Chapter 7. The differential coefficient i.e. or f means the gradient of the curve, and since the slope of the line y is this result can be obtained by inspection. Hence, in general, if y k where k is a constant, then the gradient of the line is k and or f k. Problem. Find the derivative of y 8 y f 8. Since there are no -values in the original equation, substituting C υ for still gives f C υ 8. Hence f limit υ! limit υ! { } f C υ f υ { } 8 8 υ Thus, when y 8, = The equation y 8 represents a straight horizontal line and the gradient of a horizontal line is zero, hence the result could have been determined by inspection. Finding the derivative means finding the gradient, hence, in general, for any horizontal line if y k where k is a constant then. Problem 6. ifferentiate from first principles f Substituting C υ for gives f C υ C υ C υ C υ C υ C υ C υ C υ C 6 υ C 6υ C υ { } f C υ f f limit υ! υ C 6 υ C 6υ C υ limit υ! υ { 6 υ C 6υ C υ } limit υ! υ { limit 6 C 6υ C υ } υ! Hence f./ = 6, i.e. the differential coefficient of is 6. Problem 7. Find the differential coefficient of y C and determine the gradient of the curve at y f C f C υ C υ C C υ C υ C υ C C υ C 8υ C υ f limit υ! limit υ! C C υ { } f C υ f υ C 8υ C υ C C υ C υ

21 { 8υ C υ } C υ limit υ! υ limit f8 C υ C g υ! i.e. = f./ = 8 Y At, the gradient of the curve f 8 C 9 Now try the following eercise Eercise 9 Further problems on differentiation from first principles In Problems to, differentiate from first principles.. y. y 7 7. y 8. y. y C C 6. y 7. f f 9. f 9 8. f 7. f C C. f. etermine d from first principles. Find d C from first principles 6. ifferentiation of y = a n by the general rule From differentiation by first principles, a general rule for differentiating a n emerges where a and n are any constants. This rule is: INTROUCTION TO IFFERENTIATION 79 if y = a n then = ann or, if f./ = a n then f./ = an n Each of the results obtained in worked problems to 7 may be deduced by using this general rule. When differentiating, results can be epressed in a number of ways. For eample: i if y then 6, ii if f then f 6, iii the differential coefficient of is 6, iv v a the derivative of is 6, and d 6 Problem 8. Using the general rule, differentiate the following with respect to : a y 7 b y p c y Comparing y 7 with y a n shows that a and n 7. Using the general rule, ann b y p.henceaandn ann c p y. Hence a andn ann 8 8 Problem 9. Find the differential coefficient of y C p C 7 y C p C 7 i.e. y C / C 7 8 ENGINEERING MATHEMATICS i.e. C / C 6 C C / = 6 Y Y p Problem. If f t t C p t find f t Hence f t t C p t t C t t C t f t t C t t t i.e. f.t/ i.e. Problem. with respect to Hence y t p t ifferentiate y C C C y C C C C C C C Now try the following eercise Eercise Further problems on differentiation of y = a n by the general rule In Problems to 8, determine the differential coefficients with respect to the variable.. y 7 8. y p p. y p p t t. y 6 C. y p C C p 6. y p C 7 C 7 p 9 7. y t 6t 8. y C C 6 C 9. Using the general rule for a n check the results of Problems to of Eercise 9, page 79.. ifferentiate f 6 C and find the gradient of the curve at a, and b. a b. Find the differential coefficient of y C and determine the gradient of the curve at. 6 C 6,. etermine the derivative of y C C 7 and determine the gradient of the curve at. 6 C, 9..6 ifferentiation of sine and cosine functions Figure.a shows a graph of y sin. The gradient is continually changing as the curve moves from O to A to B to C to. The gradient, given by, may be plotted in a corresponding position d below y sin, as shown in Fig..b. i At, the gradient is positive and is at its steepest. Hence is a maimum positive value.

22 INTROUCTION TO IFFERENTIATION 8 8 ENGINEERING MATHEMATICS a b + + y Figure. A p A p p y = sin q p B B d dq sin q = cos q p C C p p p q radians q radians ii Between O and A the gradient is positive but is decreasing in value until at A the gradient is zero, shown as A. iii Between A and B the gradient is negative but is increasing in value until at B the gradient is at its steepest. Hence B is a maimum negative value. iv If the gradient of y sin is further investigated between B and C and C and then the resulting graph of is seen to be a cosine d wave. Hence the rate of change of sin is cos, i.e. if y = sin q then dq = cos q It may also be shown that: if y = sin aq, = a cos aq dq where a is a constant and if y = sin.aq Y a/, = a cos.aq Y a/ dq where a and are constants. If a similar eercise is followed for y cos then the graphs of Fig..6 result, showing d to be a graph of sin, but displaced by radians. If each point on the curve y sin as shown in Fig..a were to be made negative, i.e. C is made, is made C, and so on then the graph shown in Fig..6b would result. This latter graph therefore represents the curve of sin. a b + + dq y y = cos q p p Figure.6 if p p d dq cos q = sin q p p p p q radians q radians Thus, if y = cos q, dq = sinq It may also be shown that: y = cos aq, = a sin aq dq where a is a constant y = cos.aqya/, = asin.aq Y a/ dq where a and are constants. and if Problem. ifferentiate the following with respect to the variable: a y sin b f t cost a y sin cos cos q d b f t cost f t sin t 6sint Problem. Find the differential coefficient of y 7sin cos y 7sin cos 7 cos sin cos Y sin Problem. ifferentiate the following with respect to the variable: a f sin. b f t cos t C. a Iff sin. f.q/ cos. p cos.pq./ b Iff t cos t C. f.t/ sin t C. sin.t Y./ Problem. An alternating voltage is given by: v sin t volts, where t is the time in seconds. Calculate the rate of change of voltage when a t. s and b t. s v sin t volts. The rate of change of v is given by dv dt. dv cos t cos t dt a When t. s, dv cos. cos dt cos means the cosine of radian make sure your calculator is on radians not degrees. dv Hence = 86 volts per second dt b When t. s, dv cos. cos. dt dv Hence = 8 volts per second dt Now try the following eercise Eercise Further problems on the differentiation of sine and cosine functions. ifferentiate with respect to : a y sin b y cos6 a cos b sin 6. Given f sin cos, find f 6 cos C sin. An alternating current is given by i sin t amperes, where t is the time in seconds. etermine the rate of change of current when t. seconds. 7. A/s. v sin t volts represents an alternating voltage where t is the time in seconds. At a time of ð seconds, find the rate of change of voltage. 9. V/s. If f t sin tc. cos t.7 determine f t cos t C. C 6sin t.7.7 ifferentiation of e a and ln a A graph of y e is shown in Fig..7a. The gradient of the curve at any point is given by and is continually changing. By drawing tangents to the curve at many points on the curve and measuring the gradient of the tangents, values of for corre- sponding values of may be obtained. These values are shown graphically in Fig..7b. The graph of against is identical to the original graph of y e. It follows that: if y = e, then = e y y = e a = e b Figure.7

23 It may also be shown that if y = e a, then = aea Therefore if y e 6,then 6e6 e 6 A graph of y ln is shown in Fig..8a. The gradient of the curve at any point is given by and is continually changing. By drawing tangents to the curve at many points on the curve and measuring the gradient of the tangents, values of for corresponding values of may be obtained. These values are shown graphically in Fig..8b. The graph of against is the graph of =. It follows that: if y = ln, then = y... Figure.8 6 = a 6 b It may also be shown that if y = ln a, then = y = ln INTROUCTION TO IFFERENTIATION 8 Note that in the latter epression a does not appear in the term. Thus if y ln, then Problem 6. ifferentiate the following with respect to the variable: a y e b f t e t a If y e then e 6e b If f t e t e t,then Problem 7. f.t/ e t e t e t ifferentiate y ln If y ln, then Now try the following eercise Eercise Further problems on the differentiation of e a and ln a. ifferentiate with respect to : a y e b y 7e a e b 7e. Given f ln ln, determine f. If f t lnt C, evaluate f t when t. 6. Evaluate when, given y e C8ln. Give the answer e correct to significant figures. 66 Methods of differentiation. ifferentiation of common functions The standard derivatives summarised below were derived in Chapter and are true for all real values of. y or f or f a n sin a cos a e a ln a an n a cos a a sin a ae a The differential coefficient of a sum or difference is the sum or difference of the differential coefficients of the separate terms. Thus, if f p C q r, wheref, p, q and r are functions, then f p Cq r ifferentiation of common functions is demonstrated in the following worked problems. Problem. Find the differential coefficients of: a y b y If y a n then ann a Since y, a and n thus 6 b y is rewritten in the standard an form as y and in the general rule a and n Thus 6 6 a b a Problem. ifferentiate: a y 6 b y 6 y 6 may be written as y 6,i.e.inthe general rule a 6andn. Hence 6 In general, the differential coefficient of a constant is always zero. Since y 6, in the general rule a 6and n Hence In general, the differential coefficient of k, where k is a constant, is always k. Problem. Find the derivatives of: a y p b y p y p is rewritten in the standard differential form as y / In the general rule, a andn Thus / p b y p / / in the standard differential form. In the general rule, a andn Thus / 7/ p 7/ 7

24 METHOS OF IFFERENTIATION 8 86 ENGINEERING MATHEMATICS Problem. ifferentiate: y C C p with respect to y C C p is rewritten as y C C / When differentiating a sum, each term is differentiated in turn. Thus C i.e C / C C / = Y p Problem. Find the differential coefficients of: a y sin b f t cost with respect to the variable a b When y sin then cos cos When f t cost then f t sint 6sint Problem 6. etermine the derivatives of: a y e b f c y 6ln e a When y e then e e b f e e, thus f e 6e 6 c When y 6ln then 6 e q 6 Problem 7. Find the gradient of the curve y C at the points, and, The gradient of a curve at a given point is given by the corresponding value of the derivative. Thus, since y C then the gradient C At the point,,. Thus the gradient C At the point,,. Thus the gradient C Problem 8. etermine the co-ordinates of the point on the graph y 7 C where the gradient is The gradient of the curve is given by the derivative. When y 7 C then 6 7 Since the gradient is then6 7, from which, When, y 7 C Hence the gradient is at the point, Now try the following eercise Eercise Further problems on differentiating common functions In Problems to 6 find the differential coefficients of the given functions with respect to the variable.. a b.. c a b 8.. c. a b 6 c a 8 b c. a p b p c a p p p b c p. a p b c sin a p b c 6 cos. a cos b e 6 c e a 8 sin b e 6 c e 6. a ln 9 b e e a b e C e c p c C p 7. Find the gradient of the curve y t C t t C at the points, and, 8., 6 8. Find the co-ordinates of the point on the graph y C wherethe gradient is., 9. a ifferentiate y C ln cos C sin e b Evaluate d when, correct to significant figure. a C C sin cos C 6 b. e. Evaluate ds, correct to significant dt figures, when t 6 given s sint C p t.9. ifferentiation of a product When y uv, andu and v are both functions of, then = u dv Y vdu This is known as the product rule. Problem 9. Find the differential coefficient of: y sin sin is a product of two terms and sin Let u and v sin Using the product rule: u dv du C v # # # # gives: cos C sin 6 i.e. 6 cos C 6 sin 6. cos Y sin / Note that the differential coefficient of a product is not obtained by merely differentiating each term and multiplying the two answers together. The product rule formula must be used when differentiating products. Problem. Find the rate of change of y with respect to given: y p ln The rate of change of y with respect to is given by. y p ln / ln, which is a product. Let u / and v ln Then u dv du C v # # # # / C ln / i.e. / C ln / C ln = p Y ln Problem. / ifferentiate: y cos ln Let u cos i.e. a product and v ln Then udv C vdu du where sin C cos dv and Hence cos C ln sin C cos

25 METHOS OF IFFERENTIATION ENGINEERING MATHEMATICS cos C ln cos sin i.e. = fcos Yln.cos sin /g Problem. etermine the rate of change of voltage, given v t sin t volts, when t. s Rate of change of voltage dv t cost C sin t dt t cos t C sint When t., dv. cos. C sin. dt cos. C sin. where cos. means the cosine of. radians.96 Hence dv.96 C.89 dt.8 C i.e. the rate of change of voltage when t =. s is.79 volts/s, correct to significant figures. Now try the following eercise Eercise Further problems on differentiating products In Problems to differentiate the given products with respect to the variable.. cos 6 cos sin. p ln p C ln. e t sin t e t cost C sint. e ln e C ln. e t ln t cos t { } t e t C ln t cos t ln t sin t 6. Evaluate di, correct to significant dt figures, when t., and i t sin t Evaluate dz, correct to significant dt figures, when t., given that z e t sin t.. ifferentiation of a quotient When y u,andu and v are both functions of v then du v = u dv v This is known as the quotient rule. Problem. Find the differential coefficient of: y sin sin is a quotient. Let u sin and v Note that v is always the denominator and u the numerator where and Hence du v udv v du cos cos dv cos sin cos 8 sin 8 cos sin 8 i.e. =. cos sin/ Note that the differential coefficient is not obtained by merely differentiating each term in turn and then dividing the numerator by the denominator. The quotient formula must be used when differentiating quotients. Problem. etermine the differential coefficient of: y tan a sin a y tan a. ifferentiation of tan a is cos a thus treated as a quotient with u sin a and v cos a du v udv v cos a a cos a sin a a sin a cos a a cos a C a sin a cos a a cos a C sin a cos a a cos a, since cos a C sin a see Chapter Hence = a sec a since sec a cos a see Chapter Problem. y sec a Find the derivative of: y sec a i.e. a quotient. Let u and cos a v cos a du v udv v cos a a sin a cos a a sin a sin a cos a a cos a cos a i.e. = a sec a tan a Problem 6. ifferentiate: y tet cost The function tet is a quotient, whose numerator cost is a product. Let u te t and v cost then du dt t et C e t and dv dt sin t du Hence v i.e. udv v cost tet C e t te t sint cost tet cos t C e t cos t C te t sin t cos t et t cos t C cos t C t sin t cos t = et.t cos t cos Y cos t Y t sin t/ t Problem 7. etermine the gradient of the curve y p p, at the point C Let y and v C du v udv C v C At the point C C p p,, p, hence the gradient C p p C Now try the following eercise Eercise Further problems on differentiating quotients In Problems to, differentiate the quotients with respect to the variable.... cos C p sin 6 sin C cos C p sin cos sin

26 .. ln t p t e sin e sin ln t p t f C sin cos g 6. Find the gradient of the curve y at the point, 8 7. Evaluate at., correct to significant figures, given y C ln.8. Function of a function It is often easier to make a substitution before differentiating. If y is a function of then = du du This is known as the function of a function rule or sometimes the chain rule. For eample, if y 9 then, by making the substitution u, y u 9, which is of the standard form. Hence du 9u8 and du Then du ð du 9u8 7u 8 Rewriting u as gives: = 7. /8 Since y is a function of u, andu is a function of, then y is a function of a function of. Problem 8. ifferentiate: y cos C Let u C thenycos u Hence du and du sinu Using the function of a function rule, du ð du sinu sin u METHOS OF IFFERENTIATION 89 Rewriting u as C gives: = sin. Y / Problem 9. Find the derivative of: y t t 6 Let u t t, theny u 6 Hence du dt t and dt 6u Using the function of a function rule, du ð du 6u t Rewriting u as t t gives: 6 t t t dt 8.t /.t t/ Problem. etermine the differential coefficient of: y p C y p C C / Let u C theny u / Hence du 6 C and du u / p u Using the function of a function rule, du ð du p 6 C p C u u i.e. = Y p Y Problem. Let u tan then y u Hence and Then i.e. ifferentiate: y tan du sec, from Problem, du u du ð du u sec tan sec = 6 tan sec 9 ENGINEERING MATHEMATICS Problem. Find the differential coefficient of: y t y t t.letu t, then y u Hence du dt 6t and du 8u 8 u Then dt du ð du dt 8 u Now try the following eercise 6t 8t.t / Eercise 6 Further problems on the function of a function In Problems to 8, find the differential coefficients with respect to the variable.. 6. sin 6 cos. cos cos sin. C C 6. e tc e tc 6. cot t C t cosec t C 7. 6 tan y C 8 sec y C 8. e tan sec e tan 9. ifferentiate: sin with respect to, and evaluate, correct to significant figures, when.86. Successive differentiation When a function y f is differentiated with respect to the differential coefficient is written as or f. If the epression is differentiated again, the second differential coefficient is obtained and is written as d y pronounced dee two y by dee squared or f pronounced f double dash. By successive differentiation further higher derivatives such as d y and d y may be obtained. Thus if y,, d y 6, d y 7, d y 7 and d y Problem. If f C, find f f C f C f./. 6/ Problem. If y cos sin, evaluate, in the range y,whend is zero Since y cos sin, sin cos and d y cos C sin When d y is zero, cos C sin, i.e. sin cos or sin cos Hence tan an tan or p rads in the range Problem. Given y e show that d y C 6 C 9y y e i.e. a product Hence e C e 6e C e

27 i.e. d y 6 e C e 6 C 6e 8e 6e 6e d y 8e e Substituting values into d y C 6 C 9y gives: 8e e C 6 6e C e C 9 e 8e e 6e C e C 8e Thus when y e, d y C 6 C 9y Problem 6. Evaluate d y d when given: y sec Since y sec, then sec tan from Problem d 8sec tan i.e. a product d y d 8sec sec C tan 8 sec tan 6 sec C 6 sec tan When, d y d 6 sec C 6 sec tan 6 C 6 6 Now try the following eercise METHOS OF IFFERENTIATION 9 Eercise 7 Further problems on successive differentiation. If y C C find a d y b d y a 6 C b 7 C. a Given f t t t C t p t C determine f t b Evaluate f t when t. a t C 6 t C p t b.9 In Problems and, find the second differential coefficient with respect to the variable.. a sin t C cos t b ln a sin t C cos t b. a cos b a sin cos b 8. Evaluate f when given f sec 8 6. Show that the differential equation d y C y is satisfied when y e 6 Some applications of differentiation 6. Rates of change If a quantity y depends on and varies with a quantity then the rate of change of y with respect to is. Thus, for eample, the rate of change of pressure p with height h is dp dh. A rate of change with respect to time is usually just called the rate of change, the with respect to time being assumed. Thus, for eample, a rate of change of current, i, is di and a rate of change of dt temperature,, is d, and so on. dt Problem. The length l metres of a certain metal rod at temperature C is given by: l C. C.. etermine the rate of change of length, in mm/ C, when the temperature is a C and b C The rate of change of length means dl d Since length l C. C., dl then. C.8 d a When C, dl. C.8 d. m/ C. mm= C b When C, dl. C.8 d.7 m/ C.7 mm= C Problem. The luminous intensity I candelas of a lamp at varying voltage V is given by: I ð V. etermine the voltage at which the light is increasing at a rate of.6 candelas per volt The rate of change of light with respect to voltage is given by di dv Since I ð V di, dv ð V 8 ð V When the light is increasing at.6 candelas per volt then C.6 8 ð V, from which, voltage.6 V 8 ð.7 ð C 7 volts Problem. Newtons law of cooling is given by: e kt, where the ecess of temperature at zero time is C and at time t seconds is C. etermine the rate of change of temperature after s, given that 6 C andk. The rate of change of temperature is d dt Since e kt then d dt k e kt k e kt When 6, k. and t then d dt. 6 e..8 e..9 C=s Problem. The displacement s cm of the end of a stiff spring at time t seconds is given by: s ae kt sin ft. etermine the velocity of the end of the spring after s, if a, k.9 andf Velocity v ds dt where s ae kt sin ft i.e. a product Using the product rule, ds dt ae kt f cos ft C sin ft ake kt

28 SOME APPLICATIONS OF IFFERENTIATION 9 9 ENGINEERING MATHEMATICS When a, k.9, f andt, velocity, v e.9 cos C sin.9 e.9. cos.78 sin..78. cm=s Note that cos means the cosine of radians, not degrees, and cos cos. Now try the following eercise Eercise 8 Further problems on rates of change. An alternating current, i amperes, is given by i sin ft, wheref is the frequency in hertz and t the time in seconds. etermine the rate of change of current when t ms, given that f Hz. A/s. The luminous intensity, I candelas, of a lamp is given by I 6 ð V, where V is the voltage. Find a the rate of change of luminous intensity with voltage when V volts, and b the voltage at which the light is increasing at a rate of. candelas per volt. a. cd/v b V. The voltage across the plates of a capacitor at any time t seconds is given by v V e t/cr,wherev, C and R are constants. Given V volts, C. ð 6 farads and R ð 6 ohms find a the initial rate of change of voltage, and b the rate of change of voltage after. s. a 6 V/s b. V/s. The pressure p of the atmosphere at height h above ground level is given by p p e h/c,wherep is the pressure at ground level and c is a constant. etermine the rate of change of pressure with height when p. ð Pascals and c 6. ð at metres..6 Pa/m 6. Velocity and acceleration When a car moves a distance metres in a time t seconds along a straight road, if the velocity v is constant then v m/s, i.e. the gradient of the t distance/time graph shown in Fig. 6. is constant. Figure 6. If, however, the velocity of the car is not constant then the distance/time graph will not be a straight line. It may be as shown in Fig. 6.. Figure 6. d d The average velocity over a small time υt and distance υ is given by the gradient of the chord AB, i.e. the average velocity over time υt is υ υt.as υt!, the chord AB becomes a tangent, such that at point A, the velocity is given by: v dt Hence the velocity of the car at any instant is given by the gradient of the distance/time graph. If an epression for the distance is known in terms of time t then the velocity is obtained by differentiating the epression. Figure 6. d The acceleration a of the car is defined as the rate of change of velocity. A velocity/time graph is shown in Fig. 6.. If υv is the change in v and υt the corresponding change in time, then a υv υt.as υt!, the chord C becomes a tangent, such that at point C, the acceleration is given by: a dv dt Hence the acceleration of the car at any instant is given by the gradient of the velocity/time graph. If an epression for velocity is known in terms of time t then the acceleration is obtained by differentiating the epression. Acceleration a dv dt. However, v dt Hence a d d dt dt The acceleration is given by the second differential coefficient of distance with respect to time t Summarising, if a bo moves a distance metres in a time t seconds then: i distance = f.t/ ii velocity v f.t/ or, which is the gradient of the distance/time graph dt iii acceleration a = dv = f or d dt dt,which is the gradient of the velocity/time graph. Problem. The distance metres moved by a car in a time t seconds is given by: t t C t. etermine the velocity and acceleration when a t, and b t. s d istance t t C t m. Velocity v dt 9t t C m/s Acceleration a d 8t m/s a When time t, velocity v 9 C m=s and acceleration a 8 m=s i.e. a deceleration b When time t. s, velocity v 9.. C 8. m=s and acceleration a 8. m=s Problem 6. Supplies are dropped from a helicopter and the distance fallen in a time t seconds is given by: gt,where g 9.8 m/s. etermine the velocity and acceleration of the supplies after it has fallen for seconds istance gt 9.8 t.9t m Velocity v dv 9.8 tm/s dt and acceleration a d 9.8 m/s When time t s, velocity v m=s and acceleration a = 9.8 m=s which is acceleration due to gravity. Problem 7. The distance metres travelled by a vehicle in time t seconds after the brakes are applied is given by: t t. etermine a the speed of the vehicle in km/h at the instant the brakes are applied, and b the distance the car travels before it stops a istance, t t. Hence velocity v dt t At the instant the brakes are applied, time Hence ð 6 ð 6 velocity v m/s km/h 7 km=h

29 SOME APPLICATIONS OF IFFERENTIATION 9 96 ENGINEERING MATHEMATICS b a b Note: changing from m/s to km/h merely involves multiplying by.6. When the car finally stops, the velocity is zero, i.e. v t, from which, t, giving t 6 s. Hence the distance travelled before the car stops is given by: t t m Problem 8. The angular displacement radians of a flywheel varies with time t seconds and follows the equation: 9t t. etermine a the angular velocity and acceleration of the flywheel when time, t s, and b the time when the angular acceleration is zero Angular displacement 9t t rad. Angular velocity ω d dt 8t 6t rad/s. When time t s,! 8 6 rad=s. Angular acceleration d 8 t rad/s. dt When time t s,a 8 6rad=s When the angular acceleration is zero, 8 t, from which, 8 t, giving time, t =. s Problem 9. The displacement cm of the slide valve of an engine is given by:.cost C.6sint. Evaluate the velocity in m/s when time t ms isplacement.cost C.6sint Velocity v. sin t dt C.6 cos t sin t C 8 cos t cm/s When time t ms, velocity sin ð ð C 8 cos ð ð sin.7 C 8 cos.7 sin 7 C 8 cos 7.69 C.9.7 cm/s.7 m=s Now try the following eercise Eercise 9 Further problems on velocity and acceleration. A missile fired from ground level rises metres vertically upwards in t seconds and t t. Find a the initial velocity of the missile, b the time when the height of the missile is a maimum, c the maimum height reached, d the velocity with which the missile strikes the ground. a m/s b s c m d m/s. The distance s metres travelled by a car in t seconds after the brakes are applied is given by s t.t. Find a the speed of the car in km/h when the brakes are applied, b the distance the car travels before it stops. a 9 km/h b 6. m. The equation Ct t gives the angle, in radians, through which a wheel turns in t seconds. etermine a the time the wheel takes to come to rest, b the angle turned through in the last second of movement. a s b rads. At any time t seconds the distance metres of a particle moving in a straight line from a fied point is given by: tcln t. etermine a the initial velocity and acceleration b the velocity and acceleration after. s c the time when the velocity is zero. a m/s; m/s b 6 m/s; m/s c s. The angular displacement of a rotating disc is given by: 6sin t, where t is the time in seconds. etermine a the angular velocity of the disc when t is. s, b the angular acceleration when t is. s, and c the first time when the angular velocity is zero. a ω. rad/s b.7 rad/s c t 6.8 s 6. t t C 6t C represents the distance, metres, moved by a bo in t seconds. etermine a the velocity and acceleration at the start, b the velocity and acceleration when t s, c the values of t when the bo is at rest, d the value of t when the acceleration is 7 m/s, and e the distance travelled in the third second. a 6 m/s, m/s b 7 m/s, 97 m/s c sor s d s e 7 6 m 6. Turning points In Fig. 6., the gradient or rate of change of the curve changes from positive between O and P to negative between P and Q, and then positive again between Q and R. At point P, the gradient is zero and, as increases, the gradient of the curve changes from positive just before P to negative just after. Such a point is called a maimum point and appears as the crest of a wave. At point Q, the gradient is also zero and, as increases, the gradient of the curve changes from negative just before Q to positive just after. Such a point is called a minimum point, and appears as the bottom of a valley. Points such as P and Q are given the general name of turning points. It is possible to have a turning point, the gradient on either side of which is the same. Such a point is given the special name of a point of infleion, and eamples are shown in Fig. 6.. Maimum and minimum points and points of infleion are given the general term of stationary points. Procedure for finding and distinguishing between stationary points i Given y f, determine i.e. f Positive gradient y O Figure 6. y Maimum point Figure 6. P Points of infleion Minimum point Negative gradient Q R Positive gradient Maimum point ii Let and solve for the values of iii Substitute the values of into the original equation, y f, to find the corresponding y-ordinate values. This establishes the coordinates of the stationary points. To determine the nature of the stationary points: Either iv Find d y and substitute into it the values of found in ii. If the result is: a positive the point is a minimum one, b negative the point is a maimum one, c zero the point is a point of infleion or v etermine the sign of the gradient of the curve just before and just after the stationary points. If the sign change for the gradient of the curve is: a positive to negative the point is a maimum one

30 b negative to positive the point is a minimum one c positive to positive or negative to negative the point is a point of infleion. Problem. Locate the turning point on the curve y 6 and determine its nature by eamining the sign of the gradient on either side Following the above procedure: i Since y 6, 6 6 ii At a turning point,, hence 6 6, from which,. iii When, y 6 Hence the co-ordinates of the turning point is, iv If is slightly less than, say,.9, then , i.e. negative If is slightly greater than, say,., then , i.e. positive Since the gradient of the curve is negative just before the turning point and positive just after i.e. C,, is a minimum point Problem. Find the maimum and minimum values of the curve y C by a eamining the gradient on either side of the turning points, and b determining the sign of the second derivative Since y C then For a maimum or minimum value Hence, from which, and š When, y C When, y C 7 Hence, and, 7 are the co-ordinates of the turning points. SOME APPLICATIONS OF IFFERENTIATION 97 a Considering the point, : If is slightly less than, say.9, then.9, which is negative. If is slightly more than, say., then., which is positive. Since the gradient changes from negative to positive, the point, is a minimum point. Considering the point, 7: If is slightly less than, say., then., which is positive. If is slightly more than, say.9, then.9, which is negative. Since the gradient changes from positive to negative, the point, 7 is a maimum point. b Since, then d y 6 When, d y is positive, hence, is a minimum value. When, d y is negative, hence, 7 is a maimum value. Thus the maimum value is 7 and the minimum value is. It can be seen that the second differential method of determining the nature of the turning points is, in this case, quicker than investigating the gradient. Problem. Locate the turning point on the following curve and determine whether it is a maimum or minimum point: y C e Since y C e then d e fora maimum or minimum value. Hence e, e, giving ln.86 see Chapter. When.86, y.86 C e.86. C.. 98 ENGINEERING MATHEMATICS Thus.86,. are the co-ordinates of the turning point. d y d e. When.86, d y d ec.86., which is positive, hence.86,. is a minimum point. Problem. etermine the co-ordinates of the maimum and minimum values of the graph y 6 C and distinguish between them. Sketch the graph Following the given procedure: i ii iii iv Since y 6 C then 6 At a turning point,. Hence 6, i.e. C, from which or When, y When, y 6 C 9 6 C 6 Thus the co-ordinates of the turning points are, 9 and, 6 Since 6then d y d y When,, which is negative. Hence, 9 is a maimum point. When, d y, which is positive. Hence, is a minimum point. 6 Knowing, 9 is a maimum point i.e. crest of a wave, and, is a minimum point i.e. bottom of a valley and that 6 when, y, a sketch may be drawn as shown in Fig Figure 6.6 Problem. etermine the turning points on the curve y sin cos in the range to radians, and distinguish between them. Sketch the curve over one cycle Since y sin cos then cos C sin, for a turning point, from which, cos sin and sin cos tan. Hence tan 6.87 or 6.87, since tangent is negative in the second and fourth quadrants. When 6.87, y sin 6.87 cos 6.87 When 6.87, y sin 6.87 cos 6.87

31 ð radians 8. rad ð radians 8.6 rad Hence., and.6, are the coordinates of the turning points. d y sin C cos When. rad, d y sin.ccos., which is negative. Hence., is a maimum point. When.6 rad, d y sin.6ccos.6, which is positive. Hence.6, is a minimum point. Asketchofy sin cosis shown in Fig Figure 6.7 p p p p Now try the following eercise Eercise 6 Further problems on turning points In Problems to 7, find the turning points and distinguish between them.. y C Minimum at,. 6 Maimum at, 9 SOME APPLICATIONS OF IFFERENTIATION 99. y C 6 Minimum, 88 ; Maimum, 9. y ln Minimum at.,.86. y e Maimum at.69, y t t t C Minimum at, ; Maimum at, t C Minimum at t, 6 8. etermine the maimum and minimum values on the graph y cos sin in the range to 6. Sketch the graph over one cycle showing relevant points. Maimum of at 7.8, Minimum of at Show that the curve y t C t t has a maimum value of and a minimum value of. 6. Practical problems involving maimum and minimum values There are many practical problems involving maimum and minimum values which occur in science and engineering. Usually, an equation has to be determined from given data, and rearranged where necessary, so that it contains only one variable. Some eamples are demonstrated in Problems to. Problem. A rectangular area is formed having a perimeter of cm. etermine the length and breadth of the rectangle if it is to enclose the maimum possible area ENGINEERING MATHEMATICS Let the dimensions of the rectangle be and y.then the perimeter of the rectangle is C y. Hence C y, or C y Since the rectangle is to enclose the maimum possible area, a formula for area A must be obtained in terms of one variable only. Area A y. From equation, y Hence, area A y y y y da y for a turning point, from which, y cm. d A, which is negative, giving a maimum point. When y cm, cm, from equation. Hence the length and breadth of the rectangle are each cm, i.e. a square gives the maimum possible area. When the perimeter of a rectangle is cm, the maimum possible area is ð cm. Problem 6. A rectangular sheet of metal having dimensions cm by cm has squares removed from each of the four corners and the sides bent upwards to form an open bo. etermine the maimum possible volume of the bo The squares to be removed from each corner are shown in Fig. 6.8, having sides cm. When the sides are bent upwards the dimensions of the bo will be: length cm, breadth cm and height, cm. cm Figure 6.8 cm Volume of bo, V 6 C dv 8 C for a turning point. Hence 6 C, i.e. C6 Using the quadratic formula, š cm or.7 cm. Since the breadth is cmthen 8.9 cm is not possible and is neglected. Hence.7 cm. d V 8 C. When.7, d V is negative, giving a mai- mum value. The dimensions of the bo are: length.7.6 cm, breadth cm, and height.7 cm. Maimum volume cm Problem 7. etermine the height and radius of a cylinder of volume cm which has the least surface area Let the cylinder have radius r and perpendicular height h. Volume of cylinder, V r h Surface area of cylinder, A rh C r Least surface area means minimum surface area and a formula for the surface area in terms of one variable only is required. From equation, h r Hence surface area, A r r C r C r r C r r da dr C r, for a turning point. r Hence r r and r, from which, r.69 cm. d A dr 8 C. r

32 SOME APPLICATIONS OF IFFERENTIATION ENGINEERING MATHEMATICS d A When r.69 cm, is positive, giving a dr minimum value. From equation, when r.69 cm, h 6.9 cm..69 Hence for the least surface area, a cylinder of volume cm has a radius of.69 cm and height of 6.9 cm. Problem 8. etermine the area of the largest piece of rectangular ground that can be enclosed by m of fencing, if part of an eisting straight wall is used as one side Let the dimensions of the rectangle be and y as shown in Fig. 6.9, where PQ represents the straight wall. From Fig. 6.9, C y Area of rectangle, A y P y Figure 6.9 Since the maimum area is required, a formula for area A is needed in terms of one variable only. From equation, y Hence, area A y y y y y da y, for a turning point, from which, y m. d A, which is negative, giving a maimum value. When y m, m from equation. Hence the maimum possible area y m Problem 9. An open rectangular bo with square ends is fitted with an overlapping lid which covers the top and the front face. etermine the maimum volume of the bo if 6 m of metal are used in its construction A rectangular bo having square ends of side and length y is shown in Fig. 6.. y Q Figure 6. y Surface area of bo, A, consists of two ends and five faces since the lid also covers the front face. Hence A C y 6 Since it is the maimum volume required, a formula for the volume in terms of one variable only is needed. Volume of bo, V y From equation, y Hence volume V y 6 dv 6 6 for a maimum or minimum value. Hence 6 6, giving m isnot possible, and is thus neglected. d V. When, d V value. From equation, when, y 6 is negative, giving a maimum Hence the maimum volume of the bo is given by V y m Problem. Find the diameter and height of a cylinder of maimum volume which can be cut from a sphere of radius cm A cylinder of radius r and height h is shown enclosed in a sphere of radius R cm in Fig. 6.. Volume of cylinder, V r h Using the right-angled triangle OPQ shown in Fig. 6., h r C R by Pythagoras theorem, i.e. Figure 6. r C h Since the maimum volume is required, a formula for the volume V is needed in terms of one variable only. From equation, r h Substituting into equation gives: V h h h h dv h, for a maimum or dh minimum value. Hence h, from which, h.86 cm. d V dh When h.86, d V 6h. is negative, giving a mai- dh mum value. From equation, r h.86, from which, radius r 9.8 cm iameter of cylinder r cm. Hence the cylinder having the maimum volume that can be cut from a sphere of radius cm is one in which the diameter is 9.6 cm and the height is.86 cm. Now try the following eercise Eercise 6 Further problems on practical maimum and minimum problems. The speed, v, of a car in m/s is related to time t s by the equation v Ct t. etermine the maimum speed of the car in km/h. km/h. etermine the maimum area of a rectangular piece of land that can be enclosed by m of fencing. 9 m. A shell is fired vertically upwards and its vertical height, metres, is given by: t t, where t is the time in seconds. etermine the maimum height reached. 8 m. A lidless bo with square ends is to be made from a thin sheet of metal. etermine the least area of the metal for which the volume of the bo is. m.. m. A closed cylindrical container has a surface area of cm. etermine the dimensions for maimum volume. radius.67 cm, height 9. cm 6. Calculate the height of a cylinder of maimum volume that can be cut from a cone of height cm and base radius 8 cm cm 7. The power developed in a resistor R by a battery of emf E and internal resistance r is given by P E R R C r. ifferentiate P with respect to R and show that the power is a maimum when R r. 8. Find the height and radius of a closed cylinder of volume cm which has the least surface area. height. cm, radius.7 cm

33 SOME APPLICATIONS OF IFFERENTIATION ENGINEERING MATHEMATICS 9. Resistance to motion, F, of a moving vehicle, is given by: F C. etermine the minimum value of resistance..7. An electrical voltage E is given by: E sin t C cos t volts, where t is the time in seconds. etermine the maimum value of voltage..7 volts 6. Tangents and normals Tangents The equation of the tangent to a curve y f at the point, y is given by: y y = m. / where m gradient of the curve at, y. Problem. Find the equation of the tangent to the curve y atthe point, Gradient, m At the point,, andm Hence the equation of the tangent is: y y m i.e. y i.e. y C or y= The graph of y is shown in Fig. 6.. The line AB is the tangent to the curve at the point C, i.e.,, and the equation of this line is y. Normals The normal at any point on a curve is the line that passes through the point and is at right angles to the tangent. Hence, in Fig. 6., the line C is the normal. It may be shown that if two lines are at right angles then the product of their gradients is. Thus if m is the gradient of the tangent, then the gradient of the normal is m. Hence the equation of the normal at the point, y is given by: y C A Figure 6. y y = m. / B y = Problem. Find the equation of the normal to the curve y atthe point, m from Problem, hence the equation of the normal is y y m i.e. y i.e. y C C or y = Thus the line C in Fig. 6. has the equation y Problem. etermine the equations of the tangent and normal to the curve y at the point, Gradient m of curve y is given by m At the point,, and m Equation of the tangent is: y y m i.e. i.e. y y C C or y C C or y = Equation of the normal is: y y m i.e. y i.e. y C C i.e. y C Multiplying each term by gives: y C Hence equation of the normal is: y Y Y 8 = Now try the following eercise Eercise 6 Further problems on tangents and normals For the following curves, at the points given, find a the equation of the tangent, and b the equation of the normal. y at the point, a y b y C 9. y at the point, 8 a y b y C 8. y at the point, a y C b 6y C C 7. y C at the point, a y C b y C C 7. t at the point, a 9 C t 6 b 9t 6 or 7t Small changes If y is a function of, i.e. y f, and the approimate change in y corresponding to a small change υ in is required, then: υy υ ³ and or f./ Problem. Given y, determine the approimate change in y if changes from to. Since y, then 8 Approimate change in y, υy ³ Ð υ ³ 8 υ When andυ., ³ 8. ³. Obviously, in this case, the eact value of υy may be obtained by evaluating y when., i.e. y...6 and then subtracting from it the value of y when, i.e. y, giving.6.6. Using υy Ð υ above gave., which shows that the formula gives the approimate change in y for a small change in. Problem. The time of swing T of a pendulum is given by T k p l,wherek is a constant. etermine the percentage change in the time of swing if the length of the pendulum l changes from. cm to. cm If T k p l kl /, dt then dl k l / k p l Approimate change in T, υt ³ dt k k dl υl ³ p υl ³ l p. l negative since l decreases Percentage error approimate change in T original value of T %

34 k p. l k p ð % l.. % % l..6% Hence the percentage change in the time of swing is a decrease of.6% Problem 6. A circular template has a radius of cm š.. etermine the possible error in calculating the area of the template. Find also the percentage error Area of circular template, A r, hence da dr r Approimate change in area, υa ³ da Ð υr ³ r υr dr When r cm and υr., υa. ³. cm i.e. the possible error in calculating the template area is approimately.7 cm.. Percentage error ³ %.% SOME APPLICATIONS OF IFFERENTIATION Now try the following eercise Eercise 6 Further problems on small changes. etermine the change in f if changes from. to. when a y b y a. b.8. The pressure p and volume v of a mass of gas are related by the equation pv. If the pressure increases from. to., determine the approimate change in the volume of the gas. Find also the percentage change in the volume of the gas..,.6%. etermine the approimate increase in a the volume, and b the surface area of a cube of side cm if increases from. cm to. cm. a 6 cm b cm. The radius of a sphere decreases from 6. cm to.96 cm. etermine the approimate change in a the surface area, and b the volume. a 6. cm b 8. cm 6 ENGINEERING MATHEMATICS Assignment This assignment covers the material in Chapters to 6. The marks for each question are shown in brackets at the end of each question.. ifferentiate the following with respect to the variable: a y C p b s e sin c y lnt cos t d p t t C. If f. 6 C find the coordinates at the point at which the gradient is.. The displacement s cm of the end of a stiff spring at time t seconds is given by: s ae kt sin ft. etermine the velocity and acceleration of the end of the spring after seconds if a, k.7 and f.. Find the co-ordinates of the turning points on the curve y C 6 C and distinguish between them. 9. The heat capacity C of a gas varies with absolute temperature as shown: C 6. C 7. ð. ð 6 etermine the maimum value of C and the temperature at which it occurs etermine for the curve y at the point, : a the equation of the tangent b the equation of the normal 7 7. A rectangular block of metal with a square cross-section has a total surface area of cm. Find the maimum volume of the block of metal. 7

35 Part 9 7 Integral Calculus Standard integration 7. The process of integration The process of integration reverses the process of differentiation. In differentiation, if f then f. Thus the integral of is, i.e. integration is the process of moving from f to f. By similar reasoning, the integral of t is t. Integration is a process of summation or adding parts together and an elongated S, shown as,is used to replace the words the integral of. Hence, from above, and t is t. In differentiation, the differential coefficient indicates that a function of is being differentiated with respect to, the indicating that it is with respect to. In integration the variable of integration is shown by adding dthe variable after the function to be integrated. Thus means the integral of with respect to, and t dt means the integral of t with respect to t As stated above, the differential coefficient of is, hence. However, the differential coefficient of C 7isalso. Hence is also equal to C 7. To allow for the possible presence of a constant, whenever the process of integration is performed, a constant c is added to the result. Thus C c and t dt t C c c is called the arbitrary constant of integration. 7. The general solution of integrals of the form a n The general solution of integrals of the form a n, wherea and n are constants is given by: a n ny a = n Y Y c This rule is true when n is fractional, zero, or a positive or negative integer, with the eception of n. Using this rule gives: i C ii iii C C c Y c C C C c C c Y c, and p / C C c C C c p Y c Each of these three results may be checked by differentiation. a The integral of a constant k is k C c. For eample, 8 8 C c 8 ENGINEERING MATHEMATICS b When a sum of several terms is integrated the result is the sum of the integrals of the separate terms. For eample, C C Y Y c 7. Standard integrals Since integration is the reverse process of differentiation the standard integrals listed in Table 7. may be deduced and readily checked by differentiation. Table 7. i Standard integrals a n anc n C C c ecept when n ii cos a sin a C c a iii sin a cos a C c a iv sec a tan a C c a v cosec a cot a C c a vi cosec a cot a cosec a C c a vii sec a tan a sec a C c a viii e a a ea C c i ln C c Problem. b t dt etermine: a The standard integral, a n anc n C C c a When a andnthen C C C c Y c b When a andnthen t dt tc C C c t C c t Y c Each of these results may be checked by differentiating them. Problem. etermine C 7 6 C 7 6 may be written as C 7 6, i.e. each term is integrated separately. This splitting up of terms only applies, however, for addition and subtraction. Hence C 7 6 C C 7 C C 7 6 C C C c 6 C c Y Y c Note that when an integral contains more than one term there is no need to have an arbitrary constant for each; just a single constant at the end is sufficient. a Problem. etermine a b t dt Rearranging into standard integral form gives: C C C c C c 6 Y c

36 STANAR INTEGRATION 9 ENGINEERING MATHEMATICS b Rearranging t dt gives: t C t dt t tc C C tc C C c t t C t C c t t Y t Y c Problem t p t Y c etermine C c 9 C p d t / C c a Problem 9. etermine a 7sec t dt b cosec d From Table 7.iv, 7sec t dt 7 tan t C c Now try the following eercise Eercise 6 Further problems on standard integrals etermine the following integrals:. a b 7 This problem shows that functions often have to be rearranged into the standard form of a n before it is possible to integrate them. Problem. etermine. Using the standard integral, a n when a andn gives: C C C c C c C c Y c Problem. etermine p For fractional powers it is necessary to appreciate np a m a m n p / C C c C Problem 6. p 9 dt t C c C c p Y c etermine 9 9t t C 9 p t dt dt t dt 9 C C c C C C p d p d C C d C C d a b C C d C C C C C C C Cc C C C C c C C C c p q Y p q Y p q Y c Problem 8. etermine a cos b sind From Table 7.ii, cos sin C c sin Y c From Table 7.iii, sin d cos C c cos q Y c b a b 7 tan t Y c From Table 7.v, cosec d cot C c cot q Y c Problem. etermine a e b dt et From Table 7.viii, e dt et e t dt e C c e Y c e t C c 6 e t C c 6e t Y c Problem. etermine m a b C dm m a b m C m ln Y c from Table 7.i m dm m C m m C m dm dm m C ln m C c m Y ln m Y c. a b a a C c b 7 C c 6 C c b C c. a b C d a C c b C C C c. a b a C c b C c p p. a b a p C c b p 9 C c 9 6. a p t dt 7. a cos b 7 p a p C c b t 7 p C c b 7sind a sin C c b 7 cos C c

37 8. a sec b cosec d a tan Cc b cot Cc STANAR INTEGRATION Note that the c term always cancels out when limits are applied and it need not be shown with definite integrals. ENGINEERING MATHEMATICS Problem. Evaluate: / sin b u du ln u ln ln 9. a cot t cosec t dt b sec t tan t dt a cosec t C c b sec t C c. a. a a e b a 8 e C c e b e C c u b du u ln u C c ln C c u b. C a p b b t a 8 p C 8 p C 8 7. efinite integrals C t C t C c t C t dt p C c Integrals containing an arbitrary constant c in their results are called indefinite integrals since their precise value cannot be determined without further information. efinite integrals are those in which limits are applied. If an epression is written as b a, b is called the upper limit and a the lower limit. The operation of applying the limits is defined as: b a b a The increase in the value of the integral as increases from to is written as Applying the limits gives: C c C c C c 9 C c C c 8 a Problem. b Evaluate a { } { } 6 b } } { { { f 9g 8 8 } { fg } 8 C Problem. Evaluate p d, taking positive square roots only C p d C d C d C C C C C C { C p } { { } { } 6 C 8 C C 8 8 p C p C p } sin cos { } cos cos { cos } { } cos { } cos { } { } C Problem. Evaluate costdt } { sin 6 cost dt sin t } { sin sin t Note that limits of trigonometric functions are always epressed in radians thus, for eample, sin 6 means the sine of 6 radians { } Hence costdt { } Problem 6. Evaluate a e b u du, each correct to significant figures a e e e e e Now try the following eercise Eercise 6 Further problems on definite integrals In Problems to 8, evaluate the definite integrals where necessary, correct to significant figures.. a. a. a. a. a 6. a b 7. a 6 b t dt a b b cos d b sin d cos cosec t dt C a 6 b b b cos d a b sint dt a b.8 6 sec a. b.98 sin cos a.7 b.68 e t dt b e a 9.9 b.7

38 8. a b C a.7 b The entropy change S, for an ideal gas is given by: T dt V S C v T R dv V V T STANAR INTEGRATION where T is the thermonamic temperature, V is the volume and R 8.. etermine the entropy change when a gas epands from litre to litres for a temperature rise from K to K given that: C v C 6 ð T C 8 ð 6 T.6 8 Integration using algebraic substitutions 8. Introduction Functions that require integrating are not always in the standard form shown in Chapter 7. However, it is often possible to change a function into a form which can be integrated by using either: i an algebraic substitution see Section 8., ii trigonometric substitutions see Chapters 9 and, iii partial fractions see Chapter, or iv integration by parts see Chapter. 8. Algebraic substitutions With algebraic substitutions, the substitution usually made is to let u be equal to f such that f u du is a standard integral. It is found that integrals of the forms: k f n f and k f n f where k and n are constants can both be integrated by substituting u for f. 8. Worked problems on integration using algebraic substitutions Problem. etermine: cos C 7 cos C7 is not a standard integral of the form shown in Table 7., page 8, thus an algebraic substitution is made. Let u C 7then du and rearranging gives du Hence cos C 7 cos u du cos u du, which is a standard integral sin u C c Rewriting u as C 7 gives: cos C 7 sin. Y 7/ Y c, which may be checked by differentiating it. Problem. Find: 7 may be multiplied by itself 7 times and then each term of the result integrated. However, this would be a lengthy process, and thus an algebraic substitution is made. Let u then du du and Hence 7 u 7 du u 7 du u 8 8 C c 6 u8 C c Rewriting u as gives:. / 7 = 6. /8 Y c

39 Problem. Find: Let u then du du and Hence du u u du ln u C c Problem. Evaluate: to significant figures ln. / Y c e 6, correct Hence INTEGRATION USING ALGEBRAIC SUBSTITUTIONS Problem 6. u du 8 8 u 6 C c 6 u6 C c 6 6. Y / 6 Y c Evaluate: /6 sin cos d Let u sin then du cos and d du d cos Hence sin cos d u cos du cos u du, by cancelling 6 ENGINEERING MATHEMATICS C c ln C c 6. e C e C C c In Problems 7 to, evaluate the definite integrals correct to significant figures / C 7. C. sin t C dt.98 cos.769 Hence p p u du 8 u / du u / C C c C p u du, by cancelling u/ p p u C c Y c Problem 9. Show that: tan d ln sec C c C c Let u 6 then du du 6and 6 Hence e 6 e u du 6 e u du eu C c e6 C c Thus e 6 e6 correct to significant figures. Problem. e e 9., etermine: C Let u C then du du 8 and 8 Hence C u du 8 u du, by cancelling 8 The original variable has been completely removed and the integral is now only in terms of u and is a standard integral. Thus u6 6 C c u6 C c sin 6 C c sin 6 C c /6 sin cos d sin 6 /6 sin 6 sin Now try the following eercise or.6 6 Eercise 66 Further problems on integration using algebraic substitutions In Problems to 6, integrate with respect to the variable.. sin C 9 cos C 9 C c. cos. sec t C sin C c tan t C C c 8. Further worked problems on integration using algebraic substitutions Problem 7. Find: Let u C then du Hence C C 6 and du 6 du u 6 du, by cancelling, 6 u 6 ln u C 6 ln. Y / Y c Problem 8. etermine: p Let u then du du 8 and 8 tan d then du d Hence sin d. Letu cos cos sin and d du sin sin cos d sin du u sin du ln u C c u ln cos C c ln cos C c, by the laws of logarithms Hence tan q dq = ln.sec q/ Y c, since cos cos sec 8. Change of limits When evaluating definite integrals involving substitutions it is sometimes more convenient to change the limits of the integral as shown in Problems and.

40 Problem. Evaluate: C 7, taking positive values of square roots only Let u C 7, then du du and It is possible in this case to change the limits of integration. Thus when,u C 7 and when,u C 7 9 Hence u u9 C 7 p u du 9 p u du u / du 9 Thus the limits have been changed, and it is unnecessary to change the integral back in terms of. Thus C 7 u / p u / p p Problem. Evaluate: p C, taking positive values of square roots only Let u C then du du and Hence p C du p u u / du Since u C, when, u 9andwhen, u Thus u / du u9 u u / du, INTEGRATION USING ALGEBRAIC SUBSTITUTIONS 7 i.e. the limits have been changed 9 u/ p 9 p, taking positive values of square roots only. Now try the following eercise Eercise 67 Further problems on integration using algebraic substitutions In Problems to 7, integrate with respect to the variable.. 6 C c. cos t sin t 6 cos6 t C c. sec tan sec C c or tan C c. t p t. ln 6. tan t t 9 C c ln C c ln sec t C c e t 7. p p e t C C c et C In Problems 8 to, evaluate the definite integrals correct to significant figures / e.76 sin cos d Integration using trigonometric substitutions 9. Introduction Table 9. gives a summary of the integrals that require the use of trigonometric substitutions, and their application is demonstrated in Problems to Worked problems on integration of sin,cos,tan and cot Problem. Evaluate: cos t dt Table 9. Integrals using trigonometric substitutions f f Method See problem. cos. sin C sin C c Use cos cos sin C c Use cos sin. tan tan C c Use C tan sec. cot cot C c Use cot C cosec. cos m sin n a If either m or n is odd but not both, use cos C sin, 6 b If both m and n are even, use either cos cos orcos sin 7, 8 6. sin A cos B Use sin A C B C sin A B 9 7. cos A sin B Use sin A C B sin A B 8. cos A cos B Use cos A C B C cos A B 9. sin A sin B Use cos A C B cos A B. p sin a a C c, Use a sin.. p a a sin a C p a C c substitution, 6 a C a tan C c Use a tan substitution 7 9 a

41 INTEGRATION USING TRIGONOMETRIC SUBSTITUTIONS 9 ENGINEERING MATHEMATICS Since cos t cos t from Chapter 6, then Hence cos t C cos t and cos t C cos 8t Problem. cos t dt C cos 8t dt t C sin 8t 8 sin 8 C 8 p or.78 etermine: sin C sin 8 Since cos sin from Chapter 6, then sin cos and sin cos 6 Hence sin cos 6 Problem. Find: tan sin 6 Y c 6 Since C tan sec,thentan sec and tan sec Hence tan sec tan Y c Problem. Evaluate cot d 6 Since cot C cosec,then cot cosec andcot cosec Hence 6 cot d cosec d cot 6 6 cot cot Now try the following eercise Eercise 68 Further problems on integration of sin, cos, tan and cot In Problems to, integrate with respect to the variable.. sin. cos t. tan t C sin C c sin t tan C c C c. cot t cot t C t C c In Problems to 8, evaluate the definite integrals, correct to significant figures.. 6. / / sin cos or.7 8 or / /6 tan t dt.8 cot d.6 9. Worked problems on powers of sines and cosines Problem. etermine: sin d Since cos C sin thensin cos. Hence sin d sin sin d sin cos d sin cos C cos d sin sincos C sin cos d cosq Y cos q cos q Y c Whenever a power of a cosine is multiplied by a sine of power, or vice-versa, the integral may be determined by inspection as shown. In general, cos n sin d and sin n cos d cosnc n C C c sinnc n C C c Alternatively, an algebraic substitution may be used as shown in Problem 6, chapter, page. Problem 6. sin cos Evaluate: sin cos sin cos cos sin sin cos sin cos sin cos sin sin sin sin or. Problem 7. Evaluate: cos d, correct to significant figures cos d C cos cos d d C cos C cos d C cos C C cos d C cos C sin C sin C 8 C sin 8 C cos d C sin / 8.78, correct to significant figures. Problem 8. Find: sin t cos t dt

42 sin t cos t dt sin t cos t dt cos t C cos t dt cos t C cost C cos t dt 8 C cost C cos t cos t 8 cos t cos t dt C cos t cos t cos t dt 8 C cos t C cos t 8 cos t sin t dt cos t C cos t sin t dt 8 t sin t Y sin t Y c INTEGRATION USING TRIGONOMETRIC SUBSTITUTIONS 9. Worked problems on integration of products of sines and cosines Problem 9. etermine: sin t cos t dt sin t cos t dt sin t C t C sin t t dt, from 6 of Table 9., which follows from Section 6., page, sin t C sin t dt cos t cos t Y c Problem. Find: cos sin ENGINEERING MATHEMATICS sin 7 means the sine of 7 radians.7 and sin Hence cos6 cos d.986 C , correct to decimal places Problem. Find: sin sin sin sin cos C cos, from 9 of Table 9. cos 8 cos sin 8 sin Y c or 8 In Problems to 8, evaluate the definite integrals.. 6. / / cos cos or.86 7 sin7t cos t dt.97 sin sin d.7 cos8t sin t dt Worked problems on integration using the sin q substitution Problem. etermine: p a Now try the following eercise Eercise 69 Further problems on integration of powers of sines and cosines Integrate the following with respect to the variable:. sin a cos C cos C c. cos sin sin C c. sin t cos t cos t C cos t C c cos. sin cos C cos7 C c 7. sin sin C sin 8 C c t 6. sin t cos t 8 sin t C c cos sin sin C sin, from 7 of Table 9. sin 7 sin 6 cos 7 cos Y Y c 6 7 Problem. Evaluate: cos6cos d, correct to decimal places cos6 cos d cos 6 C C cos 6 d, from 8 of Table 9. sin 7 sin cos 7 C cos d C 7 sin 7 C sin sin C sin 7 7 Now try the following eercise. sin sin 8/ Y c 6 Eercise 7 Further problems on integration of products of sines and cosines In Problems to, integrate with respect to the variable.. sin t cos t cos 7t 7. sin sin sin C. cos 6 cos sin 7 C 7. cos sin cos cos t C c sin C c sin C c cos 6 C c 6 Let a sin, then a cos and d a cos d. Hence p a a cos d a a sin a cos d a sin a cos d p a cos, since sin C cos a cos d d C c a cos Since a sin, thensin a and sin a Hence p a sin a Y c Problem. Evaluate p 9

43 INTEGRATION USING TRIGONOMETRIC SUBSTITUTIONS ENGINEERING MATHEMATICS From Problem, p 9 sin since a sin sin p or.78 Problem. a Find: Let a sin then a cos and d a cos d a Hence a a sin acos d a sin a cos d pa cos acos d a cos a cos d C cos a cos d a d since cos cos a sin C C c a C sin cos C c since from Chapter 6, sin sin cos a C sin cos C c Since a sin, thensin a and sin a Also, cos C sin, from which, cos sin a a a p a a a Thus a C sin cos a sin p a C a C c a a a sin a Y p a Y c Problem 6. Evaluate: 6 From Problem, 6 6 sin C 6 8sin C p 8 sin C 8sin 8 p or.7 Now try the following eercise Eercise 7 Further problems on integration using the sine q substitution. etermine: p dt t sin t C c. etermine:. etermine: p 9 sin C c sin C p C c 6. etermine: 9t dt 8 t sin C t p 6 9t C c. Evaluate: 6. Evaluate: p 6 or Worked problems on integration using the tan q substitution Problem 7. etermine: a C Let a tan then d a sec and a sec d Hence a C a C a tan a sec d a sec d a C tan a sec d a sec, since C tan sec a d a C c Since a tan, tan a Hence.a Y / = a tan a Y c Problem 8. From Problem 7, Evaluate: C C tan since a tan tan p or.97 8 Problem 9. Evaluate: C, correct to decimal places C / C p / C p tan p / / tan tan , correct to decimal places. Now try the following eercise Eercise 7. etermine:. etermine:. Evaluate: Further problems on integration using the tan q substitution C t dt tan C c 6 C 9 d tan C c dt.6 C t. Evaluate:.7 C

44 INTEGRATION USING TRIGONOMETRIC SUBSTITUTIONS Assignment This assignment covers the material in Chapters 7 to 9. The marks for each question are shown in brackets at the end of each question.. etermine: a p t dt b p c C d 9. Evaluate the following integrals, each correct to significant figures: a b / sint dt C C. etermine the following integrals: a 6t C 7 dt ln b c p d 9. Evaluate the following definite integrals: / a sin t C dt b e. etermine the following integrals: a cos sin b p Evaluate the following definite integrals, correct to significant figures: a b / / sin t dt cos sin d c C

45 Integration by parts. Introduction From the product rule of differentiation: d du uv v C u dv, where u and v are both functions of. Rearranging gives: u dv d uv vdu Integrating both sides with respect to gives: u dv d uv v du i.e. or u dv uv v du udv = uv v du This is known as the integration by parts formula and provides a method of integrating such products of simple functions as e, t sin t dt, e cos d and ln. Given a product of two terms to integrate the initial choice is: which part to make equal to u and which part to make equal to dv. The choice must be such that the u part becomes a constant after successive differentiation and the dv part can be integrated from standard integrals. Invariable, the following rule holds: If a product to be integrated contains an algebraic term such as, t or then this term is chosen as the u part. The one eception to this rule is when a ln term is involved; in this case ln is chosen as the u part.. Worked problems on integration by parts Problem. etermine cos From the integration by parts formula, u dv uv v du Let u, from which du, i.e. du and let dv cos, from which v cos sin. Epressions for u, du and v are now substituted into the by parts formula as shown below. u dv = u v v du cos = sin sin i.e. cos sin cos C c sin Y cos Y c This result may be checked by differentiating the right hand side, d i.e. sin C cos C c cos C sin sin C using the product rule cos, which is the function being integrated Problem. Find: te t dt Let u t, from which, du, i.e. du dt and dt let dv e t dt, from which, v e t dt et Substituting into u dv uv v du gives: te t dt t et et dt e t dt tet tet e t C c Hence t e t dt = et t Y c, which may be checked by differentiating. Problem. Evaluate sin d du Let u, from which,, i.e. du d d and let dv sin d, from which, v sin d cos Substituting into u dv uv v du gives: sin d cos cos d cos C cos d Hence cos C sin C c sin d cos C sin cos C sin C sin C C since cos and sin Problem. Evaluate: to significant figures e, correct Let u, from which du, i.e. du and let dv e, from which, v e e Substituting into u dv uv v du gives: e e e e e e e C c e C c Hence e e e Problem. INTEGRATION BY PARTS e 6 e 6.86 C..99., correct to significant figures. etermine: sin Let u, from which, du, i.e.du, and let dv sin, from which, v sin cos Substituting into u dv uv v du gives: sin cos cos cos C cos The integral, cos, is not a standard integral and it can only be determined by using the integration by parts formula again. From Problem, cos sin C cos Hence sin cos C f sin C cos gcc cos C sin C cos C c. / cos Y sin Y c In general, if the algebraic term of a product is of power n, then the integration by parts formula is applied n times. Now try the following eercise Eercise 78 Further problems on integration by parts etermine the integrals in Problems to using integration by parts.

46 6 ENGINEERING MATHEMATICS..... e e e C c e C C c sin cos C sin C c cos d sin C cos C c t e t dt et t t C C c Evaluate the integrals in Problems 6 to 9, correct to significant figures e 6.78 sin. t cos t dt.67 e.78. Further worked problems on integration by parts Problem 6. Find: ln The logarithmic function is chosen as the u part Thus when u ln, then du,i.e.du Letting dv gives v Substituting into u dv uv v du gives: ln ln ln ln C c Hence or Problem 7. ln = ln Y c.ln / Y c etermine: ln ln is the same as ln Let u ln, from which, du,i.e.du and let dv, from which, v Substituting into u dv uv v du gives: ln ln ln ln C c Hence ln =.ln / Y c 9 p Problem 8. Evaluate: ln, correct to significant figures Let u ln, from which du and let dv p, from which, v Substituting into u dv uv v du gives: p ln ln p ln p ln C c ln p C c Hence 9 p ln p ln 9

47 Areas under and between curves. Area under a curve The area shown shaded in Fig.. may be determined using approimate methods such as the trapezoidal rule, the mid-ordinate rule or Simpson s rule or, more precisely, by using integration. y o Figure. = a y d y = f = b i Let A be the area shown shaded in Fig.. and let this area be divided into a number of strips each of width υ. One such strip is shown and let the area of this strip be υa. Then: υa ³ yυ The accuracy of statement increases when the width of each strip is reduced, i.e. area A is divided into a greater number of strips. ii Area A is equal to the sum of all the strips from a to b, iii i.e. b A limit yυ υ! a From statement, υa υ ³ y In the limit, as υ approaches zero, becomes the differential coefficient da. υa υ iv υa Hence limit da υ! υ y, from statement. By integration, da y i.e. A y The ordinates a and b limit the area and such ordinate values are shown as limits. Hence A b a y Equating statements and gives: =b b Area A = limit y = y =a a b = f./ a v If the area between a curve f y, the y-ais and ordinates y p and y q is required, then area q p Thus, determining the area under a curve by integration merely involves evaluating a definite integral. There are several instances in engineering and science where the area beneath a curve needs to be accurately determined. For eample, the areas between limits of a: velocity/time graph gives distance travelled, force/distance graph gives work done, voltage/current graph gives power, and so on. Should a curve drop below the -ais, then y f becomes negative and f is negative. When determining such areas by integration, a negative sign is placed before the integral. For the curve shown in Fig.., the total shaded area is given by area E C area F C area G. y a Figure. E b By integration, total shaded area b a f F c b c f C G d d c y = f f Note that this is not the same as d f. It it usually necessary to sketch a curve in order to check whether it crosses the -ais.. Worked problems on the area under a curve Problem. etermine the area enclosed by y C, the -ais and ordinates and y C is a straight line graph as shown in Fig.., where the required area is shown shaded. y 8 6 y = + Figure. By integration, shaded area AREAS UNER AN BETWEEN CURVES 9 y C C 6 C C square units This answer may be checked since the shaded area is a trapezium. Area of trapezium sum of parallel sides C square units perpendicular distance between parallel sides Problem. The velocity v of a bo t seconds after a certain instant is: t C m/s. Find by integration how far it moves in the interval from t tots Since t C is a quadratic epression, the curve v t C is a parabola cutting the v-ais at v, as shown in Fig... The distance travelled is given by the area under the v/t curve shown shaded in Fig... By integration, shaded area i.e. v dt t C dt t C t C distance travelled = 6.67 m

48 ENGINEERING MATHEMATICS AREAS UNER AN BETWEEN CURVES v m/s v = t + t s Figure. Problem. Sketch the graph y C 6 between and and determine the area enclosed by the curve and the -ais y = + 6 y y 6 8 Shaded area y y, the minus sign before the second integral being necessary since the enclosed area is below the -ais. Hence shaded area C 6 C 6 C 6 6 { C 6 C } { } { } { 8 8 } C 8 { C 6 } { } C 6 { } { } y Figure y y = + The curve y C is shown plotted in Fig..6. a By the trapezoidal rule Area = width of interval sum of first Y last Y remaining ordinate ordinates Selecting 6 intervals each of width. gives: Area. 7 C C.7 C 6 C.7 C C square units c d By Simpson s rule, area width of first + last interval ordinates sum of even C ordinates sum of remaining C odd ordinates Selecting 6 intervals, each of width., gives: area. 7 C C.7 C.7 C.7 C 6 C 7 square units By integration, shaded area y C C 7 square units Integration gives the precise value for the area under a curve. In this case Simpson s rule is seen to be the most accurate of the three approimate methods. Problem. Find the area enclosed by the curve y sin, the-ais and the ordinates an/ Asketchofy sin is shown in Fig..7. y y =sin Figure. A table of values is produced and the graph sketched as shown in Fig.. where the area enclosed by the curve and the -ais is shown shaded. or.8 square units Problem. etermine the area enclosed by the curve y C, the -ais and ordinates anbyathe trapezoidal rule, b the mid-ordinate rule, c Simpson s rule, and d integration b By the mid-ordinate rule, area width of intervalsum of mid-ordinates. Selecting 6 intervals, each of width. gives the midordinates as shown by the broken lines in Fig..6. Thus, area. 8. C C 9 C 6. C. C 6 7. square units p/ p/ p Figure.7 Note that y sin has a period of,i.e. radians.

49 ENGINEERING MATHEMATICS Shaded area / / y sin / cos { cos } { cos } { } { } C square units Now try the following eercise Eercise 8 Further problems on area under curves Unless otherwise stated all answers are in square units.. Show by integration that the area of the triangle formed by the line y, the ordinates anandthe-ais is 6 square units.. Sketch the curve y C between an. etermine by integration the area enclosed by the curve, the -ais and ordinates an. Use an approimate method to find the area and compare your result with that obtained by integration. In Problems to 8, find the area enclosed between the given curves, the horizontal ais and the given ordinates.. y ;, 7.. y C ;, 7.. y sin;, 6. t C e t ; t, t y cost; t, t y ;,.67. Further worked problems on the area under a curve Problem 6. A gas epands according to the law pv constant. When the volume is m the pressure is kpa. Given that work done v p dv, determine the work v done as the gas epands from m to a volume of 6 m pv constant. When v m and p kpa the constant is given by ð kpa m or kj. Hence pv, or p v 6 Work done v dv ln v 6 ln 6 ln ln 6 ln 9. kj Problem 7. etermine the area enclosed by the curve y cos,the-ais and ordinates and The curve y cos / is shown in Fig..8. y y = cos q p/ p p p p Figure.8 Note that y cos has a maimum value of and period / /, i.e. rads. / / Shaded area yd cos d sin / 8sin 8sin.67 square units Problem 8. etermine the area bounded by the curve y e t/,thet-ais and ordinates t andt, correct to significant figures A table of values is produced as shown. t y e t/ Since all the values of y are positive the area required is wholly above the t-ais. Hence area y dt e t/ dt e t/ e t/ e e / square units Problem 9. Sketch the curve y C between an. Find the area enclosed by the curve, the -ais and the ordinates an. etermine also, by integration, the area enclosed by the curve and the y-ais, between the same limits A table of values is produced and the curve y C plotted as shown in Fig..9. y Shaded area y C C square units When, y C, and when, y. y A C AREAS UNER AN BETWEEN CURVES B y = + P Q Figure.9 Since y C then y and p y The area enclosed by the curve y C i.e. p y, they-ais and the ordinates y and y i.e. area ABC of Fig..9 is given by: y Area y y y / Let u y, then du and du Hence y / u / du u/ for algebraic substitutions, see Chapter 8 Since u y then y y / p 9 8 square units Check: From Fig..9, area BCPQ C area ABC C 8 square units, which is the area of rectangle ABQP. Problem. etermine the area between the curve y 8 and the -ais y 8 8 C When y, then or C or, i.e. when y, or or,which means that the curve crosses the -ais at, and

50 ENGINEERING MATHEMATICS AREAS UNER AN BETWEEN CURVES. Since the curve is a continuous function, only one other co-ordinate value needs to be calculated before a sketch of the curve can be produced. When, y 9, showing that the part of the curve between an is negative. A sketch of y 8 is shown in Fig... Another method of sketching Fig.. would have been to draw up a table of values. Figure. y Shaded area square units Now try the following eercise 8 Eercise 8 Further problems on areas under curves In Problems and, find the area enclosed between the given curves, the horizontal ais and the given ordinates.. y ;, 6 square units. y ;,. square units. The force F newtons acting on a bo at a distance metres from a fied point is given by: F C. If work done F, determine the work done when the bo moves from the position where m to that when m. 9. Nm. Find the area between the curve y and the -ais..67 square units. Sketch the curves y C and y 7 and determine the area enclosed by them..8 square units 6. etermine the area enclosed by the curves y sin and y cos and the y- ais.. square units 7. The velocity v of a vehicle t seconds after a certain instant is given by: v t C m/s. etermine how far it moves in the interval from t sto t s. m. The area between curves The area enclosed between curves y f and y f shown shaded in Fig.. is given by: y shaded area b a b a f b f./ f./ = a = b Figure. a y = f y = f f Problem. etermine the area enclosed between the curves y C an 7 At the points of intersection, the curves are equal. Thus, equating the y-values of each curve gives: C 7, from which C 6. Factorizing gives C, from which, an. By firstly determining the points of intersection the range of -values has been found. Tables of values are produced as shown below. y C y 7 7 A sketch of the two curves is shown in Fig... Figure. Shaded area y 7 6 y = + y = 7 C 7 C square units C 9 a b c Problem. a etermine the coordinates of the points of intersection of the curves y and y 8. b Sketch the curves y and y 8 on the same aes. c Calculate the area enclosed by the two curves At the points of intersection the coordinates of the curves are equal. When y then y. Hence at the points of intersection 8, by equating the y values. Thus 8, from which 8, i.e. or 8. Hence at the points of intersection or. When, y and when, y. Hence the points of intersection of the curves y = and y = 8 are, and, A sketch of y and y 8 is shown in Fig.. Figure. Shaded area f p 8 g f p 8 / g p 8 / {p p } fg square units

51 6 ENGINEERING MATHEMATICS Problem. etermine by integration the area bounded by the three straight lines y, y and y Each of the straight lines is shown sketched in Fig... Figure. Shaded area C 6 6 C 6 C C 6 square units Now try the following eercise Eercise 8 Further problems on areas between curves. etermine the coordinates of the points of intersection and the area enclosed between the parabolas y and y., and,, sq. units. etermine the area enclosed by the curve y C, the -ais and the ordinates an. Find also the area enclosed by the curve and the y-ais between the same limits. sq. units, 9 sq. units. Calculate the area enclosed between y and the -ais using an approimate method and compare your result with the true area obtained by integration. 7.8 sq. units. A gas epands according to the law pv constant. When the volume is m the pressure is kpa. Find the work done as the gas epands from m to a volume of m given that work done v p dv 69. kj. etermine the area enclosed by the three straight lines y, y and y C. sq. units v Mean and root mean square values. Mean or average values i The mean or average value of the curve shown in Fig.., between a and b, is given by: Figure. mean or average value, area under curve y length of base ii When the area under a curve may be obtained by integration then: mean or average value, b y a y b a Mean value, y y Problem. A sinusoidal voltage is given by v sin ωt volts. etermine the mean value of the voltage over half a cycle using integration Half a cycle means the limits are to radians. Mean value, v v d ωt sin ωt d ωt cos ωt cos cos C 6.66 volts iii i.e. y = b f./ b a a For a periodic function, such as a sine wave, the mean value is assumed to be the mean value over half a cycle, since the mean value over a complete cycle is zero. Note that for a sine wave, mean value ð maimum value In this case, mean value ð 6.66 V Problem. etermine, using integration, the mean value of y between and Problem. Calculate the mean value of y C in the range toby a the mid-ordinate rule and b integration

52 8 ENGINEERING MATHEMATICS a A graph of y over the required range is shown in Fig.. using the following table: X Y y Figure. y = + Using the mid-ordinate rule, mean value area under curve length of base sum of mid-ordinates number of mid-ordinates Selecting 6 intervals, each of width., the mid-ordinates are erected as shown by the broken lines in Fig.... C.7 C 6.7 C. Mean value C7. C b By integration, mean value y C C f 7 C 6 g The answer obtained by integration is eact; greater accuracy may be obtained by the midordinate rule if a larger number of intervals are selected. Problem. The number of atoms, N, remaining in a mass of material during radioactive decay after time t seconds is given by: N N e t,wheren and are constants. etermine the mean number of atoms in the mass of material for the time period t andt Mean number of atoms / N / / N dt N e t dt e e t t / dt N N e / e N e e CN e e N e.6 N Now try the following eercise Eercise 86 Further problems on mean or average values. etermine the mean value of a y p from to b y sin from to c y e t from t to t a b or.67 c Calculate the mean value of y C in the range to byathe mid-ordinate rule, and b integration. 9. The speed v of a vehicle is given by: v t C m/s, where t is the time in seconds. etermine the average value of the speed from t tots. 9m/s. Find the mean value of the curve y 6C which lies above the -ais by using an approimate method. Check the result using integration..7. The vertical height h km of a missile varies with the horizontal distance d km, and is given by h d d. etermine the mean height of the missile from d to d km..67 km 6. The velocity v of a piston moving with simple harmonic motion at any time t is given by: v c sin ωt, where c is a constant. etermine the mean velocity between t andt c ω.. Root mean square values The root mean square value of a quantity is the square root of the mean value of the squared values of the quantity taken over an interval. With reference to Fig.., the r.m.s. value of y f over the range a to b is given by: b r.m.s. value y b a a One of the principal applications of r.m.s. values is with alternating currents and voltages. The r.m.s. value of an alternating current is defined as that current which will give the same heating effect as the equivalent direct current. Problem. etermine the r.m.s. value of y between an R.m.s. value y MEAN AN ROOT MEAN SQUARE VALUES 9 p Problem 6. A sinusoidal voltage has a maimum value of V. Calculate its r.m.s. value A sinusoidal voltage v having a maimum value of V may be written as: v sin. Over the range to, r.m.s. value v d sin d sin d which is not a standard integral. It is shown in Chapter 6 that cos A sin A and this formula is used whenever sin A needs to be integrated. Rearranging cos A sin A gives sin A cos A Hence sin d cos d p 7.7 volts sin sin sin

53 6 ENGINEERING MATHEMATICS Note that for a sine wave, r.m.s. value p ð maimum value. In this case, r.m.s. value p ð 7.7 V Problem 7. In a frequency distribution the average distance from the mean, y, is related to the variable,, by the equation y. etermine, correct to significant figures, the r.m.s. deviation from the mean for values of from toc R.m.s. deviation y C C C C p , correct to significant figures. Now try the following eercise Eercise 87 Further problems on root mean square values. etermine the r.m.s. values of: a y from to b y t from t tot c y sin from to a 6.98 b.99 c p or Calculate the r.m.s. values of: a y sin from to b y C sin t from t tot c y cos from to Note that cos t C cos t, from Chapter 6. a p or.77 b. c.. The distance, p, of points from the mean value of a frequency distribution are related to the variable, q, by the equation p C q. etermine the standard q deviation i.e. the r.m.s. value, correct to significant figures, for values from q toq..8. A current, i sin t amperes is applied across an electric circuit. etermine its mean and r.m.s. values, each correct to significant figures, over the range t tot ms. 9. A,. A. A sinusoidal voltage has a peak value of V. Calculate its mean and r.m.s. values, correct to significant figures. 6 V, V 6. etermine the form factor, correct to significant figures, of a sinusoidal voltage of maimum value volts, given that r.m.s. value form factor. average value

54 Laplace transforms 6 Introduction to Laplace transforms 6. Introduction The solution of most electrical circuit problems can be reduced ultimately to the solution of differential equations. The use of Laplace transforms provides an alternative method to those discussed in Chapters 6 to for solving linear differential equations. 6. efinition of a Laplace transform The Laplace transform of the function f t is defined by the integral e st f tdt, where s is a parameter assumed to be a real number. Common notations used for the Laplace transform There are various commonly used notations for the Laplace transform of f t and these include: i L{ f t} or L{ f t} ii L f orlf iii f s orf s Also, the letter p is sometimes used instead of s as the parameter. The notation adopted in this book will be f t for the original function and L{ f t} for its Laplace transform. Hence, from above: L{ f t}= e st f tdt 6. Linearity property of the Laplace transform From equation, L{kf t} = e st kftdt = k e st f tdt i.e L{kft}=kL{ f t} where k is any constant. Similarly, L{aft + bgt} = e st aft + bgt dt = a e st f tdt + b e st gtdt i.e. L{aft + bgt} =al{ f t}+bl{gt}, where a and b are any real constants. The Laplace transform is termed a linear operator because of the properties shown in equations and. 6. Laplace transforms of elementary functions Using the definition of the Laplace transform in equation a number of elementary functions may be transformed. For eample: a f t =. From equation, e L{} = e st st dt = s = s e s e = s = provided s > s b f t = k. From equation, L{k} =kl{} Hence L{k} =k = k, from a above. s s c f t = e at where a is a real constant =. K 68 LAPLACE TRANSFORMS From equation, L{e at }= e st e at dt = e s at dt, from the laws of indices, e s at = = s a s a = s a provided s a >, i.e. s > a d f t = cos at where a is a real constant. From equation, L{cos at} = e st cos at dt e st = s a sin at s cos at + a by integration by parts twice see page, e s = s a sin a s cos a + a e s a sin s cos + a s = s provided s > + a e f t = t. From equation, te L{t} = e st st e st t dt = s s dt te st = s e st s by integration by parts, e s = e s s s e s = s since =, = provided s > s f f t = t n where n =,,,,. By a similar method to e it may be shown that L{t }= s and L{t }= s =! s. These results can be etended to n being any positive integer. Thus L{t n }= n! provided s > sn+ g f t = sinh at. From Chapter, sinh at = eat e at. Hence, { L{sinh at} =L eat } e at = L{eat } L{e at } = from equations and, s a s + a from c above, = s a s + a a = s provided s > a a A list of elementary standard Laplace transforms are summarized in Table 6.. Table 6. Elementary standard Laplace transforms Function Laplace transforms f t L{f t}= e st f tdt i s ii k k s iii e at s a iv sin at a s + a v cos at s s + a vi t s vii t! s viii t n n =,,,... i cosh at sinh at n! s n+ s s a a s a

55 6. Worked problems on standard Laplace transforms Problem. Using a standard list of Laplace transforms determine the following: a L { + t } t b L{e t e t }. a L { + t } t = L{}+L{t} L{t }, from equations and = s + s! s +, from i, vi and viii of Table 6. = s + s... s = s + s 8 s b L{e t e t }=Le t L{e t }, from equations and =, s s from iii of Table 6. = s s + = = s + s s s + s + s s Problem. Find the Laplace transforms of: a 6 sin t cos t b cosh θ sinh θ. a L{6 sin t cos t} = 6L{sin t} L{cos t} INTROUCTION TO LAPLACE TRANSFORMS 69 = 6 s + s s +, from iv and v of Table 6. = 8 s + 9 s s + b L{ cosh θ sinh θ} = L{cosh θ} L{sinh θ} s = s s from i and of Table 6. = s s s 9 Problem. Prove that a L{sin at}= a s + a b L{t }= s s c L{cosh at}= s a. a From equation, L{sin at} = e st sin at dt e st = s s sin at a cos at + a by integration by parts, = s + a e s s sin a a cos a e s sin a cos = s a + a a = s provided s > + a b From equation, L{t }= e st t dt t e st = te st s s e st s by integration by parts twice, K 6 LAPLACE TRANSFORMS = s = provided s > s c From equation, { } L{cosh at} =L eat + e at, from Chapter = L{eat }+ L{e at }, = equations and + s a s a from iii of Table 6. = s a + s + a = s + a + s a s as + a s = s provided s > a a Problem. etermine the Laplace transforms of: a sin t b cosh. a Since cos t = sin t then sin t = cost. Hence, { } L{sin t}=l cos t = L{} L{cos t} = s s s + from i and v of Table 6. = s + s ss = + ss + = ss + b Since cosh = cosh then cosh = + cosh from Chapter. Hence cosh = + cosh 6 { } Thus L{cosh } =L + cosh 6 = L{}+ L{cosh 6} = + s s s 6 = s 6 ss 6 = s 8 ss 6 Problem. Find the Laplace transform of sin ωt + α, where ω and α are constants. Using the compound angle formula for sina + B, from Chapter 8, sinωt + α may be epanded to sin ωt cos α + cos ωt sin α. Hence, L{sin ωt + α} = L{sin ωt cos α + cos ωt sin α} = cos αl{sin ωt}+sin αl{cos ωt}, = cos α ω s + ω + sin α since α is a constant s s + ω from iv and v of Table 6. = s + ω ω cos α + s sin α Now try the following eercise. Eercise Further problems on an introduction to Laplace transforms etermine the Laplace transforms in Problems to9.. a t b t + t a s b s s + s s. a t t t + b t + t a s s + s b 8 s 6 8 s + s

56 Laplace transforms. a e t b e t a. a sin t b cos t a. a 7 cosh b sinh t a s b s + s + 9 b s s + 7s s b s 9 6. a cos t b sin a s + ss b + ss + 6 INTROUCTION TO LAPLACE TRANSFORMS 6 7. a cosh t b sinh θ s a ss b 6 ss 6 8. sinat + b, where a and b are constants s a cos b + s sin b + a 9. cosωt α, where ω and α are constants s s cos α + ω sin α + ω. Show that Lcos t sin t = s s + 6 K 6 Properties of Laplace transforms 6. The Laplace transform of e at f t From Chapter 6, the definition of the Laplace transform of f t is: L{f t} = e st f tdt Thus L{e at f t} = e st e at f t dt = e s a f tdt where a is a real constant Hence the substitution of s a for s in the transform shown in equation corresponds to the multiplication of the original function f t bye at. This is known as a shift theorem. 6. Laplace transforms of the form e at f t From equation, Laplace transforms of the form e at f t may be deduced. For eample: i L{e at t n } Since L{t n }= n! from viii of Table 6., sn+ page 68. then L{e at t n n! }= from equation n+ s a above provided s > a. ii L{e at sin ωt} ω Since L{sin ωt} = s + ω from iv of Table 6., page 68. then L{e at ω sin ωt}= s a from equation provided s > + ω a. iii L{e at cosh ωt} s Since L{cosh ωt} = s ω from i of Table 6., page 68. then L{e at s a cosh ωt}= s a from equation provided s > a. ω A summary of Laplace transforms of the form e at f t is shown in Table 6.. Table 6. Laplace transforms of the form e at f t Function e at f t Laplace transform a is a real constant L{e at f t} i e at t n n! s a n+ ii e at sin ωt ω s a + ω iii e at cos ωt s a s a + ω iv e at sinh ωt ω s a ω v e at cosh ωt s a s a ω Problem. etermine a L{t e t } b L{e t cos t}. a From i of Table 6., L{t e t }=L{t e t! }= s + = s = 8 s b From iii of Table 6., L{e t cos t} =L{e t cos t} s = s +

57 PROPERTIES OF LAPLACE TRANSFORMS 6 6 LAPLACE TRANSFORMS s = s 6s s = s 6s + Problem. etermine a L{e t sin t} b L{e θ cosh θ}. a From ii of Table 6., L{e t sin t}= s + = s+ + 9 = s + s = s + s + b From v of Table 6., L{e θ cosh θ}= L{e θ s cosh θ}= s = s s s = s+ 6 s s Problem. etermine the Laplace transforms of a e t sinh t b e t cos t sin t. a From iv of Table 6., L{e t sinh t}= L{e t sinh t} = s = s + = s + 6s+9 = s + 6s + b L{e t cos t sin t} = 8L{e t cos t} L{e t sin t} = = 8s s + s + from iii and ii of Table 6. 8s 8s s + = s 6s + Problem. Show that { } L e sin = 8 s + s + s + 7. Since cos = sin, sin = cos. Hence, { } L e sin { } = L e cos = { } L e } {e L cos = s s s + from iii of Table 6. page 68 and iii of Table 6. above, s + = s + s + + = s + 6s + s + s + + = s + 6s + s + s + 7 = s + s + 7 6s + s + s + s + s + 7 = s + s + s 6s 6s s + s + s + 7 = 8 s + s + s + 7 K Now try the following eercise. Eercise Further problems on Laplace transforms of the form e at f t etermine the Laplace transforms of the following functions:. a te t b t e t a. a t e t b t e t a. a e t cos t b e t sin t s a s s + s b s + b b. a e t cos t b e t sin t a b s + s + s +. a e t sin t b et cos t s s + 6 s s + 8 s + s + 6 a s s s s + b s + s s 6s + 6. a e t sinh t b e t cosh t a ss b 7. a e t sinh t b e t cosh t 6 a s + s 8 8. a e t cos t sin t b b e t sinh t cosh t s a s b s + s s s s + s + 6s + 6s + ss + 6. The Laplace transforms of derivatives a First derivative Let the first derivative of f t bef t then, from equation, L{f t} = e st f tdt From Chapter, when integrating by parts u dv dt = uv v du dt dt dt When evaluating e st f tdt, let u = e st and dv dt = f t from which, du dt = se st and v = f tdt = f t Hence e st f tdt = e st f t f t se st dt = f + s e st f tdt = f + sl{f t} assuming e st f t ast, and f is the value of f tatt =. Hence, or L{ f t}=sl{ f t} f { } L = sl{ y} y where y is the value of y at =. b Second derivative Let the second derivative of f t bef t, then from equation, L{ f t} = e st f tdt Integrating by parts gives: e st f tdt = e st f t + s e st f tdt = f + sl{ f t}

58 assuming e st f t ast, and f is the value of f t att =. Hence { f t}= f + ss f t f, from equation, i.e. or L{ f t} = s L{ f t} sf f { d } y L = s L{ y} sy y where y is the value of at =. Equations and are important and are used in the solution of differential equations see Chapter 67 and simultaneous differential equations Chapter 68. Problem. Use the Laplace transform of the first derivative to derive: a L{k} = k s c L{e at }= s + a b L{t} = s From equation, L{ f t}=sl{ft} f. a Let f t = k, then f t = and f = k. Substituting into equation gives: L{}=sL{k} k i.e. k = sl{k} Hence L{k} = k s b Let f t = t then f t = and f =. Substituting into equation gives: L{} =sl{t} i.e. s = sl{t} Hence L{t}= s c Let f t = e at then f t = ae at and f =. Substituting into equation gives: L{ ae at }=sl{e at } PROPERTIES OF LAPLACE TRANSFORMS 6 al{e at }=sl{e at } = sl{e at }+al{e at } = s + al{e at } Hence L{e at }= s + a Problem 6. Use the Laplace transform of the second derivative to derive L{cos at} = From equation, s s + a L{ f t} =s L{ f t} sf f Let f t = cos at, then f t = asin at and f t = a cos at, f = and f = Substituting into equation gives: L{ a cos at}=s {cos at} s i.e. a L{cos at}=s L{cos at} s Hence s = s + a L{cos at} from which, L{cos at}= s s + a Now try the following eercise. Eercise Further problems on the Laplace transforms of derivatives. erive the Laplace transform of the first derivative from the definition of a Laplace transform. Hence derive the transform L{} = s. Use the Laplace transform of the first derivative to derive the transforms: a L{e at }= b L{t }= 6 s a s. erive the Laplace transform of the second derivative from the definition of a Laplace transform. Hence derive the transform a L{sin at} = s + a K 66 LAPLACE TRANSFORMS. Use the Laplace transform of the second derivative to derive the transforms: a a L{sinh at} = s a s b L{cosh at} = s a 6. The initial and final value theorems There are several Laplace transform theorems used to simplify and interpret the solution of certain problems. Two such theorems are the initial value theorem and the final value theorem. a The initial value theorem states: limit t f t = limit sl{ f t} s For eample, if f t = e t then L{e t }= s from iii of Table 6., page 68. By the initial value theorem, limit t et = limit s s s i.e. e = i.e. =, which illustrates the theorem. Problem 7. Verify the initial value theorem for the voltage function + cos t volts, and state its initial value. Let f t = + cos t L{f t} =L{ + cos t} = s + s s + 9 from ii and v of Table 6., page 68. By the initial value theorem, limit t f t = limit sl{ f t} s i.e. limit + cos t = limit s t s = limit s s + s s s s + 9 i.e. + = = + i.e. 7 = 7, which verifies the theorem in this case. The initial value of the voltage is thus 7V. Problem 8. Verify the initial value theorem for the function t and state its initial value. Let f t = t = t t + 9 Let L{ f t}=lt t + 9 = s s + 9 s from vii, vi and ii of Table 6., page 68. By the initial value theorem, 8 limit t t = limit s s s s + 9 s 8 = limit s s s + 9 i.e. = i.e. 9 = 9, which verifies the theorem in this case. The initial value of the given function is thus 9. b The final value theorem states: limit f t = limit sl{ f t} t s For eample, if f t = e t then: limit t e t = limit s s s + i.e. e = + i.e. =, which illustrates the theorem.

59 Laplace transforms Problem 9. Verify the final value theorem for the function + e t sin t cm, which represents the displacement of a particle. State its final stea value. Let f t = + e t sin t L{ f t}=l{ + e t sin t} = s + s + = s + s from ii of Table 6., page 68 and ii of Table 6. on page 6. By the final value theorem, limit f t = limit sl{ f t} t s i.e. limit + t e t sin t s = limit s = limit s s + s s + s i.e. + = + i.e. =, which verifies the theorem in this case. The final value of the displacement is thus cm. PROPERTIES OF LAPLACE TRANSFORMS 67 The initial and final value theorems are used in pulse circuit applications where the response of the circuit for small periods of time, or the behaviour immediately after the switch is closed, are of interest. The final value theorem is particularly useful in investigating the stability of systems such as in automatic aircraft-landing systems and is concerned with the stea state response for large values of time t, i.e. after all transient effects have died away. Now try the following eercise. Eercise Further problems on initial and final value theorems. State the initial value theorem. Verify the theorem for the functions a sin t b t and state their initial values. a b 6. Verify the initial value theorem for the voltage functions: a + cos t b t cos t and state their initial values. a 6 b. State the final value theorem and state a practical application where it is of use. Verify the theorem for the function + e t sin t + cos t representing a displacement and state its final value.. Verify the final value theorem for the function t e t and determine its stea state value. K 66 Inverse Laplace transforms 66. efinition of the inverse Laplace transform If the Laplace transform of a function f t isfs, i.e. L{ f t}=fs, then f t is called the inverse Laplace transform of Fs and is written as f t = L {Fs}. For eample, since L{}= { } s then L =. s Similarly, since L{sin at}= a s + a then { } L a s + a = sin at, and so on. 66. Inverse Laplace transforms of simple functions Tables of Laplace transforms, such as the tables in Chapters 6 and 6 see pages 68 and 6 may be used to find inverse Laplace transforms. However, for convenience, a summary of inverse Laplace transforms is shown in Table 66.. Problem. Find the following inverse Laplace transforms: { } { } a L s b L + 9 s a From iv of Table 66., { } L a s + a = sin at, { } { } Hence L s = L + 9 s + = { } L s + Table 66. Fs = L{ f t} i ii iii iv v vi vii viii i i ii iii iv v { } b L s Inverse Laplace transforms L {Fs}=f t s k k s e at s a a sin at s + a s cos at s + a t s! t s n! t n s n+ a sinh at s a s cosh at s a n! e at t n s a n+ ω e at sin ωt s a + ω s a e at cos ωt s a + ω ω e at sinh ωt s a ω s a e at cosh ωt s a ω = L = L s s = e t = sin t from iii of Table 66.

60 Problem. Find the following inverse Laplace transforms: { } { } 6 a L s b L s a From vii of Table 66., L { s } = t Hence L { 6 s } = L { s } = t. b From viii of Table 66., if s is to have a power of then n =. { } { }! 6s Thus L s = t i.e. L = t { } Hence L s = { } 6 L s = t. Problem. etermine { } { } 7s s a L s b L + s 6 { } { } 7s a L s = 7L s + s + = 7 cos t, from v of Table 66. { } { } s b L s = L s 6 s = cosh t, from of Table 66. Problem. Find { } { } a L s b L 7 s a From i of Table 66., { } L a s a = sinh at Thus { } { } L s = L 7 s 7 INVERSE LAPLACE TRANSFORMS 69 { = } 7 L 7 s 7 = 7 sinh 7t b From i of Table 66., { L n! s a n+ Thus { L } = e at t n } = n! eat t n s a n+ { and comparing with L s n = and a =. Hence { } { L = L s Problem. etermine { } a L s s + { } s + b L s + s + =! et t s } shows that } = et t { } { } a L s = L s + s + = e t sin t, from ii of Table 66. { } { } s + s + b L s = L + s + s + + Problem 6. etermine { } a L s + s { } s b L s s = e t cos t, from iii of Table 66. K 6 LAPLACE TRANSFORMS { } { } a L s = L + s s + = L s + = e t sinh t, from iv of Table 66. { } { } s b L s = L s s s { } s + = L s { } s = L s { } + L s = e t cosh t + L s from v of Table 66. = e t cosh t + et sinh t, from iv of Table 66. Now try the following eercise. Eercise Further problems on inverse Laplace transforms of simple functions etermine the inverse Laplace transforms of the following:. a 7 b a 7 b e t s s. a. a s + s + b s s + a e t b cos t b s + 9 a sin t b sin t. a s s + 8 b 6 s a cos t b 6t. a s b 8 s a t b t s 7 6. a b s s 8 6 a 6 cosh t b 7 sinh t 7. a s 7 b s a sinh t b et t 8. a s + b s a 6 e t t b 8 et t s + 9. a s + s + s. a s 6s + b a e t cos t b ae t cos t s + 6s + b e t sin t 7 s 8s + b 7 et sinh t s + s +. a s b + s s 8s + a e t cosh t + e t sinh t b e t cos t + et sin t 66. Inverse Laplace transforms using partial fractions Sometimes the function whose inverse is required is not recognisable as a standard type, such as those listed in Table 66.. In such cases it may be possible, by using partial fractions, to resolve the function into

61 simpler fractions which may be inverted on sight. For eample, the function, Fs = s ss cannot be inverted on sight from Table 66.. However, by using partial fractions, s ss s + which may be inverted as s + e t from i and iii of Table 6.. Partial fractions are discussed in Chapter, and a summary of the forms of partial fractions is given in Table. on page 8. { } s Problem 7. etermine L s s s s s s s s + A s + B s + As+ + Bs s s + Hence s As + + Bs. When s =, = A, from which, A =. When s =, 9 = B, from which, B =. { } s Hence L s s { L s + s + = L { s } } + L { s + = e t + e t, from iii of Table 66. { s Problem 8. Find L + s } + s + s s + s + s + s + s s + A s + B s + + C s + + s + As + + Bs s + + Cs s + + s s s + } Hence INVERSE LAPLACE TRANSFORMS 6 s + s + s + As + + Bs s + + Cs s + + s When s =, 8 = 6A, from which, A =. When s =, =, from which, =. Equating s terms gives: = A + B, from which, B =. Equating constant terms gives: = A B C, i.e. = C 9, from which, C = and C = Hence { s L + s } + s + s s + { L s + } s + s + + s + = e t + e t e t t + e t t, Problem 9. etermine { s L } + 8s s + s + from iii and i of Table 66. s + 8s s + s + A s + + Bs + C s + As + + Bs + Cs + s + s + Hence s + 8s As + + Bs + Cs +. When s =, = A, from which, A =. Equating s terms gives: = A + B, from which, B =, since A =. Equating s terms gives: 8 = B + C, from which, C =, since B =. K 6 LAPLACE TRANSFORMS { s Hence L } + 8s s + s + { L s + + s s + L { s + } } + L { s s + } L { s + = e t + cos t sin t, from iii, v and iv of Table 66. { } Problem. Find L 7s + ss + s + 7s + ss + s + A s + Bs + C s + s + As + s + + Bs + Cs ss + s + Hence 7s + As + s + + Bs + Cs. When s =, = A, from which, A =. Equating s terms gives: = A + B, from which, B =. Equating s terms gives: 7 = A + C, from which, C =. { } Hence L 7s + ss + s + { L s + s + } s + s + { } { } L + L s + s s + + { } { } s + + L + L s s + + { } { } L L s + s s + + { } + L s + + e t cos t + e t sin t from i, iii and ii of Table 66. } Now try the following eercise. Eercise 6 Further problems on inverse Laplace transforms using partial fractions Use partial fractions to find the inverse Laplace transforms of the following functions: s. s e t e t + s.... s 9s s + s s + e t e t + e t s s 9 s + s e t + e t e t t s + 6s + s + e t t t 7s + s + s + s + cos t + sin t + e t + 6s + s s 6. s s s ss + s Poles and zeros + t + sin t cos t e t cos t e t sin t It was seen in the previous section that Laplace transforms, in general, have the form f s = φs. This is θs the same form as most transfer functions for engineering systems, a transfer function being one that relates the response at a given pair of terminals to a source or stimulus at another pair of terminals. Let a function in the s domain be given by: φs f s = where φs is of less s as bs c degree than the denominator.

62 INVERSE LAPLACE TRANSFORMS 6 6 LAPLACE TRANSFORMS Poles: The values a, b, c, that makes the denominator zero, and hence f s infinite, are called the system poles of f s. If there are no repeated factors, the poles are simple poles. If there are repeated factors, the poles are multiple poles. Zeros: Values of s that make the numerator φs zero, and hence f s zero, are called the system zeros of f s. s For eample: has simple poles at s + s s = and s =+, and a zero at s = s + s + s + has a simple pole at s = and double poles at s =, and a zero at s = s + and has simple poles at ss s + s + s =, +,, and and a zero at s = Pole-zero diagram The poles and zeros of a function are values of comple frequency s and can therefore be plotted on the comple frequency or s-plane. The resulting plot is the pole-zero diagram or pole-zero map. Onthe rectangular aes, the real part is labelled the σ-ais and the imaginary part the jω-ais. The location of a pole in the s-plane is denoted by a cross and the location of a zero by a small circle o. This is demonstrated in the following eamples. From the pole-zero diagram it may be determined that the magnitude of the transfer function will be larger when it is closer to the poles and smaller when it is close to the zeros. This is important in understanding what the system does at various frequencies and is crucial in the stu of stability and control theory in general. Problem. etermine for the transfer s + function: Rs = s s + s + s + a the zero and b the poles. Show the poles and zero on a pole-zero diagram. a For the numerator to be zero, s + =. Hence, s = is a zero of Rs. b For the denominator to be zero, s = ors = or s + s + =. Using the quadratic formula. s = ± = ± ± j = = ± j Hence, poles occur at s =, s =, + j and j The pole-zero diagram is shown in Figure 66.. σ Figure 66. jω j j Problem. etermine the poles and zeros for s + s the function: Fs = s + s + s + and plot them on a pole-zero map. For the numerator to be zero, s + = and s =, hence zeros occur at s = and at s =+ Poles occur when the denominator is zero, i.e. when s + =, i.e. s =, and when s + s + =, i.e. s = ± = ± j = ± = + jor j The poles and zeros are shown on the pole-zero map of Fs in Figure 66.. It is seen from these problems that poles and zeros are always real or comple conjugate. K σ Figure 66. Now try the following eercise. Eercise 7 Further problems on poles and zeros. etermine for the transfer function: s + Rs = s s + s 8s + a the zero and b the poles. Show the poles and zeros on a pole-zero diagram. a s = b s =, s =, s = + j, s = j jω j j. etermine the poles and zeros for the function: Fs = s s + s + s and plot s + them on a pole-zero map. poles at s =, s = + j, s = j, zeros at s =+, s = s. For the function Gs = s + s + s + determine the poles and zeros and show them on a pole-zero diagram. poles at s =, s = + j, s = j, zero at s =. Find the poles and zeros for the transfer function: Hs = s s 6 ss and plot the results + in the s-plane. poles at s =, s =+j, s = j, zeros at s =, s = 6

63 Laplace transforms 67 The solution of differential equations using Laplace transforms 67. Introduction An alternative method of solving differential equations to that used in Chapters 6 to is possible by using Laplace transforms. 67. Procedure to solve differential equations by using Laplace transforms i Take the Laplace transform of both sides of the differential equation by applying the formulae for the Laplace transforms of derivatives i.e. equations and of Chapter 6 and, where necessary, using a list of standard Laplace transforms, such as Tables 6. and 6. on pages 68 and 6. ii Put in the given initial conditions, i.e. y and y. iii Rearrange the equation to make L{y} the subject. iv etermine y by using, where necessary, partial fractions, and taking the inverse of each term by using Table 66. on page Worked problems on solving differential equations using Laplace transforms Problem. Use Laplace transforms to solve the differential equation d y + y =, given that when =, y = and = 9. This is the same problem as Problem of Chapter, page 76 and a comparison of methods can be made. Using the above procedure: i L { d y } + L { } L{y}=L{} s L{y} sy y + sl{y} y L{y}=, from equations and of Chapter 6. ii y = and y = 9 Thus s L{y} s 9 + sl{y} L{y}= i.e. s L{y} 8s 8 + sl{y} L{y}= iii Rearranging gives: s + s L{y}=8s + 8 8s + 8 i.e. L{y} = s + s { } 8s + 8 iv y = L s + s 8s + 8 s + s 8s + 8 s s + A s + B s + As + + Bs s s + Hence 8s + 8 = As + + Bs. When s =,= A, from which, A =. When s =, = 7B, from which, B =. K 66 LAPLACE TRANSFORMS { } 8s + 8 Hence y = L s + s { = L s s + { } { } = L s L s + Hence y = 6e e, from iii of Table 66.. Problem. Use Laplace transforms to solve the differential equation: d y y =, given that when =, y = and = 7. This is the same as Problem of Chapter, page 77. Using the above procedure: { d } { } i L + 6L + L{y}=L{} Hence s L{y} sy y + 6sL{y} y + L{y} =, from equations and of Chapter 6. ii y = and y = 7 Thus s L{y} s 7 + 6sL{y} 8 + L{y} = iii Rearranging gives: s + 6s + L{y} =s + i.e. L{y} = { } s + iv y = L s + 6s + { } = L s + s + + { } s = L s + + } s + s + 6s + { } s + = L s + + { } + L 8 s + + = e t cos t + 8e t sin t, from iii and ii of Table 66. Hence y = e t cos t + 8 sin t Problem. Use Laplace transforms to solve the differential equation: d y = 9, given that when =, y = and =. This is the same problem as Problem of Chapter, page 8. Using the procedure: { d } { } y i L L = L{9} Hence s L{y} sy y sl{y} y = 9 s ii y = and y = Hence s L{y} sl{y} = 9 s iii Rearranging gives: s sl{y} = 9 s 9 i.e. L{y} = ss s = 9 s s { } iv y = L 9 s s 9 s s A s + B s + C s Ass + Bs + Cs s s Hence 9 Ass + Bs + Cs. When s =, 9 = B, from which, B =. When s =, 9 = 9C, from which, C =.

64 THE SOLUTION OF IFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS 67 Equating s terms gives: = A + C, from which, A =, since C =. Hence, iv y = { s } + 6s { } { L 9 s = L s s s + } L ss s s + 6s s ss s = + e, from i, vi and iii of Table 66.. A s + B s + C s + s i.e. y = e As s + Bss + Css s + ss Problem. Use Laplace transforms to solve ss s the differential equation: Hence d y 7 + y = e +, given that when s + 6s =, y = and = As s + Bss + Css s + ss When s =, = A, from which, A = 6. Using the procedure: { d } { } When s =, 8 = B, from which, B =. y i L 7L + L{y}=L{ e + } When s =, 6 = 6, from which, =. Hence s L{y} sy y Equating s terms gives: = A + B + C, from 7sL{y} y + L{y}= s + which, C =. s Hence { s } + 6s ii y = and y = L ss s Hence s L{y} = { 6 L s + s } s s 7sL{y}+ = s 6 + e e e + L{y}= ss iii s s 7s + L{y}= ss Thus y = + e e e s ss Problem. The current flowing in an electrical circuit is given by the differential equation = ss Ri + Ldi/dt = E, where E, L and R are constants. Use Laplace transforms to solve the = s + 6s ss equation for current i given that when t =, i =. s + 6s Hence L{y} = ss s 7s + Using the procedure: = s { + 6s i L{Ri}+L L di } = L{E} ss s s dt = s + 6s ss s i.e. RL{i}+LsL{i} i = E s K 68 LAPLACE TRANSFORMS ii i =, hence RL{i}+LsL{i}= E s iii Rearranging gives: R + LsL{i} = E s E i.e. L{i} = sr + Ls { } iv i = L E sr + Ls Hence When E sr + Ls A s + B R + Ls AR + Ls + Bs sr + Ls E = AR + Ls + Bs s =, E = AR, from which, A = E R When s = R L, E = B R L from which, B = EL R { } Hence L E sr + Ls { E/R = L + EL/R } s R + Ls { } E = L Rs EL RR + Ls = L E E R s R R L + s = E R L s s + R L Hence current i = E e Rt L R Now try the following eercise. Eercise 8 Further problems on solving differential equations using Laplace transforms. A first order differential equation involving current i in a series R L circuit is given by: di dt + i = E and i = at time t =. Use Laplace transforms to solve for i when a E = b E = e t and c E = sin t. a i = e t b i = e t e t c i = e t cos t + sin t In Problems to 9, use Laplace transforms to solve the given differential equations.. 9 d y dt dt and y =. + 6y =, given y = y = te t d. + =, given = and dt =. = cos t d i di. + + i =, given dt dt i = and i =. i = t e t d. dt =, given = and dt = 8. = e t e t 6. d y + y = e, given y = and y = y = e + e d y y = cos, giveny = and y =. y = cos + sin + sin

65 Laplace transforms d y 8. + y = cos sin, given y = and y = 6 THE SOLUTION OF IFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS 69 y = e e + sin d y 9. + y = e cos, given y = and y = y = e cos + sin e cos. Solve, using Laplace transforms, Problems to 9 of Eercise 88, page 77 and Problems to of Eercise 89, page 8.. Solve, using Laplace transforms, Problems to 6 of Eercise 9, page 8, Problems and 6 of Eercise 9, page 8, Problems and 7 of Eercise 9, page 87 and Problems and 6 of Eercise 9, page The solution of simultaneous differential equations using Laplace transforms K 68. Introduction It is sometimes necessary to solve simultaneous differential equations. An eample occurs when two electrical circuits are coupled magnetically where the equations relating the two currents i and i are typically: di L + M di + R i = E dt dt di L + M di + R i = dt dt where L represents inductance, R resistance, M mutual inductance and E the p.d. applied to one of the circuits. 68. Procedure to solve simultaneous differential equations using Laplace transforms i Take the Laplace transform of both sides of each simultaneous equation by applying the formulae for the Laplace transforms of derivatives i.e. equations and of Chapter 6, page 6 and using a list of standard Laplace transforms, as in Table 6., page 68 and Table 6., page 6. ii Put in the initial conditions, i.e., y,, y. iii Solve the simultaneous equations for L{y} and L{} by the normal algebraic method. iv etermine y and by using, where necessary, partial fractions, and taking the inverse of each term. 68. Worked problems on solving simultaneous differential equations by using Laplace transforms Problem. Solve the following pair of simultaneous differential equations dt + = dt y + et = given that at t =, = and y =. Using the above procedure: { i L dt { L dt } + L{} =L{} } L{y}+L{e t }= Equation becomes: sl{y} y + L{} = s from equation, page 6 and Table 6., page 68. Equation becomes: sl{} L{y} = s ii = and y = hence Equation becomes: sl{y}+l{} = s

66 THE SOLUTION OF SIMULTANEOUS IFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS 6 6 LAPLACE TRANSFORMS and equation becomes: sl{} L{y} = s or L{y}+sL{} = s iii equation and s equation gives: sl{y}+l{} = s sl{y}+s L{} = s s Adding equations and gives: s + L{} = s s s s ss = ss = s + s ss from which, L{} = s + s ss s + Using partial fractions s + s ss s + A s + B s + Cs + s + As s + + Bss + + Cs + ss = ss s + Hence s + s = As s + + Bss + + Cs + ss When s =, = A hence A = When s =, = B hence B = Equating s coefficients: = A + B + C hence C = since A = and B = Equating s coefficients: = A + C hence = since A = and C = Thus L{} = s + s ss s + = s s + s s + iv Hence { = L s s + s } s + = L { s s + s s + s + i.e. = e t + cos t sin t, from Table 66., page 68 From the second equation given in the question, dt y + et = from which, y = dt + et = d dt et + cos t sin t + e t = e t sin t cos t + e t i.e. y = e t sin t cos t Alternatively, to determine y, return to equations and Problem. Solve the following pair of simultaneous differential equations dt dt + = 6 dt dt y = given that at t =, = 8 and y =. Using the above procedure: i { L dt { L dt } L } L { dt { dt } } + L{} =L{6} } L{y} =L{ } K Equation becomes: sl{} sl{y} y + L{} = 6 s from equation, page 6, and Table 6., page 68. i.e. sl{} sl{y} + y + L{} = 6 s i.e. s + L{} sl{y} + y = 6 s Equation becomes: sl{y} y sl{} L{y} = s from equation, page 6, and Table 6., page 68, i.e. sl{y} y sl{} i.e. s L{y} y sl{} + L{y} = s + = s ii = 8 and y =, hence equation becomes s + L{} 8 sl{y} + = 6 s and equation becomes s L{y} sl{} + 8 = s i.e. s + L{} sl{y} = 6 s + 9 s + L{} sl{y} = 6 s + 9 sl{}+s L{y} = s A iii s equation and s+ equation gives: 6 ss + L{} s L{y} =s s + 9 ss + L{}+s + s L{y} = s + s i.e. ss + L{} s L{y} =6 + 9s ss + L{}+6s + s L{y} = 6s s 7 Adding equations and gives: s + s L{y} = + s s = s + s s from which, L{y} = s s ss + s Using partial fractions s s ss + s A s + B s + + C s = As + s + Bss + Css + ss + s i.e. s s = As + s + Bss + Css + When s =, = A, hence A = When s =, = C, hence C = When s =, = 6B, hence B = Thus L{y} = s s ss + s = s + s + { iv Hence y = L s + } = + e t s +

67 THE SOLUTION OF SIMULTANEOUS IFFERENTIAL EQUATIONS USING LAPLACE TRANSFORMS 6 6 LAPLACE TRANSFORMS Returning to equations A to determine L{} and hence : s equation and s gives: s s + L{} ss L{y} 6 = s s + 9 and i.e. ssl{}+ ss L{y} = s s 6s + s L{} ss L{y} 6 = + 8s 6 s 9 and s L{}+ss L{y} = s 6 Adding equations and 6 gives: s + s L{} = + 8s 6 s = s + 8s 6 s from which, L{} = 8s s 6 ss + s = 8s s 6 ss + s Using partial fractions 8s s 6 ss + s A s + B s + + C s = As + s + Bss + Css + ss + s i.e. 8s s 6 = As + s + Bss + Css + When s =, 6 = A, hence A = When s =, = C, hence C = When s =, = 6B, hence B = Thus L{} = 8s s 6 ss + s = s + s + { Hence = L s + } = + e t s + Therefore the solutions of the given simultaneous differential equations are y = + e t and = + e t These solutions may be checked by substituting the epressions for and y into the original equations. Problem. Solve the following pair of simultaneous differential equations d dt = y d y dt + y = given that at t =, =, y =, dt = and dt =. Using the procedure: i s L{} s L{} =L{y} s L{y} sy y + L{y} = L{} from equation, page 6 ii =, y =, = and y = hence s L{} s L{} =L{y} s L{y}+s + L{y} = L{} iii Rearranging gives: s L{} L{y} =s L{}+s + L{y} = s Equation s + and equation gives: s + s L{} s + L{y} = s + s L{}+s + L{y} = s 6 K Adding equations and 6 gives: s + s + L{} =s + s s i.e. s L{} =s + s = ss + from which, L{} = ss + s = s + s = s s + s = s + s iv Hence = L { s + s } i.e. = + t Returning to equations and to determine y: equation and s equation gives: s L{} L{y} =s 7 s L{}+s s + L{y} = ss 8 Equation 7 equation 8 gives: s s + L{y} = s + ss i.e. s L{y} =s + s and L{y} = s + s s = s s from which, { y = L s } s i.e. y = t Now try the following eercise. Eercise 9 Further problems on solving simultaneous differential equations using Laplace transforms Solve the following pairs of simultaneous differential equations:. dt + dt = et dt dt = given that when t =, = and y = = e t t and y = t + e t. dt y + + dt sin t = dt + y + dt et =. given that at t =, = and y = = cos t + sin t e t e t and y = e t + e t sin t d dt + = y d y dt + y = given that at t =, =, y =, dt = and dt = = cos t + cos t and y = cos t cos t

68 Laplace transforms Assignment 8 This assignment covers the material contained in Chapters 6 to 68. The marks for each question are shown in brackets at the end of each question. { s b L } + s 9 ss s + { c L s } ss + s + This page intentionally left blank. Find the Laplace transforms of the following functions: a t t + b e t sin t c cosh t d t e t e e t cos t f e t sinh t 6. Find the inverse Laplace transforms of the following functions: a b s + s c s s d + 9 s 9 e s s + f s 8s g 8 s s + 7. Use partial fractions to determine the following: a { } s L s s. In a galvanometer the deflection θ satisfies the differential equation: d θ dt + dθ dt + θ = Use Laplace transforms to solve the equation for dθ θ given that when t =, θ = and dt =. Solve the following pair of simultaneous differential equations: = + y dt + = 6y dt given that when t =, = and y =. 6. etermine the poles and zeros for the transfer function: Fs = s + s s + s and plot + s + them on a pole-zero diagram. K

69 Fourier series 69 Fourier series for periodic functions of period π L 68 FOURIER SERIES ii a, a n and b n are called the Fourier coefficients of the series and if these can be determined, the series of equation is called the Fourier series corresponding to f. iii An alternative way of writing the series is by using the a cos + b sin = c sin + α relationship introduced in Chapter 8, i.e. f = a + c sin + α + c sin + α + +c n sinn + α n, 69. Worked problems on Fourier series of periodic functions of period π Problem. Obtain a Fourier series for the periodic function f defined as: { k, when π < < f = +k, when < <π 69. Introduction Fourier series provides a method of analysing periodic functions into their constituent components. Alternating currents and voltages, displacement, velocity and acceleration of slider-crank mechanisms and acoustic waves are typical practical eamples in engineering and science where periodic functions are involved and often requiring analysis. 69. Periodic functions A function f is said to be periodic if f + T = f for all values of, where T is some positive number. T is the interval between two successive repetitions and is called the period of the functions f. For eample, y = sin is periodic in with period π since sin = sin + π = sin + π, and so on. In general, if y = sin ωt then the period of the waveform is π/ω. The function shown in Fig. 69. is also periodic of period π and is defined by: {, when π < < f =, when < <π Figure 69. f π π π π If a graph of a function has no sudden jumps or breaks it is called a continuous function, eamples being the graphs of sine and cosine functions. However, other graphs make finite jumps at a point or points in the interval. The square wave shown in Fig. 69. has finite discontinuities at = π, π, π, and so on. A great advantage of Fourier series over other series is that it can be applied to functions which are discontinuous as well as those which are continuous. 69. Fourier series i The basis of a Fourier series is that all functions of practical significance which are defined in the interval π π can be epressed in terms of a convergent trigonometric series of the form: f = a + a cos + a cos + a cos + +b sin + b sin + b sin + when a, a, a,...b, b,... are real constants, i.e. f = a + a n cos n + b n sin n n= where for the range π to π: and a = π f π π a n = π f cos n π π n =,,,... b n = π f sin n π π n =,,,... where a is a constant, c = a + b,...c n = a n + b n are the amplitudes of the various components, and phase angle α n = arctan a n b n iv For the series of equation : the term a cos + b sin orc sin + α is called the first harmonic or the fundamental, the term a cos + b sin orc sin + α is called the second harmonic, and so on. For an eact representation of a comple wave, an infinite number of terms are, in general, required. In many practical cases, however, it is sufficient to take the first few terms only see Problem. The sum of a Fourier series at a point of discontinuity is given by the arithmetic mean of the two limiting values of f as approaches the point of discontinuity from the two sides. For eample, for the waveform shown in Fig. 69., the sum of the Fourier series at the points of discontinuity i.e. at π, π,...is given by: 8 + = or Figure 69. f 8 π π/ π/ π π/ The function is periodic outside of this range with period π. The square wave function defined is shown in Fig Since f is given by two different epressions in the two halves of the range the integration is performed in two parts, one from π to and the other from to π. Figure 69. f k π π π k From Section 69.i: a = π f π π = π k + π π k = π { k π + kπ }= a is in fact the mean value of the waveform over a complete period of π and this could have been deduced on sight from Fig From Section 69.i: a n = π f cos n π π = { π } k cos n + k cos n π π { k = sin n } k sin n π + = π n π n

70 FOURIER SERIES FOR PERIOIC FUNCTIONS OF PERIO π FOURIER SERIES Hence a, a, a,... are all zero since sin = sin nπ = sin nπ =, and therefore no cosine terms will appear in the Fourier series. From Section 69.i: b n = π f sin n π π = { π } k sin n + k sin n π π { k = cos n } k cos n π + π n π n When n is odd: b n = k { π n + n = k { π n + n n } = k nπ n } Hence b = k π, b = k π, b = k, and so on. π When n is even: b n = k { π n + n n } = n Hence, from equation, the Fourier series for the function shown in Fig. 69. is given by: f = a + a n cos n + b n sin n i.e. f = k π i.e. f = k π n= = + + b n sin n n= k sin + π sin + sin k sin + sin + π + sin + Problem. For the Fourier series of Problem let k = π. Show by plotting the first three partial sums of this Fourier series that as the series is added together term by term the result approimates more and more closely to the function it represents. If k = π in the Fourier series of Problem then: f = sin + sin + sin + sin is termed the first partial sum of the Fourier series of f, sin + sin is termed the second partial sum of the Fourier series, and sin + sin + sin is termed the third partial sum, and so on. Let P = sin, P = sin + sin and P = sin + sin + sin. Graphs of P, P and P, obtained by drawing up tables of values, and adding waveforms, are shown in Figs. 69.a to c and they show that the series is convergent, i.e. continually approimating towards a definite limit as more and more partial sums are taken, and in the limit will have the sum f = π. Even with just three partial sums, the waveform is starting to approach the rectangular wave the Fourier series is representing. Problem. If in the Fourier series of Problem, k =, deduce a series for π at the point = π. If k = in the Fourier series of Problem : f = sin + sin + sin + π When = π, f =, sin = sin π =, sin = sin π =, sin = sin π =, and so on. Hence = π i.e. π = L π π/ f π a f π π/ π π P f π π/ π/ π / sin b f π π P P f π/ π π/ π / sin Figure 69. Problem. c π P f P etermine the Fourier series for the full wave rectified sine wave i = sin θ shown in Fig Figure 69. i θ i = sin / π π π i = sin θ is a periodic function of period π. Thus i = f θ = a + a n cos nθ + b n sin nθ n= In this case it is better to take the range to π instead of π to +π since the waveform is continuous between and π. a = π f θdθ = π sin θ π π dθ = cos θ π π = cos π cos π = = π π a n = π sin θ cos nθ dθ π = π { θ sin π + nθ } θ + sin nθ dθ see Chapter, page = cos θ + n π + n cos θ n n π θ

71 { = cos π + n π + n cos π n n } cos + n cos n When n is both odd and even, { a n = π + n + n } + n n { } = π + n + n { } = π + n + n Hence a = + π = π a = + π = π a = + π 7 = π π b n = π = π π sin θ sin nθ dθ = π { cos θ + n cos = π 7 = 7π and so on } θ n dθ from Chapter = sin θ n π n sin θ + n π + n FOURIER SERIES FOR PERIOIC FUNCTIONS OF PERIO π 66 { = sin π n π n sin π + n + n } sin n sin + n When n is both odd and even, b n = since sin π, sin, sin π, sin π,...are all zero. Hence the Fourier series for the rectified sine wave, i = sin θ is given by: f θ = a + a n cos nθ + b n sin nθ n= i.e. i = f θ = π cos θ cos θ π π cos θ 7π i.e. i = π cos θ cos θ cos θ 7 Now try the following eercise. Eercises Further problems on Fourier series of periodic functions of period π. etermine the Fourier series for the periodic function: {, when π < < f = +, when < <π which is periodic outside this range of period π. f = 8 sin + sin π + sin +. For the Fourier series in Problem, deduce a series for π at the point where = π π = For the waveform shown in Fig determine a the Fourier series for the function L 66 FOURIER SERIES and b the sum of the Fourier series at the points of discontinuity. a f = + cos cos π + cos b π Figure 69.6 π π f. For Problem, draw graphs of the first three partial sums of the Fourier series and show that as the series is added together term by term the result approimates more and more closely to the function it represents.. Find the term representing the third harmonic for the periodic function of period π given by: f = π π π {, when π < <, when < <π sin π 6. etermine the Fourier series for the periodic function of period π defined by:, when π <t < f t =, when < t < π π, when < t <π The function has a period of π cos t cos t f t = + cos t π + sin t + sin 6t + sin t + 7. Show that the Fourier series for the periodic function of period π defined by {, when π<θ< f θ = sin θ, when <θ<π is given by: f θ = π cos θ cos θ cos 6θ 7

72 Fourier series 7 Fourier series for a non-periodic function over range π 7. Epansion of non-periodic functions If a function f is not periodic then it cannot be epanded in a Fourier series for all values of.however, it is possible to determine a Fourier series to represent the function over any range of width π. Given a non-periodic function, a new function may be constructed by taking the values of f in the given range and then repeating them outside of the given range at intervals of π. Since this new function is, by construction, periodic with period π, it may then be epanded in a Fourier series for all values of. For eample, the function f = is not a periodic function. However, if a Fourier series for f = is required then the function is constructed outside of this range so that it is periodic with period π as shown by the broken lines in Fig. 7.. For non-periodic functions, such as f =, the sum of the Fourier series is equal to f at all points in the given range but it is not equal to f at points outside of the range. For determining a Fourier series of a non-periodic function over a range π, eactly the same formulae for the Fourier coefficients are used as in Section 69.i. 7. Worked problems on Fourier series of non-periodic functions over a range of π Problem. etermine the Fourier series to represent the function f = in the range π to +π. The function f = is not periodic. The function is shown in the range π to π in Fig. 7. and is then constructed outside of that range so that it is periodic of period π see broken lines with the resulting saw-tooth waveform. Figure 7. π Figure 7. f π f = π π π π For a Fourier series: f π π f = π π π f = a + a n cos n + b n sin n n= From Section 69.i, a = π f π π = π = π = π π π π a n = π f cos n = π cos n π π π π = sin n sin n π π n n π by parts see Chapter L 66 FOURIER SERIES = sin n cos n π + π n n π = cos nπ cos n π + π n + n b n = π f sin n = π sin n π π π π = cos n cos n π n n = π π by parts = cos n sin n π + π n n π = π cos nπ sin nπ + π n n π cos n π sin n π + n n = π cos nπ π cos nπ = cos nπ π n n n When n is odd, b n = n. Thus b =, b =, b =, and so on. When n is even, b n = n. Thus b =, b =, b 6 =, and so on. 6 Thus f = = sin sin + sin sin + sin sin i.e. = sin sin + sin sin + sin sin for values of f between π and π. For values of f outside the range π to +π the sum of the series is not equal to f. Problem. In the Fourier series of Problem, by letting = π/, deduce a series for π/. When = π/, f = π from Fig. 7.. Thus, from the Fourier series of equation : π = sin π sin π + sin π i.e. sin π + sin π 6 sin 6π + 7 π = + π = Problem. Obtain a Fourier series for the function defined by: {, when < <π f =, when π< < π. The defined function is shown in Fig. 7. between and π. The function is constructed outside of this range so that it is periodic of period π, as shown by the broken line in Fig. 7.. Figure 7. f π f = π π π π π For a Fourier series: f = a + a n cos n + b n sin n n= It is more convenient in this case to take the limits fromtoπ instead of from π to +π. The value of the Fourier coefficients are unaltered by this change of limits. Hence a = π f = π π + π π π π π = π = π = π

73 FOURIER SERIES FOR A NON-PERIOIC FUNCTION OVER RANGE π FOURIER SERIES a n = π f cos n π = π π cos n + π π = sin n cos n π + π n n from Problem, by parts = { π sin nπ cos nπ + π n n + cos } n = cos nπ πn When n is even, a n =. When n is odd, a n = πn. Hence a = π, a = π, a =, and so on π b n = π f sin n π = π π sin n π π = cos n sin n π + π n n from Problem, by parts = { π cos nπ sin nπ + π n n + sin } n = π cos nπ cos nπ = π n n Hence b = cos π =, b =, b =, and so on. Thus the Fourier series is: f = a + a n cos n + b n sin n n= i.e. f = π π cos cos π cos +sin π sin + sin i.e. f = π cos cos cos + π sin sin + sin Problem. For the Fourier series of Problem : a what is the sum of the series at the point of discontinuity i.e. at = π? b what is the amplitude and phase angle of the third harmonic? and c let =, and deduce a series for π /8. a The sum of the Fourier series at the point of discontinuity is given by the arithmetic mean of the two limiting values of f as approaches the point of discontinuity from the two sides. Hence sum of the series at = π is π = π b The third harmonic term of the Fourier series is π cos + sin This may also be written in the form c sin + α, where amplitude, c = + π =. and phase angle, α = tan π =.98 or.9 radians Hence the third harmonic is given by. sin.9 c When =, f = see Fig. 7.. Hence, from the Fourier series: = π cos + cos + cos + + π i.e. π = π L Hence π 8 = Problem. educe the Fourier series for the function f θ = θ in the range to π. f θ = θ is shown in Fig. 7. in the range to π. The function is not periodic but is constructed outside of this range so that it is periodic of period π, as shown by the broken lines. Figure 7. For a Fourier series: fθ π fθ = θ π π π π f = a + a n cos n + b n sin n n= a = π f θdθ = π θ dθ π π = θ π = 8π π π = π a n = π f θ cos nθ dθ π = π θ cos nθ dθ π = θ π sin nθ θ cos nθ sin nθ + π n n n by parts = π cos πn + π n = n cos πn = when n =,,, n Hence a =, a =, a = and so on θ b n = π f θ sin nθ dθ = π θ sin nθ dθ π π = θ π cos nθ θ sin nθ cos nθ + π n n + n by parts = π cos πn cos πn + + π n n + + cos n = π + π n n n = π n Hence b = π, b = π, b = π Thus f θ = θ = π +, and so on. n cos nθ π sin nθ n n= i.e. θ = π cos + θ + cos θ + cos θ + π sin θ + sin θ + sin θ + for values of θ between and π. Problem 6. In the Fourier series of Problem, let θ = π and determine a series for π. When θ = π, f θ = π Hence π = π + cos π + cos π + 9 cos π + cos π + 6 π sin π + sin π + sin π +

74 i.e. Hence or π π = π π = π = π = π = + + Now try the following eercise. Eercise Further problems on Fourier series of non-periodic functions over a range of π. Show that the Fourier series for the function f = over the range = to = π is given by: f = π sin + sin + sin + sin +. etermine the Fourier series for the function defined by: { t, when π <t < f t = + t, when < t <π raw a graph of the function within and outside of the given range. f t = π + cos t cos t + π cos t + + FOURIER SERIES FOR A NON-PERIOIC FUNCTION OVER RANGE π 667. etermine the Fourier series up to and including the third harmonic for the function defined by: {, when < <π f = π, when π< < π Sketch a graph of the function within and outside of the given range, assuming the period is π. f = π π cos + cos + + cos. Epand the function f θ = θ in a Fourier series in the range π <θ<π. Sketch the function within and outside of the given range. f θ = π cos θ + cos θ cos θ 6. For the Fourier series obtained in Problem, let θ = π and deduce the series for n= n π = 6 7. Show that the Fourier series for the triangular waveform shown in Fig. 7. is given by: y = 8 sin π θ sin θ + sin θ sin 7θ + 7 in the range to π. y 668 FOURIER SERIES 8. Sketch the waveform defined by: +, when π < < π f =, when < <π π etermine the Fourier series in this range. f = 8 cos π + cos + cos + cos For the Fourier series of Problem 8, deduce a series for π 8 π 8 = Find the Fourier series for the function f = + π within the range π < <π. f = π + sin sin + sin Figure 7. π π θ L

75 Fourier series 7 Even and odd functions and half-range Fourier series 67 FOURIER SERIES f π/ π π/ π/ π π/ π Hence a = 8 π, a = 8 π, a = 8, and so on. π Hence the Fourier series for the waveform of Fig. 7. is given by: f = 8 π cos cos + cos cos Even and odd functions Even functions A function y = f is said to be even if f = f for all values of. Graphs of even functions are always symmetrical about the y-ais i.e. is a mirror image. Two eamples of even functions are y = and y = cos as shown in Fig. 9., page 99. Odd functions A function y = f is said to be odd if f = f for all values of. Graphs of odd functions are always symmetrical about the origin. Two eamples of odd functions are y = and y = sin as shown in Fig. 9.6, page. Many functions are neither even nor odd, two such eamples being shown in Fig. 9.7, page. See also Problems and, page. 7. Fourier cosine and Fourier sine series a Fourier cosine series The Fourier series of an even periodic function f having period π contains cosine terms only i.e. contains no sine terms and may contain a constant term. Hence f = a + a n cos n n= where a = π f π π = π f π due to symmetry and a n = π f cos n π π = π f cos n π b Fourier sine series The Fourier series of an odd periodic function f having period π contains sine terms only i.e. contains no constant term and no cosine terms. Hence f = b n sin n where n= b n = π f sin n π π = π f sin n π Problem. etermine the Fourier series for the periodic function defined by:, when π< < π f =, when π < < π π, when < <π. and has a period of π The square wave shown in Fig. 7. is an even function since it is symmetrical about the f ais. Hence from para. a the Fourier series is given by: f = a + a n cos n n= i.e. the series contains no sine terms L Figure 7. From para. a, a = π f π = { π/ π = π π } + π/ } { π/ + π π/ = π + π π = π a n = π f cos n π = { π/ π } cos n + cos n π π/ { sin = n π/ } sin n π + π n n π/ = { sin π/n π n } sin π/n + n = sin π/n = 8 sin nπ π n πn When n is even, a n = When n is odd, a n = 8 for n =,, 9,... πn and a n = 8 for n =, 7,,... πn Problem. In the Fourier series of Problem let = and deduce a series for π/. When =, f = from Fig. 7.. Thus, from the Fourier series, = 8 cos π cos + cos cos + 7 Hence π 8 = i.e. π = Problem. Obtain the Fourier series for the square wave shown in Fig. 7.. Figure 7. f π π π π The square wave shown in Fig. 7. is an odd function since it is symmetrical about the origin. Hence, from para. b, the Fourier series is given by: f = b n sin n n=

76 The function is defined by: {, when π< < f =, when < <π From para. b, b n = π f sin n π = π sin n π = cos n π π n = cos nπ π n n = cos nπ πn When n is even, b n =. When n is odd, b n = 8 = πn πn Hence b = 8 π, b = 8 π, b = 8 π, and so on Hence the Fourier series is: f = 8 sin + π sin + sin + sin Problem. etermine the Fourier series for the function f θ = θ in the range π<θ<π. The function has a period of π. EVEN AN O FUNCTIONS AN HALF-RANGE FOURIER SERIES 67 fθ π fθ = θ π π π π Figure 7. and a n = π f θ cos nθ dθ π = π θ cosnθ dθ π = θ π sin nθ θ cos nθ sin nθ + π n n n by parts = π cos nπ + π n = cos nπ n When n is odd, a n = n. Hence a =, a =, a =, and so on. When n is even, a n = n. Hence a =, a =, and so on. Hence the Fourier series is: f θ = θ = π cos θ cos θ + cos θ cos θ + cos θ A graph of f θ = θ is shown in Fig. 7. in the range π to π with period π. The function is symmetrical about the f θ ais and is thus an even function. Thus Problem. For the Fourier series of Problem, a Fourier cosine series will result of the form: let θ = π and show that n= n = π 6 f θ = a + a n cos nθ n= From para. a, When θ = π, f θ = π see Fig. 7.. Hence from a = π f θdθ = π the Fourier series: θ dθ π π π = π = θ π cos π cos π + cos π = π π cos π + cos π θ L 67 FOURIER SERIES i.e. π π = π = π i.e. = π i.e. 6 = Hence n = π 6 n= Now try the following eercise. Eercise Further problems on Fourier cosine and Fourier sine series. etermine the Fourier series for the function defined by:, π < < π f =, π < < π π, < <π which is periodic outside of this range of period π. f = cos cos π + cos cos Obtain the Fourier series of the function defined by: t + π, π <t < f t = t π, < t <π which is periodic of period π. Sketch the given function. f t = sin t + sin t + sin t + sin t +. etermine the Fourier series defined by {, π < < f = +, < <π which is periodic of period π. f = π + π cos + cos + cos +. In the Fourier series of Problem, let = and deduce a series for π /8. π 8 = Half-range Fourier series a When a function is defined over the range say to π instead of from to π it may be epanded in a series of sine terms only or of cosine terms only. The series produced is called a half-range Fourier series. b If a half-range cosine series is required for the function f = in the range to π then an even periodic function is required. In Figure 7., f = is shown plotted from = to = π. Since an even function is symmetrical about the f ais the line AB is constructed as shown. If the triangular waveform produced is assumed to be periodic of period π outside of this range then the waveform is as shown in Fig. 7.. When a half-range cosine series is required then the Fourier coefficients a and a n are calculated B f π A f = π π π π Figure 7.

77 Figure 7. f π π f = C π π π π π Problem 6. etermine the half-range Fourier cosine series to represent the function f = in the range π. From para. b, for a half-range cosine series: f = a + a n cos n n= EVEN AN O FUNCTIONS AN HALF-RANGE FOURIER SERIES 67 as in Section 7.a, i.e. When f =, f = a + a n cos n a = π f = π π π n= where a = π = π = π f π π and a n = π a n = π f cos n f cos n π π = π cos n c If a half-range sine series is required for the π function f = in the range to π then an odd periodic function is required. In Figure 7., = 6 sin n cos n π + π n n f = is shown plotted from = to = π. by parts Since an odd function is symmetrical about the origin the line C is constructed as shown. If = 6 π sin nπ cos nπ + π n n + cos n the sawtooth waveform produced is assumed to be periodic of period π outside of this range, = 6 cos nπ + then the waveform is as shown in Fig. 7.. π n cos n When a half-range sine series is required then the Fourier coefficient b n is calculated as in = 6 cos nπ Section 7.b, i.e. πn When n is even, a n = f = b n sin n n= When n is odd, a n = 6 = πn πn where b n = π f sin n Hence a = π π, a = π, a =, and so on. π Hence the half-range Fourier cosine series is given by: f = = π π cos + cos + cos + Problem 7. Find the half-range Fourier sine series to represent the function f = in the range π. From para. c, for a half-range sine series: f = b n sin n n= When f =, b n = π f sin n = π sin n π π = 6 cos n sin n π + π n n by parts L 67 FOURIER SERIES = 6 π cos nπ + π n = 6 cos nπ n When n is odd, b n = 6 n. sin nπ n + Hence b = 6, b = 6, b = 6 and so on. When n is even, b n = 6 n. Hence b = 6, b = 6, b 6 = 6 and so on. 6 Hence the half-range Fourier sine series is given by: f = = 6 sin sin + sin sin + sin Problem 8. Epand f = cos as a half-range Fourier sine series in the range π, and sketch the function within and outside of the given range. When a half-range sine series is required then an odd function is implied, i.e. a function symmetrical about the origin. A graph of y = cos is shown in Fig. 7.6 in the range to π. For cos to be symmetrical about the origin the function is as shown by the broken lines in Fig. 7.6 outside of the given range. Figure 7.6 π f y = cos π π From para. c, for a half-range Fourier sine series: f = b n sin n n= b n = π f sin n π = π cos sin n π = π sin + n sin n π = cos + n cos n π + π + n n = cos π + n cos π n + π + n n cos + n + cos n When n is odd, b n = π + n + n + n + n = When n is even, b n = π + n n + n + n = π + n n = n + n π n = n n π n = πn Hence b = 8 π, b = 6 π, b 6 = and so on. π Hence the half-range Fourier sine series for f in the range to π is given by: f = 8 6 sin + sin π π + sin 6 + π

78 Fourier series or f = 8 π sin + sin + sin Now try the following eercise. Eercise Further problems on halfrange Fourier series. etermine the half-range sine series for the function defined by:, < < π f = π, < <π f = sin + π sin π sin 9 π sin + 8. Obtain a the half-range cosine series and b the half-range sine series for the function, < t < π f t = π, < t <π a f t = π cos t cos t + cos t EVEN AN O FUNCTIONS AN HALF-RANGE FOURIER SERIES 67 b f t = π sin t sin t + sin t + sin t sin 6t +. Find a the half-range Fourier sine series and b the half-range Fourier cosine series for the function f = sin in the range π. Sketch the function within and outside of the given range. a f = 8 sin sin π sin 7 sin 7 79 b f = cos. etermine the half-range Fourier cosine series in the range = to = π for the function defined by:, < < π f = π π, < <π f = π π cos cos cos + L 7 Fourier series over any range 7. Epansion of a periodic function of period L a A periodic function f of period L repeats itself when increases by L, i.e. f + L = f. The change from functions dealt with previously having period π to functions having period L is not difficult since it may be achieved by a change of variable. b To find a Fourier series for a function f in the range L L a new variable u is introduced such that f, as a function of u, has period π. If u = π then, when = L L, u = π and when = L, u =+π. Also, let Lu f = f = Fu. The Fourier series for π Fu is given by: Fu = a + a n cos nu + b n sin nu, n= where a = π Fudu, π π a n = π Fu cos nu du π π and b n = π Fu sin nu du π π c It is however more usual to change the formula of para. b to terms of. Since u = π L, then du = π L, and the limits of integration are L to +L instead of from π to +π. Hence the Fourier series epressed in terms of is given by: f = a + πn a n cos n= L πn + b n sin L where, in the range L to +L : and a = L a n = L b n = L L f, L L f cos L L f sin L πn L πn L The limits of integration may be replaced by any interval of length L, such as from to L. Problem. The voltage from a square wave generator is of the form: {, < t < vt =, < t < and has a period of 8 ms. Find the Fourier series for this periodic function. The square wave is shown in Fig. 7.. From para. c, the Fourier series is of the form: vt = a + a n cos n= πnt L πnt + b n sin L

79 Figure 7. vt 8 8 t ms Period L = 8 ms a = L vtdt = vtdt L L 8 = { } dt + dt = 8 8 t = a n = L πnt vt cos dt L L L = πnt vt cos dt 8 8 = { πnt cos dt } πnt + cos dt πnt = sin πn = sin πn sin πn = for n =,,,... b n = L πnt vt sin dt L L L = πnt vt sin dt 8 8 = { πnt sin dt } πnt + sin dt πnt = cos πn = cos πn cos πn FOURIER SERIES OVER ANY RANGE 677 When n is even, b n = When n is odd, b = = π π, b = = π π, b =, and so on. π Thus the Fourier series for the function vt is given by: vt = + πt sin + πt π sin + πt sin + Problem. Obtain the Fourier series for the function defined by:, when < < f =, when < <, when < < The function is periodic outside of this range of period. The function f is shown in Fig. 7. where period, L =. Since the function is symmetrical about the f ais it is an even function and the Fourier series contains no sine terms i.e. b n =. Figure 7. f L = Thus, from para. c, πn f = a + a n cos L n= a = L f = f L L L 678 FOURIER SERIES = { + + } = = = = a n = L πn f cos L L L = πn f cos = { πn cos πn + cos πn } + cos = sin πn πn = πn πn sin sin πn When n is even, a n = When n is odd, a = = π π a = = π π a = = and so on. π π Hence the Fourier series for the function f is given by: f = + π cos + cos π π π cos 7π 7 cos + Problem. etermine the Fourier series for the function f t = t in the range t = tot =. The function f t = t in the interval to is shown in Fig. 7.. Although the function is not periodic it may be constructed outside of this range so that Figure 7. ft Period L = ft = t 6 t it is periodic of period, as shown by the broken lines in Fig. 7.. From para. c, the Fourier series is given by: πnt f t = a + a n cos L n= πnt + b n sin L a = L f t = L f t L L L = t dt = t = a n = L πnt f t cos dt L L L = L πnt t cos dt L L = πnt t cos dt πnt πnt = t sin cos + πn πn by parts = sin πn cos πn + πn πn + cos πn =

80 FOURIER SERIES OVER ANY RANGE FOURIER SERIES b n = L πnt f t sin dt L L L = L πnt t sin dt L L = πnt t sin dt πnt πnt = t cos sin + πn πn by parts = cos πn sin πn + πn πn + sin πn = cos πn πn = cos πn = πn πn Hence b = π, b = π, b = and so on. π Thus the Fourier series for the function f t inthe range to is given by: f t = πt sin + πt π sin + 6πt sin + Now try the following eercise. Eercise Further problems on Fourier series over any range L. The voltage from a square wave generator is of the form: {, < t < vt =, < t < and is periodic of period. Show that the Fourier series for the function is given by: vt = + sin π πt + sin πt + πt sin +. Find the Fourier series for f = in the range = to =. f = π sin π + π sin + 6π sin +. A periodic function of period is defined by: {, < < f = +, < < Sketch the function and obtain the Fourier series for the function. f = π sin π + π sin + π sin +. etermine the Fourier series for the half wave rectified sinusoidal voltage V sin ωt defined by: V sin ωt, < t < π ω f t = π, ω < t < π ω which is periodic of period π ω f t = V π + V sin ωt V cos ωt π cos ωt cos 6ωt L 7. Half-range Fourier series for functions defined over range L a By making the substitution u = π see Section 7., the range = to = L corresponds to the range u = tou = π. Hence a function may be epanded in a series of either cosine terms or sine terms only, i.e. a half-range Fourier series. b A half-range cosine series in the range to L can be epanded as: where nπ f = a + a n cos n= L a = L f L and a n = L nπ f cos L L c A half-range sine series in the range to L can be epanded as: f = b n sin n= nπ L where b n = L nπ f sin L L Problem. etermine the half-range Fourier cosine series for the function f = in the range. Sketch the function within and outside of the given range. A half-range Fourier cosine series indicates an even function. Thus the graph of f = in the range to is shown in Fig. 7. and is etended outside of this range so as to be symmetrical about the f ais as shown by the broken lines. From para. b, for a half-range cosine series: nπ f = a + a n cos n= L f f = 6 Figure 7. a = L f = L = = a n = L nπ f cos L L = nπ cos nπ nπ sin cos = nπ + nπ sin nπ cos nπ = nπ + nπ + cos nπ cos nπ = nπ nπ = cos nπ πn When n is even, a n = a = 8 π, a = 8 π, a = 8 π and so on. Hence the half-range Fourier cosine series for f in the range to is given by: f = 8 π cos π + π cos + cos π +

81 Problem. Find the half-range Fourier sine series for the function f = in the range. Sketch the function within and outside of the given range. A half-range Fourier sine series indicates an odd function. Thus the graph of f = in the range to is shown in Fig. 7. and is etended outside of this range so as to be symmetrical about the origin, as shown by the broken lines. f f = Hence b = π = π b = = π π b = π = π FOURIER SERIES OVER ANY RANGE 68 and so on. Thus the half-range Fourier sine series in the range toisgivenby: f = π sin π π sin + π sin π sin + 68 FOURIER SERIES f t = 8 πt sin π πt sin + πt sin. Show that the half-range Fourier cosine series for the function f θ = θ in the range to is given by: f θ = 6 6 π cos πθ cos πθ + cos πθ Sketch the function within and outside of the given range. Figure 7. 6 From para. c, for a half-range sine series: f = b n sin n= nπ L b n = L nπ f sin L L = nπ sin L nπ nπ cos sin = nπ + nπ cos nπ sin nπ = nπ + nπ + sin nπ = cos nπ nπ = cos nπ nπ Now try the following eercise. Eercise Further problems on halfrange Fourier series over range L. etermine the half-range Fourier cosine series for the function f = in the range. Sketch the function within and outside of the given range. f = { π π cos + cos π + cos π } +. Find the half-range Fourier sine series for the function f = in the range. Sketch the function within and outside of the given range. f = 6 π sin π π sin + sin π sin π +. etermine the half-range Fourier sine series for the function defined by: { t, < t < f t = t, < t < L

82 Number and Algebra 6 Arithmetic and geometric progressions 6. Arithmetic progressions When a sequence has a constant difference between successive terms it is called an arithmetic progression often abbreviated to AP. Eamples include: i,, 7,,,...where the common difference is and ii a, a + d, a + d, a + d,...where the common difference is d. If the first term of an AP is a and the common difference is d then the n th term is: a + n d In eample i above, the 7th term is given by + 7 = 9, which may be readily checked. The sum S of an AP can be obtained by multiplying the average of all the terms by the number of terms. The average of all the terms = a + l, where a is the first term and l is the last term, i.e. l = a + n d, for n terms. Hence the sum of n terms, a + l S n = n i.e. = n {a + a + n d} S n = n a + n d For eample, the sum of the first 7 terms of the series,, 7,,,...is given by S 7 = 7 + 7, since a = and d = = = 7 = 7 6. Worked problems on arithmetic progressions Problem. etermine a the ninth, and b the siteenth term of the series, 7,, 7,..., 7,, 7,...is an arithmetic progression with a common difference, d,of. a The n th term of an AP is given by a + n d Since the first term a =, d = and n = 9 then the 9th term is: + 9 = + 8 = + = b The 6th term is: + 6 = + = + 7 = 77. Problem. The 6th term of an AP is 7 and the th term is 8. etermine the 9th term. The n th term of an AP is a + n d The 6th term is: a + d = 7 The th term is: a + d = 8 Equation equation gives: 7d =, from which, d = 7 =. Substituting in equation gives: a + = 7, from which, a =. Hence the 9th term is: a + n d = + 9 = + 8 = + = 6. Problem. etermine the number of the term whose value is in the series,,,7,...,,,7,... is an AP where a = and d =. A NUMBER AN ALGEBRA Hence if the n th term is then: a + n d = i.e. + n = n = = 9. n = 9 = and n = + = i.e. the th term of the AP is. Problem. Find the sum of the first terms of the series, 9,, 7,..., 9,, 7,... is an AP where a = and d =. The sum of n terms of an AP, S n = n a + n d Hence the sum of the first terms, S = + = 6 + = 6 = Problem. Find the sum of the first terms of the series.,.,.7,.,....,.,.7,.,... is an AP where a =. and d =.6. The sum of the first terms, S = a + n d =. +.6 = 7 + = 99 9 = = 99. Now try the following eercise. Eercise 8 Further problems on arithmetic progressions. Find the th term of the series 8,,, 6, Find the 7th term of the series,.7,.,., The seventh term of a series is 9 and the eleventh term is. etermine the siteenth term. 8.. Find the th term of an arithmetic progression of which the first term is. and the tenth term is 6... etermine the number of the term which is 9 in the series 7, 9.,.,.6, Find the sum of the first terms of the series, 7,,, etermine the sum of the series 6., 8., 9.,.,..., Further worked problems on arithmetic progressions Problem 6. The sum of 7 terms of an AP is and the common difference is.. etermine the first term of the series. n = 7, d =. and S 7 = Since the sum of n terms of an AP is given by S n = n a + n d, then = 7 a + 7. = 7 a + 7. Hence = a = a + 7. Thus a = 7. =.8, from which a =.8 =. i.e. the first term, a =. Problem 7. Three numbers are in arithmetic progression. Their sum is and their product is 8. etermine the three numbers. Let the three numbers be a d, a and a + d Then a d + a + a + d =, i.e. a =, from which, a = Also, aa da + d = 8, i.e. aa d = 8 Since a =, d = 8 d = 8

83 ARITHMETIC AN GEOMETRIC PROGRESSIONS NUMBER AN ALGEBRA 8 = d = d from which, d = = 9. Hence d = 9 =±. The three numbers are thus, and +, i.e., and 8. Problem 8. Find the sum of all the numbers between and 7 which are eactly divisible by. The series, 6, 9,,..., 7 is an AP whose first term a = and common difference d = The last term is a + n d = 7 i.e. + n = 7, from which n = 7 = 68 Hence n = 68 + = 69 The sum of all 69 terms is given by S 69 = n a + n d = = = = 7 Problem 9. The first, twelfth and last term of an arithmetic progression are,, and 76 respectively. etermine a the number of terms in the series, b the sum of all the terms and c the 8 th term. a Let theap be a, a + d, a + d,..., a + n d, where a = The th term is: a + d = i.e. + d =, from which, d = = 7 Hence d = 7 = The last term is a + n d i.e. + n = 76 n = 76 = 7 = 9 Hence the number of terms in the series, n = 9 + = b Sum of all the terms, S = n a + n d = + = = = 78. = 87 c The 8th term is: a + n d = + 8 = + 79 = = Now try the following eercise. Eercise 9 Further problems on arithmetic progressions. The sum of terms of an arithmetic progression is. and the common difference is. Find the first term of the series... Three numbers are in arithmetic progression. Their sum is 9 and their product is.. etermine the three numbers..,,.. Find the sum of all the numbers between and which are eactly divisible by Find the number of terms of the series, 8,,...of which the sum is. A. Insert four terms between and. to form an arithmetic progression. 8.,,., 9 6. The first, tenth and last terms of an arithmetic progression are 9,., and. respectively. Find a the number of terms, b the sum of all the terms and c the 7th term. a b 67 c. 7. On commencing employment a man is paid a salary of 7 per annum and receives annual increments of. etermine his salary in the 9th year and calculate the total he will have received in the first years., 9 8. An oil company bores a hole 8 m deep. Estimate the cost of boring if the cost is for drilling the first metre with an increase in cost of per metre for each succeeding metre Geometric progressions When a sequence has a constant ratio between successive terms it is called a geometric progression often abbreviated to GP. The constant is called the common ratio, r. Eamples include i,,, 8,...where the common ratio is and ii a, ar, ar, ar,...where the common ratio is r. If the first term of a GP is a and the common ratio is r, then the n th term is: ar n which can be readily checked from the above eamples. For eample, the 8th term of the GP,,, 8,...is 7 = 8, since a = and r =. LetaGPbea, ar, ar, ar,..., ar n then the sum of n terms, S n = a + ar + ar + ar + +ar n Multiplying throughout by r gives: rs n = ar + ar + ar + ar + +ar n + ar n + Subtracting equation from equation gives: S n rs n = a ar n i.e. S n r = a r n Thus the sum of n terms, S n = a r n which r is valid when r <. Subtracting equation from equation gives S n = ar n r which is valid when r >. For eample, the sum of the first 8 terms of the GP,,, 8, 6,... is given by S 8 = 8, since a = and r = i.e. S 8 = 6 = When the common ratio r of a GP is less than unity, the sum of n terms, S n = a rn, which may be r written as S n = a r arn r. Since r <, r n becomes less as n increases, i.e. r n asn. ar n Hence r asn. Thus S n a r as n. a The quantity is called the sum to infinity, r S, and is the limiting value of the sum of an infinite number of terms, i.e. S = a which is valid when < r <. r For eample, the sum to infinity of the GP is S =, since a = and r =, i.e. S =.

84 ARITHMETIC AN GEOMETRIC PROGRESSIONS 6 NUMBER AN ALGEBRA 6. Worked problems on geometric progressions Problem. etermine the tenth term of the series, 6,,,..., 6,,,... is a geometric progression with a common ratio r of. The n th term of a GP is ar n, where a is the first term. Hence the th term is: = 9 = = 6. Problem. Find the sum of the first 7 terms of the series,,,,,...,,,,... is a GP with a common ratio r = The sum of n terms, S n = arn r Hence S 7 = 7 87 = = 6 Problem. The first term of a geometric progression is and the fifth term is. etermine the 8 th term and the th term. The th term is given by ar =, where the first term a = Hence r = a = and r = = The 8th term is ar 7 = = 7. The th term is ar = = 9.7 Problem. Which term of the series 87, 79,,...is 9? 87, 79,,... is a GP with a common ratio r = and first term a = 87 The n th term of a GP is given by: ar n Hence 9 = 87 n from which n = 987 = 7 = 9 9 = Thus n = 9, from which, n = 9 + = i.e. 9 is the th term of the GP Problem. Find the sum of the first 9 terms of the series 7., 7.6, 6.8,... The common ratio, r = ar a = =.8 also ar ar = =.8 The sum of 9 terms, S 9 = a rn = r = =.7. Problem. Find the sum to infinity of the series,,,...,,,...is a GP of common ratio, r = The sum to infinity, S = a r = = = 9 = Now try the following eercise. Eercise Further problems on geometric progressions. Find the th term of the series,,,, etermine the sum of the first 7 terms of the series,,,6, The first term of a geometric progression is and the 6th term is 8. etermine the 8th and th terms., 96 A. Find the sum of the first 7 terms of the series,,,...correct to significant figures 8.. etermine the sum to infinity of the series,,, Find the sum to infinity of the series,, 8, Further worked problems on geometric progressions Problem 6. In a geometric progression the sith term is 8 times the third term and the sum of the seventh and eighth terms is 9. etermine a the common ratio, b the first term, and c the sum of the fifth to eleventh terms, inclusive. a Let the GP be a, ar, ar, ar,..., ar n The rd term = ar and the sith term = ar The 6th term is 8 times the rd. Hence ar = 8ar from which, r = 8, r = 8 i.e. the common ratio r =. b The sum of the 7th and 8th terms is 9. Hence ar 6 + ar 7 = 9. Since r =, then 6a + 8a = 9 9a = 9, from which, a, the first term, =. c The sum of the th to th terms inclusive is given by: S S = ar ar r r = = = = 8 6 = Problem 7. A hire tool firm finds that their net return from hiring tools is decreasing by % per annum. If their net gain on a certain tool this year is, find the possible total of all future profits from this tool assuming the tool lasts for ever. The net gain forms a series: , which is a GP with a = and r =.9. The sum to infinity, a S = r =.9 = = total future profits Problem 8. If is invested at compound interest of 8% per annum, determine a the value after years, b the time, correct to the nearest year, it takes to reach more than. a Let the GP be a, ar, ar,..., ar n The first term a = The common ratio r =.8 Hence the second term is ar =.8 = 8, which is the value after year, the third term is ar =.8 = 6.6, which is the value after years, and so on. Thus the value after years = ar =.8 =.89 b When has been reached, = ar n i.e. and =.8 n =.8 n Taking logarithms to base of both sides gives: lg = lg.8 n = n lg.8, by the laws of logarithms from which, n = lg lg.8 =. Hence it will take years to reach more than. Problem 9. A drilling machine is to have 6 speeds ranging from rev/min to 7 rev/ min. If the speeds form a geometric progression determine their values, each correct to the nearest whole number.

85 Number and Algebra Let the GP of n terms be given by a, ar, ar,..., ar n. The first term a = rev/min The 6th term is given by ar 6, which is 7 rev/min, i.e., ar = 7 from which r = 7 a = 7 = Thus the common ratio, r = =.788 The first term is a = rev/min the second term is ar =.788 = 8.9, the third term is ar =.788 = 7.7, the fourth term is ar =.788 =.89, the fifth term is ar =.788 = 6.9, the sith term is ar =.788 = 7.6 Hence, correct to the nearest whole number, the 6 speeds of the drilling machine are, 86, 8,, 6 and 7 rev/min. Now try the following eercise. Eercise Further problems on geometric progressions. In a geometric progression the th term is 9 times the rd term and the sum of the 6th and 7th terms is 9. etermine a the common ratio, b the first term and c the sum of the th to th terms inclusive. a b c 9 ARITHMETIC AN GEOMETRIC PROGRESSIONS 7. Which term of the series, 9, 7,... is 99? th. The value of a lathe originally valued at depreciates % per annum. Calculate its value after years. The machine is sold when its value is less than. After how many years is the lathe sold? 66, years. If the population of Great Britain is million and is decreasing at.% per annum, what will be the population in years time? 8.7 M. g of a radioactive substance disintegrates at a rate of % per annum. How much of the substance is left after years? 7. g 6. If is invested at compound interest of 6% per annum determine a the value after years, b the time, correct to the nearest year, it takes to reach 7. a 99. b 9 years 7. A drilling machine is to have 8 speeds ranging from rev/min to rev/min. If the speeds form a geometric progression determine their values, each correct to the nearest whole number., 9, 9, 68, 7, 8, 7, rev/min A 7 The binomial series 7. Pascal s triangle A binomial epression is one which contains two terms connected by a plus or minus sign. Thus p + q, a +, +y are eamples of binomial epressions. Epanding a + n for integer values of n from to 6 gives the results as shown at the bottom of the page. From these results the following patterns emerge: i a decreases in power moving from left to right. ii increases in power moving from left to right. iii The coefficients of each term of the epansions are symmetrical about the middle coefficient when n is even and symmetrical about the two middle coefficients when n is odd. iv The coefficients are shown separately in Table 7. and this arrangement is known as Pascal s triangle. A coefficient of a term may be obtained by adding the two adjacent coefficients immediately above in the previous row. This is shown by the triangles in Table 7., where, for eample, + =, + =, and so on. v Pascal s triangle method is used for epansions of the form a + n for integer values of n less than about 8. Table 7. as shown in below. The first and last terms of the epansion of a + 7 are a 7 and 7 respectively. The powers of a decrease and the powers of increase moving from left to right. Hence a + 7 = a 7 + 7a 6 + a + a + a + a + 7a Problem. Use the Pascal s triangle method to determine the epansion of a + 7. From Table 7., the row of Pascal s triangle corresponding to a + 6 is as shown in below. Adding adjacent coefficients gives the coefficients of a+ 7 Problem. etermine, using Pascal s triangle method, the epansion of p q. Comparing p q with a + shows that a = p and = q. a + = a + = a + a+ a + = a + a + = a + a + a + = a + a + = a + a + a + a + = a + a + = a + a + 6a + a + a + = a + a + = a + a + a + a + a + a + 6 = a + a + = a 6 + 6a + a + a + a + 6a + 6

86 THE BINOMIAL SERIES 9 6 NUMBER AN ALGEBRA Using Pascal s triangle method: a + = a + a + a + a + Hence p q = p + p q + p q + p q + p q + q i.e. p q = p p q + 7p q Now try the following eercise. 8p q + 8pq q Eercise Further problems on Pascal s triangle. Use Pascal s triangle to epand y y + y y + y y + 7y 6 y 7. Epand a + b using Pascal s triangle a + a b + 7a b 7. The binomial series + 8a b + 8ab + b The binomial series or binomial theorem is a formula for raising a binomial epression to any power without lengthy multiplication. The general binomial epansion of a + n is given by: a + n = a n + na n nn + a n! nn n + a n! + where! denotes and is termed factorial. With the binomial theorem n may be a fraction, a decimal fraction or a positive or negative integer. When n is a positive integer, the series is finite, i.e., it comes to an end; when n is a negative integer, or a fraction, the series is infinite. In the general epansion of a + n it is noted that nn n the th term is: a n. The number! is very evident in this epression. For any term in a binomial epansion, say the r th term, r is very evident. It may therefore be reasoned that the r th term of the epansion a + n is: nn n... to r terms a n r r r! If a = in the binomial epansion of a + n then: + n nn = + n +! nn n + +! which is valid for < <. When is small compared with then: + n + n 7. Worked problems on the binomial series Problem. Use the binomial series to determine the epansion of + 7. The binomial epansion is given by: a + n = a n + na n nn + a n! nn n + a n +! When a = and n = 7: + 7 = i.e. + 7 = A Epand c using the bino- c Problem. mial series. c = c + c c c + c c + c c + c c + c i.e. c = c c + c c c + c c Problem. Without fully epanding + 7, determine the fifth term. The r th term of the epansion a + n is given by: nn n... to r terms a n r r r! Substituting n = 7, a = and r = = gives: 76 7 i.e. the fifth term of + 7 = = 9 Problem 6. Find the middle term of p q In the epansion of a + there are +, i.e. terms. Hence the middle term is the sith. Using the general epression for the r th term where a = p, =, n = and r = gives: q 9876 p q = p q Hence the middle term of p is p q q Problem 7. Evaluate. 9 using the binomial theorem correct to a decimal places and b 7 significant figures. + n nn = + n +! nn n + +!. 9 = +. 9 Substituting =. and n = 9 in the general epansion for + n gives: +. 9 = = = Hence. 9 =.8, correct to decimal places =.8, correct to 7 significant figures Problem 8. Evaluate.97 6 correct to significant figures using the binomial epansion is written as. 6 Using the epansion of + n where n = 6 and =. gives:. 6 = =

87 THE BINOMIAL SERIES 6 6 NUMBER AN ALGEBRA i.e =.8, correct to significant figures Problem 9. etermine the value of.9, correct to 6 significant figures using the binomial theorem..9 may be written in the form + n as:.9 = +.9 = +.9 = = = =.8 correct to 8 significant figures Hence.9 =.8 = 8.98, correct to 6 significant figures Now try the following eercise. Eercise Further problems on the binomial series. Use the binomial theorem to epand a +. a + 8a + a + a + 6. Use the binomial theorem to epand Epand y 6 96 y + 6 y 6y + 8y. etermine the epansion of Epand p + q as far as the fifth term. p + p q + p 9 q + p 8 q + 8p 7 q 6. etermine the sith term of p + q. 79 p 8 q 7. etermine the middle term of a b 8. 7 a b 8. Use the binomial theorem to determine, correct to decimal places: a. 8 b. 7 a. b.7 9. Use the binomial theorem to determine, correct to significant figures: a.98 7 b. 9 a.868 b.. Evaluate. 6 correct to decimal places Further worked problems on the binomial series Problem. a Epand + in ascending powers of as far as the term in, using the binomial series. b State the limits of for which the epansion is valid. a Using the binomial epansion of + n, where n = and is replaced by gives: = + + A = + +! + +! = b The epansion is valid provided <, i.e. < or < < Problem. a Epand in ascending powers of as far as the term in, using the binomial theorem. b What are the limits of for which the epansion in a is true? a = = = 6 Using the epansion of + n = 6 = 6 + +! + +! = b The epansion in a is true provided <, i.e. < or < < Problem. Use the binomial theorem to epand + in ascending powers of to four terms. Give the limits of for which the epansion is valid. + = + = + = + Using the epansion of + n, + = + + / /! + / / / +! 8 = = This is valid when <, i.e. < or < < Problem. Epand in ascending t powers of t as far as the term in t. State the limits of t for which the epression is valid. t = t = + t + / / t! + / / / t +,! using the epansion for + n = + t + t + t + The epression is valid when t <, i.e. t < or < t <

88 THE BINOMIAL SERIES 6 6 NUMBER AN ALGEBRA + Problem. Simplify + given that powers of above the first may be neglected. + + = when epanded by the binomial theorem as far as the term only, = + = + when powers of higher than unity are neglected = + Problem. Epress as a power series as far as the term in. State the range of values of for which the series is convergent. + = + + = + + / / +! = + + which is valid for <, i.e. < = + / + / / +! = which is valid for Hence + = + <, i.e. < = = , neglecting terms of higher power than, = + + The series is convergent if < < Now try the following eercise. Eercise Further problems on the binomial series In problems to epand in ascending powers of as far as the term in, using the binomial theorem. State in each case the limits of for which the series is valid , < , < < < A < 6. Epand + 6 to three terms. For what values of is the epansion valid? < 7. When is very small show that: a + b + + c If is very small such that and higher powers may be neglected, determine the power + 8 series for + 9. Epress the following as power series in ascending powers of as far as the term in. State in each case the range of for which the series is valid. a b a +, < b 7, < 7. Practical problems involving the binomial theorem Binomial epansions may be used for numerical approimations, for calculations with small variations and in probability theory see Chapter 7. Problem 6. The radius of a cylinder is reduced by % and its height is increased by %. etermine the approimate percentage change in a its volume and b its curved surface area, neglecting the products of small quantities. Volume of cylinder = πr h. Let r and h be the original values of radius and height. The new values are.96r or.r and.h or +.h. a New volume = π.r +.h = πr h. +. Now. =. +. =.8, neglecting powers of small terms. Hence new volume πr h.8 +. πr h.8 +., neglecting products of small terms πr h.6 or.9πr h, i.e. 9% of the original volume Hence the volume is reduced by approimately 6%. b Curved surface area of cylinder = πrh. New surface area = π.r +.h = πrh. +. πrh. +., neglecting products of small terms πrh. or.98πrh, i.e. 98% of the original surface area Hence the curved surface area is reduced by approimately %. Problem 7. The second moment of area of a rectangle through its centroid is given by bl. etermine the approimate change in the second moment of area if b is increased by.% and l is reduced by.%.

89 THE BINOMIAL SERIES 6 66 NUMBER AN ALGEBRA New values of b and l are +.b and.l respectively. New second moment of area = +.b.l = bl +.. bl +..7, neglecting powers of small terms bl +..7, neglecting products of small terms bl. or.96bl, i.e. 96% of the original second moment of area Hence the second moment of area is reduced by approimately %. Problem 8. The resonant frequency of a vibrating shaft is given by: f = k π I, where k is the stiffness and I is the inertia of the shaft. Use the binomial theorem to determine the approimate percentage error in determining the frequency using the measured values of k and I when the measured value of k is % too large and the measured value of I is % too small. Let f, k and I be the true values of frequency, stiffness and inertia respectively. Since the measured value of stiffness, k, is % too large, then = π +. k. I = π k I +.. i.e. f = f +.. f f Neglecting the products of small terms, f +. +.f.f Thus the percentage error in f based on the measured values of k and I is approimately., i.e. % too large. Now try the following eercise. Eercise Further practical problems involving the binomial theorem. Pressure p and volume v are related by pv = c, where c is a constant. etermine the approimate percentage change in c when p is increased by % and v decreased by.%..6% decrease. Kinetic energy is given by mv. etermine the approimate change in the kinetic energy when mass m is increased by.% and the velocity v is reduced by %..% decrease A. The radius of a cone is increased by.7% and its height reduced by.9%. etermine the approimate percentage change in its volume, neglecting the products of small terms..% increase 6. The electric field strength H due to a magnet of length l and moment M at a point on its ais distance from the centre is given by H = M { l l + l Show that if l is very small compared with, then H M. 7. The shear stress τ in a shaft of diameter under a torque T is given by: τ = kt π. etermine the approimate percentage error in calculating τ if T is measured % too small and.% too large. 7.% decrease 8. The energy W stored in a flywheel is given by: W = kr N, where k is a constant, r is the radius and N the number of revolutions. etermine the approimate percentage change in W when r is increased by.% and N is decreased by %..% increase 9. In a series electrical circuit containing inductance L and capacitance C the resonant } frequency is given by: f r = π. If the LC values of L and C used in the calculation are.6% too large and.8% too small respectively, determine the approimate percentage error in the frequency..9% too small. The viscosity η of a liquid is given by: η = kr, where k is a constant. If there is νl an error in r of +%, in ν of +% and l of %, what is the resultant error in η? +7%. A magnetic pole, distance from the plane of a coil of radius r, and on the ais of the coil, is subject to a force F when a current flows in the coil. The force is given k by: F =, where k is a constant. r + Use the binomial theorem to show that when is small compared to r, then F k r k r 7. The flow of water through a pipe is given by: d H G =. If d decreases by % and H L by %, use the binomial theorem to estimate the decrease in G..% k = k = +.k The measured value of inertia, I, is % too small, hence I = 98 I =.I. An error of +.% was made when measuring the radius of a sphere. Ignoring the products of small quantities determine the approimate error in calculating a the volume, and b the surface area. a.% increase b.% increase The measured value of frequency, f = k = π I = π k I π +.k.i. The power developed by an engine is given by I = k PLAN, where k is a constant. etermine the approimate percentage change in the power when P and A are each increased by.% and L and N are each decreased by.%..% increase

90 Number and Algebra 8 Maclaurin s series 8. Introduction Some mathematical functions may be represented as power series, containing terms in ascending powers of the variable. For eample, e = + +! +! + sin =! +! 7 7! + and cosh = +! +! + as introduced in Chapter Using a series, called Maclaurin s series, mied functions containing, say, algebraic, trigonometric and eponential functions, may be epressed solely as algebraic functions, and differentiation and integration can often be more readily performed. 8. erivation of Maclaurin s theorem Let the power series for f be f = a + a + a + a + a + a + where a, a, a,...are constants. When =, f = a. ifferentiating equation with respect to gives: f = a + a + a + a ifferentiating equation with respect to gives: f = a + a + a + When =, f = a =!a, i.e. a = f! Continuing the same procedure gives a = f iv,! a = f v, and so on.! Substituting for a, a, a,...in equation gives: f = f + f + f! + f +! i.e. f = f + f +! f +! f + Equation is a mathematical statement called Maclaurin s theorem or Maclaurin s series. A 68 NUMBER AN ALGEBRA on; thus cos meets this condition. However, if f = ln, f = =, thus ln does not meet this condition. c The resultant Maclaurin s series must be convergent In general, this means that the values of the terms, or groups of terms, must get progressively smaller and the sum of the terms must reach a limiting value. For eample, the series is convergent since the value of the terms is getting smaller and the sum of the terms is approaching a limiting value of. 8. Worked problems on Maclaurin s series Problem. etermine the first four terms of the power series for cos. The values of f, f, f,... in the Maclaurin s series are obtained as follows: f = cos f = cos = f = sin f = sin = f = cos f = cos = f = sin f = sin = f iv = cos f iv = cos = f v = sin f v = sin = f vi = cos f vi = cos = Replacing with θ in the series obtained in Problem gives: cos θ = θ! + θ! θ6 6! + = θ + 6θ 6θ6 7 + i.e. cos θ = θ + θ θ6 + Problem. etermine the power series for tan as far as the term in. f = tan f = tan = f = sec f = sec = cos = f = sec sec tan = sec tan f = sec tan = f = sec sec + tan sec sec tan, by the product rule, = sec + sec tan f = sec + sec tan = Substituting these values into equation gives: f = tan = + + +!! + a + When =, f = a. ifferentiating equation with respect to gives: f = a + a + a + a + When =, f = a =!a, i.e. a = f! 8. Conditions of Maclaurin s series Maclaurin s series may be used to represent any function, say f, as a power series provided that at = the following three conditions are met: a f = For eample, for the function f = cos, f = cos =, thus cos meets the condition. However, if f = ln, f = ln =, thus ln does not meet this condition. b f, f, f,... = For eample, for the function f = cos, f = sin =, f = cos =, and so Substituting these values into equation gives: f = cos = + + +!! i.e. Problem. cos θ. +! +! ! cos =! +! 6 6! + etermine the power series for i.e. tan = + Problem. Epand ln + to five terms. f = ln + f = ln + = f = f = + + = f = + f = + = f = + f = + =

91 MACLAURIN S SERIES 69 7 NUMBER AN ALGEBRA f iv = 6 + f iv = 6 + = 6 f v = + f v = + = Substituting these values into equation gives: f = ln + = + +! !!! i.e. ln + = + + Problem. Epand ln to five terms. Replacing by in the series for ln + in Problem gives: ln = + + i.e. ln = Problem 6. etermine the power series for + ln. + ln = ln + ln by the laws of logarithms, and from Problems and, + ln = + + = i.e. ln = Problem 7. Use Maclaurin s series to find the epansion of +. f = + f = = 6 f = + f = = f = + f = = 8 f = + f = = 8 f iv = f iv = Substituting in equation gives: + = f + f +! f +! f +! f iv = !!! = This epression could have been obtained by applying the binomial theorem. Problem 8. Epand e as far as the term in. f = e f = e = f = e f = e = f = e f = e = f = 8 e f = 8 e = 8 f iv = 6 e f iv = 6 e = 6 Substituting in equation gives: e = f + f +! f = + +! f +! f iv + + +!! 8 +! 6 + i.e. e = A Problem 9. evelop a series for sinh using Maclaurin s series. f = sinh f = sinh = e e = f = cosh f = cosh = e + e = f = sinh f = sinh = f = cosh f = cosh = f iv = sinh f iv = sinh = f v = cosh f v = cosh = Substituting in equation gives: sinh = f + f +! f +! f +! f iv +! f v + = !!! + +! i.e. sinh = +! +! + as obtained in Section. Problem. Produce a power series for cos as far as the term in 6. From double angle formulae, cos A = cos A see Chapter 8. from which, cos A = + cos A and cos = + cos From Problem, cos =! +! 6 6! + hence cos =! +! 6 6! + = Thus cos = + cos = i.e. cos = Now try the following eercise. Eercise 6 Further problems on Maclaurin s series. etermine the first four terms of the power series for sin using Maclaurin s series. sin = Use Maclaurin s series to produce a power series for cosh as far as the term in Use Maclaurin s theorem to determine the first three terms of the power series for ln + e. ln etermine the power series for cos t as far as the term in t 6. 8t + t 6 t6. Epand e in a power series as far as the term in evelop, as far as the term in, the power series for sec Epand e θ cos θ as far as the term in θ using Maclaurin s series. + θ θ 8. etermine the first three terms of the series for sin by applying Maclaurin s theorem. + 6

92 MACLAURIN S SERIES 7 7 NUMBER AN ALGEBRA 9. Use Maclaurin s series to determine the epansion of + t t + 6t + 96t + 6t 8. Numerical integration using Maclaurin s series The value of many integrals cannot be determined using the various analytical methods. In Chapter, the trapezoidal, mid-ordinate and Simpson s rules are used to numerically evaluate such integrals. Another method of finding the approimate value of a definite integral is to epress the function as a power series using Maclaurin s series, and then integrating each algebraic term in turn. This is demonstrated in the following worked problems. Problem. Evaluate.. esin θ dθ, correct to significant figures. A power series for e sin θ is firstly obtained using Maclaurin s series. f θ = e sin θ f = e sin = e = f θ = cos θ e sin θ f =cos e sin =e = f θ = cos θcos θ e sin θ + e sin θ sin θ, by the product rule, = e sin θ cos θ sin θ; f = e cos sin = f θ = e sin θ cos θ sin θ cos θ + cos θ sin θcos θ e sin θ = e sin θ cos θ sin θ + cos θ sin θ f = e cos + = Hence from equation : e sin θ = f + θf + θ! f + θ! f + = + θ + θ +.. Thus e sin θ dθ = + θ + θ dθ... = + θ + θ dθ.. = θ + θ + θ. = =.98. =.77, correct to significant figures sin θ Problem. Evaluate dθ using θ Maclaurin s series, correct to significant figures. Let f θ = sin θ f = f θ = cos θ f = f θ = sin θ f = f θ = cos θ f = f iv θ = sin θ f iv = f v θ = cos θ f v = Hence from equation : sin θ = f + θf + θ! f + θ! f + θ! f iv + θ! f v + = + θ + θ θ +!! + θ θ + +!! i.e. sin θ = θ θ! + θ! Hence θ θ sin θ! + θ! θ7 7! + dθ = dθ θ θ = θ 6 + θ θ6 + dθ A = θ θ 8 + θ 6 θ7 7 + = =.96, correct to significant figures Problem. Evaluate. ln+ using Maclaurin s theorem, correct to decimal places. From Problem, ln + = + +. Hence ln +. = + +. = = = = =.9, correct to decimal places Now try the following eercise. Eercise 7 Further problems on numerical integration using Maclaurin s series. Evaluate.6. esin θ dθ, correct to decimal places, using Maclaurin s series..78. Use Maclaurin s theorem to epand cos θ and hence evaluate, correct to decimal cos θ places, θ dθ..88. etermine the value of θ cos θ dθ, correct to significant figures, using Maclaurin s series... Use Maclaurin s theorem to epand ln + as a power series. Hence evaluate, correct to decimal places,. ln Limiting values It is { sometimes } necessary to find limits of the form f lim, where f a = and ga =. a g For eample, lim { } = = and is generally referred to as indeterminate. For certain limits a knowledge of series can sometimes help. For eample, { } tan lim + lim + from Problem = lim + = lim { } = Similarly, { } sinh lim + lim! +! + from Problem 9 } = lim { +! +! + = However, a knowledge { of series does not help with } + eamples such as lim 7 + 6

93 L Hopital s rule will enable us to determine such limits when the differential coefficients of the numerator and denominator can be found. L Hopital s rule states: lim a { f g } = lim a { f } g provided g a = { f } It can happen that lim a g is still ; if so, the numerator and denominator are differentiated again and again until a non-zero value is obtained for the denominator. The following worked problems demonstrate how L Hopital s rule is used. Refer to Chapter 7 for methods of differentiation. Problem. { } + etermine lim The first step is to substitute = into both numerator and denominator. In this case we obtain.itis only when we obtain such a result that we then use L Hopital s rule. Hence applying L Hopital s rule, lim { Problem. } = lim { } + 7 i.e. both numerator and denominator have been differentiated = = { } sin etermine lim Substituting = gives { } sin lim = sin = Applying L Hopital s rule gives { } { } sin cos lim = lim Substituting = gives cos = = again MACLAURIN S SERIES 7 Applying L Hopital s rule again gives { } cos { } sin lim = lim = Problem 6. { } sin etermine lim tan Substituting = gives { } sin lim = sin tan tan = Applying L Hopital s rule gives { } { } sin cos lim = lim tan sec Substituting = gives { } cos lim sec = cos sec = = again Applying L Hopital s rule gives { } { } cos sin lim sec = lim sec sec tan { } sin = lim sec tan Substituting = gives sin sec tan = again Applying L Hopital s rule gives { } sin lim sec tan cos = lim sec sec + tan sec tan using the product rule Substituting = gives cos sec sec tan = = A 7 NUMBER AN ALGEBRA { } sin Hence lim = tan Now try the following eercise. Eercise 8 Further problems on limiting values etermine the following limiting values { } +. lim + { } sin 9. lim { } ln +. lim { } sin. lim + { } sin θ θ cos θ. lim θ θ { ln t } 6. lim t t { } sinh sin 7. lim { } sin θ 8. lim θ π ln sin θ 9. lim t { sec t t sin t }

94 Number and Algebra Assignment A This assignment covers the material contained in Chapters 6 to 8. The marks for each question are shown in brackets at the end of each question.. etermine the th term of the series.6,,.,.8,.... The sum of terms of an arithmetic progression is 86 and the common difference is. etermine the first term of the series.. An engineer earns per annum and receives annual increments of 6. etermine the salary in the 9th year and calculate the total earnings in the first years.. etermine the th term of the series.,, 6,,.... Find the sum of the first eight terms of the series,., 6.,..., correct to decimal place. 6. etermine the sum to infinity of the series,,, A machine is to have seven speeds ranging from rev/min to rev/min. If the speeds form a geometric progression, determine their value, each correct to the nearest whole number Use the binomial series to epand a b etermine the middle term of 8 y 6. Epand the following in ascending powers of t as far as the term in t a + t b t For each case, state the limits for which the epansion is valid.. When is very small show that: +. The modulus of rigidity G is given by G = R θ L where R is the radius, θ the angle of twist and L the length. Find the approimate percentage error in G when R is measured.% too large, θ is measured % too small and L is measured % too small. 7. Use Maclaurin s series to determine a power series for e cos as far as the term in.. Show, using Maclaurin s series, that the first four terms of the power series for cosh is given by: cosh = Epand the function ln + sin using Maclaurin s series and hence evaluate: ln + sin correct to significant figures.

95 Statistics and probability Presentation of statistical data. Some statistical terminology ata are obtained largely by two methods: a by counting for eample, the number of stamps sold by a post office in equal periods of time, and b by measurement for eample, the heights of a group of people. When data are obtained by counting and only whole numbers are possible, the data are called discrete. Measured data can have any value within certain limits and are called continuous see Problem. A set is a group of data and an individual value within the set is called a member of the set. Thus, if the masses of five people are measured correct to the nearest. kg and are found to be. kg, 9. kg, 6. kg, 77.8 kg and 6. kg, then the set of masses in kilograms for these five people is: {., 9., 6., 77.8, 6.} and one of the members of the set is 9.. A set containing all the members is called a population. Some members selected at random from a population are called a sample. Thus all car registration numbers form a population, but the registration numbers of, say, cars taken at random throughout the country are a sample drawn from that population. The number of times that the value of a member occurs in a set is called the frequency of that member. Thus in the set: {,,,,,,, 7, 9}, member has a frequency of three, member has a frequency of and the other members have a frequency of one. The relative frequency with which any member of a set occurs is given by the ratio: frequency of member total frequency of all members For the set: {,,,, 7,, 6,, 8}, the relative frequency of member is 9. Often, relative frequency is epressed as a percentage and the percentage relative frequency is: relative frequency %. Problem. ata are obtained on the topics given below. State whether they are discrete or continuous data. a The number of days on which rain falls in a month for each month of the year. b The mileage travelled by each of a number of salesmen. c The time that each of a batch of similar batteries lasts. d The amount of money spent by each of several families on food. a The number of days on which rain falls in a given month must be an integer value and is obtained by counting the number of days. Hence, these data are discrete. b A salesman can travel any number of miles and parts of a mile between certain limits and these data are measured. Hence the data are continuous. c The time that a battery lasts is measured and can have any value between certain limits. Hence these data are continuous. d The amount of money spent on food can only be epressed correct to the nearest pence, the amount being counted. Hence, these data are discrete. Now try the following eercise. Eercise Further problems on discrete and continuous data In Problems and, state whether data relating to the topics given are discrete or continuous.. a The amount of petrol produced daily, for each of days, by a refinery. J 8 STATISTICS AN PROBABILITY b The amount of coal produced daily by each of miners. c The number of bottles of milk delivered daily by each of milkmen. d The size of samples of rivets produced by a machine. a continuous b continuous c discrete d continuous. a The number of people visiting an ehibition on each of days. b The time taken by each of athletes to run metres. c The value of stamps sold in a day by each of post offices. d The number of defective items produced in each of one-hour periods by a machine. a discrete b continuous c discrete d discrete. Presentation of ungrouped data Ungrouped data can be presented diagrammatically in several ways and these include: a pictograms, in which pictorial symbols are used to represent quantities see Problem, b horizontal bar charts, having data represented by equally spaced horizontal rectangles see Problem, and c vertical bar charts, in which data are represented by equally spaced vertical rectangles see Problem. Trends in ungrouped data over equal periods of time can be presented diagrammatically by a percentage component bar chart. In such a chart, equally spaced rectangles of any width, but whose height corresponds to %, are constructed. The rectangles are then subdivided into values corresponding to the percentage relative frequencies of the members see Problem. A pie diagram is used to show diagrammatically the parts making up the whole. In a pie diagram, the area of a circle represents the whole, and the areas of the sectors of the circle are made proportional to the parts which make up the whole see Problem 6. Problem. The number of television sets repaired in a workshop by a technician in si, one-month periods is as shown below. Present these data as a pictogram. Month Number repaired January February 6 March April 9 May June 8 Each symbol shown in Fig.. represents two television sets repaired. Thus, in January, symbols are used to represent the sets repaired, in February, symbols are used to represent the 6 sets repaired, and so on. Figure. Problem. The distance in miles travelled by four salesmen in a week are as shown below. Salesmen P Q R S istance travelled miles 6 97 Use a horizontal bar chart to represent these data diagrammatically. Equally spaced horizontal rectangles of any width, but whose length is proportional to the distance travelled, are used. Thus, the length of the rectangle for salesman P is proportional to miles, and so on. The horizontal bar chart depicting these data is shown in Fig...

96 PRESENTATION OF STATISTICAL ATA 9 STATISTICS AN PROBABILITY Figure. Problem. The number of issues of tools or materials from a store in a factory is observed for seven, one-hour periods in a day, and the results of the survey are as follows: Period 6 7 Number of issues Present these data on a vertical bar chart. In a vertical bar chart, equally spaced vertical rectangles of any width, but whose height is proportional to the quantity being represented, are used. Thus the height of the rectangle for period is proportional to units, and so on. The vertical bar chart depicting these data is shown in Fig... Figure. Problem. The numbers of various types of dwellings sold by a company annually over a three-year period are as shown below. raw percentage component bar charts to present these data. Year Year Year -roomed bungalows 7 7 -roomed bungalows roomed houses -roomed houses roomed houses A table of percentage relative frequency values, correct to the nearest %, is the first requirement. Since, percentage relative frequency frequency of member = total frequency then for -roomed bungalows in year : percentage relative frequency = = % The percentage relative frequencies of the other types of dwellings for each of the three years are similarly calculated and the results are as shown in the table below. Year Year Year % % % -roomed bungalows 7 -roomed bungalows 9 8 -roomed houses -roomed houses 6-roomed houses 7 The percentage component bar chart is produced by constructing three equally spaced rectangles of any width, corresponding to the three years. The heights of the rectangles correspond to % relative frequency, and are subdivided into the values in the table of percentages shown above.a key is used different types of shading or different colour schemes to indicate corresponding percentage values in the rows of the table of percentages. The percentage component bar chart is shown in Fig... Problem 6. The retail price of a product costing is made up as follows: materials p, labour p, research and development p, overheads 7 p, profit 6 p. Present these data on a pie diagram. A circle of any radius is drawn, and the area of the circle represents the whole, which in this case is. The circle is subdivided into sectors so that the areas of the sectors are proportional to the parts, i.e. the parts which make up the total retail price. For the J Figure. area of a sector to be proportional to a part, the angle at the centre of the circle must be proportional to that part. The whole, or p, corresponds to 6. Therefore, p corresponds to 6 degrees, i.e. 8 p corresponds to 6 degrees, i.e. 6 and so on, giving the angles at the centre of the circle for the parts of the retail price as: 8,6,7, 6 and 8, respectively. The pie diagram is shown in Fig... Figure. Problem 7. a Using the data given in Fig.. only, calculate the amount of money paid to each salesman for travelling epenses, if they are paid an allowance of 7 p per mile. b Using the data presented in Fig.., comment on the housing trends over the threeyear period. c etermine the profit made by selling 7 units of the product shown in Fig... a By measuring the length of rectangle P the mileage covered by salesman P is equivalent to miles. Hence salesman P receives a travelling allowance of 7, i.e..8 Similarly, for salesman Q, the miles travelled are 6 and his allowance is 6 7, i.e Salesman R travels 97 miles and he receives 97 7, i.e..89 Finally, salesman S receives 7, i.e..9 b An analysis of Fig.. shows that -roomed bungalows and -roomed houses are becoming more popular, the greatest change in the three years being a % increase in the sales of -roomed bungalows. c Since.8 corresponds to p and the profit occupies 8 of the pie diagram, then the profit per unit is 8, that is, 6 p.8 The profit when selling 7 units of the product is 7 6, that is, Now try the following eercise. Eercise 6 Further problems on presentation of ungrouped data. The number of vehicles passing a stationary observer on a road in si ten-minute intervals is as shown. raw a pictogram to represent these data.

97 PRESENTATION OF STATISTICAL ATA STATISTICS AN PROBABILITY Period of Time 6 Number of Vehicles If one symbol is used to represent vehicles, working correct to the nearest vehicles, gives,,6,7,and symbols respectively.. The number of components produced by a factory in a week is as shown below: ay Number of Components Mon 8 Tues 9 Wed 8 Thur 8 Fri 8 Show these data on a pictogram. If one symbol represents components, working correct to the nearest components gives: Mon 8, Tues, Wed 9, Thurs and Fri 6.. For the data given in Problem above, draw a horizontal bar chart. 6 equally spaced horizontal rectangles, whose lengths are proportional to,, 6, 68, 9 and, respectively.. Present the data given in Problem above on a horizontal bar chart. equally spaced horizontal rectangles, whose lengths are proportional to 8, 9, 8, 8 and 8 units, respectively.. For the data given in Problem above, construct a vertical bar chart. 6 equally spaced vertical rectangles, whose heights are proportional to,, 6, 68, 9 and units, respectively. 6. epict the data given in Problem above on a vertical bar chart. equally spaced vertical rectangles, whose heights are proportional to 8, 9, 8, 8 and 8 units, respectively. 7. A factory produces three different types of components. The percentages of each of these components produced for three, onemonth periods are as shown below. Show this information on percentage component bar charts and comment on the changing trend in the percentages of the types of component produced. Month Component P Component Q Component R Three rectangles of equal height, subdivided in the percentages shown in the columns above. P increases by % at the epense of Q and R 8. A company has five distribution centres and the mass of goods in tonnes sent to each centre during four, one-week periods, is as shown. Week Centre A Centre B Centre C 8 7 Centre 97 7 Centre E 8 9 Use a percentage component bar chart to present these data and comment on any trends. Four rectangles of equal heights, subdivided as follows: week : 8%, 7%, %, %, 8% week : %, 8%, %, %, 7% week : %, %, 9%, %, % week : %, 9%, 7%, 9%, %. Little change in centres A and B,a reduction of about 8% in C,an increase of about 7% in and a reduction of about % in E. J 9. The employees in a company can be split into the following categories: managerial, supervisory 9, craftsmen, semi-skilled 67, others. Show these data on a pie diagram. A circle of any radius, subdivided into sectors having angles of 7,,, 67 and, respectively.. The way in which an apprentice spent his time over a one-month period is as follows: drawing office hours, production 6 hours, training hours, at college 8 hours. Use a pie diagram to depict this information. A circle of any radius, subdivided into sectors having angles of 7, 6,9 and 68, respectively.. a With reference to Fig.., determine the amount spent on labour and materials to produce 6 units of the product. b If in year of Fig.., % corresponds to. dwellings, how many bungalows are sold in that year. a 9, b 88. a If the company sell units per annum of the product depicted in Fig.., determine the cost of their overheads per annum. b If % of the dwellings represented in year of Fig.. corresponds to dwellings, find the total number of houses sold in that year. a 6, b 8. Presentation of grouped data When the number of members in a set is small, say ten or less, the data can be represented diagrammatically without further analysis, by means of pictograms, bar charts, percentage components bar charts or pie diagrams as shown in Section.. For sets having more than ten members, those members having similar values are grouped together in classes to form a frequency distribution. To assist in accurately counting members in the various classes, a tally diagram is used see Problems 8 and. A frequency distribution is merely a table showing classes and their corresponding frequencies see Problems 8 and. The new set of values obtained by forming a frequency distribution is called grouped data. The terms used in connection with grouped data are shown in Fig..6a. The size or range of a class is given by the upper class boundary value minus the lower class boundary value, and in Fig..6 is , i.e... The class interval for the class shown in Fig..6b is 7. to 7.6 and the class mid-point value is given by, upper class boundary value and in Fig..6 is Figure.6 + lower class boundary value , i.e. 7.. One of the principal ways of presenting grouped data diagrammatically is by using a histogram, in which the areas of vertical, adjacent rectangles are made proportional to frequencies of the classes see Problem 9. When class intervals are equal, the heights of the rectangles of a histogram are equal to the frequencies of the classes. For histograms having unequal class intervals, the area must be proportional to the frequency. Hence, if the class interval of class A is twice the class interval of class B, then for equal frequencies, the height of the rectangle representing

98 PRESENTATION OF STATISTICAL ATA STATISTICS AN PROBABILITY A is half that of B see Problem. Another method of presenting grouped data diagrammatically is by using a frequency polygon, which is the graph produced by plotting frequency against class mid-point values and joining the co-ordinates with straight lines see Problem. A cumulative frequency distribution is a table showing the cumulative frequency for each value of upper class boundary. The cumulative frequency for a particular value of upper class boundary is obtained by adding the frequency of the class to the sum of the previous frequencies. A cumulative frequency distribution is formed in Problem. The curve obtained by joining the co-ordinates of cumulative frequency vertically against upper class boundary horizontally is called an ogive or a cumulative frequency distribution curve see Problem. Problem 8. The data given below refer to the gain of each of a batch of transistors, epressed correct to the nearest whole number. Form a frequency distribution for these data having seven classes The range of the data is the value obtained by taking the value of the smallest member from that of the largest member. Inspection of the set of data shows that, range = 89 7 = 8. The size of each class is given approimately by range divided by the number of classes. Since 7 classes are required, the size of each class is 8/7, that is, approimately. To achieve seven equal classes spanning a range of values from 7 to 89, the class intervals are selected as: 7 7, 7 7, and so on. To assist with accurately determining the number in each class, a tally diagram is produced, as shown intable.a. This is obtained by listing the classes in the left-hand column, and then inspecting each of the members of the set in turn and allocating them to the appropriate classes by putting s in the appropriate rows. Every fifth allocated to the particular row is shown as an oblique line crossing the four previous s, to help with final counting. A frequency distribution for the data is shown in Table.b and lists classes and their corresponding frequencies, obtained from the tally diagram. Class mid-point value are also shown in the table, since they are used for constructing the histogram for these data see Problem 9. Table.b Table.a Class Class mid-point Frequency Problem 9. Construct a histogram for the data given in Table.b. The histogram is shown in Fig..7. The width of the rectangles correspond to the upper class boundary values minus the lower class boundary values and the heights of the rectangles correspond to the class frequencies. The easiest way to draw a histogram is to mark the class mid-point values on the horizontal scale and draw the rectangles symmetrically about the appropriate class mid-point values and touching one another. Figure.7 J Problem. The amount of money earned weekly by people working part-time in a factory, correct to the nearest, is shown below. Form a frequency distribution having 6 classes for these data Inspection of the set given shows that the majority of the members of the set lie between 8 and and that there are a much smaller number of etreme values ranging from to 7. If equal class intervals are selected, the frequency distribution obtained does not give as much information as one with unequal class intervals. Since the majority of members are between 8 and, the class intervals in this range are selected to be smaller than those outside of this range. There is no unique solution and one possible solution is shown in Table.. Problem. raw a histogram for the data given in Table. When dealing with unequal class intervals, the histogram must be drawn so that the areas, and not the heights, of the rectangles are proportional to the frequencies of the classes. The data given are shown Table. Table. 6 Class Frequency Upper class boundary Lower class boundary Class range Height of rectangle Class Frequency in columns and of Table.. Columns and give the upper and lower class boundaries, respectively. In column, the class ranges i.e. upper class boundary minus lower class boundary values are listed. The heights of the rectangles are proportional to the ratio frequency, as shown in column 6. The class range histogram is shown in Fig..8. Problem. The masses of ingots in kilograms are measured correct to the nearest. kg and the results are as shown below. Produce a frequency distribution having about 7 classes for these data and then present the grouped data as a a frequency polygon and b a histogram = 6 = = 9 = = =

99 PRESENTATION OF STATISTICAL ATA 6 STATISTICS AN PROBABILITY Table. draw a histogram is to mark class mid-point values on the horizontal scale and to draw the rectangles symmetrically about the appropriate class mid-point values and touching one another. A histogram for the data given in Table. is shown in Fig... Problem. The frequency distribution for the masses in kilograms of ingots is: Figure.8 The range of the data is the member having the largest value minus the member having the smallest value. Inspection of the set of data shows that: range = =. The size of each class is given approimately by range number of classes. Since about seven classes are required, the size of each class is./7, that is approimately., and thus the class limits are selected as 7. to 7., 7. to 7.6, 7.7 to 7.9, and so on. The class mid-point for the 7. to 7. class is , i.e. 7., for the 7. to 7.6 class is , i.e. 7., and so on. To assist with accurately determining the number in each class, a tally diagram is produced as shown in Table.. This is obtained by listing the classes in the left-hand column and then inspecting each of the members of the set of data in turn and allocating it to the appropriate class by putting a in the appropriate row. Each fifth allocated to a particular row is marked as an oblique line to help with final counting. A frequency distribution for the data is shown in Table. and lists classes and their corresponding frequencies. Class mid-points are also shown in this table, since they are used when constructing the frequency polygon and histogram. A frequency polygon is shown in Fig..9, the co-ordinates corresponding to the class midpoint/frequency values, given in Table.. The co-ordinates are joined by straight lines and the polygon is anchored-down at each end by joining to the net class mid-point value and zero frequency. A histogram is shown in Fig.., the width of a rectangle corresponding to upper class boundary value lower class boundary value and height corresponding to the class frequency. The easiest way to Table. Class Class mid-point Frequency 7. to to to to to to to Figure.9 Figure. J 7. to 7., 7. to 7.6, 7.7 to 7.9 9, 8. to 8., 8. to 8., 8.6 to 8.8, 6, 8.9 to 9., Form a cumulative frequency distribution for these data and draw the corresponding ogive. A cumulative frequency distribution is a table giving values of cumulative frequency for the value of upper class boundaries, and is shown in Table.6. Columns and show the classes and their frequencies. Column lists the upper class boundary values for the classes given in column. Column gives the cumulative frequency values for all frequencies less than the upper class boundary values given in column. Thus, for eample, for the 7.7 to 7.9 class shown in row, the cumulative frequency value is the sum of all frequencies having values of less than 7.9, i.e = 7, and so on. The ogive for the cumulative frequency distribution given in Table.6 is shown in Fig... The co-ordinates corresponding to each upper class boundary/cumulative frequency value are plotted and the co-ordinates are joined by straight lines not the best curve drawn through the co-ordinates as in eperimental work. The ogive is anchored at its start by adding the co-ordinate 7.,. Table.6 Class Frequency Upper Class Cumulative boundary frequency Less than Figure. Now try the following eercise. Eercise 7 Further problems on presentation of grouped data. The mass in kilograms, correct to the nearest one-tenth of a kilogram, of 6 bars of metal are as shown. Form a frequency distribution of about 8 classes for these data There is no unique solution, but one solution is: ; ; ; ;.. ;.. 7;..6 ;.7.8. raw a histogram for the frequency distribution given in the solution of Problem. Rectangles, touching one another, having mid-points of 9., 9., 9.7, 9.9,...and heights of,, 9, 7,...