About the HELM Project HELM (Helping Engineers Learn Mathematics) materials were the outcome of a three-year curriculum development project

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2 About the HELM Project HELM (Helping Engineers Lern Mthemtics) mterils were the outcome of three-yer curriculum development project undertken by consortium of five English universities led by Loughborough University, funded by the Higher Eduction Funding Council for Englnd under the Fund for the Development of Teching nd Lerning for the period October September 5. HELM ims to enhnce the mthemticl eduction of engineering undergrdutes through rnge of flexible lerning resources in the form of Workbooks nd web-delivered interctive segments. HELM supports two CAA regimes: n integrted web-delivered implementtion nd CD-bsed version. HELM lerning resources hve been produced primrily by tems of writers t six universities: Hull, Loughborough, Mnchester, Newcstle, Reding, Sunderlnd. HELM grtefully cknowledges the vluble support of collegues t the following universities nd colleges involved in the criticl reding, trilling, enhncement nd revision of the lerning mterils: Aston, Bournemouth & Poole College, Cmbridge, City, Glmorgn, Glsgow, Glsgow Cledonin, Glenrothes Institute of Applied Technology, Hrper Adms University College, Hertfordshire, Leicester, Liverpool, London Metropolitn, Mory College, Northumbri, Nottinghm, Nottinghm Trent, Oxford Brookes, Plymouth, Portsmouth, Queens Belfst, Robert Gordon, Royl Forest of Den College, Slford, Sligo Institute of Technology, Southmpton, Southmpton Institute, Surrey, Teesside, Ulster, University of Wles Institute Crdiff, West Kingswy College (London), West Notts College. HELM Contcts: Post: HELM, Mthemtics Eduction Centre, Loughborough University, Loughborough, LE 3TU. Emil: helm@lboro.c.uk Web: HELM Workbooks List Bsic Algebr 6 Functions of Complex Vrible Bsic Functions 7 Multiple Integrtion 3 Equtions, Inequlities & Prtil Frctions 8 Differentil Vector Clculus 4 Trigonometry 9 Integrl Vector Clculus 5 Functions nd Modelling 3 Introduction to Numericl Methods 6 Exponentil nd Logrithmic Functions 3 Numericl Methods of Approximtion 7 Mtrices 3 Numericl Initil Vlue Problems 8 Mtrix Solution of Equtions 33 Numericl Boundry Vlue Problems 9 Vectors 34 Modelling Motion Complex Numbers 35 Sets nd Probbility Differentition 36 Descriptive Sttistics Applictions of Differentition 37 Discrete Probbility Distributions 3 Integrtion 38 Continuous Probbility Distributions 4 Applictions of Integrtion 39 The Norml Distribution 5 Applictions of Integrtion 4 Smpling Distributions nd Estimtion 6 Sequences nd Series 4 Hypothesis Testing 7 Conics nd Polr Coordintes 4 Goodness of Fit nd Contingency Tbles 8 Functions of Severl Vribles 43 Regression nd Correltion 9 Differentil Equtions 44 Anlysis of Vrince Lplce Trnsforms 45 Non-prmetric Sttistics z-trnsforms 46 Relibility nd Qulity Control Eigenvlues nd Eigenvectors 47 Mthemtics nd Physics Miscellny 3 Fourier Series 48 Engineering Cse Studies 4 Fourier Trnsforms 49 Student s Guide 5 Prtil Differentil Equtions 5 Tutor s Guide Copyright Loughborough University, 6

3 Contents 4 Applictions of Integrtion 4. Integrtion s the Limit of Sum 4. The Men Vlue nd the Root-Men-Squre Vlue 4.3 Volumes of Revolution 4.4 Lengths of Curves nd Surfces of Revolution 7 Lerning outcomes In this Workbook you will lern to interpret n integrl s the limit of sum. You will lern how to pply this pproch to the mening of n integrl to clculte importnt ttributes of curve: the re under the curve, the length of curve segment, the volume nd surfce re obtined when segment of curve is rotted bout n xis. Other quntities of interest which cn lso be clculted using integrtion is the position of the centre of mss of plne lmin nd the moment of inerti of lmin bout n xis. You will lso lern how to determine the men vlue of n integl.

4 The Men Vlue nd the Root-Men-Squre Vlue 4. Introduction Currents nd voltges often vry with time nd engineers my wish to know the men vlue of such current or voltge over some prticulr time intervl. The men vlue of time-vrying function is defined in terms of n integrl. An ssocited quntity is the root-men-squre (r.m.s). For exmple, the r.m.s. vlue of current is used in the clcultion of the power dissipted by resistor. Prerequisites Before strting this Section you should... Lerning Outcomes On completion you should be ble to... be ble to clculte definite integrls be fmilir with tble of trigonometric identities clculte the men vlue of function clculte the root-men-squre vlue of function HELM (8): Workbook 4: Applictions of Integrtion

5 . Averge vlue of function Suppose time-vrying function f(t) is defined on the intervl t b. The re, A, under the grph of f(t) is given by the integrl A = f(t) b f(t) dt. This is illustrted in Figure 5. f(t) m b t b t () the re under the curve from t = to t = b Figure 5 (b) the re under the curve nd the re of the rectngle re equl On Figure 3 we hve lso drwn rectngle with bse spnning the intervl t b nd which hs the sme re s tht under the curve. Suppose the height of the rectngle is m. Then re of rectngle = re under curve m(b ) = b f(t) dt m = b The vlue of m is the men vlue of the function cross the intervl t b. b f(t) dt Key Point The men vlue of function f(t) in the intervl t b is b b f(t) dt The men vlue depends upon the intervl chosen. If the vlues of or b re chnged, then the men vlue of the function cross the intervl from to b will in generl chnge s well. Exmple Find the men vlue of f(t) =t over the intervl t 3. Solution Using Key Point with =nd b =3nd f(t) =t men vlue = b b f(t) dt = 3 3 t dt = t = 3 3 HELM (8): Section 4.: The Men Vlue nd the Root-Men-Squre Vlue

6 Volumes of Revolution 4.3 Introduction In this Section we show how the concept of integrtion s the limit of sum, introduced in Section 4., cn be used to find volumes of solids formed when curves re rotted round the x or y xis. Prerequisites Before strting this Section you should... Lerning Outcomes On completion you should be ble to... be ble to clculte definite integrls understnd integrtion s the limit of sum clculte volumes of revolution HELM (8): Workbook 4: Applictions of Integrtion

7 Lengths of Curves nd Surfces of Revolution 4.4 Introduction Integrtion cn be used to find the length of curve nd the re of the surfce generted when curve is rotted round n xis. In this Section we stte nd use formule for doing this. Prerequisites Before strting this Section you should... Lerning Outcomes On completion you should be ble to... be ble to clculte definite integrls find the length of curves find the re of the surfce generted when curve is rotted bout n xis HELM (8): Section 4.4: Lengths of Curves nd Surfces of Revolution 7

8 . The length of curve To find the length of curve in the xy plne we first divide the curve into lrge number of pieces. We mesure (or, t lest, pproximte) the length of ech piece nd then by n obvious summtion process obtin n estimte for the length of the curve. Theoreticlly, we llow the number of pieces to increse without bound, implying tht the length of ech piece will tend to zero. In this limit the summtion process becomes n integrtion process. y δx δy y(x) Figure Figure shows the portion of the curve y(x) between x = nd x = b. A smll piece of this curve hs been selected nd cn be considered s the hypotenuse of tringle with bse δx nd height δy. (Here δx nd δy re intended to be smll so tht the curved segment cn be regrded s stright segment.) δy Using Pythgors theorem, the length of the hypotenuse is: δx + δy = + δx δx By summing ll such contributions between x = nd x = b, nd letting δx we obtin n expression for the totl length of the curve: δy + δx δx x=b lim δx x= But we lredy know how to write such n expression in terms of n integrl. We obtin the following result: b x Key Point 7 Given curve with eqution y = f(x), then the length of the curve between the points where x = nd x = b is given by the formul: b dy + dx dx Becuse of the complicted form of the integrnd, nd in prticulr the squre root, integrls of this type re often difficult to clculte. In prctice, pproximte numericl methods rther thn exct methods re normlly needed to perform the integrtion. We shll first illustrte the ppliction of the formul in Key Point 7 by problem which could be clculted in much simpler wy, before looking t some hrder problems. 8 HELM (8): Workbook 4: Applictions of Integrtion

9 Exmple 5 Find the length of the curve y =3x +between x =nd x =5. Solution In this Exmple, the curve is in fct stright line segment, nd its length could be obtined using Pythgors theorem without the need for integrtion. Notice from the formul in Key Point 7 tht it is necessry to find dy, which in this cse is 3. dx Applying the formul we find length of curve = = (3) dx dx 5 = x = (5 ) = 4 =.65 to d.p. Thus the length of the curve y =3x +between the points where x =nd x =5is.65 units. Tsk Find the length of the curve y = cosh x between x =nd x =shown in the digrm. y y = cosh x x First write down dy dx : dy dx = dy dx = sinh x HELM (8): Section 4.4: Lengths of Curves nd Surfces of Revolution 9

10 Hence write down the required integrl: + sinh x dx This integrl cn be evluted by mking use of the hyperbolic identity cosh x sinh x. Write down the integrl which results fter pplying this identity: cosh x dx Perform the integrtion to find the required length: sinh x =3.63 to d.p. Thus the length of y = cosh x between x =nd x =is 3.63 units. The next Tsk is more complicted still nd requires the use of hyperbolic substitution nd knowledge of the hyperbolic identities. Tsk Find the length of the curve y = x between x =nd x =3. Given y = x then dy dx integrl required: =x. Use this result nd pply the formul in Key Point 7 to obtin the 3 +4x dx 3 HELM (8): Workbook 4: Applictions of Integrtion

11 Mke the substitution x = dx sinh u, giving du = cosh u, to obtin n integrl in terms of u: sinh 6 + sinh u cosh u du Use the hyperbolic identity cosh u sinh u to eliminte sinh u: sinh 6 cosh u du Use the hyperbolic identity cosh u (cosh u +)to rewrite the integrnd in terms of cosh u: 4 sinh 6 (cosh u +)du Finlly, perform the integrtion to complete the clcultion: 4 sinh 6 (cosh u +)du = sinh sinh u 6 + u 4 = 9.75 to d.p. Thus the length of the curve y = x between x =nd x =3is 9.75 units. HELM (8): Section 4.4: Lengths of Curves nd Surfces of Revolution 3

12 Exercises. Find the length of the line y =x +7between x =nd x =3using the technique of this Section. Verify your result from your knowledge of the stright line.. Find the length of y = x 3/ between x =nd x =5. 3. Clculte the length of the curve y =4x 3 between x =nd x =, in the first qudrnt. s The distnce is from (.9) to (3, 3) long the line. This is given using Pythgors theorem s +4 = = (first qudrnt only).. The re of surfce of revolution In Section 4. we found n expression for the volume of solid of revolution. Here we consider the more complicted problem of formulting n expression for the surfce re of solid of revolution. y(x) y ( x, y) δx δy b x Figure Figure shows the portion of the curve y(x) between x = nd x = b which is rotted round the x xis through 36. A smll disc, of thickness δx, of the solid of revolution hs been selected. Its rdius is y nd so its circumference hs length πy. (As usul we ssume δx is smll so tht the curved prt of y(x) representing the hypotenuse of the highlighted tringle cn be regrded s stright). This surfce ribbon, shown shded, hs length πy nd width (δx) +(δy) nd so its re is, to good pproximtion, πy (δx) +(δy). We now let δx to obtin the result in Key Point 8: 3 HELM (8): Workbook 4: Applictions of Integrtion

13 Key Point 8 Given curve with eqution y = f(x), then the surfce re of the solid generted by rotting tht prt of the curve between the points where x = nd x = b round the x xis is given by the formul: b dy re of surfce = πy + dx dx Tsk Find the re of the surfce generted when the prt of the curve y = x 3 between x =nd x =4is rotted round the x xis. Using Key Point 8 write down the integrl: re = b πy + dy 4 dx = πx +(3x 3 dx ) dx = 4 πx 3 +9x 4 dx Use the substitution u =+9x 4 so du dx = 36x3 to write down the integrl in terms of u: π 8 35 u du Perform the integrtion: π u 3/ HELM (8): Section 4.4: Lengths of Curves nd Surfces of Revolution 33

14 Apply the limits of integrtion to find the re: π (35) 3/ 7 Exercises. The line y = x between x =nd x =is rotted round the x xis. () Find the re of the surfce generted. (b) Verify this result by finding the curved surfce re of the corresponding cone. curved surfce re of cone of rdius r nd slnt height is πr.) (The. Find the re of the surfce generted when y = x in the intervl x is rotted bout the x xis. s. π HELM (8): Workbook 4: Applictions of Integrtion

15 . Volumes generted by rotting curves bout the x-xis Figure 8 shows grph of the function y =x for x between nd 3. 6 y y = x O 3 x Figure 8: A grph of the function y =x, for x 3 Imgine rotting the line y =x by one complete revolution (36 or π rdins) round the x-xis. The surfce so formed is the surfce of cone s shown in Figure 9. Such three-dimensionl shpe is known s solid of revolution. We now discuss how to obtin the volumes of such solids of revolution. 6 y y = x O 3 x Figure 9: When the line y =x is rotted round the xis, solid is generted Tsk Find the volume of the cone generted by rotting y =x, for x 3, round the x-xis, s shown in Figure 9. In order to find the volume of this solid we ssume tht it is composed of lots of thin circulr discs ll ligned perpendiculr to the x-xis, such s tht shown in the digrm. From the digrm below we note tht typicl disc hs rdius y, which in this cse equls x, nd thickness δx. HELM (8): Section 4.3: Volumes of Revolution

16 y 6 y = x (x, y) δx O 3 x The cone is divided into number of thin circulr discs. The volume of circulr disc is the circulr re multiplied by the thickness. Write down n expression for the volume of this typicl disc: π(x) δx =4πx δx To find the totl volume we must sum the contributions from ll discs nd find the limit of this sum s the number of discs tends to infinity nd δx tends to zero. Tht is x=3 lim δx x= 4πx δx This is the definition of definite integrl. Write down the corresponding integrl: 3 4πx dx Find the required volume by performing the integrtion: HELM (8): Workbook 4: Applictions of Integrtion

17 4πx = 36π Tsk A grph of the function y = x for x between nd 4 is shown in the digrm. The grph is rotted round the x-xis to produce the solid shown. Find its volume. y 6 y = x (x, y) O δx 4 x The solid of revolution is divided into number of thin circulr discs. As in the previous Tsk, the solid is considered to be composed of lots of circulr discs of rdius y, (which in this exmple is equl to x ), nd thickness δx. Write down the volume of ech disc: π(x ) δx = πx 4 δx Write down the expression which represents summing the volumes of ll such discs: x=4 πx 4 δx x= Write down the integrl which results from tking the limit of the sum s δx : HELM (8): Section 4.3: Volumes of Revolution 3

18 4 πx 4 dx Perform the integrtion to find the volume of the solid: 4 5 π 5 = 4.8π Tsk In generl, suppose the grph of y(x) between x = nd x = b is rotted bout the x-xis, nd the solid so formed is considered to be composed of lots of circulr discs of thickness δx. Write down n expression for the rdius of typicl disc: y Write down n expression for the volume of typicl disc: πy δx The totl volume is found by summing these individul volumes nd tking the limit s δx tends to zero: x=b lim δx x= πy δx Write down the definite integrl which this sum defines: b πy dx 4 HELM (8): Workbook 4: Applictions of Integrtion

19 Key Point 5 If the grph of y(x), between x = nd x = b, is rotted bout the x-xis the volume of the solid formed is b πy dx Exercises. Find the volume of the solid formed when tht prt of the curve between y = x between x =nd x =is rotted bout the x-xis.. The prbol y =4x for x, is rotted round the x-xis. Find the volume of the solid formed. s. 3π/5,. π.. Volumes generted by rotting curves bout the y-xis We cn obtin different solid of revolution by rotting curve round the y-xis insted of round the x-xis. See Figure. y y(x) δy (x, y) O x Figure : A solid generted by rottion round the y-xis To find the volume of this solid it is divided into number of circulr discs s before, but this time the discs re horizontl. The rdius of typicl disc is x nd its thickness is δy. The volume of the disc will be πx δy. The totl volume is found by summing these individul volumes nd tking the limit s δy. If the lower nd upper limits on y re c nd d, we obtin for the volume: y=d lim δy y=c πx δy which is the definite integrl d c πx dy HELM (8): Section 4.3: Volumes of Revolution 5

20 Key Point 6 If the grph of y(x), between y = c nd y = d, is rotted bout the y-xis the volume of the solid formed is d c πx dy Tsk Find the volume generted when the grph of y = x between x =nd x = is rotted round the y-xis. Using Key Point 6 write down the required integrl: πx dy This integrl cn be written entirely in terms of y, using the fct tht y = x to eliminte x. Do this now, nd then evlute the integrl: πx dy = πy πy dy = = π Exercises. The curve y = x for <x< is rotted bout the y-xis. Find the volume of the solid formed.. The line y = x for x is rotted round the y-xis. Find the volume of revolution. s. 5π. 6π 3. 6 HELM (8): Workbook 4: Applictions of Integrtion

21 Tsk Find the men vlue of f(t) =t over the intervl t 5. Use Key Point with =nd b =5to write down the required integrl: men vlue = 5 5 t dt Now evlute the integrl: men vlue = 5 5 t dt = 3 t = = = 3 Engineering Exmple Sonic boom Introduction Impulsive signls re described by their pek mplitudes nd their durtion. Another quntity of interest is the totl energy of the impulse. The effect of blst wve from n explosion on structures, for exmple, is relted to its totl energy. This Exmple looks t the clcultion of the energy on sonic boom. Sonic booms re cused when n ircrft trvels fster thn the speed of sound in ir. An idelized sonic-boom pressure wveform is shown in Figure 6 where the instntneous sound pressure p(t) is plotted versus time t. This wve type is often clled n N-wve becuse it resembles the shpe of the letter N. The energy in sound wve is proportionl to the squre of the sound pressure. P p(t) T t P Figure 6: An idelized sonic-boom pressure wveform HELM (8): Workbook 4: Applictions of Integrtion

22 Problem in words Clculte the energy in n idel N-wve sonic boom in terms of its pek pressure, its durtion nd the density nd sound speed in ir. Mthemticl sttement of problem Represent the positive pek pressure by P nd the durtion by T. The totl coustic energy E crried cross unit re norml to the sonic-boom wve front during time T is defined by E = <p(t) > T/ρc where ρ is the ir density, c the speed of sound nd the time verge of [p(t)] is <p(t) > = T T p(t) dt () Find n pproprite expression for p(t). () () (b) Hence show tht E cn be expressed in terms of P,T,ρnd c s E = TP 3ρc. Mthemticl nlysis () The intervl of integrtion needed to compute () is [,T]. Therefore it is necessry to find n expression for p(t) only in this intervl. Figure 6 shows tht, in this intervl, the dependence of the sound pressure p on the vrible t is liner, i.e. p(t) =t + b. From Figure 6 lso p() = P nd p(t )= P. The constnts nd b re determined from these conditions. At t =, +b = P implies tht b = P. At t = T, T + b = P implies tht = P /T. Consequently, the sound pressure in the intervl [,T] my be written p(t) = P T (b) This expression for p(t) my be used to compute the integrl () T T p(t) dt = T = T = P T T P T t + P dt = T 4P 3T t3 P T 4 3T T 3 T T + T T t + P t T =P /3. t + P. 4P t 4P T T t + P dt Hence, from Eqution (), the totl coustic energy E crried cross unit re norml to the sonicboom wve front during time T is E = TP 3ρc. Interprettion The energy in n N-wve is given by third of the sound intensity corresponding to the pek pressure multiplied by the durtion. HELM (8): Section 4.: The Men Vlue nd the Root-Men-Squre Vlue 3

23 Exercises. Clculte the men vlue of the given functions cross the specified intervl. () f(t) =+t cross [, ] (b) f(x) =x cross [, ] (c) f(t) =t cross [, ] (d) f(t) =t cross [, ] (e) f(z) =z + z cross [, 3]. Clculte the men vlue of the given functions over the specified intervl. () f(x) =x 3 cross [, 3] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(z) =z 3 cross [, ] 3. Clculte the men vlue of the following: () f(t) = sin t cross, π (b) f(t) = sin t cross [,π] (c) f(t) = sin ωt cross [,π] (d) f(t) = cos t cross, π (e) f(t) = cos t cross [,π] (f) f(t) = cos ωt cross [,π] (g) f(t) = sin ωt + cos ωt cross [, ] 4. Clculte the men vlue of the following functions: () f(t) = t +cross [, 3] (b) f(t) =e t cross [, ] (c) f(t) =+e t cross [, ] s. () (b) (c) (d) 4 (e) () (b).693 (c).948 (d) 3. () (b) (c) [ cos(πω)] π π πω (d) + sin ω cos ω (g) ω 4. () 4 (b).75 (c).75 9 π (e) (f) sin(πω) πω 4 HELM (8): Workbook 4: Applictions of Integrtion

24 . Root-men-squre vlue of function If f(t) is defined on the intervl t b, the men-squre vlue is given by the expression: b b [f(t)] dt This is simply the men vlue of [f(t)] over the given intervl. The relted quntity: the root-men-squre (r.m.s.) vlue is given by the following formul. Key Point 3 Root-Men-Squre Vlue b r.m.s vlue = [f(t)] b dt The r.m.s. vlue depends upon the intervl chosen. If the vlues of or b re chnged, then the r.m.s. vlue of the function cross the intervl from to b will in generl chnge s well. Note tht when finding n r.m.s. vlue the function must be squred before it is integrted. Exmple 3 Find the r.m.s. vlue of f(t) =t cross the intervl from t =to t =3. Solution r.m.s = b b [f(t)] dt = 3 3 [t ] dt = 3 t 4 dt = t HELM (8): Section 4.: The Men Vlue nd the Root-Men-Squre Vlue 5

25 Exmple 4 Clculte the r.m.s vlue of f(t) = sin t cross the intervl t π. Solution Here =nd b =π so r.m.s = π π sin t dt. The integrl of sin t is performed by using trigonometricl identities to rewrite it in the lterntive form ( cos t). This technique ws described in 3.7. π π ( cos t) sin t r.m.s. vlue = dt = t = π 4π 4π (π) = =.77 Thus the r.m.s vlue is.77 to 3 d.p. In the previous Exmple the mplitude of the sine wve ws, nd the r.m.s. vlue ws.77. In generl, if the mplitude of sine wve is A, its r.m.s vlue is.77a. Key Point 4 The r.m.s vlue of ny sinusoidl wveform tken cross n intervl of width equl to one period is.77 mplitude of the wveform. Engineering Exmple 3 Electrodynmic meters Introduction A dynmometer or electrodynmic meter is n nlogue instrument tht cn mesure d.c. current or.c. current up to frequency of khz. A typicl dynmometer is shown in Figure 7. It consists of circulr dynmic coil positioned in mgnetic field produced by two wound circulr sttor coils connected in series with ech other. The torque T on the moving coil depends upon the mutul inductnce between the coils given by: T = I I dm dθ 6 HELM (8): Workbook 4: Applictions of Integrtion

26 where I is the current in the fixed coil, I the current in the moving coil nd θ is the ngle between the coils. The torque is therefore proportionl to the squre of the current. If the current is lternting the moving coil is unble to follow the current nd the pointer position is relted to the men vlue of the squre of the current. The scle cn be suitbly grduted so tht the pointer position shows the squre root of this vlue, i.e. the r.m.s. current. Pointer Scle Moving coil Spring Fixed sttor coils Figure 7: An electrodynmic meter Problem in words A dynmometer is in circuit in series with 4 Ω resistor, rectifying device nd 4 V r.m.s lternting sinusoidl power supply. The rectifier resists current with resistnce of Ω in one direction nd resistnce of kω in the opposite direction. Clculte the reding indicted on the meter. Mthemticl Sttement of the problem We know from Key Point 4 in the text tht the r.m.s. vlue of ny sinusoidl wveform tken cross n intervl equl to one period is.77 mplitude of the wveform. Where.77 is n pproximtion of. This llows us to stte tht the mplitude of the sinusoidl power supply will be: V pek = V rms = V rms In this cse the r.m.s power supply is 4 V so we hve V pek = 4 = V During the prt of the cycle where the voltge of the power supply is positive the rectifier behves s resistor with resistnce of Ω nd this is combined with the 4 Ω resistnce to give resistnce of 6 Ω in totl. Using Ohm s lw V = IR I = V R As V = V pek sin(θ) where θ = ωt where ω is the ngulr frequency nd t is time we find tht during the positive prt of the cycle Irms = π sin(θ) dθ π 6 HELM (8): Section 4.: The Men Vlue nd the Root-Men-Squre Vlue 7

27 During the prt of the cycle where the voltge of the power supply is negtive the rectifier behves s resistor with resistnce of kω nd this is combined with the 4 Ω resistnce to give 4 Ω in totl. So we find tht during the negtive prt of the cycle Irms = π π π sin(θ) 4 dθ Therefore over n entire cycle Irms = π π sin(θ) 6 dθ + π π π We cn clculte this vlue to find I rms nd therefore I rms. Mthemticl nlysis Irms = π π sin(θ) 6 π Irms = sin (θ) π 36 dθ + π π π dθ + π π sin(θ) dθ sin(θ) dθ 4 sin (θ) 96 dθ Substituting the trigonometric identity sin (θ) cos(θ) we get I rms = = = π cos(θ) 4π π θ 36 sin(θ) 7 π dθ + + π π cos(θ) 96 θ 96 sin(θ) π 4π 36 + π = A 96 π π dθ I rms =.3 A to d.p. Interprettion The reding on the meter would be.3 A. 8 HELM (8): Workbook 4: Applictions of Integrtion

28 Exercises. Clculte the r.m.s vlues of the given functions cross the specified intervl. () f(t) =+t cross [, ] (b) f(x) =x cross [, ] (c) f(t) =t cross [, ] (d) f(t) =t cross [, ] (e) f(z) =z + z cross [, 3]. Clculte the r.m.s vlues of the given functions over the specified intervl. () f(x) =x 3 cross [, 3] (b) f(x) = x cross [, ] (c) f(t) = t cross [, ] (d) f(z) =z 3 cross [, ] 3. Clculte the r.m.s vlues of the following: () f(t) = sin t cross, π (b) f(t) = sin t cross [,π] (c) f(t) = sin ωt cross [,π] (d) f(t) = cos t cross, π (e) f(t) = cos t cross [,π] (f) f(t) = cos ωt cross [,π] (g) f(t) = sin ωt + cos ωt cross [, ] 4. Clculte the r.m.s vlues of the following functions: s () f(t) = t +cross [, 3] (b) f(t) =e t cross [, ] (c) f(t) =+e t cross [, ]. ().87 (b).575 (c).447 (d).7889 (e) ().4957 (b).77 (c) (d) ().77 (b).77 (c) (d).77 (e).77 (f) 4. ().58 (b).3466 (c).74 sin πω cos πω πω sin πω cos πω + πω (g) + sin ω ω HELM (8): Section 4.: The Men Vlue nd the Root-Men-Squre Vlue 9

29 Integrtion s the Limit of Sum 4. Introduction In 3, integrtion ws introduced s the reverse of differentition. A more rigorous tretment would show tht integrtion is process of dding or summtion. By viewing integrtion from this perspective it is possible to pply the techniques of integrtion to finding res, volumes, centres of grvity nd mny other importnt quntities. The content of this Section is importnt becuse it is here tht integrtion is defined more crefully. A thorough understnding of the process involved is essentil if you need to pply integrtion techniques to prcticl problems. Prerequisites Before strting this Section you should... Lerning Outcomes On completion you should be ble to... be ble to clculte definite integrls explin integrtion s the limit of sum evlute the limit of sum in simple cses HELM (8): Workbook 4: Applictions of Integrtion

30 . The limit of sum y re required y( x) b x Figure : The re under curve Consider the grph of the positive function y(x) shown in Figure. Suppose we re interested in finding the re under the grph between x = nd x = b. One wy in which this re cn be pproximted is to divide it into number of rectngles of equl width, find the re of ech rectngle, nd then dd up ll these individul rectngulr res. This is illustrted in Figure, which shows the re divided into n rectngles (with some smll discrepncies t the tops), nd Figure b which shows the dimensions of typicl rectngle which is locted t x = x k. y y n rectngles y( x) y(x k ) δx y(x) b x x k x () The re pproximted by n rectngles (b) A typicl rectngle Figure We wish to find n expression for the re under curve bsed on the sum of mny rectngles. Firstly, we note tht the distnce from x = to x = b is b. In Figure the re hs been divided into n rectngles. If n rectngles spn the distnce from to b the width of ech rectngle is b n : It is conventionl to lbel the width of ech rectngle s δx, i.e. δx = b. We lbel the x n coordintes t the left-hnd side of the rectngles s x, x up to x n (here x = nd x n+ = b). A typicl rectngle, the kth rectngle, is shown in Figure b. Note tht its height is y(x k ), so its re is y(x k ) δx. The sum of the res of ll n rectngles is then y(x )δx + y(x )δx + y(x 3 )δx + + y(x n )δx which we write concisely using sigm nottion s n y(x k )δx k= HELM (8): Section 4.: Integrtion s the Limit of Sum 3

31 This quntity gives us n estimte of the re under the curve but it is not exct. To improve the estimte we must tke lrge number of very thin rectngles. So, wht we wnt to find is the vlue of this sum when n tends to infinity nd δx tends to zero. We write this vlue s lim n n y(x k )δx k= The lower nd upper limits on the sum correspond to the first rectngle nd lst rectngle where x = nd x = b respectively nd so we cn write this limit in the equivlent form x=b lim δx x= y(x)δx Here, s the number of rectngles increses without bound we drop the subscript k from x k nd write y(x) which is the vlue of y t typicl vlue of x. If this sum cn ctully be found, it is clled the definite integrl of y(x), from x = to x = b nd it is written b () y(x)dx. You re lredy fmilir with the technique for evluting definite integrls which ws studied in Section 4.. Therefore we hve the following definition: Key Point The definite integrl b x=b y(x)dx is defined s lim y(x)δx δx x= Note tht the quntity δx represents the thickness of smll but finite rectngle. When we hve tken the limit s δx tends to zero to obtin the integrl, we write dx, which reminds us of the vrible of integrtion. This process of dividing n re into very smll regions, performing clcultion on ech region, nd then dding the results by mens of n integrl is very importnt. This will become pprent when finding volumes, centres of grvity, moments of inerti etc in the following Sections where similr procedures re followed. 4 HELM (8): Workbook 4: Applictions of Integrtion

32 Exmple The re under the grph of y = x between x =nd x =is to be found by pproximting it by lrge number of thin rectngles nd finding the limit of the sum of their res. From Eqution () this is lim δx x= y(x) δx. Write down the integrl which this sum defines nd evlute it to obtin the re under the curve. x= Solution The limit of the sum defines the integrl y(x)dx. Here y = x nd so x x 3 dx = 3 = 3 To show tht the process of tking the limit of sum ctully works we investigte the problem in detil. We use the ide of the limit of sum to find the re under the grph of y = x between x =nd x =, s illustrted in Figure 3. y y =x n rectngles x Figure 3: The re under y = x is pproximted by number of thin rectngles Tsk Refer to the digrm below to help you nswer the questions below. y y =x n rectngles x If the intervl between x =nd x =is divided into n rectngles wht is the width of ech rectngle? HELM (8): Section 4.: Integrtion s the Limit of Sum 5

33 /n Mrk this on the digrm. Wht is the x coordinte t the left-hnd side of the first rectngle? Wht is the x coordinte t the left-hnd side of the second rectngle? /n Wht is the x coordinte t the left-hnd side of the third rectngle? /n Mrk these coordintes on the digrm. Wht is the x coordinte t the left-hnd side of the kth rectngle? (k )/n Given tht y = x, wht is the y coordinte t the left-hnd side of the kth rectngle? k n The re of the kth rectngle is its height its width. Write down the re of the kth rectngle: k n n = (k ) n 3 6 HELM (8): Workbook 4: Applictions of Integrtion

34 To find the totl re A n of the n rectngles we must dd up ll these individul rectngulr res: n (k ) A n = k= n 3 This sum cn be simplified nd then clculted s follows. You will need to mke use of the formuls for the sum of the first n integers, nd the sum of the squres of the first n integers: n n =n, k = n n(n +), k = n(n + )(n +) 6 k= k= k= Then, the totl re of the rectngles is given by A n = n (k ) k= n 3 = n (k ) n 3 k= = n (k k +) n 3 k= = n 3 n k= k n k + k= n k= = n 3 n 6 (n + )(n +) n (n +)+n = (n + )(n +) n 6 = (n + )(n +) n 6 (n +)+ n = 6n n 3n + = 3 n + 6n Note tht this is formul for the exct totl re of the n rectngles. It is n estimte of the re under the grph of y = x. However, s n gets lrger, the terms n nd become smll nd will 6n eventully tend to zero. If we let n tend to infinity we obtin the exct nswer of 3. The required re is. It hs been found s the limit of sum nd of course grees with tht 3 clculted by integrtion. In the clcultions which follow in subsequent Sections the need to evlute complicted limits like this is voided by performing the integrtion using the techniques of 3. Nevertheless it will sometimes be necessry to go through the process of dividing region into smll sections, performing clcultion on ech section nd then dding the results, in order to formulte the integrl required. When numericl methods of integrtion re studied ( 3) this summtion method will prove fundmentl. HELM (8): Section 4.: Integrtion s the Limit of Sum 7

35 Engineering Exmple Pulley belt tension Problem Consider tht belt is prtilly wound round pulley so tht there is difference in the tension either side of the pulley (see Figure 4). The pulley will be sttionry s long s the friction between belt nd pulley is sufficient. The frictionl force on the pulley will depend on the extent of the contct between belt nd pulley i.e. on the ngle θ shown in Figure 4. Given tht the tensions on either side of the belt re T nd T nd tht the coefficient of friction between belt nd pulley is µ, find n expression for T in terms of T, µ nd θ. Solution Consider smll element of the belt, t ngle θ where the tension is T. Chnging the ngle by smll mount θ chnges the tension from T to T + T. θ R θ T T Figure 4 Tke moments bout the centre of the pulley, denoting the rdius of the pulley by R nd ssuming tht the frictionl force is µt per unit length. For the pulley to remin sttionry, R θµt = R(T + T ) RT or θ = T µt. Using integrtion s the limit of sum, θ = T T dt µt = T ln T = µ T µ ln T T. So T = T e µθ. 8 HELM (8): Workbook 4: Applictions of Integrtion

36 Exercises. Find the re under y = x +from x =to x = using the limit of sum.. Find the re under y =3x from x =to x =using the limit of sum. 3. Write down, but do not evlute, the integrl defined by the limit s δx, orδt of the following sums: () s. 6,. 8, x= x 3 δx, x= (b) x=4 4πx δx, x= (c) t= t 3 δt, t= (d) x= 6mx δx. x= 3. () x 3 dx, (b) 4π 4 x dx, (c) t 3 dt, (d) 6m x dx. HELM (8): Section 4.: Integrtion s the Limit of Sum 9

37 Index for Workbook 4 Are 3 Cone, 34 Curve length 8 Definite integrl 4 Dynmometer 6 Electrodynmic meter 3 Energy 3 Integrtion s summtion 3-7, 3 Length of curve 8 Limit 3, Men vlue Pulley belt 8 R.M.S. 5 Root-men-squre vlue 5- Rottion - bout x-xis - bout y-xis 5 Solid of revolution Sonic boom Surfce of revolution 3 Tension 8 Volume of revolution -6 EXERCISES 9, 4, 9, 6, 34 ENGINEERING EXAMPLES Pulley belt tension 8 Sonic boom 3 Electrodynmic meters 6

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