Engineering mathematic  a foundation for electronic, electrical, communications and systems engineers

ENGINEERING MATHEMATICS A Foundation for Electronic, Electrical, Communications and Systems Engineers FIFTH EDITION Anthony Croft • Robert Davison Martin Hargreaves • James Flint Tai ngay!!! Ban co the xoa dong chu nay!!! Engineering Mathematics At Pearson, we have a simple mission: to help people make more of their lives through learning We combine innovative learning technology with trusted content and educational expertise to provide engaging and effective learning experiences that serve people wherever and whenever they are learning From classroom to boardroom, our curriculum materials, digital learning tools and testing programmes help to educate millions of people worldwide – more than any other private enterprise Every day our work helps learning flourish, and wherever learning flourishes, so people To learn more, please visit us at www.pearson.com/uk Fifth Edition Engineering Mathematics A Foundation for Electronic, Electrical, Communications and Systems Engineers Anthony Croft Loughborough University Robert Davison Martin Hargreaves Chartered Physicist James Flint Loughborough University Harlow, England • London • New York • Boston • San Francisco • Toronto • Sydney Dubai • Singapore • Hong Kong • Tokyo • Seoul • Taipei • New Delhi Cape Town • São Paulo • Mexico City • Madrid • Amsterdam • Munich • Paris • Milan PEARSON EDUCATION LIMITED Edinburgh Gate Harlow CM20 2JE United Kingdom Tel: +44 (0)1279 623623 Web: www.pearson.com/uk First edition published under the Addison-Wesley imprint 1992 (print) Second edition published under the Addison-Wesley imprint 1996 (print) Third edition published under the Prentice Hall imprint 2001 (print) Fourth edition published 2013 (print and electronic) Fifth edition published 2017 (print and electronic) © Addison-Wesley Publishers Limited 1992, 1996 (print) © Pearson Education Limited 2001 (print) © Pearson Education Limited 2013, 2017 (print and electronic) The rights of Anthony Croft, Robert Davison, Martin Hargreaves and James Flint to be identified as authors of this work have been asserted by them in accordance with the Copyright, Designs and Patents Act 1988 The print publication is protected by copyright Prior to any prohibited 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sites ISBN: 978-1-292-14665-2 (print) 978-1-292-14667-6 (PDF) 978-1-292-14666-9 (ePub) British Library Cataloguing-in-Publication Data A catalogue record for the print edition is available from the British Library Library of Congress Cataloging-in-Publication Data Names: Croft, Tony, 1957– author Title: Engineering mathematics : a foundation for electronic, electrical, communications and systems engineers / Anthony Croft, Loughborough University, Robert Davison, De Montfort University, Martin Hargreaves, De Montfort University, James Flint, Loughborough University Description: Fifth edition | Harlow, England ; New York : Pearson, 2017 k Revised edition of: Engineering mathematics : a foundation for electronic, electrical, communications, and systems engineers / Anthony Croft, Robert Davison, Martin Hargreaves 3rd editon 2001 | Includes index Identifiers: LCCN 2017011081| ISBN 9781292146652 (Print) | ISBN 9781292146676 (PDF) | ISBN 9781292146669 (ePub) Subjects: LCSH: Engineering mathematics | Electrical engineering–Mathematics | Electronics–Mathematics Classification: LCC TA330 C76 2017 | DDC 510–dc23 LC record available at https://lccn.loc.gov/2017011081 A catalog record for the print edition is available from the Library of Congress 10 21 20 19 18 17 Print edition typeset in 10/12 Times Roman by iEnerziger Aptara® , Ltd Printed in Slovakia by Neografia NOTE THAT ANY PAGE CROSS REFERENCES REFER TO THE PRINT EDITION To Kate, Tom and Harvey A.C To Kathy R.D To my father and mother M.H To Suzanne, Alexandra and Dominic J.F Contents Chapter Chapter Chapter Preface xvii Acknowledgements xix Review of algebraic techniques 1.1 1.2 Introduction Laws of indices 1.3 1.4 Number bases Polynomial equations 11 1.5 1.6 Algebraic fractions Solution of inequalities 26 1.7 1.8 Partial fractions Summation notation 39 Review exercises 50 20 33 46 Engineering functions 54 2.1 2.2 Introduction Numbers and intervals 54 2.3 2.4 Basic concepts of functions Review of some common engineering functions and techniques Review exercises 56 55 70 113 The trigonometric functions 115 3.1 3.2 3.3 Introduction Degrees and radians The trigonometric ratios 115 3.4 3.5 The sine, cosine and tangent functions The sinc x function 120 3.6 3.7 Trigonometric identities Modelling waves using sin t and cos t 3.8 Trigonometric equations Review exercises 116 116 123 125 131 144 150 viii Contents Chapter Chapter Chapter Chapter Chapter Coordinate systems 154 4.1 Introduction 154 4.2 4.3 Cartesian coordinate system – two dimensions Cartesian coordinate system – three dimensions 154 4.4 4.5 Polar coordinates Some simple polar curves 159 4.6 4.7 Cylindrical polar coordinates Spherical polar coordinates Review exercises 166 157 163 170 173 Discrete mathematics 175 5.1 5.2 5.3 Introduction Set theory Logic 175 5.4 Boolean algebra Review exercises 185 175 183 197 Sequences and series 200 6.1 Introduction 200 6.2 6.3 Sequences Series 201 6.4 6.5 The binomial theorem Power series 214 6.6 Sequences arising from the iterative solution of non-linear equations Review exercises 209 218 219 222 Vectors 224 7.1 7.2 7.3 Introduction Vectors and scalars: basic concepts Cartesian components 224 7.4 7.5 Scalar fields and vector fields The scalar product 240 7.6 7.7 The vector product Vectors of n dimensions 246 Review exercises 255 224 232 241 253 Matrix algebra 257 8.1 8.2 257 Introduction Basic definitions 258 Contents 8.3 Addition, subtraction and multiplication 259 8.4 8.5 Using matrices in the translation and rotation of vectors Some special matrices 267 8.6 8.7 The inverse of a × matrix Determinants 274 8.8 The inverse of a × matrix 8.9 Application to the solution of simultaneous equations 8.10 Gaussian elimination Chapter 10 278 281 283 286 294 8.13 Iterative techniques for the solution of simultaneous equations 8.14 Computer solutions of matrix problems 312 307 319 321 Complex numbers 324 9.1 9.2 Introduction Complex numbers 324 9.3 9.4 Operations with complex numbers Graphical representation of complex numbers 328 9.5 9.6 9.7 Polar form of a complex number Vectors and complex numbers The exponential form of a complex number 333 9.8 9.9 Phasors De Moivre’s theorem 340 325 332 336 337 344 9.10 Loci and regions of the complex plane Review exercises 351 Differentiation 356 10.1 Introduction 356 10.2 Graphical approach to differentiation 10.3 Limits and continuity 357 10.4 Rate of change at a specific point 10.5 Rate of change at a general point 362 10.6 Existence of derivatives 10.7 Common derivatives 10.8 Differentiation as a linear operator 370 Review exercises 10 Chapter 11 271 8.11 Eigenvalues and eigenvectors 8.12 Analysis of electrical networks Review exercises Chapter ix 354 358 364 372 375 385 Techniques of differentiation 386 11.1 Introduction 386 x Contents 11.2 Rules of differentiation 386 11.3 Parametric, implicit and logarithmic differentiation 11.4 Higher derivatives 393 Review exercises 11 Chapter 12 Chapter 13 Chapter 14 Chapter 15 404 Applications of differentiation 406 12.1 Introduction 12.2 Maximum points and minimum points 406 12.3 Points of inflexion 12.4 The Newton–Raphson method for solving equations 415 12.5 Differentiation of vectors Review exercises 12 423 Integration 428 406 418 427 13.1 Introduction 428 13.2 Elementary integration 13.3 Definite and indefinite integrals Review exercises 13 429 Techniques of integration 457 14.1 Introduction 14.2 Integration by parts 14.3 Integration by substitution 457 442 453 457 463 14.4 Integration using partial fractions Review exercises 14 468 Applications of integration 471 15.1 Introduction 15.2 Average value of a function 15.3 Root mean square value of a function 471 Review exercises 15 Chapter 16 400 466 471 475 479 Further topics in integration 480 16.1 Introduction 16.2 Orthogonal functions 480 16.3 Improper integrals 16.4 Integral properties of the delta function 483 16.5 Integration of piecewise continuous functions 16.6 Integration of vectors 491 Review exercises 16 480 489 493 494 402 Chapter 11 Techniques of differentiation d3 y or y′′′ or y(3) The fourth derivative dx d5 y d4 y is written or yiv or y(4) The fifth derivative is written or yv or y(5) dx dx notation is used The third derivative is written Example 11.21 Find the first five derivatives of z(t ) = 2t + sin t Solution z′ = 6t + cos t z′′ = 12t − sin t z′′′ = 12 − cos t ziv = sin t zv = cos t Example 11.22 Calculate the values of x for which y′′ = 0, given y = x4 − x3 Solution y = x4 − x3 y′ = 4x3 − 3x2 y′′ = 12x2 − 6x Putting y′′ = gives 12x2 − 6x = and so 6x(2x − 1) = Hence x = 0, The first and second derivatives can be used to describe the nature of increasing and decreasing functions In Figure 11.2(a, b) the tangents to the curves have positive gradients, that is y′ > As can be seen, as x increases the value of the function increases Conversely, in Figure 11.2(c, d) the tangents have negative gradients (y′ < 0) and as x increases the value of the function decreases The sign of the first derivative tells us whether y is increasing or decreasing However, the curves in (a) and (b) both show y increasing but, clearly, there is a difference in the way y changes Consider again Figure 11.2(a) The tangents at A, B and C are shown As x increases the gradient of the tangent increases, that is y′ increases as x increases Since y′ increases as x increases then the derivative of y′ is positive, that is y′′ > (Compare with: y increases when its derivative is positive.) So for the curve shown in Figure 11.2(a), y′ > and y′′ > For that shown in Figure 11.2(b) the situation is different The value of y′ decreases as x increases, as can be seen by considering the gradients of the tangents at A, B and C, that is the derivative of y′ must be negative For this curve y′ > and y′′ < A function is concave down when y′ decreases and concave up when y′ increases Hence Figure 11.2(a) illustrates a concave up function; Figure 11.2(b) illustrates a concave down function The sign of the second derivative can be used to distinguish between concave up and concave down functions Consider now the functions shown in Figure 11.2(c) and Figure 11.2(d) In both (c) and (d), y is decreasing and so y′ < In (c) the gradient of the tangent becomes increasingly negative; that is, it is decreasing Hence, for the function in (c) y′′ < Conversely, for the function in (d) the gradient of the tangent is increasing as x increases, although it is always negative, that is y′′ > So for the function in (c) y′ < and y′′ < 0; that 11.4 Higher derivatives y y y C 403 y A C B B B A A B C A x (a) (b) x C (c) x x (d) Figure 11.2 (a) y is concave up (y′ > 0, y′′ > 0); (b) y is concave down (y′ > 0, y′′ < 0); (c) y is concave down (y′ < 0, y′′ < 0); (d) y is concave up (y′ < 0, y′′ > 0) is, the function is concave down For the function in (d) y′ < and y′′ > 0; that is, the function is concave up In summary, we can state: When y′ > 0, y is increasing When y′ < 0, y is decreasing When y′ is increasing the function is concave up In this case y′′ > When y′ is decreasing the function is concave down In this case y′′ < An easy way of determining the concavity of a curve is to note that as the curve is traced from left to right, an anticlockwise motion reveals that the curve is concave up A clockwise motion means that the curve is concave down As will be seen in the next chapter, higher derivatives are used to determine the location and nature of important points called maximum points, minimum points and points of inflexion EXERCISES 11.4 (a) (b) (c) (d) (e) (f) (g) (h) (i) (j) d2 y dy and given dt dt y = t2 + t y = 2t − t + y = sin 2t y = sin kt k constant y = 2e3t − t + t y= t +1 t y = cos y = et t y = sinh 4t y = sin2 t Calculate If y= 7t t3 − + 12t − Determine whether the following functions are concave up or concave down (a) y = et (b) y = t (c) y = + t − t Determine the interval on which y = t is (a) concave up, (b) concave down Evaluate y′′ at the specified value of t (a) y = cos t − t t=1 sin t + cos t (b) y = t = π/2 (c) y = (1 + t )et t=0 y = 2x3 + 3x2 − 12x + find values of x for which y′′ = dy = 3t + t, find If dt d2 y d3 y (a) (b) dt dt Find values of t at which y′′ = 0, where d2 y given xy + x2 = y2 dx2 dx x Find when x3 + = t + x2 t dt t Find 404 Chapter 11 Techniques of differentiation Solutions (a) 2t + 1, (b) 6t − 2t, 12t − (c) cos 2t, −4 sin 2t (d) k cos kt, −k2 sin kt (e) 6e3t − 2t, 18e3t − 2 ,− (f) (t + 1)2 (t + 1)3 (g) −2 sin(t/2), − cos(t/2) et (t + 1), et (t (a) concave up (b) concave up (c) concave down (h) + 2) (i) cosh 4t, 16 sinh 4t (j) sin t cos t, cos 2t − (a) 6t + (b) (a) concave up on (0, ∞) (b) concave down on (−∞, 0) (a) −3.08 (b) − (c) 10y2 − 10x2 − 10yx (2y − x)3 x + 2t + x2 t 3x2 t + t − 2xt REVIEW EXERCISES 11 (d) x2 + 3xy + y2 = Differentiate each of the following functions: (e) = 3x3 + 2xy2 − y (a) y = sin(5 + x)2 (b) y = e2 sin x (c) y = (4x + 7)5 (d) y = x4 sin 3x e4x + 11 (f) y = x2 tan x cos 3x (g) y = x2 (h) y = e−x cos 5x (e) y = x3 (i) y = ln cos 4x (j) y = sin 2t cos 2t (k) y = x +1 Find dy in each of the following cases: dx (a) y = x3 sin 2x cos x (b) y = x3 e−x tan x (c) y = xe5x sin x 5 + 3t 2−t dy d2 y and y = find and 1−t 1−t dx dx dy If x = 4(1 + cos θ ) and y = 3(θ − sin θ ) find dx d2 y dy and If x = cos2 θ and y = sin2 θ find dx dx If x = Show that y = e−4x sin 8x satisfies the equation y′′ + 8y′ + 80y = Differentiate y = xx Given y = x3 e2x find dy d2 y (a) (b) dx dx2 (c) d3 y dx3 Use logarithmic differentiation to find dx given dt (a) x = tet sin t (b) x = t e−t cos 3t (c) x = t e3t sin 4t cos 3t 10 Show that if y(t ) = A sin ωt + B cos ωt, where ω is a constant, then y′′ + ω2 y = Review exercises 11 405 Solutions (a) 2(5 + x) cos(5 + x)2 (d) − (b) cos xe2 sin x (c) 20(4x + 7)4 (d) 3x4 cos 3x + 4x3 sin 3x (e) e4x (4x3 − 3x2 + 44) (x3 + 11)2 (f) x2 sec2 x + 2x tan x −3x sin 3x − cos 3x (g) x3 (h) −e−x (5 sin 5x + cos 5x) (i) −4 tan 4x (j) cos 4t −2x (k) (x2 + 1)2 (a) cos x(2x3 cos 2x + 3x2 sin 2x) + x3 sin 2x sin x cos2 x which simplifies to (b) (c) 2x3 cos x + 6x2 sin x e−x (x3 sec2 x − x3 tan x + 3x2 tan x) e5x (5x sin x + sin x − x cos x) sin2 x (e) 2x + 3y 3x + 2y 9x2 + 2y2 − 4xy ,0 3(1 − cos θ ) − sin θ − ,0 xx (ln x + 1) (a) e2x (2x3 + 3x2 ) (b) 2xe2x (2x2 + 6x + 3) (c) 2e2x (4x3 + 18x2 + 18x + 3) (a) et (t cos t + (t + 1) sin t ) (b) −e−t ((t − 2t ) cos 3t + 3t sin 3t ) (c) t e3t cos 4t sin 3t +3+ − sin 4t cos 3t t sin 4t cos 3t ! 12 Contents 12.1 Applications of differentiation 12.1 Introduction 406 12.2 Maximum points and minimum points 406 12.3 Points of inflexion 415 12.4 The Newton Raphson method for solving equations 418 12.5 Differentiation of vectors 423 Review exercises 12 427 INTRODUCTION In this chapter the techniques of differentiation are used to solve a variety of problems It is possible to use differentiation to find the maximum or minimum values of a function For example, it is possible to find the maximum power transferred from a voltage source to a load resistor, as we shall show later in the chapter Differentiation is also used in the Newton Raphson method of solving non-linear equations Such an equation needs to be solved to calculate the steady-state values of current and voltage in a series diode resistor circuit Finally we show how vectors can be differentiated This forms an introduction to the important topic of vector calculus which is discussed in Chapter 26 12.2 MAXIMUM POINTS AND MINIMUM POINTS Consider Figure 12.1 A and B are important points on the curve At A the function stops increasing and starts to decrease At B it stops decreasing and starts to increase A is a local maximum, B is a local minimum Note that A is not the highest point on the curve, nor B the lowest point However, for that part of the curve near to A, A is the highest point The word ‘local’ is used to stress that A is maximum in its locality Similarly, B is the lowest point in its locality 12.2 Maximum points and minimum points y y A A B B x (a) 407 x (b) Figure 12.1 The function y has a local maximum at A and a local minimum at B In Figure 12.1(a) tangents drawn at A and B would be parallel to the x axis and so dy is zero However, in Figure 12.1(b) there are corners at A and B It is at these points dx dy impossible to draw tangents at these points and so does not exist at these points dx Hence, when searching for maximum and minimum points we need only examine dy dy is zero, or does not exist those points at which dx dx dy is zero are known as turning points or stationary values of the Points at which dx function At maximum and minimum points either: dy (i) does not exist, or dx dy (ii) =0 dx dy on dx either side of the point At maximum points such as A, y is increasing immediately to the dy left of the point, and decreasing immediately to the right That is, is positive immedidx dy ately to the left, and is negative immediately to the right At minimum points such as dx B, y is decreasing immediately to the left of the point, and increasing immediately to the dy dy is negative immediately to the left, and is positive immediately to the right That is, dx dx right This so-called first-derivative test enables us to distinguish maxima from minima This test can be used even when the derivative does not exist at the point in question To distinguish between maximum and minimum points we can study the sign of The first-derivative test to distinguish maxima from minima: dy dy is positive; to the right, is negative To the left of a maximum point, dx dx dy dy To the left of a minimum point, is negative; to the right, is positive dx dx 408 Chapter 12 Applications of differentiation Example 12.1 Determine the position and nature of all maximum and minimum points of the following functions: (a) y = x2 (b) y = −t + t + x2 x3 + − 2x + (c) y = (d) y = |t| Solution dy = 2x dx Recall that at maximum and minimum points either (a) If y = x2 , then by differentiation (i) dy does not exist, or dx (ii) dy = We must check both of these conditions dx The function 2x exists for all values of x, and so we move to examine any points dy = So, we have where dx dy = 2x = dx The equation 2x = has one solution, x = We conclude that a turning point exists at x = Furthermore, from the given function y = x2 , we see that when x = the value of y is also 0, so a turning point exists at the point with coordinates (0, 0) To determine whether this point is a maximum or minimum we use the first-derivative dy test and examine the sign of on either side of x = To the left of x = 0, x is dx clearly negative and so 2x is also negative To the right of x = 0, x is positive and so 2x is also positive Hence y has a minimum at x = A graph of y = x2 showing this minimum is given in Figure 12.2 (b) If y = −t + t + 1, then y′ = −2t + and this function exists for all values of t Solving y′ = we have −2t + = and so t =  2 1 The y coordinate here is − + 2 1 + = We now inspect the sign of y′ to the left and to the right of t = A little to the left, say at t = 0, we see that y′ = −2(0) + = which is positive A ′ little to the right, say at t = 1, we see that + = −1 which is negative  y = −2(1)  1 Hence there is a maximum at the point ,1 A graph of the function is shown in Figure 12.3 We conclude that there is a turning point at t = 12.2 Maximum points and minimum points y y 409 y = –t + t + y = x2 – t x Figure 12.2 The function y has a minimum at x = Figure 12.3 The function y has a maximum at t = x3 x2 + − 2x + 1, then y′ = x2 + x − and this function exists for all values of x Solving y′ = we find (c) If y = x2 + x − = (x − 1)(x + 2) = x = 1, −2 There are therefore two turning points, one at x = and one at x = −2 We consider each in turn At x = 1, we examine the sign of y′ to the left and to the right of x = A little way to the left, say at x = 0, we see that y′ = −2 which is negative A little to the right, say at x = 2, we see that y′ = 22 + − = which is positive So the point where x = is a minimum At x = −2, we examine the sign of y′ to the left and to the right of x = −2 A little way to the left, say at x = −3, we see that y′ = (−3)2 + (−3) − = which is positive A little to the right, say at x = −1, we see that y′ = (−1)2 +(−1)−2 = −2 which is negative So the point where x = −2 is a maximum A graph of the function is shown in Figure 12.4 (d) Recall that the modulus function y = |t| is defined as follows:  −t t y = |t| = t t>0 A graph of this function was given in Figure 10.13(a) and this should be looked at dy dy = −1 for t negative, and = +1 for t positive before continuing Note that dt dt The derivative is not defined at t = because of the corner there There are no points dy when = Because the derivative is not defined at t = this point requires dt dy dy further scrutiny To the left of t = 0, < 0; to the right, > and so t = is a dt dt minimum point y x3 + — x2 – 2x + y= — –2 x Figure 12.4 The function y has a maximum at x = −2 and a minimum at x = 410 Chapter 12 Applications of differentiation y y dy — >0 dx dy — 0 dx dy — implies a minimum point; y′′ < implies a maximum point If y′′ = 0, then we must return to the earlier, more basic test of examining y′ on both sides of the point In summary: The second-derivative test to distinguish maxima from minima: If y′ = and y′′ < at a point, then this indicates that the point is a maximum turning point If y′ = and y′′ > at a point, then this indicates that the point is a minimum turning point If y′ = and y′′ = at a point, the second-derivative test fails and you must use the first-derivative test Example 12.2 Use the second-derivative test to find all maximum and minimum points of the functions in Example 12.1 Solution (a) Given y = x2 then y′ = 2x and y′′ = We locate the position of maximum and minimum points by solving y′ = and so such a point exists at x = Evaluating y′′ at this point we see that y′′ (0) = which is positive Using the second-derivative test we conclude that the point is a minimum (b) Given y = −t + t + then y′ = −2t + and y′′ = −2 Solving y′ = we find   1 = −2 which is negative Using t = Evaluating y′′ at this point we find y′′ 2 the second-derivative test we conclude that t = is a maximum point x3 x2 ′ (c) Given y = + − 2x + 1, then y = x + x − and y′′ = 2x + y′ = at x = and x = −2 At x = 1, y′′ = which is positive and so the point 12.2 Maximum points and minimum points 411 is a minimum At x = −2, y′′ = −3 which is negative and so the point is a maximum.  t0 (d) y =   undefined at t = Since y′ (0) is undefined, we use the first-derivative test This was employed in Example 12.1 Engineering application 12.1 Risetime for a second-order electrical system Consider the electrical system illustrated in Figure 12.7 The input voltage, vi , is applied to terminals a b The output from the system is a voltage, vo , measured across the terminals c d The easiest way to determine the time response of this system to a particular input is to use the technique of Laplace transforms (see Chapter 21) When a step input is applied to the system, the general form of the response depends on whether a quantity called the damping ratio, ζ , is such that ζ > 1, ζ = or ζ < The quantity ζ itself depends upon the values of L, C and R This is illustrated in Figure 12.8 If the damping ratio, ζ < 1, then vo overshoots its final value and the system is said to be underdamped For this case it can be shown that   α sin(βt ) for t>0 (12.1) vo = U − Ue−αt cos(βt ) + β where U is the height of a step input applied at t = 0, and R α= 2L ωr = √ resonant frequency LC q β = ωr2 − α natural frequency (12.2) (12.3) (12.4) Engineers are often interested in knowing how quickly a system will respond to a particular input For many systems this is an important design criterion One way of characterizing the speed of response of the system is the time taken for the output to reach a certain level in response to a step input This is known as the rise time and is often defined as the time taken for the output to rise from 10% to 90% of its final value However, by looking at the underdamped response illustrated in Figure 12.8 it is clear that the time, tm , required for the output to reach its maximum value would also provide an indicator of system response time As the derivative of a function is zero at a maximum point it is possible to calculate this time Differentiating Equation (12.1) and using the product rule,    dvo α sin(βt ) d −αt cos(βt ) + U − Ue t>0 = dt dt β   d −αt d e−αt α sin(βt ) = − U (e cos(βt )) − U dt dt β ➔ 412 Chapter 12 Applications of differentiation = −U (−αe−αt cos(βt ) − e−αt β sin(βt ))   −αe−αt α sin(βt ) e−αt αβ cos(βt ) −U + β β   α sin(βt ) −αt −α cos(βt ) − β sin(βt ) − = −Ue + α cos(βt ) β   α2 sin(βt ) = Ue−αt β + β dv At a turning point o = Hence dt   −αt β + α Ue sin(βt ) = β yo(t) z1 yo C tm b t d Figure 12.7 A second-order electrical system Figure 12.8 Response of a second-order system to a step input This occurs when sin(βt ) = 0, which corresponds to t = kπ/β, k = 0, 1, It is now straightforward to calculate tm , once β has been calculated, using Equations (12.2), (12.3) and (12.4) for particular values of R, L and C You may like to show that the turning point corresponding to k = is a maximum by calculating d2 vo and carrying out the second-derivative test dt It is possible to check whether or not a system is underdamped using the following formulae: R ζ = damping ratio (12.5) Rc r L critical resistance (12.6) Rc = C Let us look at a specific case with typical values L = 40 mH, C = µF, R = 200  Using Equations (12.5) and (12.6), we find s r L × 10−2 =2 = 400 Rc = C × 10−6 ζ = R 200 = 0.5 = Rc 400 12.2 Maximum points and minimum points 413 and therefore the system is underdamped because ζ < Also, = 5000 ωr = √ LC 200 R = = 2500 2L × × 10−2 q p β = ωr2 − α = 50002 − 25002 = 4330 α= Finally, π = 7.26 ì 104 = 726 às 4330 We conclude for this case that the risetime is 726 µs tm = Engineering application 12.2 Maximum power transfer Consider the circuit of Figure 12.9 in which a non-ideal voltage source is connected to a variable load resistor with resistance RL The source voltage is V and its internal resistance is RS Calculate the value of RL which results in the maximum power being transferred from the voltage source to the load resistor This is an essential piece of information for engineers involved in the design of power systems Often an important design consideration is to transfer the maximum amount from the power source to the point where the power is being consumed i RS Non-ideal voltage source RL + V – Figure 12.9 Maximum power transfer occurs when RL = RS Solution Let i be the current flowing in the circuit Using Kirchhoff’s voltage law and Ohm’s law gives V = i(RS + RL ) Let P be the power developed in the load resistor Then, P = i2 RL = V RL (RS + RL )2 ➔ 414 Chapter 12 Applications of differentiation Clearly P depends on the value of RL Differentiating w.r.t RL and using the quotient rule, we obtain dP 1(RS + RL )2 − RL 2(RS + RL ) = V2 dRL (RS + RL )4 = V2 = V2 Equating V2 that is, (RS + RL ) − 2RL (RS + RL )3 RS − RL (RS + RL )3 dP to zero to obtain the turning point gives dRL RS − RL =0 (RS + RL )3 RL = RS So a turning point occurs when the load resistance equals the source resistance We need to check if this is a maximum turning point, so R − RL dP = V2 S dRL (RS + RL )3 d2 P −1(RS + RL )3 − (RS − RL )3(RS + RL )2 = V2 dRL (RS + RL )6 = V2 = V2 −(RS + RL ) − 3(RS − RL ) (RS + RL )4 2RL − 4RS (RS + RL )4 = 2V (RL − 2RS ) (RS + RL )4 When RL = RS , this expression is negative and so the turning point is a maximum Therefore, maximum power transfer occurs when the load resistance equals the source resistance EXERCISES 12.2 Locate the position of any turning points of the following functions and determine whether they are maxima or minima (a) y = x2 − x + (b) y = 2x2 + 3x + (c) y = x − (d) y = + x − 2x2 (e) y = x − 12x (f) y = + 3x Locate and identify all turning points of x3 − 3x2 + 8x + (b) y = te−t (a) y = (c) y = x4 − 2x2 12.3 Points of inflexion 415 Solutions (a) (b) (c) (d) (e) (f)   , , minimum (−0.75, −0.125), minimum none (0.25, 1.125), maximum (2, −16) minimum, (−2, 16) maximum none     23 19 2, maximum, 4, minimum 3 (b) (1, 0.368), maximum (a) (c) (0, 0) maximum, (1, −1) minimum, (−1, −1) minimum Technical Computing Exercises 12.2 Computer languages such as MATLAB® are matrix orientated and not always provide the ability to differentiate functions Others such as Wolfram Mathematica and Maplesoft Maple have this capability by default If you are attempting the following exercises in MATLAB® you may require the Symbolic Math Toolbox which is an add-on for the main program 12.3 (a) Use a technical computing language to find y′ and y′′ when y = e−0.2t cos t (b) Solve y′ = and hence locate any turning points in the interval [0, 6] and determine their type (c) Plot a graph of y and check the position of the turning points with the results obtained in part (b) POINTS OF INFLEXION Recall from Section 11.4 that when the gradient of a curve, that is y′ , is increasing, the second derivative y′′ is positive and the curve is said to be concave up When the gradient is decreasing the second derivative y′′ is negative and the curve is said to be concave down A point at which the concavity of a curve changes from concave up to concave down or vice versa is called a point of inflexion A point of inflexion is a point on a curve where the concavity changes from concave up to concave down or vice versa It follows that y′′ = at such a point or, in exceptional cases, y′′ does not exist Figure 12.10(a) shows a graph for which a point of inflexion occurs at the point marked A Note that at this point the gradient of the graph is zero Figure 12.10(b) shows a graph with points of inflexion occurring at A and B Note that at these points the gradient of the graph is not zero To locate a point of inflexion we must look for a point where y′′ = or does not exist We must then examine the concavity of the curve on either side of such a point 416 Chapter 12 Applications of differentiation y y A A x (a) B x (b) Figure 12.10 (a) There is a point of inflexion at A; (b) there are points of inflexion at A and B y y y = x4 y = x3 y'' > y'' = y' < 0, y'' > y' > 0, y'' > x x y'' < Figure 12.11 The second derivative, y′′ , changes sign at x = Figure 12.12 The derivative, y′ , changes sign at x = 0, but y′′ remains positive Example 12.3 Locate any points of inflexion of the curve y = x3 Solution Given y = x3 , then y′ = 3x2 and y′′ = 6x Points of inflexion can only occur where y′′ = or does not exist Clearly y′′ exists for all x and is zero when x = It is possible that a point of inflexion occurs when x = but we must examine the concavity of the curve on either side To the left of x = 0, x is negative and so y′′ is negative Hence to the left, the curve is concave down To the right of x = 0, x is positive and so y′′ is positive Hence to the right, the curve is concave up Thus the concavity changes at x = We conclude that x = is a point of inflexion A graph is shown in Figure 12.11 Note that at this point of inflexion y′ = too A common error is to state that if y′ = y′′ = then there is a point of inflexion This is not always true; consider the next example Example 12.4 Locate all maximum points, minimum points and points of inflexion of y = x4 Solution y′ = 4x3 y′′ = 12x2