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 Learn Mathematics) materials were the outcome of a three-ear curriculum development project undertaken b a consortium of five English universities led b Loughborough Universit, funded b the Higher Education Funding Council for England under the Fund for the Development of Teaching and Learning for the period October 00 September 005. HELM aims to enhance the mathematical education of engineering undergraduates through a range of fleible learning resources in the form of Workbooks and web-delivered interactive segments. HELM supports two CAA regimes: an integrated web-delivered implementation and a CD-based version. HELM learning resources have been produced primaril b teams of writers at si universities: Hull, Loughborough, Manchester, Newcastle, Reading, Sunderland. HELM gratefull acknowledges the valuable support of colleagues at the following universities and colleges involved in the critical reading, trialling, enhancement and revision of the learning materials: Aston, Bournemouth & Poole College, Cambridge, Cit, Glamorgan, Glasgow, Glasgow Caledonian, Glenrothes Institute of Applied Technolog, Harper Adams Universit College, Hertfordshire, Leicester, Liverpool, London Metropolitan, Mora College, Northumbria, Nottingham, Nottingham Trent, Oford Brookes, Plmouth, Portsmouth, Queens Belfast, Robert Gordon, Roal Forest of Dean College, Salford, Sligo Institute of Technolog, Southampton, Southampton Institute, Surre, Teesside, Ulster, Universit of Wales Institute Cardiff, West Kingswa College (London), West Notts College. HELM Contacts: Post: HELM, Mathematics Education Centre, Loughborough Universit, Loughborough, LE11 3TU. helm@lboro.ac.uk Web: HELM Workbooks List 1 Basic Algebra 6 Functions of a Comple Variable Basic Functions 7 Multiple Integration 3 Equations, Inequalities & Partial Fractions 8 Differential Vector Calculus 4 Trigonometr 9 Integral Vector Calculus 5 Functions and Modelling 30 Introduction to Numerical Methods 6 Eponential and Logarithmic Functions 31 Numerical Methods of Approimation 7 Matrices 3 Numerical Initial Value Problems 8 Matri Solution of Equations 33 Numerical Boundar Value Problems 9 Vectors 34 Modelling Motion 10 Comple Numbers 35 Sets and Probabilit 11 Differentiation 36 Descriptive Statistics 1 Applications of Differentiation 37 Discrete Probabilit Distributions 13 Integration 38 Continuous Probabilit Distributions 14 Applications of Integration 1 39 The Normal Distribution 15 Applications of Integration 40 Sampling Distributions and Estimation 16 Sequences and Series 41 Hpothesis Testing 17 Conics and Polar Coordinates 4 Goodness of Fit and Contingenc Tables 18 Functions of Several Variables 43 Regression and Correlation 19 Differential Equations 44 Analsis of Variance 0 Laplace Transforms 45 Non-parametric Statistics 1 z-transforms 46 Reliabilit and Qualit Control Eigenvalues and Eigenvectors 47 Mathematics and Phsics Miscellan 3 Fourier Series 48 Engineering Case Studies 4 Fourier Transforms 49 Student s Guide 5 Partial Differential Equations 50 Tutor s Guide Copright Loughborough Universit, 006

3 Contents 1 Applications of Differentiation 1.1 Tangents and Normals 1. Maima and Minima The Newton-Raphson Method Curvature Differentiation of Vectors Case Stud: Comple Impedance 60 Learning outcomes In this Workbook ou will learn to appl our knowledge of differentiation to solve some basic problems connected with curves. First ou will learn how to obtain the equation of the tangent line and the normal line to an point of interest on a curve. Secondl, ou will learn how to find the positions of maima and minima on a given curve. Thirdl, ou will learn how, given an approimate position of the root of a function, a better estimate of the position can be obtained using the Newton-Raphson technique. Lastl ou will learn how to characterise how sharpl a curve is turning b calculating its curvature.

4 Maima and Minima 1. Introduction In this Section we analse curves in the local neighbourhood of a stationar point and, from this analsis, deduce necessar conditions satisfied b local maima and local minima. Locating the maima and minima of a function is an important task which arises often in applications of mathematics to problems in engineering and science. It is a task which can often be carried out using onl a knowledge of the derivatives of the function concerned. The problem breaks into two parts finding the stationar points of the given functions distinguishing whether these stationar points are maima, minima or, eceptionall, points of inflection. This Section ends with maimum and minimum problems from engineering contets. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... be able to obtain first and second derivatives of simple functions be able to find the roots of simple equations eplain the difference between local and global maima and minima describe how a tangent line changes near a maimum or a minimum locate the position of stationar points use knowledge of the second derivative to distinguish between maima and minima 14 HELM (008): Workbook 1: Applications of Differentiation

5 1. Maima and minima Consider the curve = f() a b shown in Figure 7: f(a) f(b) a 0 1 b Figure 7 B inspection we see that there is no -value greater than that at = a (i.e. f(a)) and there is no value smaller than that at = b (i.e. f(b)). However, the points on the curve at 0 and 1 merit comment. It is clear that in the near neighbourhood of 0 all the -values are greater than the -value at 0 and, similarl, in the near neighbourhood of 1 all the -values are less than the -value at 1. We sa f() has a global maimum at = a and a global minimum at = b but also has a local minimum at = 0 and a local maimum at = 1. Our primar purpose in this Section is to see how we might locate the position of the local maima and the local minima for a smooth function f(). A stationar point on a curve is one at which the derivative has a zero value. In Figure 8 we have sketched a curve with a maimum and a curve with a minimum. 0 0 Figure 8 B drawing tangent lines to these curves in the near neighbourhood of the local maimum and the local minimum it is obvious that at these points the tangent line is parallel to the -ais so that d =0 0 HELM (008): Section 1.: Maima and Minima 15

6 The Newton-Raphson Method 1.3 Introduction This Section is concerned with the problem of root location ; i.e. finding those values of which satisf an equation of the form f() =0. An initial estimate of the root is found (for eample b drawing a graph of the function). This estimate is then improved using a technique known as the Newton-Raphson method, which is based upon a knowledge of the tangent to the curve near the root. It is an iterative method in that it can be used repeatedl to continuall improve the accurac of the root. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... be able to differentiate simple functions be able to sketch graphs distinguish between simple and multiple roots estimate the root of an equation b drawing a graph emplo the Newton-Raphson method to improve the accurac of a root 38 HELM (008): Workbook 1: Applications of Differentiation

7 Curvature 1.4 Introduction Curvature is a measure of how sharpl a curve is turning. At a particular point along the curve a tangent line can be drawn; this tangent line making an angle ψ with the positive -ais. Curvature is then defined as the magnitude of the rate of change of ψ with respect to the measure of length on the curve - the arc length s. That is Curvature = dψ ds In this Section we eamine the concept of curvature and, from its definition, obtain more useful epressions for curvature when the equation of the curve is epressed either in Cartesian form = f() or in parametric form = (t) = (t). We show that a circle has a constant value for the curvature, which is to be epected, as the tangent line to a circle turns equall quickl irrespective of the position on the circle. For all curves, ecept circles, other than a circle, the curvature will depend upon position, changing its value as the curve twists and turns. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... HELM (008): Section 1.4: Curvature understand the geometrical interpretation of the derivative be able to differentiate standard functions be able to use the parametric description of a curve understand the concept of curvature calculate curvature when the curve is defined in Cartesian form or in parametric form 47

8 1. Curvature Curvature is a measure of how quickl a tangent line turns as the contact point moves along a curve. For eample, consider a simple parabola, with equation =. Its graph is shown in Figure 7. R P Q Figure 7 It is obvious, geometricall, that the tangent lines to this curve turn more quickl between P and Q than between Q and R. It is the purpose of this Section to give, a quantitative measure of this rate of turning. If we change from a parabola to a circle, (centred on the origin, of radius 1), we can again consider how quickl the tangent lines turn as we move along the curve. See Figure 8. It is immediatel clear that the tangent lines to a circle turn equall quickl no matter where located on the circle. Figure 8 However, if we consider two circles with the same centre but different radii, as in Figure 9, it is again obvious that the smaller circle bends more tightl than the larger circle and we sa it has a larger curvature. Athletes who run the 00 metres find it easier to run in the outside lanes (where the curve turns less sharpl) than in the inside lanes. Q Q P ψ P ψ Figure 9 On the two circle diagram (Figure 9) we have drawn tangent lines at P and P ; both lines make an angle ψ (greek letter psi) with the positive -ais. We need to measure how quickl the angle 48 HELM (008): Workbook 1: Applications of Differentiation

9 Differentiation of Vectors 1.5 Introduction The area of mathematics known as vector calculus is used to model mathematicall a vast range of engineering phenomena including electrostatics, electromagnetic fields, air flow around aircraft and heat flow in nuclear reactors. In this Section we introduce briefl the differential calculus of vectors. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... have a knowledge of vectors, in Cartesian form be able to calculate the scalar and vector products of two vectors be able to differentiate and integrate scalar functions differentiate vectors 54 HELM (008): Workbook 1: Applications of Differentiation

10 Case Stud: Comple Impedance 1.6 Electronic Filters Electronic filters are used widel, for eample in audio equipment to correct for imperfections in microphones or loudspeakers, or to introduce special effects. The purpose of a filter is to produce an alternating current (a.c.) output voltage that varies with the frequenc of the input voltage. A filter must have at least one component which has an impedance that varies with frequenc. The impedance is given b the time dependent ratio of voltage across the component to current through the component. This means that a filter must contain at least one inductance or capacitance. An inductor consists of a large number of coils of wire. When the current i flowing through an inductor changes, the associated magnetic field changes and produces a voltage v across the inductor which is proportional to the rate of change of the current. The constant of proportionalit (inductance) is given the smbol L. In electronics, it is usual to use lower case smbols for the time varing quantities. The standard representations for a.c. electronic signals are v = V 0 e jωt and i = I 0 e jωt where V 0 is the (real) amplitude of the a.c. voltage and I 0 is the (real) amplitude of the a.c. current and j = 1. v v i i L (a) C (b) Figure 33: (a) an inductor (b) a capacitor An inductor (see Figure 33) gives rise to an a.c. voltage v = L di dt = jωli Hence v/i = jwl is the impedance of the inductor. The purel imaginar quantit, jwl, is called the reactance of the inductor. Usuall a coil of wire forming an inductor also has resistance but this can be designed or assumed to be negligible. A capacitor consists of two conducting plates separated b a thin insulator. The charge (q) on the plates is proportional to the voltage (v) between the plates. The constant of proportionalit (capacitance) is given the smbol C. So q = Cv. The current (i) into the capacitor is equal to the rate of change of the charge on the capacitor i.e. 60 HELM (008): Workbook 1: Applications of Differentiation

11 i = dq dt = C dv dt = jωcv. Hence, for a capacitor, the impedance Z c = v/i =1/jwC. This purel imaginar quantit is also a reactance. Because of Ohm s law (v = ir), a resistance R provides a constant (real) contribution of R to the impedance of a circuit. If two resistors R 1 and R are in series the same current passes through both of them and the combined resistance is R 1 + R. In the circuit shown in Figure 34 (consider the left-hand representation of this circuit first but note that the right-hand version is equivalent), the input voltage across both resistors and the output voltage across R are related b v in = i(r 1 + R ) and v out = ir so Such a circuit is called a potential divider. v out v in = R R 1 + R. R 1 v in R 1 v in R v out R v out Figure 34: Two representations of a potential divider circuit Now consider this circuit with the resistor R replaced b a capacitor C as in Figure 35. R v in C v out Figure 35: Low pass filter circuit containing a resistor and a capacitor If R 1 is replaced b R and R b Z C =1/jwC, in the relevant epression for the potential divider circuit, then v out v in = 1/jωC R +1/jωC = 1 1+jωRC The square of the magnitude of the voltage ratio is given b multipling the eisting comple epression b its comple conjugate, i.e. v out 1 v in = (1 + jωrc)(1 jωrc) = 1 (1 + ω R C ) HELM (008): Section 1.6: Case Stud: Comple Impedance 61

12 Figure 36 shows a plot of the magnitude of the voltage ratio as a function of ω, i.e. the frequenc response for R = 10 Ω and C =1 µf (i.e F). Note that the magnitude of the output voltage is close to that of the input voltage at low frequencies but decreases rapidl as frequenc increases. This is an ideal low pass filter response. Output voltage/input voltage Angular frequenc rad/s Engineering problem stated in words Figure 36: Frequenc response of a low pass filter R L v in C v out Figure 37: An LC filter circuit Plot the frequenc response of the LC filter circuit shown in Figure 37 if R = 10 Ω, L =0.1 mh (i.e H) and C = 1 µf. After plotting the response for two values of R below 10 Ω, comment on the wa in which the response varies as R varies. Identif the frequenc for which the response is maimum. Engineering problem epressed mathematicall (a) Noting that the resistor and inductor are in series, replace R 1 b (R + jwl) and R b v out 1/jwC in the equation = v in R 1 + R (b) Derive an epression for v out v in (c) Hence plot v out v in as a function of ω for R = 10 Ω. R 6 HELM (008): Workbook 1: Applications of Differentiation

13 (d) Plot v out v in for two further values of R<10 Ω (e.g. 5Ωand Ω). (e) Find an epression for the value of ω = ω res at which v out v in is maimum. Mathematical analsis (a) The substitutions R 1 (R + jwl) and R 1/jwC in the equation v out R v out = ield = v in R 1 + R v in (b) Multipling b the comple conjugate of the denominator v out v in = (c) See the solid line in Figure 38. 1/jωC R + jωl +1/jωC = 1 (1 ω LC + jωrc) 1 (1 ω LC + jωrc)(1 ω LC jωrc) = 1 (1 ω LC) + ω R C Output voltage/input voltage R = Ω R = 5Ω R = 10Ω Angular frequenc rad/s Figure 38: Frequenc response of LC filter (d) See the other broken lines in Figure 38. There is a peak in the voltage output, which can eceed the voltage input b a considerable amount. It is particularl noticeable for small values of the resistance and decreases as the resistance increases. (e) v out v in will be maimum when the first term in the denominator is zero (the other term is alwas positive for ω>0) i.e. when ω = ω res = 1 LC or f res = ω res π = 1 π LC The corresponding frequenc is known as the resonant frequenc of the circuit. Additional comment The resonant behaviour depicted in Figure 38 is found in certain vibrating sstems as well as electronic circuits. This gives rise to an electrical analog for such mechanical sstems and will be eplored further after 19 on differential equations. HELM (008): Section 1.6: Case Stud: Comple Impedance 63

14 1. Differentiation of vectors Consider Figure 31. P r C Figure 31 If r represents the position vector of an object which is moving along a curve C, then the position vector will be dependent upon the time, t. We write r = r(t) to show the dependence upon time. Suppose that the object is at the point P, with position vector r at time t and at the point Q, with position vector r(t + δt), at the later time t + δt, as shown in Figure 3. r(t) P r(t + δt) P Q Q Figure 3 Then PQrepresents the displacement vector of the object during the interval of time δt. The length of the displacement vector represents the distance travelled, and its direction gives the direction of motion. The average velocit during the time from t to t + δt is defined as the displacement vector divided b the time interval δt, that is, average velocit = PQ δt = r(t + δt) r(t) δt If we now take the limit as the interval of time δt tends to zero then the epression on the right hand side is the derivative of r with respect to t. Not surprisingl we refer to this derivative as the instantaneous velocit, v. B its ver construction we see that the velocit vector is alwas tangential to the curve as the object moves along it. We have: v = lim δt 0 r(t + δt) r(t) δt = dr dt HELM (008): Section 1.5: Differentiation of Vectors 55

15 Now, since the and coordinates of the object depend upon time, we can write the position vector r in Cartesian coordinates as: r(t) = (t)i + (t)j Therefore, so that, r(t + δt) =(t + δt)i + (t + δt)j (t + δt)i + (t + δt)j (t)i (t)j v(t) = lim δt 0 δt (t + δt) (t) (t + δt) (t) = lim i + j δt 0 δt δt = d dt i + d dt j This is often abbreviated to v =ṙ = ẋi +ẏj, using notation for derivatives with respect to time. So we see that the velocit vector is the derivative of the position vector with respect to time. This result generalizes in an obvious wa to three dimensions as summarized in the following Ke Point. Given r(t) = (t)i + (t)j + z(t)k then the velocit vector is Ke Point 8 v =ṙ(t) =ẋ(t)i +ẏ(t)j +ż(t)k The magnitude of the velocit vector gives the speed of the object. We can define the acceleration vector in a similar wa, as the rate of change (i.e. the derivative) of the velocit with respect to the time: a = dv dt = d r = r =ẍi +ÿj + zk dt 56 HELM (008): Workbook 1: Applications of Differentiation

16 Eample 6 If w =3t i + cos tj, find (a) dw (b) dw dt dt (c) d w dt Solution (a) If w =3t i + cos tj, then differentiation with respect to t ields: dw dt (b) dw dt = (6t) +( sin t) = 36t + 4 sin t (c) d w dt =6i 4cos tj =6ti sin tj It is possible to differentiate more complicated epressions involving vectors provided certain rules are adhered to as summarized in the following Ke Point. Ke Point 9 If w and z are vectors and c is a scalar, all these being functions of time t, then: d dw (w + z) = dt dt + dz dt d (cw) dt = cdw dt + dc dt w d (w z) dt = dz w dt + dw dt z d (w z) dt = dz w dt + dw dt z HELM (008): Section 1.5: Differentiation of Vectors 57

17 Eample 7 If w =3ti t j and z =t i +3j, verif the result d dz (w z) =w dt dt + dw dt z Solution w z = (3ti t j) (t i +3j) =6t 3 3t. Therefore Also so d dt (w z) =18t 6t (1) dw dt =3i tj and dz dt =4ti w dz dt + z dw dt = (3ti t j) (4ti) + (t i +3j) (3i tj) = 1t +6t 6t = 18t 6t () We have verified d dz (w z) =w dt dt + dw z since (1) is the same as (). dt Eample 8 If w =3ti t j and z =t i +3j, verif the result d dz (w z) =w dt dt + dw dt z Solution w z = w dz dt i j k 3t t 0 t 3 0 = i j k 3t t 0 4t 0 0 = (9t +t4 )k impling d dt (w z) = (9 + 8t3 )k (1) =4t3 k () dw dt z = i j k 3 t 0 t 3 0 = (9 + 4t3 )k (3) We can see that (1) is the same as () + (3) as required. 58 HELM (008): Workbook 1: Applications of Differentiation

18 1. If r =3ti +t j + t 3 k, find (a) dr dt Eercises. Given B = te t i + cos tjfind (a) db dt 3. If r =4t i +tj 7k evaluate r and dr dt (b) d r dt when t =1. 4. If w = t 3 i 7tk and z = ( + t)i + t j k (b) d B dt (a) find w z, 5. Given r = sin ti+ cos tj (b) find dw dt, dz (c) find dt, (d) show that d dz (w z) =w dt dt + dw dt z (a) find ṙ, (b) find r, (c) find r (d) Show that the position vector r and velocit vector ṙ are perpendicular. s 1. (a) 3i +4tj +3t k (b) 4j +6tk. (a) ( te t + e t )i sin tj (b) e t (t )i cos tj 3. 4i +j 7k, 8i +j 4. (a) t(t 3 +t + 14) (b) 3t i 7k (c) i +tj 5. (a) cos ti sin tj (b) sin ti cos tj (c) 1 (d) Follows b showing r ṙ =0. HELM (008): Section 1.5: Differentiation of Vectors 59

19 ψ changes as we move along the curve. As we move from P to Q (inner circle), or from P to Q (outer circle), the angle ψ changes b the same amount. However, the distance traversed on the inner circle is less than the distance traversed on the outer circle. This suggests that a measure of curvature is: curvature is the magnitude of the rate of change of ψ with respect to the distance moved along the curve. We shall denote the curvature b the Greek letter κ (kappa). So κ = dψ ds where s is the measure of arc-length along a curve. This rather odd-looking derivative needs converting to involve the variable if the equation of the curve is given in the usual form = f(). As a preliminar we note that dψ ds = dψ d ds d We now obtain epressions for the derivatives dψ d Consider Figure 30 below. and ds d in terms of the derivatives of f(). δ δs δ ψ Figure 30 Small increments in the - and -directions have been denoted b δ and δ respectivel. hpotenuse on this small triangle is δs which is the change in arc-length along the curve. From Pthagoras theorem: so δs = δ + δ δs =1+ δ δ δ so that δs δ = 1+ δ δ In the limit as the increments get smaller and smaller, we write this relation in derivative form: ds d d = 1+ d The HELM (008): Section 1.4: Curvature 49

20 However, as = f() is the equation of the curve we obtain ds d = 1+ = (1 + [f ()] ) 1/ d We also know the relation between the angle ψ and the derivative d : d = tan ψ so differentiating again: d f d = sec ψ dψ d = (1 + tan ψ) dψ d = (1 + [f ()] ) dψ d Inverting this relation: dψ d = f () (1 + [f ()] ) and so, finall, the curvature is given b κ = dψ ds = dψ ds d d = f () (1 + [f ()] ) 3/ Ke Point 6 Curvature At each point on a curve, with equation = f(), the tangent line turns at a certain rate. A measure of this rate of turning is the curvature κ defined b κ = f () (1 + [f ()] ) 3/ 50 HELM (008): Workbook 1: Applications of Differentiation

21 Task Obtain the curvature of the parabola =. First calculate the derivatives of f(): f() = f() = d = d = d f d = Now find an epression for the curvature: κ = κ = f () [1 + [f ())] ] 3/ = [1 + 4 ] 3/ d f d = Finall, plot the curvature κ as a function of : κ κ The figure above supports what we have alread argued: Close to =0the parabola turns sharpl (near =0the curvature κ is relativel, large). Further awa from =0the curve is more gentle (in these regions κ is small). In general, the curvature κ is a function of position. However, from what we have said earlier, we epect the curvature to be a constant for a given circle but to increase as the radius of the circle decreases. This can now be checked directl. HELM (008): Section 1.4: Curvature 51

22 Eample 5 Find the curvature of =(a ) 1/ (this is the equation of the upper half of a circle centred at the origin of radius a). Solution Here f() =(a ) 1 d = (a ) 1 d f d = a (a ) 3 1+[f ()] =1+ a = r a a (a κ = ) 3/ a 3/ = 1 a a For a circle of radius a, the curvature is constant, with value 1 a. The value of κ (at an particular point on the curve, i.e. at a particular value of ) indicates how sharpl the curve is turning. What this result states is that, for a circle, the curvature is inversel related to the radius. The bigger the radius, the smaller the curvature; precisel what we predicted.. Curvature for parametricall defined curves An epression for the curvature is also available if the curve is described parametricall: = g(t) = h(t) t 0 t t 1 We remember the basic formulae connecting derivatives d d = ẏ ẋ d ẋÿ ẏẍ = d ẋ 3 where, as usual ẋ d dt, ẍ d dt etc. Then κ = f () {1+[f ()] } 3/ = ẋÿ ẏẍ ẏ 3/ ẋ 1+ 3 ẋ = ẋÿ ẏẍ [ẋ +ẏ ] 3/ 5 HELM (008): Workbook 1: Applications of Differentiation

23 Ke Point 7 The formula for curvature in parametric form is κ = ẋÿ ẏẍ [ẋ +ẏ ] 3/ Task An ellipse is described parametricall b the equations = cos t = sin t 0 t π Obtain an epression for the curvature κ and find where the curvature is a maimum or a minimum. First find ẋ, ẏ, ẍ, ÿ: ẋ = ẏ = ẍ = ÿ = ẋ = sin t ẏ =cost ẍ = cos t ÿ = sin t Now find κ: κ = κ = ẋÿ ẏẍ [ẋ +ẏ ] 3/ = sin t + cos t [4 sin t + cos t] 3/ = [1 + 3 sin t] 3/ Find maimum and minimum values of κ b inspection of the epression for κ: ma κ = min κ = Denominator is ma when t = π/. This gives minimum value of κ =1/4, Denominator is min when t =0. This gives maimum value of κ =. minimum value of κ 1 1 maimum value of κ HELM (008): Section 1.4: Curvature 53

24 1. The Newton-Raphson method We first remind the reader of some basic notation: If f() is a given function the value of for which f() =0is called a root of the equation or zero of the function. We also distinguish between various tpes of roots: simple roots and multiple roots. Figures 1-3 illustrate some common eamples. = f() = ( 1) = ( ) simple root double root triple root Figure 1 Figure Figure 3 More precisel; a root 0 is said to be: a simple root if f( 0 )=0 and d =0. 0 a double root if f( 0 )=0, d f d =0 and 0 d =0, 0 and so on. In this Section we shall concentrate on the location of simple roots of a given function f(). Task Given graphs of the functions (a) f() = , (b) f() = 1 + sin classif the roots into simple or multiple. (a) f() = : The negative root is: and the positive root is: = The negative root is simple and the positive root is double. (b) f() = 1 + sin : Each root is a root Each root is a double root. HELM (008): Section 1.3: The Newton-Raphson Method 39

25 . Finding roots of the equation f()=0 A first investigation into the roots of f() might be graphical. Such an analsis will suppl information as to the approimate location of the roots. Task Sketch the function f() = + ln > 0 and estimate the value of the root. 1 An estimate of the root is: 1 A simple root is located near 1.5 One method of obtaining a better approimation is to halve the interval 1 into and 1.5 and test the sign of the function at the end-points of these new regions. We find f() 1 < < 0 > 0 so a root must lie between =1.5 and =because the sign of f() changes between these values and f() is a continuous curve. We can repeat this procedure and divide the interval (1.5, ) into the two new intervals (1.5, 1.75) and (1.75, ) and test again. This time we find f() 1.5 < > 0.0 > 0 40 HELM (008): Workbook 1: Applications of Differentiation

26 so a root lies in the interval (1.5, 1.75). It is obvious that proceeding in this wa will give a smaller and smaller interval in which the root must lie. But can we do better than this rather laborious bisection procedure? In fact there are man was to improve this numerical search for the root. In this Section we eamine one of the best methods: the Newton-Raphson method. To derive the method we eamine the general characteristics of a curve in the neighbourhood of a simple root. Consider Figure 4 showing a function f() with a simple root at = whose value is required. Initial analsis has indicated that the root is approimatel located at = 0. The aim is to provide a better estimate to the location of the root. = f() 0 Figure 4 The basic premise of the Newton-Raphson method is the assumption that the curve in the close neighbourhood of the simple root at is approimatel a straight line. Hence if we draw the tangent to the curve at 0, this tangent will intersect the -ais at a point closer to than is 0 : see Figure 5. R =f( 0 ) P 1 θ Q 0 From the geometr of this diagram we see that 1 = 0 PQ But from the right-angled triangle P QR we have RQ PQ = tan θ = f ( 0 ) and so PQ = RQ f ( 0 ) = f( 0) f ( 0 ) Figure 5 1 = 0 f( 0) f ( 0 ) If f() has a simple root near 0 then a closer estimate to the root is 1 where 1 = 0 f( 0) f ( 0 ) This formula can be used iterativel to get closer and closer to the root, as summarised in Ke Point 5: HELM (008): Section 1.3: The Newton-Raphson Method 41

27 Ke Point 5 Newton-Raphson Method If f() has a simple root near n then a closer estimate to the root is n+1 where n+1 = n f( n) f ( n ) This is the Newton-Raphson iterative formula. The iteration is begun with an initial estimate of the root, 0, and continued to find 1,,... until a suitabl accurate estimate of the position of the root is obtained. This is judged b the convergence of 1,,... to a fied value. Eample 4 f() = + ln has a root near =1.5. Use the Newton-Raphson method to obtain a better estimate. Solution Here 0 =1.5, f(1.5) = ln(1.5) = f () =1+ 1 f (1.5) = = 5 3 Hence using the formula: 1 =1.5 ( ) (1.6667) = The Newton-Raphson formula can be used again: this time beginning with as our estimate: = 1 f( 1) f ( 1 ) = f(1.5567) f (1.5567) { ln(1.5567)} = = { } {1.644} = This is in fact the correct value of the root to 4 d.p., which calculating 3 would confirm. 4 HELM (008): Workbook 1: Applications of Differentiation

28 Task The function f() = tan has a simple root near =4.5. Use one iteration of the Newton-Raphson method to find a more accurate value for the root. First find d : d = d =1 sec = tan Now use the formula 1 = 0 f( 0 )/f ( 0 ) with 0 =4.5 to obtain 1 : f(4.5) = 4.5 tan(4.5) = f (4.5) = 1 sec (4.5) = tan (4.5) = 1 =4.5 f(4.5) f (4.5) = f(4.5) = , f (4.5) = = = As the value of 1 has changed little from 0 =4.5 we can epect the root to be 4.49 to 3 d.p. Task Sketch the function f() = 3 +3and confirm that there is a simple root between = and = 1. Use 0 = as an initial estimate to obtain the value to d.p. First sketch f() = 3 +3and identif a root: HELM (008): Section 1.3: The Newton-Raphson Method 43

29 Clearl a simple root lies between = and = 1. Now use one iteration of Newton-Raphson to improve the estimate of the root using 0 = : f() = f () = 0 = 1 = 0 f( 0) f ( 0 ) = f() = 3 +3, f () =3 1 0 = 1 = { 8++3} = = 1.77 Now repeat this process for a second iteration using 1 = 1.77: = 1 f( 1 )/f ( 1 )= = 1.77 { (1.77) }/{3(1.77) 1} = {(0.44)/(7.948) = Repeat for a third iteration and state the root to d.p.: 3 = f( )/f ( )= 3 = { (1.674) }/{3(1.674) 1} = {0.017}/{7.407} = 1.67 We conclude the value of the simple root is 1.67 correct to d.p. 44 HELM (008): Workbook 1: Applications of Differentiation

30 Engineering Eample 5 Buckling of a strut The equation governing the buckling load P of a strut with one end fied and the other end simpl P supported is given b tan µl = µl where µ =, L is the length of the strut and EI is the EI fleural rigidit of the strut. For safe design it is important that the load applied to the strut is less than the lowest buckling load. This equation has no eact solution and we must therefore use the method described in this Workbook to find the lowest buckling loadp. deflected shape P L P Figure 6 We let µl = and so we need to solve the equation tan =. Before starting to appl the Newton- Raphson iteration we must first obtain an approimate solution b plotting graphs of = tan and = using the same aes. = tan = 0 π/ π 3π/ From the graph it can be seen that the solution is near to but below =3π/ ( 4.7). We therefore start the Newton-Raphson iteration with a value 0 =4.5. The equation is rewritten as tan =0. Let f() = tan then f () = sec 1 = tan The Newton-Raphson iteration is so 1 =4.5 tan(4.5) 4.5 tan 4.5 Rounding to 4 sig.fig. and iterating: =4.494 tan(4.494) tan n+1 = n tan n n tan n, 0 =4.5 = = = to 7 sig.fig. So we conclude that the value of is to 4 sig.fig. As = µl = = to 7 sig.fig. re-arrangement, that the smallest buckling load is given b P = 0.19 EI L. P/EI L we find, after HELM (008): Section 1.3: The Newton-Raphson Method 45

31 Eercises 1. B sketching the function f() = 1 sin show that there is a simple root near =. Use two iterations of the Newton-Raphson method to obtain a better estimate of the root.. Obtain an estimation accurate to d.p. of the point of intersection of the curves = 1 and =cos. s 1. 0 =, 1 =1.936, = The curves intersect when 1 cos =0. Solve this using the Newton-Raphson method with initial estimate (sa) 0 =1.. The point of intersection is (1.834, ) to 6 significant figures. 46 HELM (008): Workbook 1: Applications of Differentiation

32 Points on the curve = f() at which d Ke Point 3 =0are called stationar points of the function. However, be careful! A stationar point is not necessaril a local maimum or minimum of the function but ma be an eceptional point called a point of inflection, illustrated in Figure 9. 0 Figure 9 Eample Sketch the curve =( ) +and locate the stationar points on the curve. Solution Here f() =( ) +so = ( ). d At a stationar point =0so we have ( ) = 0 so =. We conclude that this function d has just one stationar point located at =(where =). B sketching the curve = f() it is clear that this stationar point is a local minimum. Figure HELM (008): Workbook 1: Applications of Differentiation

33 Task Locate the position of the stationar points of f() = First find d : d = d =3 3 6 Now locate the stationar points b solving d =0: 3 3 6=3( + 1)( ) = 0 so = 1 or =. When = 1, f() = 13.5 and when =, f() =0, so the stationar points are ( 1, 13.5) and (, 0). We have, in the figure, sketched the curve which confirms our deductions. ( 1, 13.5).5 HELM (008): Section 1.: Maima and Minima 17

34 Task Sketch the curve = cos 0.1 3π and on it locate the position 4 of the global maimum, global minimum and an local maima or minima. 0.1 π/4 π/ 3π/4 global maimum 0.1 π/4 π/ 3π/4 local maimum local minimum and global minimum. Distinguishing between local maima and minima We might ask if it is possible to predict when a stationar point is a local maimum, a local minimum or a point of inflection without the necessit of drawing the curve. To do this we highlight the general characteristics of curves in the neighbourhood of local maima and minima. For eample: at a local maimum (located at 0 sa) Figure 11 describes the situation: f() to the left of the maimum d > 0 0 to the right of the maimum d < 0 Figure 11 If we draw a graph of the derivative against then, near a local maimum, it must take one d of two basic shapes described in Figure 1: 18 HELM (008): Workbook 1: Applications of Differentiation

35 d 0 α or d α = (a) (b) In case (a) d tan α<0 d d 0 Figure 1 whilst in case (b) d =0 d d 0 We reach the conclusion that at a stationar point which is a maimum the value of the second derivative d f is either negative or zero. d Near a local minimum the above graphs are inverted. Figure 13 shows a local minimum. f() to the left of the minimum d < 0 0 to the right of the minimum d > 0 Figure 13 Figure 134 shows the two possible graphs of the derivative: d 0 β or d 0 (a) (b) Here, for case (a) Figure 14 d = tan β>0 whilst in (b) d d 0 d =0. d d 0 In this case we conclude that at a stationar point which is a minimum the value of the second derivative d f is either positive or zero. d HELM (008): Section 1.: Maima and Minima 19

36 For the third possibilit for a stationar point - a point of inflection - the graph of f() against and of against take one of two forms as shown in Figure 15. d f() f() d 0 d to the left of 0 to the right of 0 d > 0 d > 0 to the left of 0 to the right of 0 d < 0 d < 0 For either of these cases d =0 d d 0 Figure 15 The sketches and analsis of the shape of a curve = f() in the near neighbourhood of stationar points allow us to make the following important deduction: Ke Point 4 If 0 locates a stationar point of f(), sothat d =0, then the stationar point 0 d f is a local minimum if d > 0 0 d f is a local maimum if d < 0 0 d f is inconclusive if d =0 0 0 HELM (008): Workbook 1: Applications of Differentiation

37 Eample 3 Find the stationar points of the function f() = 3 6. Are these stationar points local maima or local minima? Solution d =3 6. At a stationar point d =0so 3 6=0, impling = ±. Thus f() has stationar points at = and =. To decide if these are maima or minima we eamine the value of the second derivative of f() at the stationar points. d f d =6 so d f d =6 > 0. Hence = locates a local minimum. = Similarl d f d = 6 < 0. Hence = locates a local maimum. = A sketch of the curve confirms this analsis: f() Figure 16 Task For the function f() = cos, 0.1 6, find the positions of an local minima or maima and distinguish between them. Calculate the first derivative and locate stationar points: d = Stationar points are located at: HELM (008): Section 1.: Maima and Minima 1

38 = sin. d Hence stationar points are at values of in the range specified for which sin =0i.e. at = π or =π or =3π (making sure is within the range 0.1 6) Stationar points at = π,= π, = 3π Now calculate the second derivative: d f d = d f = 4 cos d Finall: evaluate the second derivative at each stationar points and draw appropriate conclusions: d f d = = π d f d = =π d f = d = 3π d f d = 4 cos π =4> 0 = π locates a local minimum. = π d f d = 4 cos π = 4 < 0 = π locates a local maimum. =π d f = 4 cos 3π =4> 0 = 3π locates a local minimum. d = 3π f() 0.1π/4 π/ 3π/4 3π/ 6 HELM (008): Workbook 1: Applications of Differentiation

39 Task Determine the local maima and/or minima of the function = First obtain the positions of the stationar points: f() = Thus d =0when: d = d =43 = (4 1) d =0when =0or when =1/4 Now obtain the value of the second derivatives at the stationar points: d f d = d f d = =0 d f d = =1/4 d f d = d f 1 d =0, which is inconclusive. =0 d f d = 1 =1/ = 1 4 > 0 Hence = 1 locates a local minimum. 4 Using this analsis we cannot decide whether the stationar point at =0is a local maimum, minimum or a point of inflection. However, just to the left of =0the value of (which equals d (4 1)) is negative whilst just to the right of =0the value of is negative again. Hence d the stationar point at =0is a point of inflection. This is confirmed b sketching the curve as in Figure 17. f() /4 Figure 17 HELM (008): Section 1.: Maima and Minima 3

40 Task A materials store is to be constructed net to a 3 metre high stone wall (shown as OA in the cross section in the diagram). The roof (AB) and front (BC) are to be constructed from corrugated metal sheeting. Onl 6 metre length sheets are available. Each of them is to be cut into two parts such that one part is used for the roof and the other is used for the front. Find the dimensions, of the store that result in the maimum cross-sectional area. Hence determine the maimum cross-sectional area. A Stone 3 m Wall B O C 4 HELM (008): Workbook 1: Applications of Differentiation

41 Note that the store has the shape of a trapezium. So the cross-sectional area (A) of the store is given b the formula: Area = average length of parallel sides distance between parallel sides: A = 1 ( +3) (1) The lengths and are related through the fact that AB + BC = 6, where BC = and AB = + (3 ). Hence + (3 ) + =6. This equation can be rearranged in the following wa: + (3 ) =6 +(3 ) = (6 ) i.e = which implies that +6 = 7 () It is necessar to eliminate either or from (1) and () to obtain an equation in a single variable. Using instead of as the variable will avoid having square roots appearing in the epression for the cross-sectional area. Hence from Equation () 7 = (3) 6 Substituting for from Equation (3) into Equation (1) gives A = = = (4) To find turning points, we evaluate da from Equation (4) to get d da d = 1 1 (45 3 ) Solving the equation da d =0gives 1 1 (45 3 )= =0 Hence = ± 15 = ± Onl >0 is of interest, so = 15 = (6) gives the required turning point. Check: Differentiating Equation (5) and using the positive solution (6) gives d A d = 6 1 = = < 0 Since the second derivative is negative then the cross-sectional area is a maimum. This is the onl turning point identified for A>0and it is identified as a maimum. To find the corresponding value of, substitute = into Equation (3) to get = = So the values of and that ield the maimum cross-sectional area are m and m respectivel. To find the maimum cross-sectional area, substitute for = into Equation (5) to get A = 1 ( )=9.685 So the maimum cross-sectional area of the store is 9.68 m to d.p. (5) HELM (008): Section 1.: Maima and Minima 5

42 Task Equivalent resistance in an electrical circuit Current distributes itself in the wires of an electrical circuit so as to minimise the total power consumption i.e. the rate at which heat is produced. The power (p) dissipated in an electrical circuit is given b the product of voltage (v) and current (i) flowing in the circuit, i.e. p = vi. The voltage across a resistor is the product of current and resistance (r). This means that the power dissipated in a resistor is given b p = i r. Suppose that an electrical circuit contains three resistors r 1,r,r 3 and i 1 represents the current flowing through both r 1 and r, and that (i i 1 ) represents the current flowing through r 3 (see diagram): R 1 R i 1 R 3 i i 1 i (a) Write down an epression for the power dissipated in the circuit: p = i 1r 1 + i 1r +(i i 1 ) r 3 (b) Show that the power dissipated is a minimum when i 1 = r 3 r 1 + r + r 3 i : 6 HELM (008): Workbook 1: Applications of Differentiation

43 Differentiate result (a) with respect to i 1 : dp di 1 = i 1 r 1 +i 1 r +(i i 1 )( 1)r 3 = i 1 (r 1 + r + r 3 ) ir 3 This is zero when r 3 i 1 = i. r 1 + r + r 3 To check if this represents a minimum, differentiate again: d p di 1 = (r 1 + r + r 3 ) This is positive, so the previous result represents a minimum. (c) If R is the equivalent resistance of the circuit, i.e. of r 1,r and r 3, for minimum power dissipation and the corresponding voltage V across the circuit is given b V = ir = i 1 (r 1 + r ), show that R = (r 1 + r )r 3 r 1 + r + r 3. Substituting for i 1 in ir = i 1 (r 1 + r ) gives So ir = r 3(r 1 + r ) r 1 + r + r 3 i. R = (r 1 + r )r 3 r 1 + r + r 3. Note In this problem R 1 and R could be replaced b a single resistor. However, treating them as separate allows the possibilit of considering more general situations (variable resistors or temperature dependent resistors). HELM (008): Section 1.: Maima and Minima 7

44 Engineering Eample 1 Water wheel efficienc Introduction A water wheel is constructed with smmetrical curved vanes of angle of curvature θ. Assuming that friction can be taken as negligible, the efficienc, η, i.e. the ratio of output power to input power, is calculated as (V v)(1 + cos θ)v η = V where V is the velocit of the jet of water as it strikes the vane, v is the velocit of the vane in the direction of the jet and θ is constant. Find the ratio, v/v, which gives maimum efficienc and find the maimum efficienc. Mathematical statement of the problem We need to epress the efficienc in terms of a single variable so that we can find the maimum value. (V v)(1 + cos θ)v Efficienc = = 1 v v (1 + cos θ). V V V Let η = Efficienc and = v then η =(1 )(1 + cos θ). V We must find the value of which maimises η and we must find the maimum value of η. To do this we differentiate η with respect to and solve dη =0in order to find the stationar points. d Mathematical analsis Now η =(1 )(1 + cos θ) = ( )(1 + cos θ) So dη d = ( 4)(1 + cos θ) Now dη d =0 4 =0 = 1 and the value of η when = 1 is 1 η = 1 1 (1 + cos θ) = 1 (1 + cos θ). This is clearl a maimum not a minimum, but to check we calculate d η = 4(1 + cos θ) which is d negative which provides confirmation. Interpretation Maimum efficienc occurs when v V = 1 and the maimum efficienc is given b η = 1 (1 + cos θ). 8 HELM (008): Workbook 1: Applications of Differentiation

45 Engineering Eample Refraction The problem A light ra is travelling in a medium (A) at speed c A. The ra encounters an interface with a medium (B) where the velocit of light is c B. Between two fied points P in media A and Q in media B, find the path through the interface point O that minimizes the time of light travel (see Figure 18). Epress the result in terms of the angles of incidence and refraction at the interface and the velocities of light in the two media. P d Medium (A) Medium (B) a θ A O θ B b Q Figure 18: Geometr of light ras at an interface The solution The light ra path is shown as POQ in the above figure where O is a point with variable horizontal position. The points P and Q are fied and their positions are determined b the constants a, b, d indicated in the figure. The total path length can be decomposed as PO+ OQ so the total time of travel T () is given b T () =PO/c A + OQ/c B. (1) Epressing the distances PO and OQ in terms of the fied coordinates a, b, d, and in terms of the unknown position, Equation (1) becomes a + T () = b +(d ) + () c A c B It is assumed that the minimum of the travel time is given b the stationar point of T () such that dt d =0. Using the chain rule in ( 1 c A a ) to compute (3) given () leads to d c B b +(d ) =0. After simplification and rearrangement c A a + = d c B b +(d ). (3) HELM (008): Section 1.: Maima and Minima 9

46 Using the definitions sin θ A = sin θ A c A = sin θ B c B. a + and sin θ B = d this can be written as b +(d ) Note that θ A andθ B are the incidence angles measured from the interface normal as shown in the figure. Equation (4) can be epressed as sin θ A sin θ B = c A c B which is the well-known law of refraction for geometrical optics and applies to man other kinds of waves. The ratio c A is a constant called the refractive inde of medium (B) with respect to medium (A). c B (4) 30 HELM (008): Workbook 1: Applications of Differentiation

47 Engineering Eample 3 Fluid power transmission Introduction Power transmitted through fluid-filled pipes is the basis of hdraulic braking sstems and other hdraulic control sstems. Suppose that power associated with a piston motion at one end of a pipeline is transmitted b a fluid of densit ρ moving with positive velocit V along a clindrical pipeline of constant cross-sectional area A. Assuming that the loss of power is mainl attributable to friction and that the friction coefficient f can be taken to be a constant, then the power transmitted, P is given b P = ρga(hv cv 3 ), where g is the acceleration due to gravit and h is the head which is the height of the fluid above some reference level (= the potential energ per unit weight of the fluid). The constant c = 4fl gd where l is the length of the pipe and d is the diameter of the pipe. The power transmission efficienc is the ratio of power output to power input. Problem in words Assuming that the head of the fluid, h, is a constant find the value of the fluid velocit, V, which gives a maimum value for the output power P. Given that the input power is P i = ρgav h, find the maimum power transmission efficienc obtainable. Mathematical statement of the problem We are given that P = ρga(hv cv 3 ) and we want to find its maimum value and hence maimum efficienc. To find stationar points for P we solve dp dv =0. To classif the stationar points we can differentiate again to find the value of d P at each stationar dv point and if this is negative then we have found a local maimum point. The maimum efficienc is given b the ratio P/P i at this value of V and where P i = ρgav h. Finall we should check that this is the onl maimum in the range of P that is of interest. Mathematical analsis P = ρga(hv cv 3 ) dp dv = ρga(h 3cV ) dp dv =0gives ρga(h 3cV )=0 V = h 3c V = ± h 3c and as V is positive V = h 3c. HELM (008): Section 1.: Maima and Minima 31

48 To show this is a maimum we differentiate dp dv again giving d P = ρga( 6cV ). Clearl this is dv h negative, or zero if V =0. Thus V = gives a local maimum value for P. 3c We note that P =0when E = ρga(hv cv 3 )=0, i.e. when hv cv 3 =0,soV =0or h h h V =. So the maimum at V = is the onl ma in this range between 0 and V = C 3C C. The efficienc E, is given b (input power/output power), so here E = ρga(hv cv 3 ) ρgav h =1 cv h h h At V = 3c then V = h c 3c and therefore E =1 3c c =1 1 3 = 3 or 66 3 %. Interpretation h Maimum power transmitted through the fluid when the velocit V = and the maimum 3c efficienc is 66 %. Note that this result is independent of the friction and the maimum efficienc 3 is independent of the velocit and (static) pressure in the pipe. P (V ) h = 3 m h = m Figure 19: Graphs of transmitted power as a function of fluid velocit for two values of the head Figure 19 shows the maima in the power transmission for two different values of the head in an oil filled pipe (oil densit 1100 kg m 3 ) of inner diameter 0.01 m and coefficient of friction 0.01 and pipe length 1 m. 3 HELM (008): Workbook 1: Applications of Differentiation

49 Engineering Eample 4 Crank used to drive a piston Introduction A crank is used to drive a piston as in Figure 0. v p l C r θ v c a c = ω r a p Figure 0: Crank used to drive a piston Problem The angular velocit of the crankshaft is the rate of change of the angle θ, ω = dθ/dt. The piston moves horizontall with velocit v p and acceleration a p ; r is the length of the crank and l is the length of the connecting rod. The crankpin performs circular motion with a velocit of v c and centripetal acceleration of ω r. The acceleration a p of the piston varies with θ and is related b a p = ω r cos θ + r cos θ l Find the maimum and minimum values of the acceleration a p when r = 150 mm and l = 375 mm. Mathematical statement of the problem We need to find the stationar values of a p = ω r cos θ + r cos θ when r = 150 mm and l = 375 l mm. We do this b solving da p =0and then analsing the stationar points to decide whether the dθ are a maimum, minimum or point of infleion. Mathematical analsis. a p = ω r cos θ r cos θ + l so da p dθ = ω r sin θ r sin θ. l To find the maimum and minimum acceleration we need to solve da p dθ =0 r sin θ ω r sin θ =0. l sin θ + r l sin θ =0 sin θ + 4r l sin θ cos θ =0 HELM (008): Section 1.: Maima and Minima 33

50 sin θ 1+ 4r cos θ =0 l sin θ =0 or cos θ = l 4r sin θ =0or cos θ = 5 8 CASE 1: sin θ =0 and as r = 150 mm and l = 375 mm If sin θ =0then θ =0or θ = π. If θ =0then cos θ =cosθ =1 so a p = ω r cos θ r cos θ + = ω r 1+ r = ω r 1+ = 7 l l 5 5 ω r If θ = π then cos θ = 1, cos θ =1so a p = ω r cos θ r cos θ + = ω r l 1+ r = ω r 1+ = 3 l 5 5 ω r In order to classif the stationar points, we differentiate da p with respect to θ to find the second dθ derivative: d a p dθ = 4r cos θ ω r cos θ = ω 4r cos θ r cos θ + l l At θ =0we get d a p dθ = ω r 1+ 4r which is negative. l So θ =0gives a maimum value and a p = 7 5 ω r is the value at the maimum. At θ = π we get d a p dθ = ω r 1+ 4 = ω r l So θ = π gives a maimum value and a p = 3 5 ω r CASE : cos θ = 5 8 If cos θ = 5 8 then cos θ = cos θ 1= a p = ω r cos θ + At cos θ = 5 8 we get d a p dθ r cos θ = ω r 5 l = ω r cos θ 3 which is negative so cos θ = r cos θ = ω r l So cos θ = 5 8 gives a minimum value and a p = ω r Thus the values of a p at the stationar points are:- = ω r r l 7 5 ω r (maimum), 3 5 ω r (maimum) and ω r (minimum). 7 which is positive HELM (008): Workbook 1: Applications of Differentiation

51 So the overall maimum value is 1.4ω r =0.1ω and the minimum value is 0.715ω r = ω where we have substituted r = 150 mm (= 0.15 m) and l = 375 mm (= m). Interpretation The maimum acceleration occurs when θ =0and a p =0.1ω. The minimum acceleration occurs when cos θ = 5 8 and a p = ω. Eercises 1. Locate the stationar points of the following functions and distinguish among them as maima, minima and points of inflection. (a) f() = ln. (b) f() = 3 (c) f() = ( 1) ( + 1)( ) [Remember d d (ln ) = 1 ] 1 <<. A perturbation in the temperature of a stream leaving a chemical reactor follows a decaing sinusoidal variation, according to T (t) =5ep( at) sin(ωt) where a and ω are positive constants. (a) Sketch the variation of temperature with time. (b) B eamining the behaviour of dt, show that the maimum temperatures occur at times dt of tan 1 ( ω a )+πn /ω. HELM (008): Section 1.: Maima and Minima 35

52 s 1. (a) d =1 1 d f d = 1 =0when =1 d f d =1> 0 =1 =1, =1locates a local minimum. f() 1 (b) d =3 =0when =0 However, d d f =6 =0when =0 d > 0 on either side of =0so (0, 0) is a point of inflection. f() (c) d = ( + 1)( ) ( 1)( 1) ( + 1)( ) This is zero when ( + 1)( ) ( 1)( 1) = 0 i.e. +3=0 However, this equation has no real roots (since b < 4ac) and so f() has no stationar points. The graph of this function confirms this: f() 1 1 Nevertheless f() does have a point of inflection at =1, =0as the graph shows, although at that point d d =0. 36 HELM (008): Workbook 1: Applications of Differentiation

53 . (a) T t (b) dt dt =0implies tan ωt = ω,sotan ωt > 0 and a ω ωt = tan 1 + kπ, k integer a Eamination of d T dt reveals that onl even values of k give d T dt setting k =n gives the required answer. < 0 for a maimum so HELM (008): Section 1.: Maima and Minima 37

54 Tangents and Normals 1.1 Introduction In this Section we see how the equations of the tangent line and the normal line at a particular point on the curve = f() can be obtained. The equations of tangent and normal lines are often written as = m + c, = n + d respectivel. We shall show that the product of their gradients m and n is such that mn is 1 which is the condition for perpendicularit. Prerequisites Before starting this Section ou should... Learning Outcomes On completion ou should be able to... be able to differentiate standard functions understand the geometrical interpretation of a derivative know the trigonometric epansions of sin(a + B), cos(a + B) obtain the condition that two given lines are perpendicular obtain the equation of the tangent line to a curve obtain the equation of the normal line to a curve HELM (008): Workbook 1: Applications of Differentiation

55 1. Perpendicular lines One form for the equation of a straight line is = m + c where m and c are constants. We remember that m is the gradient of the line and its value is the tangent of the angle θ that the line makes with the positive -ais. The constant c is the value obtained where the line intersects the -ais. See Figure 1: c = m + c m = tan θ θ Figure 1 If we have a second line, with equation = n + d then, unless m = n, the two lines will intersect at one point. These are drawn together in Figure. The second line makes an angle ψ with the positive -ais. c = m + c θ = n + d n = tan ψ ψ Figure A simple question to ask is what is the relation between m and n if the lines are perpendicular? If the lines are perpendicular, as shown in Figure 3, the angles θ and ψ must satisf the relation: ψ θ = 90 c θ ψ Figure 3 HELM (008): Section 1.1: Tangents and Normals 3

56 This is true since the angles in a triangle add up to 180. According to the figure the three angles are 90, θ and 180 ψ. Therefore 180 = 90 + θ + (180 ψ) impling ψ θ = 90 In this special case that the lines are perpendicular or normal to each other the relation between the gradients m and n is easil obtained. In this deduction we use the following basic trigonometric relations and identities: sin(a B) sin A cos B cos A sin B cos(a B) cos A cos B + sin A sin B Now tan A sin A cos A sin 90 = 1 cos 90 =0 m = tan θ = tan(ψ 90 o ) (see Figure 3) = sin(ψ 90o ) cos(ψ 90 o ) = cos ψ sin ψ = 1 tan ψ = 1 n So mn = 1 Ke Point 1 Two straight lines = m + c, = n + d are perpendicular if m = 1 n or equivalentl mn = 1 This result assumes that neither of the lines are parallel to the -ais or to the -ais, as in such cases one gradient will be zero and the other infinite. Eercise Which of the following pairs of lines are perpendicular? (a) = +1, = +1 (b) + 1=0, + =0 (c) =8 +3, = (a) perpendicular (b) not perpendicular (c) perpendicular 4 HELM (008): Workbook 1: Applications of Differentiation

57 . Tangents and normals to a curve As we know, the relationship between an independent variable and a dependent variable is denoted b = f() As we also know, the geometrical interpretation of this relation takes the form of a curve in an plane as illustrated in Figure 4. = f() Figure 4 We know how to calculate a value of given a value of. We can either do this graphicall (which is inaccurate) or else use the function itself. So, at an value of 0 the corresponding value is 0 where 0 = f( 0 ) Let us eamine the curve in the neighbourhood of the point ( 0, 0 ). constructions of interest the tangent line at ( 0, 0 ) the normal line at ( 0, 0 ) These are shown in Figure 5. There are two important tangent line 0 θ 0 ψ normal line Figure 5 We note the geometricall obvious fact: the tangent and normal lines at an given point on a curve are perpendicular to each other. HELM (008): Section 1.1: Tangents and Normals 5

58 Task The curve = is drawn below. On this graph draw the tangent line and the normal line at the point ( 0 =1, 0 = 1): 1 1 tangent line 1 θ 1 ψ normal line From our graph, estimate the values of θ and ψ in degrees. (You will need a protractor.) θ ψ θ 63.4 o ψ o Returning to the curve = f() :we know, from the geometrical interpretation of the derivative that d = tan θ 0 (the notation d means evaluate 0 d at the value = 0) Here θ is the angle the tangent line to the curve = f() makes with the positive -ais. This is highlighted in Figure 6: = f() d = tan θ 0 θ 0 Figure 6 6 HELM (008): Workbook 1: Applications of Differentiation

59 3. The tangent line to a curve Let the equation of the tangent line to the curve = f() at the point ( 0, 0 ) be: = m + c where m and c are constants to be found. The line just touches the curve = f() at the point ( 0, 0 ) so, at this point both must have the same value for the derivative. That is: m = d 0 Since we know (in an particular case) f() and the value 0 we can readil calculate the value for m. The value of c is found b using the fact that the tangent line and the curve pass through the same point ( 0, 0 ). 0 = m 0 + c and 0 = f( 0 ) Thus m 0 + c = f( 0 ) leading to c = f( 0 ) m 0 Ke Point The equation of the tangent line to the curve = f() at the point ( 0, 0 ) is = m + c where m = d and c = f( 0 ) m 0 0 Alternativel, the equation is 0 = m( 0 ) where m = d and 0 = f( 0 ) 0 Eample 1 Find the equation of the tangent line to the curve = at the point (1,1). Solution Method 1 Here f() = and 0 =1thus d = m = d = 0 Also c = f( 0 ) m 0 = f(1) m =1 = 1. The tangent line has equation = 1. Method 0 = f( 0 )=f(1) = 1 =1 The tangent line has equation 1=( 1) = 1 HELM (008): Section 1.1: Tangents and Normals 7

60 Task Find the equation of the tangent line to the curve = e at the point =0.The curve and the line are displaed in the following figure: tangent line First specif 0 and f: 0 = f() = 0 =0 f() =e Now obtain the values of d and f( 0 ) m 0 : 0 d = 0 f( 0 ) m 0 = d = e d =1 and f (0) 1(0) = e 0 0=1 0 Now obtain the equation of the tangent line: = = +1 8 HELM (008): Workbook 1: Applications of Differentiation

61 Task Find the equation of the tangent line to the curve = sin 3 at the point = π 4 and find where the tangent line intersects the -ais. See the following figure: tangent line First specif 0 and f: 0 = f() = 0 = π 4 f() = sin 3 Now obtain the values of d 0 and f( 0 ) m 0 correct to d.p.: d = f( 0 ) m 0 = 0 d = 3 cos 3 d π 4 and f mπ 4 = sin 3π 4 Now obtain the equation of the tangent line: = π 4 = 3 cos 3π 4 = 3 =.1 3 π 4 = π 4 =.37 to d.p. = (4 + 3π) so = (to d.p.) HELM (008): Section 1.1: Tangents and Normals 9

62 Where does the line intersect the -ais? = When = = 0 =1.1 to d.p. 4. The normal line to a curve We have alread noted that, at an point ( 0, 0 ) on a curve = f(), the tangent and normal lines are perpendicular. Thus if the equations of the tangent and normal lines are, respectivel = m + c = n + d then m = 1 n or, equivalentl n = 1 m. We have also noted, for the tangent line m = d 0 so n can easil be obtained. To find d, we again use the fact that the normal line = n + d and the curve have a point in common: 0 = n 0 + d and 0 = f( 0 ) so n 0 + d = f( 0 ) leading to d = f( 0 ) n 0. Task Find the equation of the normal line to curve = sin 3 at the point = π 4. [The equation of the tangent line was found in the previous Task.] First find the value of m: m = d π 4 = m = 3 Hence find the value of n: n = nm = 1 n = 3 10 HELM (008): Workbook 1: Applications of Differentiation

63 The equation of the normal line is = + d. Now find the value of d to d.p.. (Remember the 3 normal line must pass through the curve at the point = π 4.) π + d = sin π d = 1 3 π Now obtain the equation of the normal line to d.p.: = = The curve and the normal line are shown in the following figure: normal line Task Find the equation of the normal line to the curve = 3 at =1. First find f(), 0, d,m,n: 0 f() = 3, 0 =1, 1 d =3 =3 m =3and n = HELM (008): Section 1.1: Tangents and Normals 11

64 Now use the propert that the normal line = n + d and the curve = 3 pass through the point (1, 1) to find d and so obtain the equation of the normal line: d = = 1 = n + d d = 1 n = = 4. Thus the equation of the normal line is 3 = The curve and the normal line through (1, 1) are shown below: 3 normal line Eercises 1. Find the equations of the tangent and normal lines to the following curves at the points indicated (a) = 4 +, (1, 3) (b) = 1, (, ) What would be obtained if the point was (1, 0)? (c) = 1/, (1, 1). Find the value of a if the two curves = e and = e a are to intersect at right-angles. 1 HELM (008): Workbook 1: Applications of Differentiation

65 s 1. (a) f() = 4 + d =43 +4, d =8 =1 tangent line =8 + c. This passes through (1, 3) so =8 5 normal line = d. This passes through (1, 3) so = (b) f() = 1 d = 1 d = 1 = tangent line = + c. This passes through, so = + normal line = + d. This passes through, so =. At (1, 0) the tangent line is =1and the gradient is infinite (the line is vertical), and the normal line is =0. (c) f() = 1 d = 1 1 d = 1 =1 tangent line: = 1 + c. This passes through (1, 1) so = normal line: = + d. This passes through (1, 1) so = +3.. The curves will intersect at right-angles if their tangent lines, at the point of intersection, are perpendicular. Point of intersection: e = e a i.e. = a =0 (a = 1 not sensible) The tangent line to = e a is = m + c where The tangent line to = e is = k + g where m = ae a =0 = a k = e =0 = 1 These two lines are perpendicular if a( 1) = 1 i.e. a = 1. = e = e HELM (008): Section 1.1: Tangents and Normals 13

66 Inde for Workbook 1 Acceleration vector 56 Angular velocit 33 Buckling load 45 Centripetal acceleration 33 Circle 48, 5 Comple independence Crank 33 Curvature Efficienc 8 Electric circuit 6 Electronic filters, 60 Ellipse 53 Fluid power transmission 31 Global maimum 15 Hdraulic brakes 31 Impedance Inflection 3 Instantaneous velocit 55 Iterative formula 4 Light ras 9 Local maimum 15 Materials store 4 Maima and minima Multiple roots 39 Newton-Raphson method Normal to curve 4, 10-1 Parabola 48, 51 Perpendicularit 4 Piston 33 Power 8, 31 Power dissipation 6 Pthagoras theorem 49 Refraction 9 Resistance 6 Root location Solving equations 40 Stationar point Straight line 3 Strut 45 Target to curve 5-9 Vector differentiation Velocit vector 56 Water wheel 8 Zero 39 EXERCISES 4, 1, 35, 46, 59 ENGINEERING EXAMPLES 1 Water wheel efficienc 8 Refraction 9 3 Fluid power transmission 31 4 Crank used to drive a piston 33 5 Buckling of a strut 44

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