12.2. Maxima and Minima. Introduction. Prerequisites. Learning Outcomes

Transcription

1 Maima and Minima 1. Introduction In this Section we anayse curves in the oca neighbourhood of a stationary point and, from this anaysis, deduce necessary conditions satisfied by oca maima and oca minima. Locating the maima and minima of a function is an important task which arises often in appications of mathematics to probems in engineering and science. It is a task which can often be carried out using ony a knowedge of the derivatives of the function concerned. The probem breaks into two parts finding the stationary points of the given functions distinguishing whether these stationary points are maima, minima or, eceptionay, points of infection. This Section ends with maimum and minimum probems from engineering contets. Prerequisites Before starting this Section you shoud... Learning Outcomes On competion you shoud be abe to... be abe to obtain first and second derivatives of simpe functions be abe to find the roots of simpe equations epain the difference between oca and goba maima and minima describe how a tangent ine changes near a maimum or a minimum ocate the position of stationary points use knowedge of the second derivative to distinguish between maima and minima 14 HELM 006): Workbook 1: Appications of Differentiation

2 1. Maima and minima Consider the curve y = f) a b shown in Figure 7: fa) y fb) a 0 1 b Figure 7 By inspection we see that there is no y-vaue greater than that at = a i.e. fa)) and there is no vaue smaer than that at = b i.e. fb)). However, the points on the curve at 0 and 1 merit comment. It is cear that in the near neighbourhood of 0 a the y-vaues are greater than the y-vaue at 0 and, simiary, in the near neighbourhood of 1 a the y-vaues are ess than the y-vaue at 1. We say f) has a goba maimum at = a and a goba minimum at = b but aso has a oca minimum at = 0 and a oca maimum at = 1. Our primary purpose in this Section is to see how we might ocate the position of the oca maima and the oca minima for a smooth function f). A stationary point on a curve is one at which the derivative has a zero vaue. In Figure 8 we have sketched a curve with a maimum and a curve with a minimum. y y 0 0 Figure 8 By drawing tangent ines to these curves in the near neighbourhood of the oca maimum and the oca minimum it is obvious that at these points the tangent ine is parae to the -ais so that d = 0 0 HELM 006): Section 1.: Maima and Minima 15

3 Points on the curve y = f) at which d Key Point 3 = 0 are caed stationary points of the function. However, be carefu! A stationary point is not necessariy a oca maimum or minimum of the function but may be an eceptiona point caed a point of infection, iustrated in Figure 9. y 0 Figure 9 Eampe Sketch the curve y = ) + and ocate the stationary points on the curve. Soution Here f) = ) + so = ). d At a stationary point = 0 so we have ) = 0 so =. We concude that this function d has just one stationary point ocated at = where y = ). By sketching the curve y = f) it is cear that this stationary point is a oca minimum. y Figure HELM 006): Workbook 1: Appications of Differentiation

4 Task Locate the position of the stationary points of f) = First find d : Your soution d = Answer d = Now ocate the stationary points by soving d = 0: Your soution Answer = 3 + 1) ) = 0 so = 1 or =. When = 1, f) = 13.5 and when =, f) = 0, so the stationary points are 1, 13.5) and, 0). We have, in the figure, sketched the curve which confirms our deductions. 1, 13.5) y.5 HELM 006): Section 1.: Maima and Minima 17

5 Task Sketch the curve y = cos 0.1 3π and on it ocate the position 4 of the goba maimum, goba minimum and any oca maima or minima. Your soution y 0.1 π/4 π/ 3π/4 Answer y goba maimum 0.1 π/4 π/ 3π/4 oca maimum oca minimum and goba minimum. Distinguishing between oca maima and minima We might ask if it is possibe to predict when a stationary point is a oca maimum, a oca minimum or a point of infection without the necessity of drawing the curve. To do this we highight the genera characteristics of curves in the neighbourhood of oca maima and minima. For eampe: at a oca maimum ocated at 0 say) Figure 11 describes the situation: f) to the eft 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 oca maimum, it must take one d of two basic shapes described in Figure 1: 18 HELM 006): Workbook 1: Appications of Differentiation

6 d 0 α or d α = a) b) In case a) d tan α < 0 d d) 0 Figure 1 whist in case b) d = 0 d d) 0 We reach the concusion that at a stationary point which is a maimum the vaue of the second derivative d f is either negative or zero. d Near a oca minimum the above graphs are inverted. Figure 13 shows a oca minimum. f) to the eft of the minimum d < 0 0 to the right of the minimum d > 0 Figure 13 Figure 134 shows the two possibe graphs of the derivative: d 0 β or d 0 a) b) Here, for case a) Figure 14 d = tan β > 0 whist in b) d d) 0 d = 0. d d) 0 In this case we concude that at a stationary point which is a minimum the vaue of the second derivative d f is either positive or zero. d HELM 006): Section 1.: Maima and Minima 19

7 For the third possibiity for a stationary point - a point of infection - 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 eft of 0 to the right of 0 d > 0 d > 0 to the eft 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 anaysis of the shape of a curve y = f) in the near neighbourhood of stationary points aow us to make the foowing important deduction: Key Point 4 If 0 ocates a stationary point of f), so that d = 0, then the stationary point 0 d f is a oca minimum if d > 0 0 d f is a oca maimum if d < 0 0 d f is inconcusive if d = HELM 006): Workbook 1: Appications of Differentiation

8 Eampe 3 Find the stationary points of the function f) = 3 6. Are these stationary points oca maima or oca minima? Soution d = 3 6. At a stationary point d = 0 so 3 6 = 0, impying = ±. Thus f) has stationary points at = and =. To decide if these are maima or minima we eamine the vaue of the second derivative of f) at the stationary points. d f d = 6 so d f d = 6 > 0. Hence = ocates a oca minimum. = Simiary d f d = 6 < 0. Hence = ocates a oca maimum. = A sketch of the curve confirms this anaysis: f) Figure 16 Task For the function f) = cos, 0.1 6, find the positions of any oca minima or maima and distinguish between them. Cacuate the first derivative and ocate stationary points: Your soution d = Stationary points are ocated at: HELM 006): Section 1.: Maima and Minima 1

9 Answer = sin. d Hence stationary points are at vaues of in the range specified for which sin = 0 i.e. at = π or = π or = 3π making sure is within the range 0.1 6) Stationary points at = π, = π, = 3π Now cacuate the second derivative: Your soution d f d = Answer d f = 4 cos d Finay: evauate the second derivative at each stationary points and draw appropriate concusions: Your soution d f d = = π d f d = =π d f = d = 3π Answer d f d = 4 cos π = 4 > 0 = π ocates a oca minimum. = π d f d = 4 cos π = 4 < 0 = π ocates a oca maimum. =π d f = 4 cos 3π = 4 > 0 = 3π ocates a oca minimum. d = 3π f) 0.1π/4 π/ 3π/4 3π/ 6 HELM 006): Workbook 1: Appications of Differentiation

10 Task Determine the oca maima and/or minima of the function y = First obtain the positions of the stationary points: Your soution f) = Thus d = 0 when: d = Answer d = 43 = 4 1) d = 0 when = 0 or when = 1/4 Now obtain the vaue of the second derivatives at the stationary points: Your soution d f d = d f d = =0 d f d = =1/4 Answer d f d = d f 1 d = 0, which is inconcusive. =0 d f d = 1 =1/ = 1 4 > 0 Hence = 1 ocates a oca minimum. 4 Using this anaysis we cannot decide whether the stationary point at = 0 is a oca maimum, minimum or a point of infection. However, just to the eft of = 0 the vaue of which equas d 4 1)) is negative whist just to the right of = 0 the vaue of is negative again. Hence d the stationary point at = 0 is a point of infection. This is confirmed by sketching the curve as in Figure 17. f) /4 Figure 17 HELM 006): Section 1.: Maima and Minima 3

11 Task A materias store is to be constructed net to a 3 metre high stone wa shown as OA in the cross section in the diagram). The roof AB) and front BC) are to be constructed from corrugated meta sheeting. Ony 6 metre ength sheets are avaiabe. 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, y of the store that resut in the maimum cross-sectiona area. Hence determine the maimum cross-sectiona area. A Stone 3 m Wa B y O C Your soution 4 HELM 006): Workbook 1: Appications of Differentiation

12 Answer Note that the store has the shape of a trapezium. So the cross-sectiona area A) of the store is given by the formua: Area = average ength of parae sides distance between parae sides: A = 1 y + 3) 1) The engths and y are reated through the fact that AB + BC = 6, where BC = y and AB = + 3 y). Hence + 3 y) + y = 6. This equation can be rearranged in the foowing way: + 3 y) = 6 y + 3 y) = 6 y) i.e y + y = 36 1y + y which impies that + 6y = 7 ) It is necessary to eiminate either or y from 1) and ) to obtain an equation in a singe variabe. Using y instead of as the variabe wi avoid having square roots appearing in the epression for the cross-sectiona area. Hence from Equation ) 7 y = 3) 6 Substituting for y from Equation 3) into Equation 1) gives A = 1 ) = 1 ) = 1 ) ) To find turning points, we evauate da from Equation 4) to get d da d = ) Soving the equation da d = 0 gives ) = = 0 Hence = ± 15 = ± Ony > 0 is of interest, so = 15 = ) gives the required turning point. Check: Differentiating Equation 5) and using the positive soution 6) gives d A d = 6 1 = = < 0 Since the second derivative is negative then the cross-sectiona area is a maimum. This is the ony turning point identified for A > 0 and it is identified as a maimum. To find the corresponding vaue of y, substitute = into Equation 3) to get y = = So the vaues of and y that yied the maimum cross-sectiona area are m and m respectivey. To find the maimum cross-sectiona area, substitute for = into Equation 5) to get A = ) = So the maimum cross-sectiona area of the store is 9.68 m to d.p. 5) HELM 006): Section 1.: Maima and Minima 5

13 Task Equivaent resistance in an eectrica circuit Current distributes itsef in the wires of an eectrica circuit so as to minimise the tota power consumption i.e. the rate at which heat is produced. The power p) dissipated in an eectrica circuit is given by the product of votage v) and current i) fowing in the circuit, i.e. p = vi. The votage across a resistor is the product of current and resistance r). This means that the power dissipated in a resistor is given by p = i r. Suppose that an eectrica circuit contains three resistors r 1, r, r 3 and i 1 represents the current fowing through both r 1 and r, and that i i 1 ) represents the current fowing 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: Your soution Answer p = i 1r 1 + i 1r + i i 1 ) r 3 b) Show that the power dissipated is a minimum when i 1 = Your soution r 3 r 1 + r + r 3 i : 6 HELM 006): Workbook 1: Appications of Differentiation

14 Answer Differentiate resut 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 resut represents a minimum. c) If R is the equivaent resistance of the circuit, i.e. of r 1, r and r 3, for minimum power dissipation and the corresponding votage V across the circuit is given by V = ir = i 1 r 1 + r ), show that R = r 1 + r )r 3 r 1 + r + r 3. Your soution Answer Substituting for i 1 in ir = i 1 r 1 + r ) gives So ir = r 3r 1 + r ) r 1 + r + r 3 i. R = r 1 + r )r 3 r 1 + r + r 3. Note In this probem R 1 and R coud be repaced by a singe resistor. However, treating them as separate aows the possibiity of considering more genera situations variabe resistors or temperature dependent resistors). HELM 006): Section 1.: Maima and Minima 7

15 Engineering Eampe 1 Water whee efficiency Introduction A water whee is constructed with symmetrica curved vanes of ange of curvature θ. Assuming that friction can be taken as negigibe, the efficiency, η, i.e. the ratio of output power to input power, is cacuated as V v)1 + cos θ)v η = V where V is the veocity of the jet of water as it strikes the vane, v is the veocity of the vane in the direction of the jet and θ is constant. Find the ratio, v/v, which gives maimum efficiency and find the maimum efficiency. Mathematica statement of the probem We need to epress the efficiency in terms of a singe variabe so that we can find the maimum vaue. V v)1 + cos θ)v Efficiency = = 1 v ) v 1 + cos θ). V V V Let η = Efficiency and = v then η = 1 )1 + cos θ). V We must find the vaue of which maimises η and we must find the maimum vaue of η. To do this we differentiate η with respect to and sove dη = 0 in order to find the stationary points. d Mathematica anaysis 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 vaue of η when = 1 is ) 1 η = 1 1 ) 1 + cos θ) = cos θ). This is ceary a maimum not a minimum, but to check we cacuate d η = 41 + cos θ) which is d negative which provides confirmation. Interpretation Maimum efficiency occurs when v V = 1 and the maimum efficiency is given by η = cos θ). 8 HELM 006): Workbook 1: Appications of Differentiation

16 Engineering Eampe Refraction The probem A ight ray is traveing in a medium A) at speed c A. The ray encounters an interface with a medium B) where the veocity of ight 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 ight trave see Figure 18). Epress the resut in terms of the anges of incidence and refraction at the interface and the veocities of ight in the two media. P d Medium A) Medium B) a θ A O θ B b Q Figure 18: Geometry of ight rays at an interface The soution The ight ray path is shown as P OQ in the above figure where O is a point with variabe horizonta position. The points P and Q are fied and their positions are determined by the constants a, b, d indicated in the figure. The tota path ength can be decomposed as P O + OQ so the tota time of trave T ) is given by T ) = P O/c A + OQ/c B. 1) Epressing the distances P O 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 trave time is given by the stationary point of T ) such that dt d = 0. Using the chain rue in 1 c A a ) to compute 3) given ) eads to d c B b + d ) = 0. After simpification and rearrangement c A a + = d c B b + d ). 3) HELM 006): Section 1.: Maima and Minima 9

17 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 anges measured from the interface norma as shown in the figure. Equation 4) can be epressed as sin θ A sin θ B = c A c B which is the we-known aw of refraction for geometrica optics and appies to many other kinds of waves. The ratio c A is a constant caed the refractive inde of medium B) with respect to medium A). c B 4) 30 HELM 006): Workbook 1: Appications of Differentiation

18 Engineering Eampe 3 Fuid power transmission Introduction Power transmitted through fuid-fied pipes is the basis of hydrauic braking systems and other hydrauic contro systems. Suppose that power associated with a piston motion at one end of a pipeine is transmitted by a fuid of density ρ moving with positive veocity V aong a cyindrica pipeine of constant cross-sectiona area A. Assuming that the oss of power is mainy attributabe to friction and that the friction coefficient f can be taken to be a constant, then the power transmitted, P is given by P = ρgahv cv 3 ), where g is the acceeration due to gravity and h is the head which is the height of the fuid above some reference eve = the potentia energy per unit weight of the fuid). The constant c = 4f gd where is the ength of the pipe and d is the diameter of the pipe. The power transmission efficiency is the ratio of power output to power input. Probem in words Assuming that the head of the fuid, h, is a constant find the vaue of the fuid veocity, V, which gives a maimum vaue for the output power P. Given that the input power is P i = ρgav h, find the maimum power transmission efficiency obtainabe. Mathematica statement of the probem We are given that P = ρgahv cv 3 ) and we want to find its maimum vaue and hence maimum efficiency. To find stationary points for P we sove dp dv = 0. To cassify the stationary points we can differentiate again to find the vaue of d P at each stationary dv point and if this is negative then we have found a oca maimum point. The maimum efficiency is given by the ratio P/P i at this vaue of V and where P i = ρgav h. Finay we shoud check that this is the ony maimum in the range of P that is of interest. Mathematica anaysis P = ρgahv cv 3 ) dp dv = ρgah 3cV ) dp dv = 0 gives ρgah 3cV ) = 0 V = h 3c V = ± h 3c and as V is positive V = h 3c. HELM 006): Section 1.: Maima and Minima 31

19 To show this is a maimum we differentiate dp dv again giving d P = ρga 6cV ). Ceary this is dv h negative, or zero if V = 0. Thus V = gives a oca maimum vaue for P. 3c We note that P = 0 when E = ρgahv cv 3 ) = 0, i.e. when hv cv 3 = 0, so V = 0 or h h h V =. So the maimum at V = is the ony ma in this range between 0 and V = C 3C C. The efficiency E, is given by input power/output power), so here E = ρgahv cv 3 ) ρgav h = 1 cv h h h At V = 3c then V = h c 3c and therefore E = 1 3c c = = 3 or 66 3 %. Interpretation h Maimum power transmitted through the fuid when the veocity V = and the maimum 3c efficiency is 66 %. Note that this resut is independent of the friction and the maimum efficiency 3 is independent of the veocity and static) pressure in the pipe. P V ) h = 3 m h = m Figure 19: Graphs of transmitted power as a function of fuid veocity for two vaues of the head Figure 19 shows the maima in the power transmission for two different vaues of the head in an oi fied pipe oi density 1100 kg m 3 ) of inner diameter 0.01 m and coefficient of friction 0.01 and pipe ength 1 m. 3 HELM 006): Workbook 1: Appications of Differentiation

20 Engineering Eampe 4 Crank used to drive a piston Introduction A crank is used to drive a piston as in Figure 0. v p C r θ v c a c = ω r a p Figure 0: Crank used to drive a piston Probem The anguar veocity of the crankshaft is the rate of change of the ange θ, ω = dθ/dt. The piston moves horizontay with veocity v p and acceeration a p ; r is the ength of the crank and is the ength of the connecting rod. The crankpin performs circuar motion with a veocity of v c and centripeta acceeration of ω r. The acceeration a p of the piston varies with θ and is reated by a p = ω r cos θ + r cos θ ) Find the maimum and minimum vaues of the acceeration a p when r = 150 mm and = 375 mm. Mathematica statement of the probem We need to find the stationary vaues of a p = ω r cos θ + ) r cos θ when r = 150 mm and = 375 mm. We do this by soving da p = 0 and then anaysing the stationary points to decide whether they dθ are a maimum, minimum or point of infeion. Mathematica anaysis. ) a p = ω r cos θ r cos θ + so da p dθ = ω r sin θ ) r sin θ. To find the maimum and minimum acceeration we need to sove ) da p dθ = 0 r sin θ ω r sin θ = 0. sin θ + r sin θ = 0 sin θ + 4r sin θ cos θ = 0 HELM 006): Section 1.: Maima and Minima 33

21 sin θ 1 + 4r ) cos θ = 0 sin θ = 0 or cos θ = 4r sin θ = 0 or cos θ = 5 8 CASE 1: sin θ = 0 and as r = 150 mm and = 375 mm If sin θ = 0 then θ = 0 or θ = π. If θ = 0 then cos θ = cos θ = 1 ) so a p = ω r cos θ r cos θ + = ω r 1 + r ) = ω r 1 + ) = ω r If θ = π then cos θ = 1, cos θ = 1 so ) a p = ω r cos θ r cos θ + = ω r 1 + r ) = ω r 1 + ) = ω r In order to cassify the stationary 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 + At θ = 0 we get d a p dθ = ω r 1 + 4r ) which is negative. So θ = 0 gives a maimum vaue and a p = 7 5 ω r is the vaue at the maimum. At θ = π we get d a p dθ = ω r ) = ω r So θ = π gives a maimum vaue 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 = ω r cos θ ) 3 which is negative. 5 ) 5 1 so cos θ = ) ) 4r cos θ = ω r So cos θ = 5 8 gives a minimum vaue and a p = ω r Thus the vaues of a p at the stationary points are:- = ω r r 7 5 ω r maimum), 3 5 ω r maimum) and ω r minimum). ) 7 which is positive HELM 006): Workbook 1: Appications of Differentiation

22 So the overa maimum vaue is 1.4ω r = 0.1ω and the minimum vaue is 0.715ω r = ω where we have substituted r = 150 mm = 0.15 m) and = 375 mm = m). Interpretation The maimum acceeration occurs when θ = 0 and a p = 0.1ω. The minimum acceeration occurs when cos θ = 5 8 and a p = ω. Eercises 1. Locate the stationary points of the foowing functions and distinguish among them as maima, minima and points of infection. a) f) = n. b) f) = 3 c) f) = 1) + 1) ) [Remember d d n ) = 1 ] 1 < <. A perturbation in the temperature of a stream eaving a chemica reactor foows a decaying sinusoida variation, according to T t) = 5ep at) sinωt) where a and ω are positive constants. a) Sketch the variation of temperature with time. b) By eamining the behaviour of dt, show that the maimum temperatures occur at times dt of tan 1 ω ) a ) + πn /ω. HELM 006): Section 1.: Maima and Minima 35

23 Answers 1. a) d = 1 1 = 0 when = 1 d f d = 1 d f d = 1 > 0 =1 = 1, y = 1 ocates a oca minimum. f) 1 b) d = 3 = 0 when = 0 However, d d f = 6 = 0 when = 0 d > 0 on either side of = 0 so 0, 0) is a point of infection. f) c) d = + 1) ) 1) 1) + 1) ) This is zero when + 1) ) 1) 1) = 0 i.e. + 3 = 0 However, this equation has no rea roots since b < 4ac) and so f) has no stationary points. The graph of this function confirms this: f) 1 1 Nevertheess f) does have a point of infection at = 1, y = 0 as the graph shows, athough at that point dy d HELM 006): Workbook 1: Appications of Differentiation

24 Answer. a) T t b) dt dt = 0 impies tan ωt = ω, so tan ωt > 0 and a ω ) ωt = tan 1 + kπ, k integer a Eamination of d T dt reveas that ony even vaues of k give d T dt setting k = n gives the required answer. < 0 for a maimum so HELM 006): Section 1.: Maima and Minima 37