Chapter 8A. Recall. where. DeMoivre: then. Now lets do something NEW some groovy algebra. therefore, So clearly, we get:

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1 Chapter 8A We are revisiting this so we can use Euler Form later in the Chapter to assist us Integrate different functions (amongst other fun things). But first lets go over some old ground: Recall where z = a + bi z = r = a ( + b ( θ = tan -. b a DeMoivre: If z = r cos θ + i r sin θ = r cis θ then z 4 = r 4 cos nθ + i r 4 sin nθ = r 4 cis nθ Now lets do something NEW some groovy algebra z 4 = z 4 r 4 cis nθ = r cis θ 4 r 4 cis nθ = r 4 cis θ 4 so cis nθ = cis θ 4 therefore, cos nθ + i sin nθ = cos θ + i sin θ 4 So clearly, we get: cos nθ = Re cos θ + i sin θ 4 sin nθ = Im cos θ + i sin θ 4 Why do we care because it allows us to get rid of the coefficient in front of Theta! (In a different way to how we did it last term with those lovely Trig rules.) 1

2 Another rule that can help is the Multiple angle formula (pge 330) cos nθ = z4 + z -4 2 sin nθ = z4 z -4 2i yes you can just remember this these! Both of these get used in Proofs as well as you will see over the page! 2

3 eg. Q1c Prove, sin 4x = 4 sin x cos > x 4 cos x sin > x LHS = sin 4x = Im cos x + i sin x? (use Ti-Nspire to expand anything above a cubic) = Im cos? x + 4i sin x cos > x + 6 i ( sin ( x cos ( x + 4 i > sin > x cos x + i? sin? x = Im cos? x + sin? x 6 sin ( x cos ( x + 4 sin x cos > x 4 sin > x cos x i = 4 sin x cos > x 4 sin > x cos x = RHS *** Strategy start with the side that has a coefficient in front of Theta. eg. Q2a. Prove, 2 sin 2x cos x = sin 3x + sin x LHS = 2 sin 2x cos x = 2 z( z -( 2i z + z-. 2 = 2 z( z -( z + z -. 4i = z> + z. z -. z -> 2i = z> z -> 2i + z. z -. 2i = sin 3x + sin x = RHS *** Did you get that? Push it into the double angle form juggle a bit get it back into the double angle form and then pop it back to standard trig form *** *** Strategy start on the side that is Factored do that last question again, but start on the right hand side. Although you can pop it into double angle form, because there are plusses everywhere, we don t have any index laws that we can use, so we don t get too far starting on the factored side we can expand brackets and apply index laws to get our wanted outcome! 3

4 eg. Q3C Prove, sin? x = (cos 4x 4 cos 2x + 3) Ė Given, sin x = HI -H JI (K LHS becomes LHS = HI -H JI (K? = z. z -.? 2? i? = z? 4z ( 4z -( + z -? = z? + z -? 4z ( 4z -( = z? + z -? 16 = 1 8 z? + z -? 2 + 4z( 4z -( z( 4z -( = 1 8 = 1 8 z? + z -? 2 4(z( + z -( ) cos 4x + 4 cos 2x + 3 = RHS So, it s a matter of understanding when to put in in what form and yes, you are correct if you are thinking that comes down to PRACTICE :-) *** if there is a single term with a coefficient in front of Theta use DeMoivre *** where there is a factor from of trig, use double angle formula *** Anything with a Power in it use the double angle formula Don t look at my solution first Prove: cos 6A = 32 cos P A 48 cos? A + 18 cos ( A 1 Hence, use this result to solve 32x P 48x? + 18x ( = >, (giving 6 answers). ( 4

5 Solution: cos 6A = Re cos A + i sin A P = Re cos P A + 6 cos Q A i sin A + 15 cos? A i sin A ( + 20 cos > A i sin A > + 15 cos ( A i sin A? + 6 cos A I sin A Q + i sin A P = cos P A 15 cos? A sin ( A + 15 cos ( A sin? A sin P A = cos P A 15 cos? A (1 cos ( A) + 15 cos ( A 1 cos ( A ( 1 cos > A > = cos P A 15 cos? A + 15 cos P A + 15 cos ( A 1 2 cos ( A + cos? A 1 3 cos ( A + 3 cos? A cos P A = cos P A 15 cos? A + 15 cos P A + 15 cos ( A 30 cos? A + 15 cos P A cos ( A 3 cos? A + cos P A cos 6A = 32 cos P A 48 cos? A + 18 cos ( A 1 QED Now, set x = cos A, substitute and solve; 32x P 48x? + 18x ( = cos P A 48 cos? A + 18 cos ( A = cos P A 48 cos? A + 18 cos ( A 1 = 1 2 cos 6A = 1 2 and we have; x = cos A = cos π 18 6A = π 3, 5π 3, 7π 3, 11π 3, 13π 3, 17π 3 A = π 18, 5π 18, 7π 18, 11π 18, 13π 18, 17π 18, cos 5π 18, cos 7π 18, cos 11π 18, cos 13π 18, cos 17π 18 5

6 Chapter 8B Euler s formula and Integration Recall: General / Cartesian Form Polar Form z = a + bi z = r(cos θ + i sin θ) = r cis θ Introducing EULER who tells us (page 333): e K[ = cis θ since z = r cis θ then z = re K[ (you don t really need to understand that series in the text book on page 333!, this is just another way of displaying a complex number) so now we have EULER Form z = re K[ Hence: r 4 = a + bi 4 = r ] cos nθ + i r 4 sin nθ = r 4 cis nθ = r 4 e K4[ Now, here is where Euler can help with integration f x sin ax dx = Im f x e K`a dx f x cos ax dx = Re f x e K`a dx No need to be able to prove where these formula came from just know them! How about this Sine has an i in it, so if the integrand has Sine in it, the solution is the i maginary part of the integral? But first, lets just do some converting between forms: 6

7 First thing is first getting into and out of Euler Form. From Euler form it just pops out Eg. Q1a e (bk becomes cis 2π becomes cos 2π + i sin 2π becomes 1 + 0i so e (bk = 1 what a fantastic equation that is it incorporates so many identities e, π, i and 1 and they are all in a relationship together awesome! Similarly e bk = 1 Go from General, into Polar, and then from Polar, to Euler! Eg. Q2a 1 + i becomes 2 cis 3π 4 θ becomes 2 e cd e K so 1 + i = 2 e cd e K 7

8 The trick in Integrating is practicing so you get used to when to swap from one form to another This is a condensed setting out to fit it on one page Eg. Q 4a e a sin 2x dx the trig goes to Euler form = Im e a e (Ka dx then join the bases = Im e af(ka dx then factorise that index = Im e (.f(k)a dx now its easy to Integrate = Im g (Ihij)k.f(K rationalise denominator = Im g Ihij k (.-(K) Q simplify and expand the index portion and expand further to get = Im = Im. Q eaf(ka 1 2i. Q ea e (Ka 1 2i use Euler to get rid of one of the e s = Im. Q ea cos 2x + i sin 2x 1 2i expand = Im. Q ea cos 2x + 2 sin 2x + sin 2x 2cos 2x i = Im. Q ea cos 2x + 2 sin 2x +. Q ea sin 2x 2 cos 2x i extract either Re or Im as needed to get your result e a sin 2x dx = 1 5 ea sin 2x 2 cos 2x + C This gets a quicker answer than what we did last Term? Or else, we just like doing the same thing different ways J TASK: Do this again and prove to yourself that it doesn t matter which way you arrange the indicies above e. That is from line 4 from above, solve for; e ((Kf.)a dx 8

9 eg. Q4d e -a cos 2x dx = Re e -a e K(a dx = Re e (ak-a dx = Re e ((K-.)a dx = Re e ((K-.)a 2i 1 = Re = Re e (Ka e -a 2i 1 2i + 1 2i + 1 e (Ka e -a 2i = Re 1 5 e-a e (Ka 2i + e (Ka = Re 1 5 e-a cis 2x 2i + cis 2x = Re = Re = Re 1 5 e-a cos 2x + i sin 2x 2i + cos 2x + i sin 2x 1 5 e-a 2 cos 2x i 2 sin 2x + cos 2x + i sin 2x 1 5 e-a 2 sin 2x + cos 2x + (2 cos 2x + sin 2x)i = 1 5 e-a cos 2x 2 sin 2x + C or e -a cos 2x dx = 1 5 e-a 2 sin 2x cos 2x + C How awesome is Question 5 tying together the multiple angle trig laws from last term, and linking it to these current multiple angle rules! EVERYTHING in maths agrees with each other J 9

10 Chapter 8C I don t think so 10

11 Chapter 8D Damped Functions *** I won t be asking about Limits, but they aren t that hard to work out anyway! Damped functions are VERY similar to all the periodic functions you have learnt in Maths B all the same principles apply. The book has then take the form: but my preference is for this form: y = e`a sin bx y = Ae`a sin bx *** My task will show you why I prefer this form! *** You need to be able to find minimum and maximum values for any given domain, so ensure you recall co-terminal angles and general minima/maxima scenarios that we have done through ALL of your learning so far (Maths B & C). Don t be fooled the new integration rules in this chapter are the SAME as in Chapter 8B you do NOT have to remember these specific integration forms, as you can always just work from the general form from Chapter 8B e`a cos bx = Re e a `fnk dx e`a sin bx = Im e a `fnk dx In fact do NOT remember these as rules. If asked in your exam, please work from First principles, that is, follow the process from Chapter 8B and manually convert the trig term, into Euler form, manually combine the like e terms, and factorise the exponent and that gets you to the above rules. Panel will be Sooooo impressed with how well I have taught you I mean they will be impressed with how thoroughly you know the curriculum J I have NO Idea why we are now introducing New titles for derivatives??? I have never seen these before, but whatever J y = y o = dy dt y = y oo = d( y dt ( 11

12 Ex 8D Q3 a i q e -a sin x dx r q = Im e -a e Ka dx r q = Im e (-.fk)a dx r = Im = Im e (-.fk)a 1 + i e -a (cos x + i sin x) 1 + i q r q 1 i 1 i r = Im e -a q ( cos x i cos x i sin x + sin x) 2 r = = e -a q ( cos x sin x) 2 r 1 2 e-a (cos x + sin x) q r = 1 2 e-q (cos C + sin C) = 1 2 e-q cos C + sin C er (cos 0 + sin 0) = e-q sin C + cos C TASK: Determine the damped function, where B = 1, that goes through points b, and >b, ( ( Try to get it in the form y = e`a sin x Then try y = Ae`a sin x How did you go the function was y = 2e a sin x can you think of why it is hard without the A? TASK: Make up some of your own questions like this for practice. If you aren t careful, you will make up a question that is Not solvable J 12

13 Chapter 8E Real Life Applications Damped functions have applications in many different things car suspensions, bridges, swings... I could contrive many models to fit this, so it is just about looking at the context of the question, and applying the same process to any situation. Electrical circuits alternate (we have a 240 AC system), so we have this periodic nature of electricity that can be modeled by trig functions, so we will include electrical circuits in this chapter. Physics may have you up to speed, but if not, think about getting a battery. The bigger the Charge of the battery, then the bigger the Current it puts out. So there is a relationship between them. Charge = q Current = I Just as the gradient of a distance graph maps to velocity, Similarly, the Gradient of a Charge graph, gives Current. So the Integral of a Current graph, works backward to obtain the Charge. Let s say the CURRENT can be modeled by I = Ae`w cos bt If we want to find the Charge at any time (t), we could say; I = q o = Ae`w cos bt So, q = I dt = Ae`w cos bt dt That Integral looks familiar (I hope so anyway J ). So in reality, this chapter does not introduce anything new, it is merely applying current knowledge. (did you like my pun there?) 13

14 Chapter 8F Malthusian Don t be fooled this is just a fancy name for what we have done in Maths B J If the rate of population growth is directly proportional to the population, then we can say: then, dp dt P dp dt = kp But now we are in Maths C, we need to Integrate to arrive at the Population function! ** We did this in the last chapter, so move through quickly as it gets a lot harder!) Via algebra, integrate both sides. { dp = k dt 1 P dp = k dt ln P = kt + C P = e }wfq P = e }w e q and as when t = 0, P = x q, we get, P t = P r e }w This chapter shouldn t pose too much trouble! But you must be able to show the Integration to arrive at the population function!! Some models will differ slightly, but it is just about following the process, and working out the different Integration to come up with the correct model. 14

15 Eg. Q2 dt = k(90 N) 1 = k dt 90 N 1 90 N 1 90 N = k dt = k dt ln(90 N) = kt ln 90 N = kt + C 90 N = e -}wfq N = 90 Ke -}w at t = 0, N = = 90 K, so K = 70 at t = 10, N = = 90 70e -.r} e -.r} = k = ln Q 10 N t = 90 70e I Iƒ ] w a) set t = 25 N 25 = 90 70e I Iƒ ] (Q = b) to find 70% potential, must find max potential, so, set t = N = 90 70e I Iƒ ] ˆ = 90 Because ln Q is negative, as t gets bigger, e}w gets smaller, and approaches zero at the limit of t thus, 70% potential is 63 units thus, solve for t as follows 63 = 90 70e I Iƒ ] w e I Iƒ ] w = ln 5 27 t = ln 7 70 t = days 15

16 Chapter 8G Verhulst / Logistic Model We learnt in Maths B that populations tend to grow in an exponential way, but this is clearly NOT the case. (the assignment question about when there will only be standing room on the planet is clearly silly!) There are many different population growth models and this chapter investigates just one of them; a Logistic Model. Common sense tells us that the rate of change of population is influenced by the actual population. As we have more potential to breed, our rate of growth increases, however, there simply has to be a limit to how big the population gets (sufficient food, air, room). So, as we get closer and closer to our population limit (carrying capacity), the rate of growth actually starts to slow. For this model, we say: dt and we Integrate (refer page 370) to get; = an bn( N t = ` n N r N r + ` n N r e -`w Note: The limiting population is defined by: Note: The maximal population (when population is increasing at its maximum ` rate), is defined by: (n ` n Be clear on this terminology and remember these Notes! This chapter has a range of skills : Straight recall, by implementing the Logistic equation directly, Verifying this equation using differentiation (verification is in your rubric J ) Using complex integration techniques and log s to arrive at the Logistic equation How about refining a mathematical model, from a Malthusian to a Verhulst model. 16

17 Eg. Q1a Given becomes, Š Šw = 0.2N N( a = 0.2, b = , N r = 100 N t = ` n N r N r + ` n N r e -`w N t = r.( r.rrr? r.( 100 e-r.(w r.rrr? N t = e -r.(w N t = e -r.(w and we have, 500 N t = 1 + 4e -r.(w From here we can find the population at any time t. It may be beneficial to revisit this chapter from the beginning again and try and understand what is happening. Rate of change (judged against time) is affected by the population. However the initial relationship does not give us a function of time!... (it s just a differential equation) We are unable to find the population or the rate of change of population given a time period! That does tend to limit our model. So we need a t in there somewhere? Once we get the Population function, we know the relationship between Population and Time, we can not only find the population at any time t, but we can also find the rate of change of population (population growth) at any time t also! Let us proceed to VERIFY our above solution through differentiation 17

18 Eg. 2b Task: Prove Š Šw = 0.2N N( This is a little like proof by induction technique, not in the process, but if you have done it a few times, you know which form each side should take, that is, the aim is to get the first section to a single denominator (you will see) That is, as we know, N = Qrr.f?g Jƒ.iŒ we can sub it into the given equation We need to prove; dt = e -r.(w e -r.(w ( dt = e -r.(w e -r.(w ( need common denominator to add fractions, dt = e-r.(w 1 + 4e-r.(w 1 + 4e -r.(w e -r.(w ( So, simplified, we need to prove; dt = e-r.(w e -r.(w ( dt = e-r.(w e -r.(w ( dt = 400e-0.2t 1 + 4e-0.2t 2 ** This is our Target ** Now, back to the given function to differentiate it Via the Chain rule, Set; 500 N t = 1 + 4e -r.(w -r.(w -. N t = e N = 500 u -. du = 500 u ( where, u = 1 + 4e -r.(w du dt = 4 e-r.(w 5 18

19 Now, dt = du. du dt dt = 500 u ( 4 e-r.(w 5 dt = 400e-0.2t 1 + 4e-0.2t 2 Q.E.D. So, just like in proof by induction, you need to have an initial aim for your first expression This is because it is very difficult to work backwards from that last line, to get all the way back to our initial equation J 19

20 Next to get the function of t through integration: Q1c dt dt = 0.2N N( = N(500 N) N(500 N) = dt 2500 N(500 N) = dt before we Integrate both sides, we will need to use the partial fractions technique on the LHS. I will skip this process and get you to revise separately. 5 N N = 1 dt 5 N N = 1 dt 5 N N = 1 dt 5 1 N N = 1 dt 5 ln N 5 ln 500 N = t + C 5 ln N ln 500 N = t + C N 5 ln = t + C 500 N N ln 500 N = 0.2t + C e r.(wfq = e q e r.(w = N 500 N N 500 N we know, at t = 0, N = 100, e q e r.( r = e q = 1 4 So, 20

21 1 N 4 er.(w = 500 N e r.(w = 4N 500 N 500e r.(w Ne r.(w = 4N 4N + Ne r.(w = 500e r.(w N(4 + e r.(w ) = 500e r.(w N = 500er.(w 4 + e r.(w N = 500er.(w e-r.(w 4 + er.(w e -r.(w N = 500 4e -r.(w N t = 1 + 4e -r.(w Q.E.D. Nice hey!!! Don t forget our Limit theory. In these models the implementation of limits won t be difficult, as the logistic model reaches a horizontal asymptote as x 21