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# Calculus II 22M 026

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This 10 page Class Notes was uploaded by Virgil Wyman on Friday October 23, 2015. The Class Notes belongs to 22M 026 at University of Iowa taught by Staff in Fall. Since its upload, it has received 28 views. For similar materials see /class/227997/22m-026-university-of-iowa in Mathematics (M) at University of Iowa.

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Date Created: 10/23/15

CALC ll 22M026AAA FINAL EXAM REVIEW Chapter 9 Differential Equations 0 Euler7s Method 7 Euler7s Method approximates the solution to a differential equation The idea behind it is that even if we cant solve for y knowing y gives us the slope of the tangent line to y at any point The equation used in Euler7s method is a series of steps where we use the slope to approximate a little piece of the curve7 then move to the top of that piece and repeat until were at the x value we want 7 Given y Fzy yz0 yo h stepsz ze7 we have x h xwl yn ywl hFxn1yn1 7 Example Use Euler7s Method with stepsize 02 to estimate 11147 given y Fy x 7 xy and y1 0 0 1 yo 0 2 14 yg yl 02Fx1y1 02 0212 712 gtk 02 0392 So our answer is y14 0392 since were at the desired X value7 z 14 7 Suggested Problems 92 217 227 237 28ac o Separable Equations 7 If we have y gfy7 ie we can separate the X and y components of the right hand side7 then we have a separable equation 7 We can solve these very intuitively7 by rearranging the equation 1 7 dy g x dx f y and then integrating both sides 7 Remember7 you7re integrating so there7s always a 0 in your answer If you7re given an initial condition you can use it to nd 0 7 If you can7 its nice to solve your nal equation for y 7 Example Solve zy y y2 with initial condition y1 71 Rearrange to y yty and then to 22172 dy idx Integrate f 22172 dy fidz i 7 lny lny 71 ln C This step used partial fractions Solve lnx C i K gtk z FindK if KH gt K2 So the complete solution rearranged is y 7 Suggested Problems 93 67 97 107 167 177 187 237 307 367 397 42 0 Population Growth Natural Growth models the situation when growth or decay of a population is proportional to the size of the population at any given time The differential equation where Pt is the size of the population k a constant is dP 7 kP dt The natural growth equation is separable since we can rearrange it to dP Hg The solution to the natural growth equation with initial condition P0 P0 is Pt Poe Logistic Growth is similar to natural growth but its equation also takes into account carrying capacity This is the idea that if the population gets too big it may not be able to sustain itself The differential equation where Pt is the size of the population k and M are constants is dPi dt 7 P kP 1 7 7 lt M M represents the carrying capacity Notice that if P is very small compared to M then kP and were basically in a natural growth situation The solution to the logistic growth equation with initial condition P0 P0 is M 7 P0 Plttgt P0 M 1 A6461 where A Suggested Problems 94 6 8 13 14 0 Linear Equations If we have a differential equation in the form Pz y Qz this is a linear equation because it is linear in y To solve a linear equation we rst calculate the integrating factor z ef PM We then multiply the entire linear equation by This turns the left hand side of our equation into the derivative of the product z y So overall at this step we have d EUW y 96 QW From here we integrate both sides giving 95 y Iz Qzdz We can solve this for y by dividing by z if we like 2 i We may also have an initial condition given with our problem as usual7 we would use this to solve for the constant at the end Err ample Solve 7 2M 3t26t2 with initial condition 00 5 7 2 Here7 Pt 7211 So t 6f tht e 642 Thus multiplying by t turns our equation into e t2 gtk 1 3t2 gtk 6 2 gtk 642 i e42 1 f3t2dt i e t2gtmt30 i 1 et2t3 C Finally7 the initial condition 00 5 gives us 5 600 C 1 gtk C so 0 5 7 Suggested Problems 95 67 87 137 147 167 197 207 26 Chapter 11 Sequences and Series 0 Sequences A sequence can be thought of in basic terms as just an ordered list of numbers Written a1 a2 a3 an or sometimes an Sequences can be given in various ways 7 as a list a function of n recursively etc FORMAL DEFINITION We say the limit of a sequence an is L or lim an L W700 if given any 6 gt 0 there exists some integer N so that for any n gt N we have 1a 7 Ll lt 8 lnformally what this means is The limit of an is L if we can always go far enough out in our sequence so that everything past that point is really really close to L It will really help you if you can nd some way to put this idea into words that make sense to you We say the limit of a sequence an is 00 if given M as big a number as we want we can always nd an integer N so that the terms past aN are larger than M A sequence may also not have a limit 7 remember problem 43 from section 111 If a sequence an is given as a function fn and this function is something that7s continuous over the reals then we can take the limit as we usually would lim an lim fn The Squeeze Theorem is just what it sounds like if we can squeeze our given sequence between two other sequences and we know those two sequences converge to L then we know our orignal sequence converges to L as well If f is a continuous function then we can pull it inside and outside of a limit For instance cos1n cos cos0 1 A sequence an is increasing if we always have an 3 an It is decreasing if we always have an 2 an Monotonic just means increasing or decreasing A sequence an is bounded above if there is some M the upper bound that our ans are always less than or equal to It is bounded below if there is some in the lower bound that our ans are always greater than or equal to We use the word bounded to just mean bounded above and below Don7t overthink this de nition Bounded means exactly what you might picture that there are high and low bounds that our sequence cannot cross THE MONOTONE CONVERGENCE THEOREM If a sequence is bounded and mono tonic then it converges This can be really useful Its also pretty intuitive 7 say our sequence is bounded below and is also decreasing Then it has to keep getting smaller but it also can7t cross that lower bound So it gets trapped in between those two things and converges 4 Suggested Problems 111 2 17 21 22 25 33 36 38 44 56 65 0 Series A series is how we deal with adding an in nite number of terms The notation is 00 2 an n1 As usual when we deal with a process involving in nity we HAVE to think of it as a limiting process To do this for a series Ea we rst de ne the partial sums 5n a1 a2 an 21 a We then de ne co 2 n1 So to deal with a series straight from the de nition we usually have to see if we can nd a pattern the partial sums follow and then take the limit of those values Again the big thing to remember when dealing with series is A SERIES EQUALS THE LIMIT OF ITS PARTIAL SUMS lim 5 Hoe The Test for Divergence of a series 2 an If limH00 an 31 0 or this limit does not exist then the series 2a is divergent BIG NOTE This doesnt go the other way 7 there are de nitely series out there whose terms do converge to zero but the series itself is divergent One particularly useful series is the geometric series 2f1ar 1 converges to a if lrl lt 1 It diverges otherwise This series lir Dealing with series as the limit of partial sums can be time consuming Fortu nately we have quite a few tests at our disposal which will help us nd convergence or divergence much more quickly gtk Integral Test If fn is a continuous positive decreasing function on 1oo then 221 fn converges if and only if flee fn dn Useful because if we can apply it we get a de nite answer about convergence 9 The Comparison Test Given 2 an and 2 bn with positive terms and an 3 bn for all n we have 1 2b converges 2a converges and 2 2a diverges 2b diverges In other words 1 if the bigger sum converges then the smaller one does and 2 if the smaller sum diverges then the bigger one does 96 The Limit Comparison Teret For series 2a and 2b with positive terms we know that if limH00 if c with 0 lt e lt 00 then both series have the same behavior 7 either they both converge or both diverge The way to use this test is to nd a series whose behavior we do know geometric series are nice and then apply this test to compare it to our unknown series gtk The Alternating Series Test Given an alternating series 2f171 1bn if we have 1 bn1 3 bn for all n and 2 limH00 bn 0 then we know the series converges gtk The Ratio Test Given a series Ea we let L lirnH00 17 lf L lt 1 then the series is absolutely convergent lf L gt 1 then the series is divergent lf L 1 then the Ratio Test doesn7t tell us anything 7 we need to do another test to know anything about convergence gtk The Root Test Given a series Ea we let L lirnH00 m lf L lt 1 then the series is absolutely convergent lf L gt 1 then the series is divergent lf L 1 then the Root Test doesn7t tell us anything 7 we need to do another test to know anything about convergence Its very straightforward to know if you should use the root test 7 if your given series involves an n h root over everything this is the test for you If not use a different test gtk The Ratio and Root tests are nice because they are very easy to apply in most cases because they are simple limits and dont require you to nd any other series for comparison Just be careful that if you do get L 1 in either test remember that this gives no result and apply another test 7 Suggested Problems All 117 exercises 7 its important not only to know the various tests but also to know when each test works best 0 Power Series 7 A power series centered at a is a series which looks like 2 cnz 7 a The cns are some sequence of coef cients 7 in a speci c problem these will be given explicitly 7 The interval of convergence of the power series 20 cnz 7 a is the set of values about for which the series converges The radius of convergence R is half the length of the interval its radius about a R can be 0 a nite positive number or 00 which means it converges for all To nd an interval of convergence we simply use one of our convergence tests 7 the ratio and root tests are very useful most of the time since the L lt 1 requirernent gives an interval There7s one more step once we7ve found our interval 7 since the ratio and root tests give inconclusive results at L 1 we have to test the endpoints of our interval separately before knowing whether to include them 7 The Taylor Series expansion of a function f about a is the series ma 2f xia n1 This series within its interval of convergence actually EQUALS the value of fx strange as that may seem Taylor series are nice because they often give us a way to differentiate or integrate functions which we cant with our usual techniques Try not to overthink Taylor Series 7 the terms are just a formula you plug your function and a value into and not anything to be intimidated by 7 The Maclaum39n Series expansion of a function f is just the Taylor series ex pansion about a 0 That is it looks like 00 f 0 n 0 n z Suggested Problems118 7 11 12 13 15 21 24 26 32 119 7 910131516 1110 7 8910151633344955 0 Miscellaneous 7 Remember that even though the test covers just chapters 9 and 11 you still have to be able to do any integrals that come up in your work A few that may be good to memorize are i arcsinE 0 a2 uz T a d 1 7u 7 arctanE C a a a2u2 du ln uu27a2 0 7702 l l Dont let yourself be overwhelmed by all this 7 breaking the material up into little sections and studying just one at at time should help it be manageable You can totally do it You guys have been an amazing class to teach this semester and I wish all of you the best of luck for nals week and for your next semester Don7t hesitate to email me in the future if you have any questions about further math classes or just want to say hi quot Annie CALC ll 22M026AAA MIDTERM 11 REVIEW 0 Parametric Equations 7 Know the difference between the parametric equations z ft y gt and the parametric curve they de ne the set of points satisfying the two equations 7 Be able to graph a given set of parametric equations 7 Eliminate the parameter to nd cartesian coordinates if possible 7 Suggested Problems 101 4 87 167 267 347 427 48 o Calculus on Parametric Curves tangents g as long as 31 0 Use this tangent slope value just like you would normally ie slope of tangent line7 increasingdecreasing test7 etc tangent line is horizontal when 07 31 0 tangent line is vertical when 07 31 0 7 areas For x W 9 9097 A fgwmdt usual area under a curve parametric version where a and b are in terms of x and we nd a and B using a ay7 b g arclength For x it7 y g1t7 b 19 L 1ltgt2dz f t29 t2dt usual arclength formula parametric version where a and b are in terms of x and we nd a and B using a ay7 b g surface area of curve rotated about the X axis For x it7 y g1t7 S 27w f t29 t2dt where a and B are in terms of t Note that this is directly related to the arclength formula at a given t we are rotating that point on the arc7 and the formula inside the integral is taking the circumference of the rotated circle 7 Suggested Problems 102 4 97 127 207 367 467 56b7 607 667 697 72 0 Polar Coordinates Be able to plot a point given in polar coordinates Conversions polar a Cartesian and Cartesian a polar remember there is am biguity converting in this direction x rcost9 y rsin6 r2 2 y2 tan0 a s Graphing It will be useful to know some basic trig values nd a way 306090 454590 triangles unit circle etc that you can easily remember them It also may be good to know xi 14 and 3 17 Suggested Problems 103 2 4 6 11 14 28 36 46 48 60 84ac o Calculus with Polar Coordinates tangents Given 7 f0 we can nd X and y as parametric equations of 0 z f0cost9 y f0sin6 dyi Then we can calculate dm dy sin0rcos6 dx cos077 sint9 The last equals sign comes straight from the product rule on the equations for z and y You may nd this derivation easier to remember than outright memorizing the end formula 7 areas Given 7 f0 1 1 b 1 A 7f62d6 42cm 1 2 1 2 Note a and b are in terms of 0 The above formula comes from the idea A 720 for the area of a sector of a circle 7 For curves that have symmetry you may want to just nd the area of one piece and multiply appropriately wGiVenrf9v 7 dr L 2 72cm a1r d6 This is directly related to the usual arclength formula with the substitutions z f0 cost and y f0 sin0 7 Suggested Problems 104 6 16 20 24 34 40 44 48 56 o Conic Sections 7 parabola vertex 11 x 7 12 4py 7 b focus ab 10 directrix y b 7 p vertex 11 y 7 b2 4px 7 a focus a p b directriX z a 7 p 7 ellipse In all cases we have center h Let 02 la2 7 bzl If a 2 b the foci are h i c k the major vertices are h i ak lf b 2 a the foci are h k i c the major vertices are h k i b 7 hyperbola 96 02 71 02 2 bz y 7 k 7 z 7 h b2 12 CASE 1 1 CASE 2 1 In all cases we have asymptotes y 7 k ibx 7 h Let 02 a2 b2 ln CASE 1 the foci are h i c k the vertices are h i a ln CASE 2 the foci are hk i c the vertices are hk i b 7 Suggested Problems 105 7 8 10 16 24 34 42 47 50 52 56 That7s it If you can study hard this weekend and work lots of problems you should be well prepared for the midterm Good luck quot Annie

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