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# Class Note for MATH 254 at UA

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Date Created: 02/06/15

MATH 254 Lectures 710i Applications and numerical simulations of rst order odels 0 Population growth 0 Radioactive decay 0 Cooling of bodies 0 Mixing of salt solutions 0 LR and RC electrical circuits 0 Mechanics of falling bodies 0 Improvements to Euler method for numerical calculations 0 Higher order numerical methods Reference NSS Chapter 3 Read all of it 0 Population growth In Lectures 34 pg 7 we introduced the subject of population dynamics Use methods introduced on pg 17 of Lectures 34 to analyze what happens if you add a constant growth term h number added h gt 0 or subtracted h lt 0 per unit time to the logistic equation d j h ky17 yK For h S 07 draw the graph of h kyl 7 yK7 say when k K 17 for various h values h 07 701 7025 705 What happens at h 7 h is called a harvesting termi o Radioactive decay See pg 57 Lectures 34i 0 Cooling of bodies We introduced Newton s law of cooling in Lectures 12 Be able to do example 2 pg 2 o Mixing of salt solutions We discussed such examples in Lectures 1 27 3 4 Understand derivation of equa tions and be able to do Q1 pg 11 Lectures 1 2 Example 17 pg2 Lectures 34 0 LR and RC electrical circuits Read Chap 55 R i W Circuit units resistance ohms VA VB 2 Bi P B capacitance farads VA VB inductance henrys VA VB L M applied emf volts VA VB 2 E givenquot 7 A 3B q Charge coulombs i current amps coulombssecond Rules Kirchoff s laws 1 Conservation of electrical charge The amount of current entering any node A is equal to the amount leaving 139 2391 i2 t C 1 II The voltage drop around any closed circuit is zero a RC circuit 139 A jW B R 99 i C I G 39 F VA VBVB VFVF VGVG VAR 0gq Et0 Therefore for an RC circuit the charge 4 on C obeys dq 1 RE 5g 7 Et 41 b LR circuit A M B L 2 R G F VA7VBV37VFVF7VGV37VAL0RiiE0 Therefore the current satis es the ode dz Ldt Rz Et 42 c LRC circuit A W B L C R G F VA7VBV37VFVF7VGV37VALqCRiiE0i Therefore the charge qt satis es dq l L7 RE 54 Et 4 3 Qili A constant emf of V0 volts E V0 is suddenly at t 0 secs applied to an RC circuit With R 1 ohm and C faradi If the initial charge q0 on the capacitor is zero nd qti What is tlirn qt equal to Explain Qi2i A constant emf of V0 volts E V0 is suddenly applied to an LR circuit With L 2 henry and R l ohrni If the initial current is zero nd What is tlirn equal to Explain o Mechanics of falling bodies Newt on s Second Law Let position be yt7 velocity vt and acceleration at fill y 0 T F l l y positive down mg Where m is the mass assumed to be constant7 g is gravity and F is the friction that acts in a direction opposite to motion m has units of slugs or kgsi gt g is ftsec2 or rnsec2 and F is pounds or newtonsi For slow ow7 F kv for fast ow F In I Will tell you the meaning of slow and fast in class The units of F are pounds or in MKS newtons and the units of k are poundfsseconds O newtonsseconds eet meters For slow ow d1 kv i 7 7 4 4 dt 9 m Initial state speci ed 110 vol To recover position yt7 rst solve for vt then integrate v to nd Equation 44 is both linear and separablei Solve as linear d1 kv 9 Integrating Factor IF 6 dt veec vv0att0 voc lt4vsgt What does the solution 45 tell us 1 As if gt 00 v gt El called the terminal velocity 2 If v0 lt k l vt increases till equilibrium 1 If 110 gt aka vt decreases till equilibrium 1 To nd yt we integrate equation 45 dy mg mg 1 e m ET 0 7 at 2 ma e yt C k kk voe Ifyyoatt0then yty Tr T 1 6 5quot mgk terminal velocity t W 9 g OY h xmf t yo 9 391 Wquot 50quot p k I t Example 1 A ball is thrown upwards at a velocity of 20m sec from a platform 30m above the ground Assuming there is nofriction determine the maximum height the ball reaches and the time it hits the ground Given v0 20msec and k 0 no friction Then mgz I m dt 9 10 20 measure yt positive up and in meters measure t in secs measure m positive in kgs measure 9 981msecs2 positive up and in meters y0 39 T 30m i ground vt gt v0 y 15 vot Maximum height is achieved at t 39 and y 4 204 m Maximum height above ground is 504 m y 30 when 2 vot 30O t v0 V113 609 vex123 609 g g z 53 seconds 110 30 t 110 t Example 2 A parachutist jumps at 5000ft Suppose the drag constant k changes from 075 0 lt t lt 10 to 12 t gt 10 after 10 seconds when the skydiver opens his chute Describe the fall yt measured in ft ve downwards t in secs mg 180 lbs g 32 feetsecs2 m 18032 0lttlt10 k0751 A 2 15 tgt10 k12 SIX 0 lt t lt 10 vt 2401 ef yt 2402 12 52400 e 1110 2401 e z 176 ftsec y10 z 1080 ft Distance to ground 3920 ft t gt 10 vt 15 2401 e 15e t m 32 t 10 yt 15t 10 076 151 e 1 10 ie yt 10 1080 so y 5000 when 3920 15t 10 N 751 egg 10 Where N 75 means approximately 1532 x 161 Solve iteratively 3920 t 10 z m 60 15 2 secs Correction of the order of 1 Why The sky diver reaches 90 terminal velocity of L5 ft sec after about t 10 1 2 seconds Therefore he falls the remaining 4000 ft at almost constant terminal velocity 15 ft sec 3 qe W LV39U M we t z l l l X0530 t The position of the center of mass of the ball is wt The velocity of the center of mass of the ball is ut vt The force due to gravity is 0 mg The friction force is b m w WWW wmwwt g is gravity 1 is the radius of the ball b is a dimensionless unitless constant 2 7 mb 2 2 mass length masslen th Why a xu 39u v has d1mens10ns of length time force The equations of motion are d2z du 12 7 7 7 7 2 2 mdt2 mdt ma u 1 u 46 d2y d1 12 7 7 7 7 7 2 2 mdtQ mdt mg ma u 1 1 47 Special Case Suppose that uo 0 Thus ut 0 for all time7 because if ut 0 07 then du 7 t 0 0 dtlt gt and so ut 0 dt 07 and so on Then 47 becomes dvi 1 dt7 9 avvi lvlv7ifvgt0andlvl7vifvlt0i We will imagine that we throw the ball directly upwards so that v0 gt 0 Then d1 12 2 779731 lt0 48 so that vt decreases until it is zero7 when the ball achieves its highest point Say this occurs at time t tli After that the ball falls down so that v lt 0 and d1 12 2 7 7 7 t t i 49 dt 9 av 7 gt 1 Describe the trajectory Take yo 0 For t lt t1 we use 48 d1 12 g v2 7gdti lntegrating both sides and rearranging gives arctan 7v 7 7t constanti ag a Att0vv0sothat arctanlt ivtgt 7H b7gtarctanlt 1110 ag a ag vt E tan 7xb7gt arctan 4 1110 i b a ag 9 01 The ball reaches zero velocity and maximum height7 yt1 yl at time t1 7 arctan 7110 i by V us Check units has units length t1mei eng l time 1 length 1 length X t lengthi W 11116 b a7gv0 has un1ts To nd yt7 integrate dy w i Mi dt 7 vt 7 b tan arctan lt agv0gt 7 a t 7 cos arctan lt 1110 7 bit l ag v a y0 0 implies constant is cos arctan 7110 i V as yt loge constanti 7 loge Thus we have a b by a 1 ya 7 loge COS arCtan UO 7 7t 7 7 loge cos arctan 7m i b ag a b ag t 7 7 E1 1 i y 1 7 M 7 7 b Oge cos arc an agvo a b 2 7 a b 2 Eloge 4 1 110 7 loge lt1 110 i For t gt t17 from 47 and 497 we have dy 7 d1 7 b 2 E 7 vt dt 7 79 av 4 10 with initial condition yt1 yl vt1 0 Then d1 12 W gdt l l by dv 77 72l7dt J gzyg v a by multiplying through by 2 and taking the difference of two squaresi lnte grating both sides and taking exponentials gives 7 1 b b 9 W consti exp 2 t i Since vt1 0 at t t17 we nd the constant to be exp72 t1i Thus exp lt2 bgt 7 t1gt 1 411 since sinht9 ee 7 6 9 7 e29 71 cosht9 7 e9 e 9 7 629 1 To nd yt7 integrate vt to nd W 710ge cosh t7t1lgt i 412 Remark 1 As t7t1 becomes large with respect to the time tanh t7t1 7gt 1 and vt 7gt 7 called the terminal velocity Question Could you have deduced the terminal velocity from the differential equation 4 10 Problem A ball is thrown directly into the air with speed v0 lOOmseci at time t 0 Given that gravity 9 lOmsec2 an approximation to 9 81m8602 and that the friction force 3 is equal to m l nd how high yl the ball rises and at what time t1 it reaches that heighti Find its velocity vt and position yt also for t gt tli For0lttltt17 107 v27 v d Fortgtt17 710 TIOOU27 dig v vt1 0 yt1 yr 11 o Improvements to Euler method Read Chapter 357 36 Basic ideal Recall that in deriving the Euler method in Lectures 567 we approxi mated T1 m yltzgtgtdz To by the area of the rectangle 107910 951 950 We could have also approximated it by fltIlvyIl 11 i 10 A better approximation would be the trapezoidal area f107y10 fltzhyltz1gtgtgtltz1 7 m lfwe did the last7 then 11710 h7 yzl yzofzo yzofzlyzl or7 in general7 where 9717 as W yn gm y mm W 413 However7 this is an implicit equation for yn1i We could solve it by a Newton iteration method but it is easier to simply approximate yn1 on the RHS of 413 by yn h rm The improved Euler algorithm is then h yn1 yn l 9n l h yn l hflt n7 414 which is an explicit formula for yn1i 0 Example Suppose y so that the solutions of d y with y0 l is 69 Then the improved Euler algorithm reads h yn1 yn yn yn1 h ynl h h22 with solution h2 yn yolt1hh22 1h 3 since yo 1 See Table 34 on pg 130 for tabulation of errors The cumulative error in improved Euler is of order h2 as compared with order h for Euleri More sophisticated methods see Chap 36 can also be constructed which have even smaller cumulative errors 12 Homework cf Lecture 4 Reconsider 47 with u 0 do I a 9 EM 415 Equation 415 is a rst order autonomous ode Let us graph fv g quot39 We see the terminal velocity V lab is an attracting and stable equilibrium The homework for this section is the HW set from the book the examples done Q1 Q2 on pg 3 on RC LR circuits and the following SECOND TRIAL EXAM 1 The rst one much more dif cult was given out in Week 3 Do it as if you are doing an exam You should be able to do within an hour 13 Q 1 Classify the order and whether linear or nonlinear autonomous or nonautonomous Equation Order Lor NL Aor NA d2y t 2 L 1 dt2y 7 NA 611 2 2 1 y 3 Jia3ysinm d3 d3 3 4 0 dx2y y 5 5amp4 dt y dy 2 6 7t dt d4y 7 dLZIJ4 da dac 25 8 dt2dt x O dza dx 9 Wdt25mcos5t d2y dy 14 Q2 Solve the initial value problem and draw the graph of the solution over 0 g a lt 00 5010 d cc 1 y dcc x24y 1c24 y 01 15 Q3 Consider 1 2 3 1 y2 y 00 lt y lt 00 a Draw the graph of 1 y2 y versus 3 00 lt y lt oo The equilibrium solutions are b 39 y c Are they asymptotically stable AS or unstable U y y d What is the basin of attraction of the asymptotically stable equilibrium solution 16 Q4 Solve y 1 y2 y 310 000 Hint You will find that 4 y n 3 17 P a 0 up At time t 0 minutes a tank initially containing 50 gallons of pure water is filled with a brine solution with a concentration of lb of salt per gallon at a rate of 4 gallons per minute It is drained at the same rate Let 23t be the number of lbs of salt in the tank at time t minutes Write a differential equation for 9ct What is 950 Solve for at 56t To what value does the concentration in the tank tend as t gt 39oo lim W t 00 50 18

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