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by: Burnice Ratke


Burnice Ratke
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Class Notes
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This 14 page Class Notes was uploaded by Burnice Ratke on Tuesday October 20, 2015. The Class Notes belongs to MATH122 at San Diego State University taught by J.Mahaffy in Fall. Since its upload, it has received 12 views. For similar materials see /class/225264/math122-san-diego-state-university in Mathematics (M) at San Diego State University.

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Date Created: 10/20/15
Spring 2003 Math 122 Integration by Substitution 1 Solutions for the integrals b For the integral below we make the substitution u 2x2 7 3 so du 4x dag It follows that xmd isz734xd 0 2273C d For the integral below we make the substitution u x2 4x 7 5 so du 200 4 dag It follows that 7172 1 2 73 mdaamp 40075 2004dw u gdu 1 711172 C 1 7 C 4 x2 1 4m 7 52 f For the integral below we make the substitution u 7052 7 205 so du 7200 7 2dao It follows that m221md 7 eim2 2m7271d 7e C 767124721 C h For the integral below we make the substitution u cos2x so du 72sin2x day It follows that sin2ao 1 1 d 77 7 72 2 d Cos2w w 2Cos2x smlt w 1 1 7 7d 211 M l1 1 c 2 n ul 1 7lnlcos2wl0 i For the integral below we make the substitution u 00 so du 0f day It follows that Ldx 2 e li d 2 2 e du 7 2euC 25 C 2 Solutions for the differential equations a The differential equation below has only a function of t on the right so it can be solved by simply integrating the right hand side i dt yt t t21dt t t21 The integral is solved with the substitution u t2 1 so du 2tdt txt21dt t212tdt 1 5 udu 1 C 1 32amp2 1 0 Thus 1 W 7 gltt2 1 C From the initial condition 310 5 we have 5 C or C Thus the solution is given by t2 1 7 32 E W 7 3 1 3 d The differential equation below is a separable differential equation that has already been separated dy 7 1 7 t y t2 7 2t 2339 The integral in t uses the substitution u t2 7 2t 1 2 so du 2t 7 2 dt so from the separable variables techniques 17t d dt y y t272t23 y2t 7 71172 C 1 2 t 7 C y 2t272t22 Thus 1 Mi m From the initial condition 311 1 we have 1 4 C or C Thus the solution is given by 0 07 1 1 y 2t272t22 239 e The differential equation below is a separable differential equation which gives the fol lowing two integrals t2 2 75 d1 1 dt yey2dy t2dt The integral in 3 uses the substitution u yz so du 2y dy so 1 E5122ydy t2dt 1 1 Eeudu gt3C 171713 75 igtC 2 2 22 7 7 3 6 7 3t 01 2 2 3 y t ln gt 01 Thus 2 yt i ln 3253 01 From the initial condition 310 O we have 0 lnCl or 01 1 Thus the solution is given by yt ln 253 1 g The differential equation below is a separable differential equation which gives the fol lowing two integrals Q A oo V l l 0 R oo m Q 111053 C lele Thus 1 W 3011003 C From the initial condition 311 72 we have 72 ln13 C or C 72 Thus the solution is given by 111003 7 2 2115 3 From the lecture notes the equation governing the motion of an object subject to gravity is given by du ngZ mi R27 which when using the relation that 34 11 gives d1 9R2 Udm T xR239 Using separation of variables we see dm 7 2 7 udu7 9R R2 W06 9R2 0 2 xR With the initial condition 0R V0 we can readily solve for the constant and 2C V02 7 29R SO 2ng 797 306 V2 7 2 R 7 71000088 W 95R 0 g x6378 Thus with the data that R 6378 km 9 00098 kmsecz and V0 5 kmsec The velocity is zero at the maximum height so we solve I797 306 0 77 71000088 90 6378 This equation gives the maximum height achieved by the object is 1594 km 5 a This problem satis es the properties of logistic growth and it can be rewritten in the following form dP P a 03P 1 7 which we saw in the lecture notes could be separated into the following integral form 1 1 7 idP 7 dP 703 dt P 300 W 7 1 Following the notes these integrals are evaluated to give Pt ln W 71 7lantl 7 703t C If A 50 then some algebra gives the general solution 300 P t 1 7 30014505 With the initial condition that P0 1 the population of game sh is given by 300 P 1 299570v3t39 From the form of the equation above it is easy to see that the carrying capacity for the game sh is 300 thousand 90 of the carrying capacity is 270 thousand so we solve 300 10 270 1 299 05 7 1 29957031 or 1 5 9 Thus 503 i 9299 2691 or t 1730 ln2691 263 yr to reach 90 of this carrying capacity b The differential equation dP a 7 70001032 7 30013 20 000 can be readily separated to give dP 70001 dt P2 7 300P 20 000 From the formula given 001 001 7 7 7 dP 700012 C P 7 200 P 7 100 001111 lPt 7 200 7 001111 lPt 7 100 7 700012 C Pt 7 200 1 7012 C n Pt7100l 1 W 200 70 1t01 70 01 P t7100 5 5 With the initial condition PO 300 A Thus implicitly Pt 200 7 1 7001 Pt 7 100 25 39 This is solved to give the solution to the differential equation with harvesting 400 7 100570 Pm 2 7 5701 39 c By letting t a 00 in the equation above one easily nds that the limiting population for this differential equation with harvesting is 200 thousand game sh The graph can be found in the short solutions 7 a To nd the maximum growth rate we differentiate kt so by the quotient rule t2117t2t 17232 k t 012 0127 t2 1 22 1 Setting the derivative equal to zero gives t i1 and 01 0712 006 Thus the maximum growth rate kt is 006 occurring at t 1 The graph of this function is available on the short solutions b The differential equation given by Q 012 t 23 dt 2amp2 1V is a separable equation that can be solved from the integrals V 23dV 012 t2 1 1tdt 3V13 006 t2 1 12tdt V13t 0021r1du V13t 002111 lul C 00211105 1 C where we took u t2 1 and du 2t dt Since VO 17 1 002ln1 C or C 1 By cubing both sides we see that the solution to the differential equation is Vt 1 00211105 13 c To double its size the cell must satisfy 2 1 002 lnt2 13 or 1002 lnt21 213 Thus7 lnt2 1 50 213 71 or t2 exp 50 213 71 71 By taking the square root7 we nd that it takes 664 min for the cell to double in volume Fall 2000 Selected Solutions Linear Differential Equations 1 b Solve d Ti 012 7 2 20 5 Factor the coef cient leading 2t to give 2 t 012t 7 20 Make the substitution wt 2t 7 20 so w0 20 7 20 715 and ut 2 t This gives the problem 10 01w w0 715 This has the solution wt 71550 225 7 20 so 225 20 71550 d Solve dh a 5 7 0211 110 10 Factor the coef cient leading 712 to give hx 702hx 7 25 Make the substitution 712 7 25 so 20 710 7 25 715 and 2 m h z This gives the problem 2 7022 20 715 This has the solution 22 7155022 may 7 25 so 1135 25 71557022 e Solve d y 7 002 2 50 d y y This is like Malthusian growth except that the initial condition begins at t 2 The general solution is Mt 2405002 with 342 50 2905004 Thus yo 505 03904 This gives the solution Mt 50570046002 506002072 3 a The differential equation for the population of Canada is C t 12105 00 24070000 which by letting t 0 corresponds to 1980 has the solution Ct 24070 0005 From the population in 1990 we have 26 620 000 C10 26 620000 24 070 00051021 or 510 m It follows that k1 1170 ln2662024070 001007 yr l To nd doubling time we solve Ctd 224 070000 24 070 000 or 5 2 It follows that the doubling time td killn2 688 yr b A similar argument gives the population of Kenya Kt 16 681 0005 with the rate constant k2 17101n2422916681 003733 yr 1 and doubling time td k721n2 186 yr c The populations in 2000 are given by C20 24 070 0005201 29440150 and K20 16 681 000520112 35192401 The populations are equal when C t Kt or am 7 24 070 Int 7 16275 240700006 7 16 681 0006 or Skit 7 167 681 Thus 5012411 3amp3 so kg 7 k1t ln g It follows that 1 24070 t W n 2135 years This would be in the middle of 1993 5 a The radioactive decay problem is This has the solution St 205 Since the half life is 28 years we have 528 10 2052812 so am 2 Thus 28k ln2 or k ln2 002476 yr l After 10 years 510 2054011 156 mg b For 7 mg remaining St 205 7 or 5 207 lt readily follows that t ln207k 424 yr 7 a From the notes we have Ca QTPsys 7 Pdla and Pdm Psyse TCDRS Thus Psys 7 Pdia Psys 7 eiTCaRS QTCa SO P QT 9219 Ca 17 e7TCaRs39 By letting Q 56 litersmin T 170 minbeats Ca 00016 litersmm Hg and RS 176 mm Hglitermin and substituting into the formula above we nd the systolic pressure as Psys 12571 mm Hg It follows with the formula for diastolic pressure that Pdm 756 mm Hg7 respectively Thus7 the 20 decrease in compliance causes minimal changes in the blood pressure 9 a The differential equation is given by T t 7kTt 7 207 T0 100 Let 2t Tt 7 207 then 20 100 7 20 80 and 2 t T t Thus7 we need to solve the initial value problem 2t 7102t7 20 80 This has the solution 225 805 Tt 7 20 so Tt 20 805 b Since T10 80 20 8051012 60 8051M or 510k 8060 43 It follows that the rate constant k ln4310 2 002877 To nd when it reaches 30 C7 we solve Tt 30 20 805 or 10 805 or ekt 8 This gives it ln8k 723 min 11 a Let at be the amount of pollutant7 and the concentration Ct is the concentration of pol lutant in ppb The change in amount the amount entering the amount leaving The change in amount7 iit7 has units massday The amount entering is 1 207 000 ppb m3day massday7 while the amount leaving is fct 40000t ppb m3day massday Thus7 05 20000 7 4000205 To make the concentration equation7 we use that Ct atV at200000 Since ct at2000007 we can write the equation above as ct 017 0020t with 00 0 To solve this initial value problem we write 6 7002c 7 5 with 30 0 We make the substitution 2t Ct 7 57 so 20 00 7 5 75 and 2 t 05 The concentration equation becomes 270022 with 2075 which has the solution 2t 7557002 Ct 7 5 Thus7 the solution is Ct 5 7 554102 b We solve Ct 5 7 55 03902t 4 or 55 03902t 1 Thus7 5002 5 or 00 ln5 Hence7 the concentration reaches 4 ppb at t 501n5 805 days c The limiting concentration is limtH00 Ct 5 ppb This easily follows because as t 7 007 7002 e 7gt 0 Spring 2003 Optimization 1 Find the area of the largest rectangle with a base on the ao axis and the upper vertices on the parabola y 12 7 002 Give the dimensions of this rectangle 2 A rectangular study plot is bounded on one side by a river7 and the other three sides are to be blocked off by a fence Find the dimensions of the plot that maximizes the area enclosed with 20 meters of fence 3 An open box with its base haVing a length twice its Width is to be constructed with 600 in2 of material Find its dimensions that maximize the volume 4 Find the dimensions of an open rectangular box with a square base that holds 32 in3 and is constructed with the least building material possible 5 Find the dimensions of a right circular cylindrical can with both a top and a that holds one liter 1000 cm3 and is constructed with the least amount of material possible 6 The strength of a rectangular beam is proportional to the product of its Width and the square of its depth Find the dimensions of the strongest beam that can be cut from a circular log with a radius of r 7 A catalyst for a chemical reaction is a substance that controls the rate of the chemical reaction Without changing the catalyst itself An autocatalytic reaction is one Whose product is a catalyst for its own formation A XLX The rate of this reaction 11 dxdt is given by the formula 11 ka 7 m Where a is the initial concentration of the substance A7 00 is the concentration of the product X7 and k is the rate constant of the reaction Find the concentration x that produces the maximum rate of reaction 8 Nutrients in low concentrations inhibit growth of an organism7 but high concentrations are often toxic Let c be the concentration of a particular nutrient in molesliter and P be the population density of an organism in numbercmZ Suppose that it is found that the effect of this nutrient causes the population to grow according to the equation 10000 P 7 C 1 100C a Find the concentration of the nutrient that yields the largest population density of this organism and What the population density of this organism is at this optimal concentration b Sketch a graph of the population density of this organism as a function of the concentra tion of the nutrient 9 One question for shery management is how to control shing to optimize pro ts for the shermen We will soon study the continuous logistic growth equation for populations One differential equation describing the population dynamics for a population of sh F with har vesting is given by the equation dF F 7 F 1 7 7 7 F dt T lt K a where r is the growth rate of this species of sh at low density K is the carrying capacity of this population and x is the harvesting effort of the shermen We will show that the non zero equilibrium of this equation is given by T 7 06 F5 K7 7 One formula for pro tability is computed by the equation P mFe PW Km in 7 Find the maximum pro t possible with this dynamics What is the equilibrium population at this optimal pro tability Also determine the maximum possible sh population for this model and at what harvesting level this occurs Clearly this is a grossly oversimpli ed model but can give some estimates for long range management 10 From Semelparous organisms breed only once during their lifetime Examples of this type of reproduction strategy can be found with Paci c salmon and bamboo The per capita rate of increase r can be thought of as a measure of reproductive tness The greater r the more offspring an individual produces The intrinsic rate of increase is typically a function of age 00 Models for agestructured populations of semelparous organisms predict that the intrinsic rate of increase as a function of x is given by 1Hll06m00l x where 05 is the probability of surviving to age as and mx is the number of female births at age 00 Suppose that and 8N Eaaa xaa ufaa agaaa REE 32sigma g aaama isiaa gian Asi a aim 33mga iiaga mams s gima wi m r x s 5 5 235a ms aag efegm a s wiaa 233iiz az mw ggE55533 E5533wasEkam a s afamni sma lt E ags xsm 5 233 g 3 4 a a E 4 3 5 32 g g a 32 n na u kzww 15 2 E A 53235 q a g 22


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