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# Calculus MAT 021B

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This 82 page Class Notes was uploaded by Otilia Murray I on Tuesday September 8, 2015. The Class Notes belongs to MAT 021B at University of California - Davis taught by Qinglan Xia in Fall. Since its upload, it has received 11 views. For similar materials see /class/187344/mat-021b-university-of-california-davis in Mathematics (M) at University of California - Davis.

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Date Created: 09/08/15

72 Exponential Growth and Decay The Law of Exponential Change In modeling many realworld situations a quantity y increases or decreases at a rate proportional to its size at a given time Examples of such quantities a the size of a population a funds earning interest in a bank account a the amount of a decaying radioactive material The Law of Exponential Change In modeling many realworld situations a quantity y increases or decreases at a rate proportional to its size at a given time Examples of such quantities a the size of a population a funds earning interest in a bank account a the amount of a decaying radioactive material Given the amount 340 at time t 250 Want the amount of the quantity at time t Mathematically we need to solve dy k 0 d y M yo where k is the rate constant Mathematically we need to solve Z i ky M 90 where k is the rate constant Now let us solve this initial value problem ldy E x ytgtdt kdt gt lny ktC ktC O kt i lyl 8 8 gt y i806 C is an arbitrary constant gt y A6 A is an arbitrary constant Substitute 310 340 we have A 340 Therefore 24 yoekt 39I It Lil ol39 Expununlial 39hangi v y quot Growlh k gt 0 Decay k lt 0 The number k is he ran nnslant of Ihe equation Remark If the derivative is proportional to itself then it must be yoekt This model have MANY APPLICATIONS in realworld Unlimited Population Growth Under ideal conditions unlimited space adequate food supply immunity to disease etc the rate of increase of a population P at time t is proportional to the size of the population at time 75 That is P t kPlttgt for some constant k gt 0 called the growth constant Thus It grows exponentially fast Example The world s population at 1980 is 45 billion and at 1990 is 53 billion 0 Determine the growth constant and the approximate population at any time t Example The world s population at 1980 is 45 billion and at 1990 is 53 billion 0 Determine the growth constant and the approximate population at any time 75 133 POekt 130 45 H10 53 53 45610k implies that 1 53 k 1 x 00164 164 10 1 45 Therefore 4580016415 0 How long will it take for the world s population to reach 9 billion 9 45800164 ie the year 1980 42 2022 a In the year 1998 the world population is estimated to be 59 billion What does the model tells us 9 45800164 ie the year 1980 42 2022 a In the year 1998 the world population is estimated to be 59 billion What does the model tells us P18 4580016418 45802952 m 6045 Difference is 6045 59 0145 Radioactive decay Radioactive substances decay at a rate proportional to the amount of the sub stance present Let Alttgt be the amount of the radioactive substance at time t Then A t kAt for some decay constant k gt 0 Thus At Anne kt Radioactive decay Radioactive substances decay at a rate proportional to the amount of the sub stance present Let Alttgt be the amount of the radioactive substance at time t Then A t kAt for some decay constant k gt 0 Thus At Anne kt Halflife is the time required for half of the radioactive substance to decay Alttgt Alt0gt 1 kt i8 12 kt 1n2andtn Thus the Halflife is 1 Example A year ago there were 5 grams of a radioactive substance Now there are 4 grams How much will remain 3 years from now What is its halflife Example A year ago there were 5 grams of a radioactive substance Now there are 4 grams How much will remain 3 years from now What is its halflife Given R 1 5 190 4 Want R6 5 48k 1nE 4 5 5 44 256 4 31n14 R6 8 lt4 53 125 The halflife is 2 n 5 31063 year 1 Continuously Compounded Interest If you invest an amount A0 of money at a xed annual interest rate r and if interest is added to your account it times a year the formula for the amount of money you have at the end of 75 years is A A01kt The interest might be added compounded bankers say a monthly k 12 a weekly k 52 a daily k 365 or even more frequently say by the hour or by the minute By taking the limit as interest is compounded more and more often we arrive at the following formula for the amount after 1 years 11m At 11m A01kt k gtoo k k gtoo 7i 7 11m AUG x a by settmg x U gt0 k 1 A011m 1 55m U gt0 A08 lim At lim 1401 mt k gtoo k k gtoo t lim A0l my by setting x U gt0 k 1 A011m 1 5cm U gt0 A08 Interest paid according to the formula At Age is said to be compounded continuously The number 7 is called the continu ous interest rate Example Suppose that 1000 is deposited in Wells Fargo that pay 5 com pounded continuously How much Will be in the account after 3 years and What is the interest earned during this period What happens after 30 years Example Suppose that 1000 is deposited in Wells Fargo that pay 5 com pounded continuously How much Will be in the account after 3 years and What is the interest earned during this period What happens after 30 years Here A0 1000 7 005 So AC 10008005t After 3 years Alt3gt 10008015 w 116183 Interest earned is 16183 After 30 years Alt30gt 1000615 448169 Heat Transfer Newton s Law of cooling When an object eg a cup of hot soup is hotter than the surrounding medium it will cool to the temperature of the surrounding medium Observation Newton s Law of Cooling The rate at which an object s temperature is changing at any given time is roughly proportional to the difference between its temperature and the temperature of the surrounding medium This observation is described by the differential equation dH k H H dt lt S where o H is the temperature of the object at time t o H S is the constant surrounding temperature LetyH HSthen dy dH kH H k The solution is W yea kt Substitute H H S for y we have H HS H0 HSekt This is the equation for Newton s Law of Cooling Example A hardboiled egg at 9800 is put in a sink of 1800 water After 5 minutes the egg s temperature is 3800 Assuming that the water has not warmed appreciably how much long will it take the egg to reach 2000 LetyH HSthen dy dH kH H k The solution is W yea kt Substitute H H S for y we have H HS H0 rage kt This is the equation for Newton s Law of Cooling Example A hardboiled egg at 9800 is put in a sink of 1800 water After 5 minutes the egg s temperature is 3800 Assuming that the water has not warmed appreciably how much long will it take the egg to reach 2000 Here the equation for Newton s Law of Cooling is H l8 98 18V 808 Whent 5 H 38 so 38 18 806 71 Logarithm De ned as an Integral The General Exponential Function a3 1 lnx dt 12 and e to be the inverse of In x How to de ne a and logg for a positive number a Last time we de ned The General Exponential Function a1 1 1 In x gdt 1 and e to be the inverse of In as How to de ne a and log for a positive number 11 Since Last time we de ned we can think of ax elnayc exlna DEFINITXON Geneval Exponential Functions a exlnu When a e the de nition gives 11 61m 6x1 e l ex Derivative of a may exlnay exlnaolt1nagtl ax Ina Thus ax axlna Derivative of a may exlnay 81an E may ax Ina Thus ax axlna By applying the Chain Rule we get a more general form If a gt O and u is a differentiable function of x then a is a differentiable function of x and u u du a a ln dx dx The integral equivalent of this last result is U audua C lna Logarithms with Base a For any positive number a 7 1 loga x is the inverse function of ax logax agt1 2 x yogagtlt alt1 FIGURE 74 The graph 01 2quot and its inverse log x 2 4 a a Since logg and a are inverses of one another we have thm x x gt 0 logamx x all x Since logg and a are inverses of one another we have Note that Thus Therefore athm x x gt 0 logamx x all x y10ga ltgt xay gt lnxylna lnx gt y Ina 1 lnr Ogo lna d d lnx 1 1 dxlt0gagt dxlt Ina xlna YABLE 72 Rules fur base a laganlhms Forany numbersx gt 0 and y Prnduct Rule lagxy log x Ioguy N Qumient Rule 10g log x 710eg 5quot Recipmcalkule log logoy 5 Power Rule log xquot y log Example Find 10 I Example Find d 10g105x2 2x 100 10x 2 in 10 5x2 2x 100 x5 m d 10g105x2 2x 100 Example Find the derivative 0f10g5 Example Find d 10g105x2 2x 100 dx d 2 2 1 5 2 100 o dx Og10lt 5 56 gt 11110 5x2 2100 15 Example Fmd the derlvatlve 0f 10g5 10 x 5 d g5 xx 2 d 10g5x 5 10g5 xx 2 d 1 Emma 5gt 510g5ltx 2gt1 1 1 1 J 1115x5 230 2 Example Evaluate 10g10 556 LE Example Evaluate l x 010 55 dx LI xl Og10xdx 1 lxcm x V ln lO 1 du by setting to ln x V ln lO 2 SVlnlO 2 lnx 32C 3Vl1qult gt u32C Example Evaluate 1 gwnt sec2 tdt 0 Let u tan t then du sec2 tdt So 1n3 31n339 Example Evaluate 2em 3088m dx 1n7r 2 2 0 Example Evaluate 2em 3088m dx m 2 2 0 2 2 Let u e then du e 2xdx So m 2 2 2xex COSlt d 0 7T cosudu l sinu jjf sin 1 Example Evaluate 10 loglolt1055gt 1 dyc LE 10 Example Evaluate 2em 3088m dx m 2 2 0 2 2 Let u e then du e 2xdx So m 2 2 2xex COSlt d 0 7T cosudu l sinu jjf sin 1 Example Evaluate 10 loglolt1055gt 1 dyc LE 10 Let u 10g1010x 1 10g10 x then du 110cm So L ln 1010g10lt1050gtd L x 10 2 u1n10du O 21n10 Example Simplify 4310g4 e In 1 Let u log1010x l log10 x then du 1111 lodx So 1010g10lt1050gtd L x 10 2 uln lOdu 0 21nlO Example Simplify 43log4 6 ln 1 311133 43log4 eln5L 4log4e ele 3 LI Practice Simplify 54log5 3 log 1 Examp a y 379 What is M Example y xx what is yquot Method 1 Take Logarithm on both sides lny harm xlnx Take derivative With respect to x y x In x x Therefore 3 ylt11nxgt x1lnx Example y xx what is yquot Method 1 Take Logarithm on both sides lny harm xlnx Take derivative With respect to x y x lnx x Therefore 3 ylt11nxgt x1lnx Method 2 y 51 1D U exlnx Thus mind y e xlnxexlnx1lnxx1lnx Practice Find y for o y lt2 1gtx2 Irlncy a sin x008 m 71 e Logarithm De ned as an Integral What is 6E 111 x We have seen and used the logarithmic function ln x and exponential func tion 8 many times But what are they How did we describe em Intuitive and informal o e is some constant e 27182821828 0 6quot when n is an integer eg 82 e 8 etc 0 of with 7 being a rational number e g 8 82 a When x is irrational the precise meaning of e is not so clear Then lnx is de ned as the inverse of 85 Also we claimed without proof that 8 ex and ln x In this chapter we will give a rigorous approach two the de nitions and prop erties of these functions and we study a wide range of applied problems in which they play a role De nition of the Natural Logarithrn Function DEFINITION The Natural Lagarithm function x1 nX7dl xgt0 To understand this function we want to study its 0 domain 0 range 0 graph monotonicity concavity etc o derivative 0 inverse 0 other properties Domain oflnx dt T L r I I X7 Irolt rlt 1IJInnK W 1 give he magma ofmis ma rgt Linen m f M gins Ius Area x GUM 741 The gm h of V In and u do ogar hm rises above the mm as x moves rmm I m Ihe righL and It falls below he ls a5 moves from to the Is The function is not de ned for x g 0 By using rectangles to obtain nite approximations of the area under the P gra h of y f and over the interval between 1 and x we can approximate the values of the function ln 95 TABLE 7 1 Typical 2place F H x values a x lnx 0 undefined 005 0 05 069 I 0 2 069 3 I 10 4 I 39 10 What is e DEFINITION The Number 2 1116 1 What is e DEFINITION The Number 2 11 e 1 That is e is the point on the x axis for which the area under the graph of y f and above the interval 17 6 equals 1 e 2 718281828 The Derivative of y ln 1 By the fundamental theorem of calculus d d ml 1 l dt h 0 dxltnwgt d1t xwenxgt The Derivative of y ln 1 By the fundamental theorem of calculus d d ml 1 l dt h 0 dxltnwgt d1t xwenxgt If u is differentiable and positive then by the Chain Rule 1 du u d lt1nultxgtgt dx The Derivative of y ln 1 By the fundamental theorem of calculus d d ml 1 l dt h 0 dxltnwgt d1t xwenxgt If u is differentiable and positive then by the Chain Rule ilt1nultxgtgt l du u dx eg d l l h Cm lnbx bx b x W en bx gt O In particular if b l and x lt 0 then d l ln x E when x lt O Thus d 1 g in 5 when y O This gives 1 dx 1nx 0 x and IllLE Il UCU ultgtdyc 1C excosxd x esmx Example Find Thus d 1 g in 5 when y O This gives 1 dx 1nx 0 x and IllLE Il UCU ultgtdyc 1C excosxd x esmx ecosx dx 1nex smx C em smx Example Find Example Find sec x dx 1n sec tan x Example Find sec x dx 1n sec tan x Let u 1n secx tan x then see tan sec sec x tan x sec xdx 2x Thus sec x ri r 1n sec x tan x d U 2 0 2x1nsecx tanx C Familiar algebraic properties of In I 1 lnbxlnblnx z Inglnbilnx 3 1171 4 lnx r In x 139 any rational number Familiar algebraic properties of In I 1 lnbxlnblnx z Inglnbilnx 3 1171 4 lnx I39lnx rany mlionalnnmber Sketch of proof Let u sz then d uI b72771 739 111110 7 7 bl 7 y d1 d Md 1 l nu 39 1171 gyiTny lnuy ilnu1 Tiny lnbarilnb Tina 111032 lnbrlnaj The graph and range of In 1 a In x is an increasing function because d 1 1 0 dx n x gt a the graph of In x is concave down because d2 1 o 11mgH00 lnx 00 because 11m ln2n 11m nln2 00 n gtOO n gtOO and In x is increasing we have a limxgm lnx 00 because 1 11m lnx 11m ln 11m lnt oo I gtO t gtoo t t gtoo Thus the range ofln x is 00 co The graph of y 111 56 looks like The graph of yogx The inverse function of in l and the Number 5 The function in x is an increasing function ofz With 0 domain 0 oo 0 range 700 00 So it has an inverse 111 1 x With 0 domain 700 00 0 range 0 oo DEFINITION Yhe Natural Exponentiai runm39an For every real numberx ex Inquot x The graph ofy hf x I e is the graph of lnx re ected across the line y x neunz 73 The graphs ofquot 1n mud y ln39 expxTl1e numbere is 1nquot1 exp In 0nd Equmium for u and in 39 2quotquot v allx gt o In a x allx Example Find d iehmsina rl dx lm Lrw Equmium for u and in 39 allx gt o allx Example Find d iehmsma rl dx d 1nsin x1 d 7 7 1 e dxsm x cosx Example Find d i 1110330273 cos 2x Innr50 Equalium for w and n 39 2quotquot x allx gt o In x x allx Example Find d iehmsm x1 x d I d 76mm 3H1 sinx 1 cos an Example Find 01 2 dx lnex 73 cos 230 d 2 d a lnex 3 COS2 g 7 3 cos 2x 2x 6 sin 295 The Derivative and Integral of 113 maimed that The Derivative and Integral of 6 We claimed that lt gtl 81 Why Since 8 is the inverse of ln x we have lnex x By the Chain Rule 1 d x 3 gt Z 1 Therefore d ex ex Moreover we have edx eC If u is any differentiable function of x then by the Chain rule d u udu 8 d Thus e du e C Example Suppose ag 8 381n22x21 What is f x If u is any differentiable function of x then by the Chain rule d u udu 8 8 dx Thus e du e C Example Suppose ag 8 381n22x21 What is f x 2 2 flltgt 3Sln2221 Sin2lt22 8 381 2x 1 6 sinlt2x2 1 3082x2 1 lt2x2 1 12 sinlt4x2 2e38m2221 Example Solve the initial value problem 2 M sec2 x dx2 with y0 O and y 0 l Example Solve the initial value problem 2 sec2 x x with y0 O and y 0 l d y tan x Cl dx yO 1 implies l 01 Thus d g tanx ly0 0 y lncosxxC 340 Oimplies y lncosx x yev mvzd a an integral mm D M Fun um 17 Lasmme we de ned 2 1 1m 7 rd 1 t and e2 to be the mvase of In e How ya de ne d2 and log for aposmve number m Smce we can Lhmk of He gear gem DEFINITION General Exponential Functions quot gt n and 39 ad en en a e the de muon gwes d2 221m 221M e a Lemme w n e2 e In a a In a Thu w a 1nd By applymg Lhe Cham Rule we get a more gmeml form If a gt 0 and u is a differentiable function of x then a is a differentiable function of x and The integral equivalent of this last result is Logaritth with Ease u For any positive number a y l loga x is the inverse function of am Al logax agt1 2 xx x r2 u 4 ylogax alt1 FIGURE 74 The gmph on quot and its inverse log x 2 4 a 3 Since log and am are inverses of one another we have 010am x x gt 0 logaaz x7 all Note that y loga m ltgt a 0y Lax ylna 111x gt 1 Ina Thus 111 x logafr Ina Therefore i d 11105 1 d z aoga i gna i xlna Example Find Example Find the derivative oflog5 5 Vi TABLE 72 Rules fur base a lugarithms Forany numbers x gt 0 and y gt 0 l PmducIRule log lngux 10g 2 QumientRule x lagav 10ng 7 log 3 RecipmcaIRule 10g i Iogy 4 Power Rule log rquot y logax dz logm5x2 7209 100 Example Example Example Example Evaluate Evaluate Evaluate Evaluate 10g1095dx f I 1 4 fant sec2 tdt 0 3 m 2 2 261 cosex dm 0 95 10 l 10 0g10lt 95 d9 10 Example Simplify 43log4 6 ln 1 Practice Simplify 54log5 3 log3 1 Example 3 05 What is 3 Practice Find 3 for o y 2x 1V2 mlnxi 0 sin 950

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