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# Math Modeling Precalculus I MATH 130

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This 24 page Class Notes was uploaded by Desmond Cormier on Monday October 12, 2015. The Class Notes belongs to MATH 130 at Hollins University taught by Julie Clark in Fall. Since its upload, it has received 33 views. For similar materials see /class/222135/math-130-hollins-university in Mathematics (M) at Hollins University.

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Date Created: 10/12/15

Questions 7 Consider the following difference equation C 13C Co17 a Find the corresponding functional equation b Find C12 3 In a continuous model for geometric growth the variable h starts out at 1095 at time 0 and decays by a factor of 078 over the next 3 days What is the fuctional equation that gives ht Where t is measured in days h 1095 x 78 gt00 4 A geometric growth model is used to describe the amount of pollution in a lake The difference equation for this model is PM 85P P0 421 Where t time in years a Find the corresponding functional equation P 421 x 85 gt30 b What is the level of pollution 3 months after the start of the d 1 m0 6 P3 421x85 404 4 A geometric growth model is used to describe the amount of pollution in a lake The difference equation for this model is PM 85P P0 421 Where t time in years c What is the level of pollution 18 months after the start of the model P15 421X851395 33 600 Modeling Radioactive decay m Pour the MampM39s Carbon14 atoms into your pan Count them and make sure you have 200 Take any extra out Record 200 as the first value in your table Mix them well Take out all the ones that have the quotMquot showing These have decayed Count them R1 Record C1 200 R1 Replace the removed MampM39s with an equal number of Reeses Pieces nitrogen atoms Mix well Pour the mixture back into the pan Remove the candy with the quotMquot showing Count them R2 and record C2 C1 R2 Replace wtih R2 Reeses Pieces Repeat until you have no quotM39s showing s m M a 314 Atoms 114 Atoms C14 Atoms C44 Atoms Homework Questions 4 Find the di erence equation for this sequence of numbers 4 6 9 13 I8 24 differences d1 2 d2 3 d3 4 d4 5 second differences e1 1 e2 1 e3 1 constant e 1 An1 An2n A0 4 2 In a quadratic growth model the initial term is Bo 10 and the difference equation is BM Bn 2 5n b What are the parameters 1 and e in the difference equation above 2 a Use the difference equation to find B 132 133 and B4 El e 5 10 25gtlt0 102 1022 5gtlt1 99 99 25gtlt2 91 91 25gtlt3 78 78 25gtlt4 6 n 0 1 2 3 4 5 3 In a quadratic growth model the initial term is yo 100 and the difference equation is ym 1 yn 5 n b What are the parameters 1 and e in the difference equation above d e 1 IE a Use the difference equation to find y 1 yz y3 and y n n El 0 100 1 1005095 2 955191 3 915288 4 885386 5 865487 Reminder of where we were on Monday El Quadratic Difference Equations The Dark Knight http lquotltv39 hnxnf mmnin r 3 d chartc Second IAm1 2A 599143n A1 15841 Notice that the graph is NOT LINEAR And the gross starts to 39 Gross in millions 1 20000 15000 10000 5000 000 7 again in week 6 Week Since Release o Gross quot mam 39 39 pr 2 I r39 In C I Second Mpdel Gross D39 lance I Terence radiation E 7 a 1T l The Dark Knight Grossin millions Week Since Release The general form for a quadratic difference equation AM Axden where d 4 240 and e constant second di erence Of course the tedious part of these quadratic difference E equations is that we 39re back to having to go one step at a time We have to nd A 1 before we can nd A 2 before we can nd A 3 before we can nd A4 etc Clearly we need a W But before we can nd one we need to practice a little notation E E Find the sum of the rst ve integers 12345 15 Now nd 5x512 15 Find the sum of the rst nine integers I23456789 45 Nowind 9X 9I2 7 45 QQEE Now take a guess about the value of the sum of the rst 125 integers 125gtlt1262 125x62 7875 And now I23 949950 950gtlt9512 125x62 451 725 And we can generalize The sum of the rst n integers must be 123 n nxn12 Why does this formula work Gauss discovered it when he was a 5year old student I23I00 SI23I00 SSI23100 zzgzgg QEQEE SS1 2 3100 SS101 101 101 101 2S100gtlt101 S 100X1012 Notation El 2 the sum of E 7 Sum up add up all the numbers integers k E 2k starting at 3 and ending with 7 23 In other words Add 34567 25 ilk 25 k3 6 2 5k k1 This time we want to add up the values 5 times k starting at kI and ending with k 6 There are two ways we can do this E 1 5123456 5D6D725D21105 E 2 5x1 5x2 5gtlt3 5gtlt4 5x5 5X6 E 51015202530 105 You Will practice this tonight as you finish the worksheet Pleuse be cureful note thutpurts e amp f ure different you should not get the sume unswerz im 3 in 3 i2 i2 The sume would be true prurts g amp h 12 12 Zak 20 22k 20 k1 k1 E E Why am 1 making you learn all this sigma notation E It39s very handy when we need to write the sum of a large set of El numbers or when we don39t yet want to specify a stopping point n This would mean start at I add up the integers and E k stop when you get to the nth integer I 39ll tell you later k1 what n is Thatis123n E n This would mean add up numbers of the form 5k4 2 5k 4 starting with k1 and stop when you get to kn k1 That is 545x245x34 5Xn4 E 9141924 5n4 Crucial point we will want for the next class The sum of the first n integers can be written using sigma notation as ZkZM k1 2

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