Mathematics for Queensland, Year 12 Worked Solutions to Exercise 5L

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1 Mathematics for Queeslad, Year 12 Worked Solutios to Exercise 5L Hits Mathematics for Queeslad, Year 12 Mathematics B, A Graphics Calculator Approach Kiddy Bolger, Rex Boggs, Rhoda Faragher, Joh Belward 1. There is t a algebraic solutio to this problem. It eeds to be tackled usig either graphs or a table. Your choice! 2. Fid a expressio for the value of the Mackay property, i terms of t. Also fid a expressio for the value of the Noosa property, i terms of t. You are iterested i whe these two fuctio iterest. 3. The aswer is ot 3 cm! With some thought, it should be clear that each cetimeter of charcoal removes 33% of the remaiig quatity of chemicals, ot 33% of the origial quatity of chemicals. 4. If there are N germs iitially, ad ½ are killed by disifectat, ad the the umber of germs remaiig icreases by 1/3, what is the expressio for the umber of germs at the start of day 1? 5. If you double the umber of speakers, you double the itesity of the soud. Remember - a soud with a itesity of 1 db has times the itesity of a soud with a itesity of db. 6. Let the curret oil reserves equal uits. If the aalysts say that the oil will last for years at the curret rate of cosumptio, the they are assumig that we use 1 uit each year. But if cosumptio is icreasig at 4% per year, the i the 1 st year we use 1 uit of oil. But how much do we use i the ext year? Ad the ext? Use a spreadsheet to track the umber of uits of oil each year ad the total amout of oil used, ad to determie whe all uits have bee used. 7. Start by fidig a mathematical model for the amout A of space juk beig added each year t from 1991 to 2. Takig 199 to be year, the weight of space juk whe t = is 1.8 millio kg. Hece the amout of space juk added each year is assumed to be a expoetial fuctio that passes through (1,.8) ad (, 1.2). Oce you have foud this fuctio, use a spreadsheet to track the weight of space juk beig added each year, ad the total weight of space juk each year. 8. Here solve for meas to fid a algebraic expressio for i terms of a, b ad m. Use logs to get oto the groud. Brig all terms ivolvig to the same side of the equatio.

2 Mathematics for Queeslad, Year 12 Mathematics B, A Graphics Calculator Approach Kiddy Bolger, Rex Boggs, Rhoda Faragher, Joh Belward Mathematics for Queeslad, Year 12 Worked Solutios to Exercise 5L 1. As there is t a aalytical (i.e. algebraic) solutio to this equatio, we will solve it by sketchig graphs ad fidig the poits of itersectio. Settig Y1 = e x ad Y2 = x 2 ad selectig (Zoom, Decimal) o a TI-83 brigs up the screeshot alogside. It shows 2 poits of itersectio. Usig (Calc, Itersect) shows us that 2 solutios are: x.77, y.58 x= 2, y = 4 By chagig the widow, we fid a 3 rd solutio: x = 4, y = 16 Have we foud all of the solutios? We have, because every expoetial fuctio evetually grows faster tha ay power fuctio. This meas that e x > x 2 for x > 4 ad hece there are o more poits of itersectio. 2. Here is a algebraic solutio. A graphical solutio is equally valid. Mackay: A= 4(1.4) Gold Coast: A= 25(1.6) 4(1.4) = 25(1.6) {Equate the 2 RHS expressios} 4(1.4) = 25(1.6) {Divide by } 1.6(1.4) = (1.6) {Divide by 25, ad covert to a decimal} log[1.6(1.4) ] = log[(1.6) ] {Take log of both sides} log1.6+ log1.4= log1.6 {Apply log laws} log1.6= log1.6 log1.4 {Brig terms with together} (log1.6 log1.4) = log1.6 {Factor out the commo factor of } log1.6 = log1.6 log1.4 {Solve for } 24.7years {Evaluate} 3. The aswer is ot 3 cm! With some thought, it should be clear that each cetimeter of charcoal removes 33% of the remaiig quatity of chemicals, ot 33% of the origial quatity of chemicals. Let the origial mass of chemicals = M. After 1 cm: M = M x.67 {If 33% is removed, 67% remais} After 2 cm: M = M x.67 x.67 {Now 67% of 67% of M remais} After 3 cm: M = M x.67 x.67 x.67 {Etc} After cm: M = M x.67 {The patter should be clear}

3 Mathematics for Queeslad, Year 12 Mathematics B, A Graphics Calculator Approach Kiddy Bolger, Rex Boggs, Rhoda Faragher, Joh Belward Whe 99% is removed:.1m = M.67 {Whe 99% is removed, 1% remais}.1=.67 {Divide both sides by M } log.1= log.67 {Take the log of both sides} log.1= log.67 {Apply log law 3} log.1 = log.67 {Solve for } = 5.75cm {Evaluate} About 6 cm of charcoal are eeded to remove 99% of the chemicals. 4. Let N be the umber of germs i the room at time t =. 1 After the disifectat is applied, the umber of germs N = 2 N. At the start of day t = 1, the umber of germs has grow to N N = o + No = No + No = No Each day the umber of germs is 2 3 of the umber of germs o the previous day. Hece o day t, t 2 N = N o 3 Whe N =.1 N, 1 {Whe oly of the germs remai}.1 2 = 3 N N o t t {Substitute} 2.1= 3 {Divide by N } log.1 t = 2 log( 3) {Solve for t} t 37.8 {Evaluate} The umber of germs reduces to less tha.1 of the origial umber after 38 days. 5. Give: I B= log I 1 For B =, I = log I {Substitute} I = log I {Divide by } I = I {Write as a idex expressio} I = I {Solve for I}

4 Mathematics for Queeslad, Year 12 Mathematics B, A Graphics Calculator Approach Kiddy Bolger, Rex Boggs, Rhoda Faragher, Joh Belward For 2 speakers I = 2 I {2 speakers = 2 x the itesity} Hece B 2 = log I ( ) I {Substitute ito 1} = log 2 {Divide out I } 3 {Evaluate} The itesity of 2 speakers is oly 3 db. 6. a. Let the curret reserves of oil equal uits. If the aalysts say that the oil will last for years at the curret rate of cosumptio, the they are assumig that we use 1 uit each year. But if cosumptio is icreasig at 4% per year, we have this model istead: Year Oil Used Total Oil Used = = = etc We eed to determie how may years are eeded for the total oil used to exceed. I the ext chapter you will lear how to do this algebraically. For ow, we will use a spreadsheet. The formula i C7 is: C7 = (1.4)^(B6) The formula i D7 is: D7 = D6 + C7 These formulas are copied dow the colums util the Total Used >. Perhaps surprisigly, the year supply of oil will oly last 42 years if cosumptio icreases by 4% per year. B C D 4 Year Oil Used Total Used

5 Mathematics for Queeslad, Year 12 Mathematics B, A Graphics Calculator Approach Kiddy Bolger, Rex Boggs, Rhoda Faragher, Joh Belward b. The mathematics is the idetical, albeit with a 6% icrease rather tha a 4% icrease. B C D 4 Year Oil Used Total Used Note: This spreadsheet is available from the mathematics-for-queeslad.com website. It oly takes 71 years to use up the year supply of oil if oil cosumptio icreases by 6% per year! 7. We first eed to fid a mathematical model for the amout A of space juk beig added each year t from 1991 to 2. Takig 199 to be year, the weight of space juk whe t = is 1.8 millio kg. The amout of space juk added each year is assumed to be a expoetial fuctio that passes through (1,.8) ad (, 1.2). Usig ExpReg o a TI-83, we fid that the equatio is give by: A =.7648 x 1.46 t. Oe way model this problem is with a spreadsheet, as a spreadsheet is a effective way to calculate the weight of space juk added ad the total amout of space juk each year. The formula i B6 is: B6 =.7648 x 1.46^A6 The formula i C6 is: C6 = C5+B6 These formulas are copied dow the colums util the Year = 15 (i.e. 25). A B C 4 Year Additioal Total Note: This spreadsheet is available from the mathematics-for-queeslad.com website.

6 Mathematics for Queeslad, Year 12 Mathematics B, A Graphics Calculator Approach Kiddy Bolger, Rex Boggs, Rhoda Faragher, Joh Belward a. A mathematical model ca be a equatio, a graph or a table. I this problem, the table give above is our model. b. The model shows that there will be 18.6 millio kg of space juk i 25. c. The model shows that there was over 5 millio kg of space juk i Here solve for meas to fid a algebraic expressio for i terms of a, b ad m. m a = b + loga m = logb {Take log of both sides} log a= ( + m)logb {Apply log law 3} loga = logb+ mlogb {Expad} loga logb= mlogb {Gather terms with together} (loga log b) = mlogb {Factorise} mlogb = {Solve for } loga logb