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(a) At what interest rate must the GIC be earning if your investment will double in value after 30 years? State your answeras a percent to 2 decimal places.A(t) =A0ert)A= 12000ertNow,A(30) = 24000)12000e30r= 24000)e30r= 2)30r= ln 2)r=130ln 2:= 0:0231 = 2:31%(b) To 2 decimal places, determine the value of your investment on November 1st, 2016 and the rate at which it is changing.Include units in your answer.A(t) = 12000e0:0231t)A0(t) = 12000(0:0231e0:0231t) = 277:20e0:0231t)On November 1st, 2016 (15years after initial inversment), its value will beA(15) = 12000e0:0231(15):= $16969:31and it will be changing at a rate ofA0(15) = 277:20e0:0231(15):= $391:99per year.2.[4marks]In baseball, a batter stands 2 feet from home plate as a pitcher throws the ball towards home plate with a velocityof 150 ft/s (or 164 km/h).Suppose:i)a(t)represents the batter±s "angle of gaze" (in radians) between the ball and home plate at timetandii)b(t)represents the distance (in ft) between the ball and home plate at timet(see diagram below).Determine the rate at which the batter±s "angle of gaze" must change (i.e.a0(t)) inorder to "keep his eyes on the ball" as it crosses home plate.tan(a(t)) =b(t)2)sec2(a(t))°a0(t) =b0(t)2)a0(t) =b0(t)2 sec2(a(t))As the ball crosses home plate,b0(t) = 150ft/s anda(t) = 0,soa0(t) =1502 sec2(0)= 75rads/sec.ORa(t) = tan°1°b(t)2±)a0(t) =11 +²b(t)2³2°°12b0(t)±=2b0(t)4 + (b(t))2As the ball crosses home plate,b0(t) = 150ft/s andb(t) = 0, soa0(t) =2(150)4= 75rads/sec.
3.[4marks]During the winter months, humans do not actually feel the temperature of the air directly.What we feel is thetemperature of our skin which is determined not only by the temperature of the air (inoC) but also the speedx(inkm=h)at which the air is moving.This e/ect is called wind chillT.When the air temperature is±10oCand the air speedx²5km=h, Environment Canada has found that the rate of changeof wind chill isT0(x) =0:16T(x)±1:10480x:(a) Given thatT(15) =±16:75oC(i.e. the wind chill is±16:75oCwhen the air speed is15km=hand the air temperatureis±10oC);°nd the linear approximationL(x)ofT(x)ata= 15km=h:[Note: Coe¢ cients can be rounded to 3 decimals.]L(x) =T(15) +T0(15)(x±15)=T(15) +0:16T(15)±1:1048015(x±15)=±16:75 +0:16(±16:75)±1:1048015(x±15)=±0:252x±12:965(b) Use the result in part (a) to approximate wind chill (to 2 decimals) when the air speed isx= 20km=h:Compare theresult to the actual air temperature of±10oC.T(20)³L(20) =±0:252(20)±12:965 =±18:00oCWhen the air speed is20km=hand the air temperature is "only"±10oC;the wind chill is approximately±18oC(i.e. itfeels8oCcolder).4.[7marks]Open the °le namedlab5mapletemplate.mw(found in MyLearningSpace) in Maple.Use the template tocomplete the following exercise.Consider the functionf(x) = 49 ln°28±3xx±.(a) De°nefin the Maple template.(b) Determine each of the following approximations offabouta= 7.[Note: In Maple,f(a); f0(a), andf00(a)are evaluated by enteringf(a),D(f)(a), andD(D(f))(a), respectively.]i. Linear Approximation.:L(x) =±28x+ 196ii.Quadratic Approximation:Q(x) =±4x2+ 28x(c) Plot each of the approximations in part (b) along withf(x)forx2[5;9]:plot([f(x),L(x),Q(x)],x=5..9,colour=[black,red,green])(d) Use each approximation in part (b) to estimate the value off(6