"6#*&%#cmG\"\"\"%\"sG !\"\"" }{TEXT -1 14 " evaluates to " }{XPPEDIT 18 0 "m/100/s;" "6#*(% \"mG\"\"\"\"$+\"!\"\"%\"sGF'" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "convert(50*km/h, units, 'cm/s');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#*&%#cmG\"\"\"%\"sG!\"\"#\"&+D\"\"\"* " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#*&%#cmG\"\"\"%\"sG!\"\"$\"+*)))))) Q\"!\"'" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 79 "How many seconds doe s it take an object, released from rest, to fall 20 meters?" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "From the equation " }{XPPEDIT 18 0 "d = g [n]*t^2/2;" "6#/%\"dG*(&%\"gG6#%\"nG\"\"\"*$%\"tG\"\"#F*F-!\"\"" } {TEXT -1 12 ", we derive:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "elapsed_time := sqrt( 2 * dist / accel );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-elapsed_timeG*&-%%sqrtG6#\"\"#\"\"\"-F'6#*&%%distGF* %&accelG!\"\"F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "eval(ela psed_time, [dist=20*m, accel=gn]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# ,$*,-%%sqrtG6#\"\"#\"\"\"-F&6#\"#?F)-F&6#\"&++#F)-F&6#\"'Lh>F)-%%UnitG 6#7#%\"sGF)#F)F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%) ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"sG$\"+w*>'>?!\"* " }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 39 "What is the rest energy of \+ an electron?" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Assume the mass of an electron is " }{XPPEDIT 18 0 "9.11 * 10^(-31)" "6#*&-%&FloatG6$\"$ 6*!\"#\"\"\")\"#5,$\"#J!\"\"F)" }{TEXT -1 11 " kilograms." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "mass := 9.11e-31 * kg;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%massG,$-%%UnitG6#7#%#kgG$\"$6*!#L" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "Approximate the speed of light by \+ " }{XPPEDIT 18 0 "Float(300,-2)*10^8;" "6#*&-%&FloatG6$\"$+$,$\"\"#!\" \"\"\"\"*$\"#5\"\")F+" }{TEXT -1 19 " meters per second." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "c := 3.00e8 * m/s;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"cG,$-%%UnitG6#7#*&%\"mG\"\"\"%\"sG!\"\"$\"$+$ \"\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 18 "Using the formula " } {XPPEDIT 18 0 "E = m*c^2;" "6#/%\"EG*&%\"mG\"\"\"*$%\"cG\"\"#F'" } {TEXT -1 54 ", you can approximate the rest energy of the electron." } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "E := mass*c^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"EG,$-%%UnitG6#7#%\"JG$\")++*>)!#@" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 122 "A more precise answer can be foun d by converting the known unit the electron mass (em) to joules using \+ energy conversions." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "conv ert(em, units, J, energy);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%Uni tG6#7#%\"JG$\"+o66(=)!#B" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 77 "App roximately what volume does 1,000,000,000 US dollars worth of gold occ upy?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "cost := 268.00*USD/o z[troy]; # April 12th, 2001" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%cos tG,$-%%UnitG6#7#*&%$USDG\"\"\"%#kgG!\"\"$\"+!3+kh)!\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "density := 19.3 * g/cm^3;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(densityG,$-%%UnitG6#7#*&%#kgG\"\"\"%\"mG! \"$$\"++++I>!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "mass := 10^9*USD/cost;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%massG,$-%%UnitG6 #7#%#kgG$\"+$\\x0;\"!\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "volume := mass/density;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'volumeG ,$-%%UnitG6#7#*$)%\"mG\"\"$\"\"\"$\"+')[N8g!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "What is the length of one side of a cube with this v olume?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "length_side := ro ot[3](volume);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,length_sideG,$-%% UnitG6#7#%\"mG$\"+&yn%==!\"*" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 131 "An Su-27 Flanker can travel at mach 1.1 at sea level. How fast i s this in miles per hour, miles per second, and meters per second?" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "convert(1.1*M, units, mph); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%$mphG$\"+PR+c\")!\" (" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "convert(1.1*M, units, \+ 'mi/s');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#*&%#miG\"\" \"%\"sG!\"\"$\"+\\mblA!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "convert(1.1*M, units, m/s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$- %%UnitG6#7#*&%\"mG\"\"\"%\"sG!\"\"$\"'1YO!\"$" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 53 "Approximately how many meters are there in 3.5 miles ?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "convert(3.5*mi, units, \+ m);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"mG$\"++SqKc!\" '" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "round(%);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"mG\"%Lc" }}}}{SECT 1 {PARA 4 " " 0 "" {TEXT -1 69 "Given 50 US gallons of water, how many 750 mL bott les could you fill?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "50*ga l[US_liquid] / (750*mL);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6##\"*\"\\Dx :\"'+]i" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "floor(%);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#\"$_#" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 115 "Given nylon with a linear mass density of 20 deniers, wh at length of thread is used in an object weighing 12 grams?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "12*g / (20*deniers);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"mG\"%+a" }}}}{SECT 1 {PARA 4 " " 0 "" {TEXT -1 55 "What is the volume in cubic inches of a 2 liter en gine." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "convert(2*L, units, inches^3);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#*$)%#inG \"\"$\"\"\"#\"*+++]#\"($Q[?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#*$)%#inG \"\"$\"\"\"$\"+#)[Z?7!\"(" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 70 "Fo r a given phenomena with a frequency of 1.420,405,761 GHz, find the:" }}{PARA 0 "" 0 "" {TEXT -1 13 " 1. period," }}{PARA 0 "" 0 "" {TEXT -1 36 " 2. number of cycles per year, and" }}{PARA 0 "" 0 "" {TEXT -1 56 " 3. number of cycles since the beginning of the earth." }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "First you must convert the frequen cy from GHz to Hz (cycles per second)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "frequency := 1.420405761 * GHz;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%*frequencyG,$-%%UnitG6#7#%$GHzG$\"+hdS?9!\"*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "The period is:" }{MPLTEXT 1 0 0 " " }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "1/freque ncy;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"sG$\"+#zT-/(! #>" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "convert(%, units, nan oseconds);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%#nsG$\"+# zT-/(!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "The number of cycles \+ per year is:" }{MPLTEXT 1 0 0 "" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 34 "convert(frequency, units, '1/yr');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#*&\"\"\"F)%#yrG!\"\"$\"+9Rm\" \\%\"\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 57 "The number of cycles \+ since the beginning of the earth is:" }{MPLTEXT 1 0 0 "" }{TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "frequency * 10000000000 *yr;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+9Rm\"\\%\"#<" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 67 "Given inductance and capacitance, find th e resistance in microohms." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "The \+ following formula relates the resistance to the inductance and capaci tance." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "resistance := sqr t(inductance / capacitance);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+res istanceG*$-%%sqrtG6#*&%+inductanceG\"\"\"%,capacitanceG!\"\"F+" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 75 "Use an inductance of 124 nanohenri es and a capacitance of 3.52 microfarads." }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 66 "eval(resistance, [inductance = 124.*nH, capacitance = 3.52 * uF]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&-%%sqrtG6#\"%+5 \"\"\"-%%UnitG6#7#%&OmegaGF)$\"+anDNf!#7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%Unit G6#7#%&OmegaG$\"+%)H*o(=!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "convert(%, units, uOmega);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-% %UnitG6#7#%'uOmegaG$\"+%)H*o(=!\"%" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 62 "Given a molar energy, find the mass energy in Btu's per pound. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "molar_energy := 523.432* Btu/mol(carbon);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-molar_energyG,$ -%%UnitG6#7#*&%\"JG\"\"\"-%$molG6#%'carbonG!\"\"$\"+wm!)=b!\"%" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "carbon_mass := 12*kg/mol(car bon);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,carbon_massG,$-%%UnitG6#7# *&%#kgG\"\"\"-%$molG6#%'carbonG!\"\"\"#7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "mass_energy := molar_energy / carbon_mass;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%,mass_energyG,$-%%UnitG6#7#*&%\"mG\"\"#%\" sG!\"#$\"+jb+*f%!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "con vert(%, units, 'Btu/lb');" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%Unit G6#7#*&%$BtuG\"\"\"%#lbG!\"\"$\"+y'R&y>!\")" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 212 "Given a distance function, find the speed by differen tiation, speed at 2.5 seconds by evaluation, and distance traveled bet ween 1 and 2.5 seconds by integration and by subtraction of the distan ce function values." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "Note: t is \+ a unit of mass, the tonne. Therefore, t1 is used to represent the time variable." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "distance := 5 /(1+4*exp(-3*t1))*m;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)distanceG,$ *&,&\"\"\"F(*&\"\"%F(-%$expG6#,$%#t1G!\"$F(F(!\"\"-%%UnitG6#7#%\"mGF( \"\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "To find the speed functi on, differentiate the distance function." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "speed := diff(distance, t1*s);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&speedG,$*(,&\"\"\"F(*&\"\"%F(-%$expG6#,$%#t1G!\"$F(F (!\"#F+F(-%%UnitG6#7#*&%\"mGF(%\"sG!\"\"F(\"#g" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "To find the speed at 2.5 seconds, evaluate the speed function at t1=2.5." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eva l(speed, t1=2.5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#*&% \"mG\"\"\"%\"sG!\"\"$\"+'\\rQI$!#6" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "By using a definite integral, you can determine the distance tr aveled from the speed function." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "int(speed, t1*s = 1..2.5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# ,$-%%UnitG6#7#%\"mG$\"+)zlL>)!#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 81 "The distance traveled can also be calculated directly from the dis tance function." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "eval(dis tance, t1=2.5) - eval(distance, t1=1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&,&$\"+LF'*))\\!\"*\"\"\"*&\"\"&F(,&F(F(*&\"\"%F(-%$expG6#!\"$F (F(!\"\"F2F(-%%UnitG6#7#%\"mGF(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"m G$\"*%eO$>)!\"*" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 85 "Find the min imum and maximum length of 1.2 yards, 1 meter, 3.2 feet, and 0.6 fatho ms." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "min(1.2*yd, m, 3.2*ft , 0.6*fathom);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"mG$ \"+++g`(*!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "max(1.2*yd, m, 3.2*ft, 0.7*fathom);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG 6#7#%\"mG$\"+++;!G\"!\"*" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 98 "Giv en a torque of 3 newton meters, how much energy is required to move a \+ lever through 10 degrees?" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "The e nergy required is the product of the torque and the angle in radians. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "energy := torque * angl e;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%'energyG*&%'torqueG\"\"\"%&ang leGF'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "eval(energy, [torq ue = 3*N*m(radius), angle = 10*deg]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%#PiG\"\"\"-%%UnitG6#7#%\"JGF&#F&\"\"'" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-% %UnitG6#7#%\"JG$\"+ex)fB&!#5" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 227 "The Hyper-X can travel at speeds up to 7200 miles per hour. How \+ long would it take to circle the earth at maximum speed (assuming it c ould carry sufficient fuel)? How far does it travel in a 10 second fl ight at maximum speed?" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "To find \+ the time to circle the earth, divide the distance by the speed." }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 227 "The meter was originally defined \+ as 1/10,000,000 th the distance from the North Pole to the Equator on \+ the meridian passing through Paris. Therefore, 40,000 kilometers is a good approximation of the circumference of the earth." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "40000 * km / (7200 * mph);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"sG#\"*++Dc\"\"&tD\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "convert(%, units, h);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"hG#\"'D1R\"'dJ6" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#,$-%%UnitG6#7#%\"hG$\"+z@1_M!\"*" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 62 "To find the distance traveled, multiply the speed \+ by the time." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "7200*mph * \+ 10*s;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"mG#\"'sY!)\" #D" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%\"mG$\"++!)o=K!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "convert(%, units, km);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%#kmG$\"++!)o=K!\")" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 122 "Given 1032 UK gallons of oil, how many c ylindrical cans with a height of 1.2 feet and diameter of 0.9 feet cou ld you fill?" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "The volume of a cy linder is " }{XPPEDIT 18 0 "h*Pi*(d/2)^2" "6#*(%\"hG\"\"\"%#PiGF%*&%\" dGF%\"\"#!\"\"F)" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "1032*gal[UK] / (1.2*ft) * Pi*((0.9*ft))/2^2;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&%#PiG\"\"\"-%%UnitG6#7#*$)%\"mG\" \"$F&F&$\"+]To'z)!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "eval f(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#*$)%\"mG\"\"$\" \"\"$\"+J)fNw#!\"*" }}}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 160 "Given an power gain from 332 microwatts to 23 milliwatts, what is the gain in \+ decibels? What would the decibel gain be if the increase were a volta ge increase?" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "A gain is a quotie nt of the final value divided by the initial value." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "gain := 23*mW / (332*uW(base));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%gainG,$-%%UnitG6#7#*&%\"WG\"\"\"-%\"WG6#% %baseG!\"\"#\"%]d\"#$)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 62 "To dete rmine the decibel gain, first take the ln of the gain. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "ln(gain);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&-%#lnG6##\"%]d\"#$)\"\"\"-%%UnitG6#7#%#NpGF+#F+\"\" #" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "convert(%, units, dB); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*(-%#lnG6##\"%]d\"#$)\"\"\"-F&6# \"#5!\"\"-%%UnitG6#7#%#dBGF+F." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%UnitG6#7#%#dB G$\"+_(*eS=!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 171 "Power is prop ortional to the square of the voltage. Therefore, the decibel increase corresponding to the voltage gain should be a factor of 2 times that \+ of the power gain." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "volta ge_gain := 23*mV / (332*uV(base));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6# >%-voltage_gainG,$-%%UnitG6#7#*&%\"VG\"\"\"-%\"VG6#%%baseG!\"\"#\"%]d \"#$)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "ln(voltage_gain); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&-%#lnG6##\"%]d\"#$)\"\"\"-%%Unit G6#7#%#NpGF*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "convert(%, \+ units, dB);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*(-%#lnG6##\"%]d\"#$) \"\"\"-F&6#\"#5!\"\"-%%UnitG6#7#%#dBGF+\"#?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$-%%U nitG6#7#%#dBG$\"+/&z6o$!\")" }}}}}{EXCHG {PARA 0 "" 0 "" {HYPERLNK 17 "Return to Index for Example Worksheets." 2 "examples/index" "" } {TEXT -1 0 "" }}}}{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 } or="[51,255,0]" pen-height="24.0" pen-width="24.0" pen-opacity="0.8"/> Taylor polynomials in several variables Worksheet by Mike May, S.J. - maymk@slu.edu Revised by Russell Blyth - blythrd@slu.edu restart; with(plots): with(Student[MultivariateCalculus]):
<Text-field style="Heading 1" layout="Heading 1">Review of Taylor polynomials in one variable</Text-field> Recall from calculus of one variable that we can approximate a nice function at a point x=a by using Taylor polynomials. We get the nth degree part of the approximation near x=a by evaluating the nth derivative at that point and multiplying by NiMqJiksJiUieEciIiIlImFHISIiJSJuR0YnLSUqZmFjdG9yaWFsRzYjRipGKQ==. We can compute the Taylor polynomial using the TaylorApproximation command from the Student[MultivariableCalculus] package . It is instructive to note that the difference between the nth degree Taylor polynomial and the (n-1)st degree Taylor polynomial is the term we just described. TaylorDegree := 10; NthTermOfTaylor := TaylorApproximation(f(x),[x]=[a],TaylorDegree)- TaylorApproximation(f(x),[x]=[a],TaylorDegree-1); Consider an example: f := x -> sin(x) + cos(2*x); a:= Pi/3; Now we compute the Taylor polynomials up to degree four for our function f at the given point a. Deg0 := TaylorApproximation(f(x),[x]=[a],0); Deg1 := TaylorApproximation(f(x),[x]=[a],1); Deg2 := TaylorApproximation(f(x),[x]=[a],2); Deg3 := TaylorApproximation(f(x),[x]=[a],3); Deg4 := TaylorApproximation(f(x),[x]=[a],4); Visually we notic{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "Hyperlink" -1 17 "" 0 1 0 128 128 1 2 0 1 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "Times" 1 18 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }{CSTYLE " " -1 257 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }{CSTYLE "" -1 258 " Times" 0 12 0 0 0 1 1 0 2 2 2 2 1 1 1 1 }{CSTYLE "" -1 259 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }{CSTYLE "" -1 260 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }{CSTYLE "" -1 261 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }{CSTYLE "" -1 262 "Times" 1 12 0 0 0 1 0 1 2 2 2 2 1 1 1 1 }{CSTYLE "" -1 263 "Times" 1 12 0 0 0 1 0 1 2 2 2 2 1 1 1 1 } {CSTYLE "" -1 264 "Times" 1 12 0 0 0 1 0 1 2 2 2 2 1 1 1 1 }{CSTYLE " " -1 265 "Times" 1 12 0 0 0 1 1 2 2 2 2 2 1 1 1 1 }{CSTYLE "" -1 266 " " 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times " 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 } {PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 3" -1 5 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple P lot" -1 13 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 }{PSTYLE "Maple Output" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Map le Output" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT 257 33 "A Demonstration of the P rocedure " }{TEXT 256 8 "minimize" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "The procedure " }{TEXT 266 8 "minimize" }{TEXT -1 201 " returns the infimum of an expression or function over a given domain. It is also possible to have minimize return both the infimum and the location where the infimum is either approached or achieved. " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "resta rt;" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 25 "The Infimum of a Function " }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "A function " }{XPPEDIT 18 0 "f " "6#%\"fG" }{TEXT -1 84 " on a given domain D is said to be bounded b elow on D if there exists a real number " }{XPPEDIT 18 0 "beta" "6#%%b etaG" }{TEXT -1 11 " such that " }{XPPEDIT 18 0 "beta <= f(x)" "6#1%%b etaG-%\"fG6#%\"xG" }{TEXT -1 10 " for each " }{XPPEDIT 18 0 "x" "6#%\" xG" }{TEXT -1 18 " in D. If such a " }{XPPEDIT 18 0 "beta" "6#%%betaG " }{TEXT -1 43 " exists then it is called a lower bound of " } {XPPEDIT 18 0 "f" "6#%\"fG" }{TEXT -1 4 " on " }{XPPEDIT 18 0 "D" "6#% \"DG" }{TEXT -1 106 ". For example, the function cosh(x) is bounded b elow by 0, (or any other number less than or equal to 1.)" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "plot( cosh(x), x =-3..3, y=-1..5 );" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7en7$$!\"$\"\"!$\"3`wxd*>mn+\"!#;7$$!3%**** *\\P&3Y$H!#<$\"3yM\"fK4WOV*F17$$!3*)*****\\2<#pGF1$\"3we%3#Q#z*R))F17$ $!3')**\\78.K7GF1$\"3?p%3'=VGa$)F17$$!3#)***\\7bBav#F1$\"3))3FoJ:k&*yF 17$$!3'***\\(=>P9p#F1$\"3t'pM#>MK5uF17$$!36++]K3XFEF1$\"3Dq?V2aNbpF17$ $!3v******H./jDF1$\"3I\\U%4>Qh_'F17$$!3%)****\\F)H')\\#F1$\"3q1yc4a+Ch F17$$!3\"****\\i3@/P#F1$\"3%fOzUcpwR&F17$$!3:++Dr^b^AF1$\"3\"*py'Q!*oQ ![F17$$!3#****\\7Sw%G@F1$\"36_e`#4P0E%F17$$!3*****\\7;)=,?F1$\"3GV0pUw ]mPF17$$!3.++DO\"3V(=F1$\"3NI!*GL^)[L$F17$$!3#******\\V'zV Z$p%HF17$$!3)*****\\d;%)G;F1$\"3gn\\t#Qhqk#F17$$!3!******\\!)H%*\\\"F1 $\"3wF!*3$*e>^BF17$$!3.+++vl[p8F1$\"3#\\IyA3%y$4#F17$$!3!******\\>iUC \"F1$\"3t!HKKTj#z=F17$$!3-++DhkaI6F1$\"3Lb^6*p-,r\"F17$$!3s******\\XF` **!#=$\"3k!p$>zigP:F17$$!3u*******>#z2))F[r$\"3]*p>^W>OT\"F17$$!3R++]7 RKvuF[r$\"3a()*z$o!eEH\"F17$$!3s,+++P'eH'F[r$\"3R6$fJ7B[?\"F17$$!3q)** *\\7*3=+&F[r$\"3dipBX-sF6F17$$!3[)***\\PFcpPF[r$\"3\"\\1$e9L*=2\"F17$$ !3:)****\\7VQ[#F[r$\"3[JQO.j+J5F17$$!32)***\\i6:.8F[r$\"31IwyPI]35F17$ $!3Wb+++v`hH!#?$\"3F(Qb`Q/++\"F17$$\"3]****\\(QIKH\"F[r$\"3i&QS^)QP35F 17$$\"38****\\7:xWCF[r$\"3]`r@$oL+.\"F17$$\"3E,++vuY)o$F[r$\"3%=#\\Nx' )zo5F17$$\"3*y******4FL(\\F[r$\"3OVYZi+CE6F17$$\"3A)****\\d6.B'F[r$\"3 4Pe..QW+7F17$$\"3s****\\(o3lW(F[r$\"3+j_81JI!H\"F17$$\"35*****\\A))oz) F[r$\"3sj68/3`79F17$$\"3e******Hk-,5F1$\"3w67l4xGW:F17$$\"35+++D-eI6F1 $\"3&RL9(Q&\\,r\"F17$$\"3t***\\(=_(zC\"F1$\"3!\\'3D5U=&)=F17$$\"3M+++b *=jP\"F1$\"3@CJAK:S1@F17$$\"3g***\\(3/3(\\\"F1$\"34[yt1I?YBF17$$\"33++ vB4JB;F1$\"3Oj%fx\\YNj#F17$$\"3u*****\\KCnu\"F1$\"3MIiv0P1bHF17$$\"3r* **\\(=n#f(=F1$\"3*3Ny6!*Q+M$F17$$\"3P+++!)RO+?F1$\"39%eu\\/;Nw$F17$$\" 30++]_!>w7#F1$\"3+4J:Kz)pD%F17$$\"3N++v)Q?QD#F1$\"3'e**eXmCX\"[F17$$\" 3G+++5jypBF1$\"3ApDW`OI%R&F17$$\"3<++]Ujp-DF1$\"3]U&)f9_i[hF17$$\"33++ D,X8iDF1$\"3-**)ei6*H?lF17$$\"3++++gEd@EF1$\"3l\"z^3L:]\"pF17$$\"31+]P Mh%\\o#F1$\"3a%y5'[$=GO(F17$$\"39++v3'>$[FF1$\"3_zyG&z,-%yF17$$\"39+++ 5h(*3GF1$\"3?Oe?IMfE$)F17$$\"37++D6EjpGF1$\"3?p[Ua(HO%))F17$$\"31+]i0j \"[$HF1$\"3Y6,CkKfN%*F17$$\"\"$F*F+-%'COLOURG6&%$RGBG$\"#5!\"\"$F*F*Fh ]l-%+AXESLABELSG6$Q\"x6\"Q\"yF]^l-%%VIEWG6$;F(F_]l;$Fg]lF*$\"\"&F*" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "If a function " }{XPPEDIT 18 0 "f " "6#%\"fG" }{TEXT -1 87 " is bounded below on a domain D and if there exists an alpha which is a lower bound of " }{TEXT 268 1 "f" }{TEXT -1 88 " on D with the additional property that for any other lower bou nd gamma it follows that " }{XPPEDIT 18 0 "gamma <= alpha" "6#1%&gamma G%&alphaG" }{TEXT -1 30 " then alpha is said to be the " }{TEXT 265 7 "infimum" }{TEXT -1 17 " of the function " }{XPPEDIT 18 0 "f" "6#%\"fG " }{TEXT -1 49 " on D. For example, the infimum of the function " } {XPPEDIT 18 0 "cosh(x)" "6#-%%coshG6#%\"xG" }{TEXT -1 59 " over the re al line is 1. Another example is the function " }{XPPEDIT 18 0 "exp(x )" "6#-%$expG6#%\"xG" }{TEXT -1 83 " on the real line which has an inf imum of 0 but there does not exist a real number " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 11 " such that " }{XPPEDIT 18 0 "exp(x) = 0" "6#/- %$expG6#%\"xG\"\"!" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "plot( exp(x), x=-5..2 );" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6%-%'CURVESG6$7W7$$!\"&\"\"!$\"3+na 3**p%zt'!#?7$$!3]LL$3#*>u%[!#<$\"3&Gxx9Y,'[yF-7$$!3!pm\"z43m9ZF1$\"3g8 8>\"o0H'*)F-7$$!3hLLe/$f`c%F1$\"31CC8cRhS5!#>7$$!3PLL3K\"o]T%F1$\"3a+z f7IP47F>7$$!3%pm\"Hn7\\lUF1$\"3&RGs0c'\\/9F>7$$!3WL$ekO9o7%F1$\"3Q,kTn %>Mh\"F>7$$!3p**\\7oCA$)RF1$\"3wL#4vB_D'=F>7$$!3;L$e9_>Z$QF1$\"3JyO=.) 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