.MCAD 306000000 Z  docDocument<mcObjectI (d2_graph_format graphData axisFormatLLtrace2D      dim_formatCmasslengthtimecharge temperature luminosity substanceUUUUNumericalFormat@dii shpRectEW 7mcDocumentObjectStateJ mcPageModel:????mcHeaderFooter99 ComputeEngine=BuiltInsB  SerialAnyvalH@!@H @"@H#@HMbP?$@H units_classA TextState0 TextStyle/@ Times New Roman0,0,0 Normalfont_style_list> font_style? VariablesTimes New Roman? ConstantsTimes New Roman? TextTimes New Roman? Greek VariablesSymbol? User^1Arial? User^2 Courier New? User^3System? User^4Script? User^5Roman? User^6Modern? User^7Times New Roman? SymbolsSymbol? Current Selection FontArial? Undefined Font? HeaderArial? FooterArial?Rotated Math FontTimes New RomanW TextRegion docRegion7shpBoxD 2R *EE CharacterMapRangeMap,@bENVIRONMENTAL PHYSICS COMPUTER LAB. #1: TUTORIAL AND EXPONENTIAL CONSUMPTION MODEL  ChrPropMap(UTb RangeElem-T  ChrPropData) RangeData.~ - )~ - )  ParPropMap*b-b ParPropData+EmbedMap"-LinkMap b-bLinkData!@NormalTimes New Roman @Dix$$;We learn first how to do simple computations with MathCad.(;-;)oArial0,0,0*;-;+"- ;-;!@NormalTimes New Roman @DH (1) Type:3*4= The answer is: 12.( - )Arial* - +" -  !- "!@NormalTimes New Roman #eqRegion3@D$tree1 p%1$&1@%'1t&3(1&4)1%*1+@)Serial_DisplayNodeF+1) _n_u_l_l_,@D 2H"(2) Type 24-35= The answer is -11("--".),Arial*"/-"0+"1- "2-"3!@NormalTimes New Roman 43@D8151 p61571@681t72491735:16;1+@:@F<1: _n_u_l_l_=@D H<(3) Type: \3= (this is square root of 3). The answer is 1.73(<>-<?)=Arial*<@@-<@A+"@B- <@C-<@D!@NormalTimes New Roman @E3@D#% @F1 p@G1@F@H1{@@G@I1@H3@J1@G@K1+@@J@F@L1@J _n_u_l_l_@M@D1C@H@F(4) Type 2.3^4.5= (this is 2.3 to the power 4.5). The answer is 42.44(F@N-F@O)@MArial*F@P-F@Q+"@R- F@S-F@T!@NormalTimes New Roman @U3@DPmh@V1 p@W1@V@X1@@W@Y1t@X2.3@Z1@X4.5@[1@W@\1+@@[@F@]1@[ _n_u_l_l_@^@Dy'H'(5) Type: (3-4)*4-6= The answer is -10('@_-'@`)@^Arial*'@a-'@b+"@c- '@d-'@e!@NormalTimes New Roman @f3@DwB  @g1 p@h1@g@i1@@h@j1@@i@k1p@@j@l1@k@m1t@l3@n1@l4@o1@j4@p1@i6@q1@h@r1+@@q@F@s1@q _n_u_l_l_@t@DH9(6) Type: 7/(2.3-8.6)space bar+3=. The answer is:1.889 (9@u-9@v)@tArial*9@w-9@x+"@y- 9@z-9@{!@NormalTimes New Roman @|3@DX@}1 p@~1@}@1@@~@1@@@1t@7@1p@@1@@1t@2.3@1@8.6@1@3@1@~@1+@@@F@1@ _n_u_l_l_@@D H (7) Type: ln(e)= The answer is 1( @- @)@oArial0,0,0* @- @+"@-  @- @!@NormalTimes New Roman @3@DX @1 p@1@@1@@@1d@ln@1p@@1@e@1@@1+@@@F@1@ _n_u_l_l_@@D!?30@77@M(8) Type: log(10^3.9)= The answer is 3.9 since log is logarithm in basis 10.(M@-M@)@oArial0,0,0*M@-M@+"@- M@-M@!@NormalTimes New Roman @3@DP@]X@1 p@1@@1@@@1d@log@1p@@1@@1t@10@1@3.9@1@@1+@@@F@1@ _n_u_l_l_@@Di{x@&(9) Type: ln(10)= The answer is 2.30 (%&@-@)@oArial0,0,0@-%@)@oArial0,0,0@@-@)@@@@*&@-&@+"@- &@-&@!@NormalTimes New Roman @3@DPg}x@1 p@1@@1@@@1d@ln@1p@@1@10@1@@1+@@@F@1@@@D(@  '(10) Type: log(e)= The answer is 0.43 ('@-'@)@oArial0,0,0*'@-'@+"@- '@-'@!@NormalTimes New Roman @3@DP@1 p@1@@1@@@1d@log@1p@@1@e@1@@1+@@@F@1@@@D8)+8(@copyright @ Miron Kaufman, 1998(@-@)@0,0,128*@-@+"@- @-@!@NormalTimes New Roman @@DIHXH@6@6@Next we learn how to define a function. Type x:1;50. It means: assign to x values from 1 to 50. : colon is assign ; semicolon is until(f@-@)@Arial@-f@)@oArial0,0,0@@-@)@kArial0,0,0@@-@)@mArial0,0,0@@-@)@kArial0,0,0@@-@)@oArial0,0,0@@- @)@iArial0,0,0@@-@)@iArial0,0,0@@-@)@oArial0,0,0@@@*@-@+"@- @-@!@NormalTimes New Roman @3@DT@1 p@1 @@1d@x@1@@1t@1@1@50@@DN(HFF,We next define the function f(x) = (1+1/x)^x(,A-,A)@Arial*,A-,A+"A- ,A-,A!@NormalTimes New Roman A3@D#3'A1 pA 1 AA 1@A A 1dA fA 1pA A 1A xA1A A1p@AA1AA1tA1A1AA1tA1A1AxA1AxA@DA;SP&@33*To get the function values just type f(x)=(*A-*A)AArial**A-*A+"A- *A-*A!@NormalTimes New Roman A3@DgBj xA1 pA 1AA!1@A A"1dA!fA#1pA!A$1A#xA%@Dh%h$$@pWe graph the function f(x). Note that as x gets bigger and bigger f(x) approaches the natural number e = 2.718.(pA&-pA')A%oArial0,0,0*pA(-pA)+"A*- pA+-pA,!@NormalTimes New Roman A-3@D`)`XA.1 pA/1A.A01@A/A11@A0A21@A1A31fA2 2.691588A41A22A51A1A61dA5 _n_u_l_l_A71A5eA81A0A91dA8fA:1pA8A;1A:xA<1A/A=1@A1@A=A?1fA>50A@1A>1AA1A=AB1dAA _n_u_l_l_AC1AA _n_u_l_l_AD1A<xAE)N^     AF@D` 8-`Y$$@fWe can graph also by typing the function in the y placeholder as you can see below for ln(x) versus x.(fAG-fAH)AFArial*fAI-fAJ+"AK- fAL-fAM!@NormalTimes New Roman AN3@Dh@]|X_AO1 pAP1 AOAQ1dAPxAR1APAS1 @ARAT1@ASAU1tAT10AV1KATAW1AV6AX1AS0.001001AY1AR10AZ@D8I48X`66@WThis means the first value of x is 0.000001, the second is 0.001001 and the last is 10.(WA[-WA\)AZArial*WA]-WA^+"A_- WA`-WAa!@NormalTimes New Roman Ab3@DhAf hbAc1 pAd1AcAe1@AdAf1@AeAg1@AfAh1tAg5Ai1KAgAj1Ai 13.815511Ak1AfAl1@AkAm1AkAn1AeAo1dAnlnAp1pAnAq1ApxAr1AdAs1@ArAt1@AsAu1fAt 9.999001Av1AtAw1dAv1Ax1AvAy1dAx10Az1KAxA{1Az6A|1AsA}1@A|A~1A|A1ArxA)NN     A@DP A P aHcopyright @ Miron Kaufman, 1998(A-A)A0,0,128*A-A+"A- A-A!@NormalTimes New Roman A@D C  83Z3ZA^This program illustrates the constant rate of growth (exponential) model. We will also demonstrate how to use semilogarithmic plot to analyze exponential dependence. Here we use a vector. j is the range variable- corresponding to time. Type n[j+1:n[j*C. Here C is the constant multiplying factor and its logarithm is the growth constant k = ln(C).(^A-^A)AoArial0,0,0*^A-^A+"A- ^A-^A!@NormalTimes New Roman A3@D( m5 ?0 A1 pA1 AA1dACA1A1.05A3@D 5 0 A1 pA1 AA1dAkA1AA1dAlnA1pAA1ACA3@DP 5 b0 A1 pA1AA1dAkA1AA1+@A@FA1A _n_u_l_l_A3@D O he 0` A1 pA1 AA1dAjA1AA1tA0A1A99A3@DO k ` A1 pA1 AA1@AA1dAnA1A0A1A1A3@D o U A1 pA1 AA1@AA1eA,nA1AA1dAjA1A1A1AA1@AA1dAnA1AjA1ACA3@D A A1 pA1AA1@AA1@AA1@AA1fA 125.239293A1A1A1AA1dA _n_u_l_l_A1A _n_u_l_l_A1AA1dAnA1AjA1AA1@AA1@AA1fA99A1A0A1AA1dA _n_u_l_l_A1A _n_u_l_l_A1AjA6NN     A3@D A A1 pA1AA1@AA1@AA1@AA1fA 125.239293A1A1A1AA1dA _n_u_l_l_A1A _n_u_l_l_A1AA1dAnA1AjA1AA1@AA1@AA1fA99A1A0A1AA1dA _n_u_l_l_A1A _n_u_l_l_A1AjA6NO     A@D Hcopyright Miron Kaufman, 1998(A-A)A0,0,128*A-A+"A- A-A!@NormalTimes New Roman A@D '  d366@We now use real data to show the applicability of the constant growth rate model. First we look at US annual natural gas consumption (in quads). (A-A)AoArial0,0,0A-A)AAAA*A-A+"A- B-B!@NormalTimes New Roman B3@D& F B1 pB1 BB1dBYearB1pBB10BB10ABB 10ABB 10AB B 10AB B 10AB B 10AB B10AB B1@BB1B1970B1B 1960B1B 1950B1B 1940B1B 1930B1B 1920B1B1910B1B1900B3@D& 1 B1 pB1 BB1dBGasB1pBB10BB10ABB10ABB 10ABB!10AB B"10AB!B#10AB"B$10AB#B%1@B$B&1B$22.0B'1B#11.0B(1B"6.3B)1B!2.6B*1B 2.0B+1B0.80B,1B0.51B-1B0.25B.3@D50B/1 pB01 B/B11dB0nB21B0B31tB20B41B27B53@DG1_HB61 pB71B6B81@B7B91@B8B:1@B9B;1fB:22B<1B:0.25B=1B9B>1dB= _n_u_l_l_B?1B= _n_u_l_l_B@1B8BA1dB@GasBB1B@nBC1B7BD1@BCBE1@BDBF1tBE1980BG1BE1890BH1BDBI1dBH _n_u_l_l_BJ1BH _n_u_l_l_BK1BCBL1dBKYearBM1BKnBN9NN     BO@D Hcopyright @ Miron Kaufman, 1998(BP-BQ)BO0,0,128*BR-BS+"BT- BU-BV!@NormalTimes New Roman BW@D1C@0)We fit the exponential model to the data.()*)BX-)BY+"BZ- )B[-)B\!@NormalTimes New Roman B]3@D@WshB^1 pB_1 B^B`1@B_Ba1dB`lnGasBb1B`nBc1B_Bd1dBclnBe1pBcBf1BeBg1dBfGasBh1BfnBi3@D@TBj1 pBk1 BjBl1dBkkBm1BkBn1dBmslopeBo1pBmBp1 BoBq1dBpYearBr1BplnGasBs3@D(:Bt1 pBu1BtBv1dBukBw1BuBx1+@Bw@FBy1Bw _n_u_l_l_Bz3@D@ TB{1 pB|1 B{B}1dB|bB~1B|B1dB~ interceptB1pB~B1 BB1dBYearB1BlnGasB3@D(:B1 pB1BB1dBbB1BB1+@B@FB1B _n_u_l_l_B3@D?B1 pB1BB1@BB1@BB1@BB1fB 3.091042B1KBB1B 1.386294B1BB1dB _n_u_l_l_B1B _n_u_l_l_B1 BB1@BB1dBlnGasB1BnB1BB15@BB1dBkB1BB1dBYearB1BnB1BbB1BB1@BB1@BB1tB1980B1B1890B1BB1dB _n_u_l_l_B1B _n_u_l_l_B1BB1dBYearB1BnB5NN     B@DG(@??AAThe growth constant k is the slope of the semilogarithmic graph: k = 0.063. Then the constant multiplying factor is  So the annual growth rate is 6.5%. The doubling time is:  i.e. it takes about 11 years for the US annual gas consumption to double. We estimate that the year 2000 US gas consumption will be: quads.(;tAB-tB)BoArial0,0,0B-B)B/Arial0,0,0BB-B)BoArial0,0,0BB-B)B/Arial0,0,0BB-9B)BoArial0,0,0BB-B)B/Arial0,0,0BB-B)B@ Arial0,0,0BB-B)BoArial0,0,0BB-B)B/Arial0,0,0BB-B)BoArial0,0,0BBB*AB-AB+"B-B EmbedData#t5EmbedObj$B EmbedObjPtr%B3@D3+hHJCB1 pB1 BB1dBCB1BB1dBeB1BkB-B#x\$B%B3@Du2HCB1 pB1BB1dBCB1BB1+@B@FB1B _n_u_l_l_BB-B#\0$B%B3@DIHxleB1 pB1 BB1dBT.2B1BB1@BB1dBlnB1pBB1B2B1BkBB-B#:$B%B3@D5B1 pB1BB1@BB1dBeB1BB1@BB1dBkB1B2000B1BbB1BB1+@B@FB1B _n_u_l_l_BBB AB-AB!@NormalTimes New Roman B@DASPHcopyright Miron Kaufman, 1998(B-B)B0,0,128*B-B+"B- B-B!@NormalTimes New Roman C@Dasp8@CSecond we will look at the annual oil consumption (in quads) in US.(CC-CC)CoArial0,0,0*CC-CC+"C- CC-CC!@NormalTimes New Roman C3@D{n6C 1 pC 1 C C 1dC YearC 1pC C 10C C10AC C10ACC10ACC10ACC10ACC10ACC10ACC1@CC1C1970C1C1965C1C1960C1C1955C1C1950C1C1940C1C1930C1C 1920C3@DSnC1 pC 1 CC!1dC OilC"1pC C#10C"C$10AC#C%10AC$C&10AC%C'10AC&C(10AC'C)10AC(C*10AC)C+1@C*C,1C*29.0C-1C)21.0C.1C(18.0C/1C'16.0C01C&12.5C11C%7.3C21C$5.1C31C#2.5C43@D0jC51 pC61 C5C71@C6C81dC7lnOilC91C7nC:1C6C;1dC:lnC<1pC:C=1C1dC=OilC?1C=nC@3@D{CA1 pCB1 CACC1dCBkCD1CBCE1dCDslopeCF1pCDCG1 CFCH1dCGYearCI1CGlnOilCJ3@D1CK1 pCL1CKCM1dCLkCN1CLCO1+@CN@FCP1CN _n_u_l_l_CQ3@DCR1 pCS1 CRCT1dCSbCU1CSCV1dCU interceptCW1pCUCX1 CWCY1dCXYearCZ1CXlnOilC[3@DAC\1 pC]1C\C^1dC]bC_1C]C`1+@C_@FCa1C_ _n_u_l_l_Cb3@D>]Cc1 pCd1CcCe1@CdCf1@CeCg1@CfCh1fCg 29.469663Ci1Cg2.5Cj1CfCk1dCj _n_u_l_l_Cl1Cj _n_u_l_l_Cm1 CeCn1@CmCo1dCnOilCp1CnnCq1CmCr1dCqeCs1pCqCt1CsCu1@CtCv1dCukCw1CuCx1dCwYearCy1CwnCz1CtbC{1CdC|1@C{C}1@C|C~1tC}1980C1C}1910C1C|C1dC _n_u_l_l_C1C _n_u_l_l_C1C{C1dCYearC1CnC1NN     C@D6a7..ADThe growth constant k is the slope of the semilogarithmic graph: k = 0.047. Then the constant multiplying factor is:  So the annual growth rate is 4.8%. The doubling time is: i.e. it takes about 15 years for the US annual oil consumption to double. We estimate that the year 2000 US oil consumption will be: quads.(CuDC-uC)CoArial0,0,0C-C)C/Arial0,0,0CC-C)CoArial0,0,0CC-C)C@ Arial0,0,0CC-C)C?CC-C)C?CC-9C)CoArial0,0,0CC-C)C/Arial0,0,0CC-C)C/Arial0,0,0CC-C)CoArial0,0,0CC-C)C/Arial0,0,0CC-C)CoArial0,0,0CC-C)CCCC*DC-DC+"C-C#u5$C%C3@D=C1 pC1 CC1dCCC1CC1dCeC1CkC-C#z $C%C3@DNXQfC1 pC1CCC-C#{\$C%C3@DXmC1 pC1CC1dCCC1CC1+@C@FC1CCC-C#\0$C%C3@DE2hC1 pC1 CC1dCT.2C1CC1@CC1dClnC1pCC1C2C1CkCC-C#q$C%C3@D*C1 pC1CC1dCT.2C1CC1+@C@FC1C _n_u_l_l_CC-C#=$C%C3@DD`a\C1 pC1CC1@CC1dCeC1CC1@CC1dCkC1C2000C1CbC1CC1+@C@FC1C _n_u_l_l_CCC DC-DC!@NormalTimes New Roman C@DXyEXHcopyright @Miron Kaufman, 1998(C-C)C0,0,128*C-C+"C- C-C!@NormalTimes New Roman C@D-l8%6%6@The table below shows the annual world energy consumption in quads for the 1980-1996 period. The data is available at the following web site of the Department of Energy: http://www.eia.doe.gov/(C-C)CoArial0,0,0*C-C+"C- C-C!@NormalTimes New Roman C3@DHXC1 pC1 CC1dCjC1CC1tC0C1C16C3@DH7SzHC1 pD1 CD1@DD1eD6YearD1DjD1DD1dDjD1D1980D3@D@BD1 pD 1 DD 1dD  worldenergyD 1pD D 10D D 10AD D10AD D10ADD10ADD10ADD10ADD10ADD10ADD10ADD10ADD10ADD10ADD10ADD10ADD10ADD10ADD1@DD1D375.069D1D365.126D 1D354.411D!1D350.855D"1D346.041D#1D344.345D$1D343.975D%1D339.783D&1D334.425D'1D322.271D(1D313.466D)1D306.499D*1D297.282D+1D283.247D,1D278.612D-1D 279.579D.1D 281.891D/3@D;0D01 pD11 D0D21@D1D31dD2 lnworldenergyD41D2jD51D1D61dD5lnD71pD5D81D7D91dD8 worldenergyD:1D8jD;3@D?UPD<1 pD=1 D1dD=kD?1D=D@1dD?slopeDA1pD?DB1 DADC1dDBYearDD1DB lnworldenergyDE3@DX?UjPDF1 pDG1DFDH1dDGkDI1DGDJ1+@DI@FDK1DIDL3@D_upDM1 pDN1 DMDO1dDNbDP1DNDQ1dDP interceptDR1pDPDS1 DRDT1dDSYearDU1DS lnworldenergyDV3@DX_ujpDW1 pDX1DWDY1dDXbDZ1DXD[1+@DZ@FD\1DZD]@D8//@AEstimate the world energy consumption in the year 2000: quads.(;7AD^-7D_)D]oArial0,0,0D`-Da)D]D^Db-Dc)D]?D`Dd-De)D]@ Arial0,0,0DbDdD^*ADf-ADg+"Dh-Di#:$Dj%Dk3@DkDl1 pDm1DlDn1@DmDo1dDneDp1DnDq1@DpDr1dDqkDs1Dq2000Dt1DpbDu1DmDv1+@Du@FDw1Du ADx-ADy!@NormalTimes New Roman Dz@DD9@<H<H@A measure of the goodness of the linear fit is the correlation coefficient r. If r is close to 1 or -1 the data show high linear correlation, i.e. the linear fit is good. The Mathcad function corr(X,Y) gives the correlation of the X and Y data values.(D{-D|)DzoArial0,0,0*D}-D~+"D- D-D!@NormalTimes New Roman D3@Do*D1 pD1 DD1dDrD1DD1dDcorrD1pDD1 DD1dDYearD1D lnworldenergyD3@D`opD1 pD1DD1dDrD1DD1+@D@FD1DD3@DA4D1 pD1DD1@DD1@DD1@DD1fD 5.931917D1D 5.623565D1DD1@DD1DD1 DD1@DD1dD lnworldenergyD1DjD1DD1@DD1dDkD1DD1dDYearD1DjD1DbD1DD1@DD1@DD1BDD1dD1.996D1DD1dD10D1D3D1DD1dD1.98D1DD1dD10D1D3D1DD1@DD1DD1DD1dDYearD1DjD2'NN     D@D@ @$$copyright @Miron Kaufman, 1998(*D-D+"D- D-D!@NormalTimes New Roman