00020001010008main.ACT0002020007CMV.EAC0100000021d3  CMV.EAC main.ACT$'+~E '9[485Geometric Interpretation of Cauchy Mean Value Theorem\= AuthorTuProfessor Wei-Chi Yang Radford University 9 e- l: wyang@r2.eduE!URL: http://www.#/5C Objectvs:[T(1) would like to g a g+i+how/() Theorem is stated.[&9(2) We would briefly describe how CMV8 proved and usAI[nm.bN (B!) Suppose the function f:[a,b]eRYg re continuouswthat e]G ir restriLs to (N)<differeAable. Mr, assumat g(x)0 for all x in 2N. =nre@ a point x+rt whichV   f(b)-f(a) gg(a) = f(x)g.[' ; Remark: [AU(1) Simply put, if we apthe Mean Value Theorem ongraph of a polar equation; RbutMwriteM( in<arametric form,vn will obta'CMV. T(2) We usY following exae to demonstra~motivproÈ\J techniqbd here c bia arbitraryw' holds.u[ $An Example: We consider a polar equation,/ of r=n in theterval n[-,0] and write Md it2_parametric form5: (now useb variable Bsteadfuture  discuss.)[yL[x(t),y]=[t*cost*sing f%]. let A=(), B=(0 ԎVC /2). Fit ograph$t whe&slope;tangent line is same as0AC. [[MIWKey: We claim that if we rotate AC to AB (a horizontal line segment),8en= e t which ]Rmakesslope of&ang9 (for r=t)dbesame a50willZW'tfdc hin curve. Sokey io find equation>\Motiv& => play anim7Έ`f pWx! `"P Q3S5@ 4! 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Note thate parametric equation for,t is asllows:[-[x(t),y]=[t*cos t*sing!f&].T2. Drag and dropoints A=[-,0],B=[0&C (/2)] intograph. )nlineAC͏Oy-0=m(x+) or y=(1/2)*C/2. We write this in parametric equation as follows:[-[x(t),y]=[t*cos ]()f]5&Step 3. Selects e point D f r=t. [/V/ 4. Constru2a perpendicular linessingroughMand)to AB;PintersI segments AC,/G n graph resp ively at G,FjEU3Step 5. Animate D along the r=t by properly selectiange of parameter for aT ion. [Us6. ColP and dratx-valuXdistancE into polar plot wesgetthin curve S.[yRy7. To (graph S, key is notiqG=EFsi G=f(t)-GF Ԉ+y2(m*x+b)%1@*gd-7  =t*sin(t)-((1/2)*(t*cos)+)[L**In other words,  heights (y values ofin graph) should be/ same as[Wf{yc. T_efore, we write S1 7[x0,y]=[,]xNote.$ Ifndragpartric equation 8BR[v],<]], with t in [0,2], backtoB_1mgeometry strip above, we will get the graph of S.[:Step 8. To find where S has a horizontal tangentYnote Cdy xU=t(dt a{Q =0 implies thatT f(t)-(m*g+b)A)(((t*cos-=0,$fo(eneed to solve WYfor t so that f(t)=m*g  (()/() =(b)-f(a g ).This is Texactly w`the Cauchy mean valueorem states. Now itEandard exercise to [,solve . We demonstratT as follows:[R define ft*sinR done0g0cos0( kt  W==G1O2W ,t,0,2,3)Rt="  2.425497143BT(Q R&,[MWe find t to be about Q (when m=(1/2),a=-, a3b=0 respectively.)]Note. MSince we are coll: ng the 'dista ' f(t)-yve dur# animation,DLchose this example inRifiedtervalcau/b is non-negWve.\`Trathree graphs.Ј#*'NFinaForm0$NGraph2DT 3h LISTSYSt4< Modify P<STATCALC d< \x S;equence,xShee] \|ؒ olveEq_wrX(UpdtupFLG1p(<LisDPicܐViewWinL_osvevdxt_((H2m` y<|<c ,Hyt3 7 Ć$K( ;O  $0<FÆT`lx !"#%̒"؍%  &(,'(< )P *dg+x, 8-D .P 0\ 1h 2t3 4 5, E@ FT Hh I|Ȇ JԆ K/L͆ MyN O( PQQ( R4 S@L ] X ^_dahblp,|͑@ ΑTFБhב| ؑ ّć   ۆfRܐ FinancialFormat  E system]Setu^a _LIST`,bT@ @@ @xĮ\Motiv& => play anim7Έ`f pWx! `"P Q3S5@ 4! DVvLr  A  L C uxy [ @  N tcos( )0-0.5+%sin%-1.570796326795d  oo9HE@d6kvYYHHI. qrQcU"UB`r%hFTB SewG Y1H!  +  tcos( )J--0.5+%sin% -1.570796327](10q Ht @  HG@6P(xrUp  ͎UBG!5A Bq%@Ibb.`5:H %Ae5+} RQCșW2g@   *H < 2-[, 8#8 <5<0#hF@6SewGY #B'ܚBH B}4No CXCX  SaSbŘN !@   W2gf@3Ę'H 9D&6\vcyg YsW62 qrQ  "((`0 `v$vi2`sqaP<efSHW`F$IU8$AP0$3E#`%@%5p .%v spwgidaaRXF `DU8P 6R0(I#$ uF@0pCp<eH fi0YS8` i $Rv0$DTYcRYa QS`T"#xYAPYaaS l`(57 61YW 1U 7v t (FQr0y9V<2( P4Tw`1H  &c 4IPCH 6Yd 5%Gp 1u@#bv $Q`03A0<v'WPHS2x 'C($isklffRH1@VIECy92xS'H2B<C&QB`5yEfDfX4lhd`aT`$b1H!+0H 6FedXH Y rETx%i3N"T((` `^`t`gyv`s`iq eSYe@WFIY`$I99S0A2)H<3%B0H% 6PT00 %klyli0YpvuIhh8`a(RTDGx69s(2g %xahUX` `DYys`Y&,Y1i,Y7`YV Yv YG8PYvyY4y#lS4xex&7vDY"PP +(/(vH "u 4b'p)Q'V`$8'@0rP<'B xR0Tar#q P0lvypw$a5 (6 &Pa0 4 9 7aE  2gY $!!Fl Q5@ˆx@TV(0DRpriRuR 3q VR"  \VXBFEXV `$8bUFvpAe u60  G$A H#!+.@  4"3#$thm (10qyX$H VH@6A0B& `X66I BsqS %uBIDGp6 Y$CdB&ۈɈ'z"@(N1Š.LE)1Ę 6!   3( [Q&SStep 1. We construct r=n. Note thate parametric equation for,t is asllows:[-[x(t),y]=[t*cos t*sing!f&].T2. Drag and dropoints A=[-,0],B=[0&C (/2)] intograph. )nlineAC͏Oy-0=m(x+) or y=(1/2)*C/2. We write this in parametric equation as follows:[-[x(t),y]=[t*cos ]()f]5&Step 3. Selects e point D f r=t. [/V/ 4. Constru2a perpendicular linessingroughMand)to AB;PintersI segments AC,/G n graph resp ively at G,FjEU3Step 5. Animate D along the r=t by properly selectiange of parameter for aT ion. [Us6. ColP and dratx-valuXdistancE into polar plot wesgetthin curve S.[yRy7. To (graph S, key is notiqG=EFsi G=f(t)-GF Ԉ+y2(m*x+b)%1@*gd-7  =t*sin(t)-((1/2)*(t*cos)+)[L**In other words,  heights (y values ofin graph) should be/ same as[Wf{yc. T_efore, we write S1 7[x0,y]=[,]xNote.$ Ifndragpartric equation 8BR[v],<]], with t in [0,2], backtoB_1mgeometry strip above, we will get the graph of S.[:Step 8. To find where S has a horizontal tangentYnote Cdy xU=t(dt a{Q =0 implies thatT f(t)-(m*g+b)A)(((t*cos-=0,$fo(eneed to solve WYfor t so that f(t)=m*g  (()/() =(b)-f(a g ).This is Texactly w`the Cauchy mean valueorem states. Now itEandard exercise to [,solve . We demonstratT as follows:[R define ft*sinR done0g0cos0( kt  W==G1O2W ,t,0,2,3)Rt="  2.425497143BT(Q R&,[MWe find t to be about Q (when m=(1/2),a=-, a3b=0 respectively.)]Note. 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