ES.182A Topic 30 Notes Jeremy Orloff

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1 ES82A opic 3 Notes Jerem Orloff 3 Non-independent vribles: chin rule Recll the chin rule: If w = f, ; nd = r, t, = r, t then = + r t r t r t = + t t t r nfortuntel, sometimes there re more complicted reltions between the vribles Well see tht through series of emples tht will get incresingl fnc Remember tht, s long s we keep the roles of the vribles stright, the multivrible chin rule is rell just the single vrible chin rule When in doubt, go bck to the tngent pproimtion formul r r 3 Creful nottion o hndle more complicted situtions we need nottion tht full specifies the role of ll the vribles So, is the prtil of w with respect ot with held constnt his shows eplicitl tht we re thinking of nd s independent vribles Likewise,, is the prtil of w with respect ot with nd held constnt his shows eplicitl tht we re thinking of,, s independent vribles As shorthnd we m write w, for 32 Chnge Derivtives mesure rtes of chnge So, before strting in on the chin rule let s look crefull t the tngent pproimtion formul nd chnge We will do this with n emple which will repper in wht follows Emple 3 se w = , look t vrious chnges in,, Emphsie we don t cre wht cused these chnges this is how w chnges Constrin = Drw pictures of constrined chnge,

2 3 NON-INDEENDEN ARIABLES: CHAIN RLE 2 33 hree methods Emple 32 Given w = constrined b the reltion = compute nswer: Note: this problem is es enough tht we could substitute for in the formul for w nd compute the prtil directl But, tht wouldn t help us find hrder derivtives where substitution is not fesible he problem tell us tht we re to consider nd s independent Clerl, chnge in or cuses chnge in So, is not independent We ll compute the derivtive using three different methods You will see similr terms nd lgebr ppering in ll three methods Method : Implicit differentition his problem is simple enough tht implicit differentition will work his mens we should differentite the formul for w with respect to, but we hve to remember tht is function of nd is constnt We get o find = we differentite the constrint reltion, ie = 2 = Method 2 sing the tngent pproimtion formul o strt with, we ignore the reltionship between,, nd write the tngent pproimtion formul for w s in our erlier emple w = + + = ,,, his ss how w chnges s,, chnge Remember it onl cres tht nd re chnging, it doesn t cre if the chnge in hppens to be cused b the chnge in Divide this eqution b, holding constnt We cn do this, becuse nd re independent w = In the first term, clerl = In the second =, since is not chnging In the third term, = 2

3 3 NON-INDEENDEN ARIABLES: CHAIN RLE 3 he lst eqution ws found b differentiting the constrint eqution = We hve found w If we let go to, we get our nswer for : = Sme s Method Method 3: otl differentils: = If we wnt, we cn substitute he totl differentil of function w = f,, is just the infinitesiml version of the usul pproimtion formul he totl differentil of w with respect to,, is dw = d+ d+ d = w d+w d+w d = 2 d+2 d+2 d, In the second formul we use the shorthnd w for, when there is no chnce of confusion, We will use this nottion, When w is function of,, we will cll w, w, w the forml derivtives of w, ie we ignore n reltions between the vribles nd just formll differentite the formul Bck to nswering the question: the trick is to eliminte d from the totl differentil We do this b tking the totl differentil of the constrint: d = 2 d + 2 d Net we substitute for d in the formul for dw: dw = 2 d + 2 d + 2 d = 2 d + 2 d + 22 d + 2 d = d d his lst epression is the totl differentil of w with respect to, So the coefficients d nd d re the prtils with respect to nd respectivel, ie = 2 + 4, = his is the sme nswer we got with the first two methods We lso found brgin! Emple 33 For w = with = find implicit differentition nd the method of totl differentils Note: this time we re holding constnt, so s chnges must lso chnge in the using the method of

4 3 NON-INDEENDEN ARIABLES: CHAIN RLE 4 nswer: Now nd re the independent vribles, nd is n intermedite vrible We go through the methods with little comment Method of implicit differentition = Remember = We need to find implicitl his is much esier thn solving for nd differentiting tht = = Now using this in Eqution : = o do this we differentite the constrint = Not surprising: constnt is constnt w = is constnt! Method of totl differentils otl differentil of w: We hve to eliminte d otl differentil of constrint: Substitute for d: dw = 2 d + 2 d + 2 d d = 2 d + 2 d d = 2 d d dw = 2 d d d + 2 d = 2 2 d d = d d Coefficients re the prtils: =, = + 2 Emple 34 Let w = 3 2 t, = t Find using implicit differentition, nswer: rible: ;, re constnts; t is function of So, o find, t = t we differentite = t implicitl with respect to :, = t,, t =,

5 3 NON-INDEENDEN ARIABLES: CHAIN RLE 5 Substitute this into the formul for t = 3 2 2, :, Emple 35 Let w = 3 2 t, = t Find totl differentils, = / = 3 2,,,, nswer: he independent vribles re,, ; t nd w depend on,, otl differentil of w: otl differentil of constrint: Solve for dt: Substitute in dw: hus, dt = d + d t d dw = 3 2 d + 3 d 2t d 2 dt d + d = t d + dt dw = 3 2 d + 3 d 2t d 2 d + d t d = 3 2 d + 3 d + 2t + t d = 3 2, = 3, = 2t + t,,, using, 34 he Jcobin In order to do the most complicted problems we go bck to the simplest Suppose w is function of, nd, re functions of, b We know the chin rule ss = + b b b = + b b b For lgebric purposes let s write this s mtri eqution [ [ b b b b b b he two-b-two mtri in this eqution is clled the Jcobin mtri of the chnge of vrible between, b nd, We used nd b s vribles to void n nottionl confusion in the net section, where we will use Eqution 2 to orgnie our lgebr 2

6 3 NON-INDEENDEN ARIABLES: CHAIN RLE 6 35 hermodnmics nd the profusion of smbols here re mn quntities tht pl role in thermodnmics: p,,,, S, H pressure, volume, temperture, internl energ, entrop, enthlp In generl, n two cn be mde independent vribles nd then the others re dependent We cn write thermodnmic lws in terms of derivtives with respect to the independent vribles For emple, when p, re independent, we hve the lw: + p + p = L Emple 36 Epress this lw when nd re the independent vribles nswer: here s no hiding the fct tht this tkes some lgebr he derivtives,, re written in terms of derivtives with the p old independent vribles p, We need to write them in terms of the new independent vribles, o do this we use Eqution 2: we replce, b the old independent vribles p, nd, b b the new vribles, his gives o write bove becomes [ 3 in terms of the new independent vribles we let w = he eqution [ he derivtives on the left re es to compute, the re nd respectivel eqution is [ We solve this with the inverse of the Jcobin mtri [ [ = D [ = D So the Here D is the determinnt It s es to compute, but we won t need to do tht We hve found tht = D Now let w =, Eqution 3 becomes [

7 3 NON-INDEENDEN ARIABLES: CHAIN RLE 7 Agin, the derivtives on the left re es to compute, we get [ [ = D We re red to replce ll the terms in the lw L b terms using the new independent vribles he lw now reds: + p = D D D Multipling the eqution b D gives our finl nswer + p = Emple 37 Epress the lw L when nd re the independent vribles nswer: his is similr to the previous emple he new wrinkle is tht is both n old nd new independent vrible his cuses no difficulties In fct, it will mke some of the derivtives es to compute As before, we need to epress,, in terms of derivtives with the p new independent vribles, In Eqution 2 we replce, b the old independent vribles p, nd, b b the new vribles, his gives [ 4 In the lst eqution we used tht =, = Since we know tht we ll need the inverse of the Jcobin, let s give it nme nd compute it J J = D [ As before, D is the determinnt nd we don t need to compute it Let w =, nd solve: [ = J [ = J = D [ [ = D [ We didn t bother writing down the unneeded second entr to the solution = D We hve

8 3 NON-INDEENDEN ARIABLES: CHAIN RLE 8 Let w = nd solve [ = J Agin, we simplified: =, = J [ = = D [ [ = D We re red to replce ll the terms in the lw L b terms using the new independent vribles he lw now reds: + p D D D = Multipling the eqution b D gives our finl nswer + p =

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