You have met function of a single variable f(x), and calculated the properties of these curves such as

Transcription

1 Chaper 5 Parial Derivaive You have me funcion of a ingle variable f(, and calculaed he properie of hee curve uch a df d. Here we have a fir look a hee idea applied o a funcion of wo variable f(,. Graphicall hi i ea o repreen a he variaion of he heigh of a urface above a plane a a poin wih coordinae (,. 5.1 Fir Parial derivaive The lope of he funcion f(, in general will depend on he direcion in which i i calculaed. We will ar here b calculaing he lope in he and direcion. To ake he parial derivaive wih repec o we mu keep fied and var b a mall amoun, δ. Thi i denoed b = lim δ 0 f( + δ, f(,. (5.1 δ Similarl he parial derivaive wih repec o i calculaed b keeping fied and varing = lim δ 0 f(, + δ f(,. (5.2 δ Thee definiion are much he ame a ordinar derivaive obained b varing one variable and keeping all oher conan. For eample conider he funcion We can define higher derivaive in he ame wa f(, = (5.3 = (5.4 = (5.5 = 2 f 2 (5.6 = 2 f 2 (5.7 = 2 f = 2 f 48 (5.8 (5.9

2 CHAPTER 5. PARTIAL DERIVATIVES 49 Noe ha provided he parial derivaive are coninuou a he poin in queion hen = 2 f i.e. he order in which we ake he derivaive doe no affec he reul. In our earlier eample 5.2 Talor Theorem (5.10 = 2 (5.11 = 2. (5.12 You have me he idea of a erie epanion for a funcion of a ingle variable. If we epand f( abou he poin 0 he reul i a Talor erie f( f( 0 + f ( 0 ( 0 + f ( ( = f (n ( 0 ( 0 n. (5.14 n! n=0 We can performa imilar epanion for a funcion of wo variable abou he poin ( 0, 0 f(, = f( 0, [ 2 ] f 2! 2 ( f + 2 f 2 ( (5.15 where = ( 0 and = ( 0. Thi epreion can be epreed more uccincl in he noaion of vecor calculu a ou will ee ne ear. 5.3 Toal differenial f(, = f( 0, 0 + ( n f (5.16 If we allow a mall change in boh and hen we can obain he oal change in he funcion f(, denoe b df a follow n=1 df = f( + δ, + δ f(, (5.17 = d + d (5.18 where we have negleced he higher order erm (δ 2 ec. in he approimaion. Thi epreion i known a he oal differenial of f. For eample conider he funcion f(, = in df = din + dco (5.19 Now uppoe ha = ( i.e. i a funcion of. The oal derivaive of f wih repec o i hen df d = + d d (5.20

3 CHAPTER 5. PARTIAL DERIVATIVES 50 Anoher wa o ee hi i a follow. If we make a mall change in f(,, where = ( hen we have: ( f( + δ, ( + δ f + δ, ( + d d δ + O(δ2 Eac oal differenial f(, ( + df d = + d d ( + d d δ + O(δ 2 The eample in he la ecion i known a an eac differenial, becaue if we ar from din + dco hen we can find he funcion f(, = in ha will produce hi differenial. Anoher eample i he differenial d + d, which mu reul from a funcion f(, = + K. Some differenial are no eac, for eample d + d. (5.21 Here he fir par inegrae o 1 2 2, bu he econd par inegrae o In general if we have a differenial hen o for i o be eac, we require 5.4 Reciproci and cclic relaion A(, d + B(, d (5.22 = A (5.23 = B (5.24 A = B. (5.25 Suppoe we have a relaion beween hree variable z = z(,. The oal differenial of hi epreion i ( ( dz = d + d. (5.26 Bu we could alo wrie = (, z d = ( d + z If we ubiue hi epreion ino our fir reul for dz we obain ( ( [( ( dz = d + d + z ( ( ( ( dz = d + d + z ( dz. (5.27 dz ] ( (5.28 dz (5.29

4 CHAPTER 5. PARTIAL DERIVATIVES 51 If we e d = 0 hen we obain which i he reciproci relaion. ( ( = 1/ Alernaivel if we e dz = 0 hen we obain ( 1 = Thi i he cclic relaion. ( z ( (5.30 ( Chain Rule Suppoe ha boh = ( and = (, i.e. boh and depend on a parameer. If we have a funcion f(, hen we can calculae i derivaive wih repec o b uing he chain rule df d = d d + d d (5.32 Thi i known a he chain rule, and i paricularl ueful if we are calculaing he derivaive of f along a paricular pah ha i defined paramericall (i.e. in erm of. Anoher wa o ee hi i b uing alor heorem δf = f( + δ, + δ f(, δ + δ + (O(δ2, δ 2, δδ (5.33 If we ubiue δ = d d δ + O(δ2, and imilarl for δ hen we obain δf ( d d δ + ( d d δ + O(δ 2. (5.34 If we divide hrough b δ and ake he limi δ 0 hen we obain he chain rule reul. 5.6 Change of variable I i ofen convenien o change he variable ha we ue o epre a paricular funcion. Suppoe we have f(, bu wan o work in erm of new variable and, ha are relaed b = (, and = (,. The derivaive of f wih repec o our new variable can be epreed a follow ( ( ( ( ( = + (5.35 ( = ( + ( ( (. (5.36 We can how ha hi i rue b uing Talor heorem again. We can obain Eq. (5.35 a follow (neglecing erm O(δ 2 ec. hroughou. δf = f( + δ, + δ f(, (5.37 = f(( + δ,, ( + δ, f(, (5.38 f( + δ, + δ f(, (5.39 ( ( ( ( δ + δ (5.40

5 CHAPTER 5. PARTIAL DERIVATIVES 52 which produce Eq.(5.35 in he limi δ 0. For eample uppoe we wih o calculae ( ( and, bu have f(, where ha = and = +. We impl appl he reul aed above ( = 1 (5.41 ( = 1 (5.42 ( ( ( = + (5.43 ( = 1 (5.44 ( = 1 (5.45 ( ( ( = +. (5.46 Alernaivel, uppoe we have a funcion in Careian coordinae f(, bu wih o work in( polar coordinae, where = r coθ and = r in θ. In erm of hee new variable we can calculae r θ. ( ( ( ( ( = + (5.47 r θ r θ r θ ( ( = coθ + in θ (5.48 ( and imilarl for θ. r 5.7 Anali of aionar poin of a funcion of wo variable For a funcion of a ingle variable a aionar poin occur when df d naure of hi poin b eamining he econd derivaive d2 f d. 2 (a d2 f d 2 > 0 he lope i increaing o we have a minimum. (b d2 f d 2 < 0 he lope i decreaing o we have a maimum. (c d2 f d 2 = 0 poin of infleion. For a funcion of wo or more variable aionar poin ill occur when = 0. We can characerie he = 0 and = 0. (5.49 However, he naure of hee aionar poin i a lile more varied. We can have maima and minima a before, bu in addiion o hee we can alo have a addle poin, where we have a maimum

6 CHAPTER 5. PARTIAL DERIVATIVES 53 in one direcion and a minimum in anoher. If we anale he econd erm in he Talor erie we can pin down he naure of he aionar ( poin. We will do hi uing he mari mehod from earlier in he coure. If we wrie = hen [ 2 ] f 2 ( f + 2 f 2 ( 2 = T H (5.50 where H = ( 2 2 = ( f f f f (5.51 i he mari of econd derivaive known a he Heian mari, and we have inroduced a hor hand for he econd derivaive. Our funcion hould increae in all direcion from a minimum, o T H > 0. If hi i rue hen H hould alwa have poiive eigenvalue. Similarl if we are a a minimum hen H hould alwa have negaive eigenvalue. If we are a a addle poin hen we hould have one poiive and one negaive eigenvalue. Equivalenl (i Minima f 2 < f f and boh f, f > 0 (boh eigenvalue poiive. (i Maima f 2 < f f and boh f, f < 0 (boh eigenvalue negaive. (i Saddle f 2 > f f (eigenvalue of mied ign. If f = f = 0 or if he eigenvalue are equal o zero hen furher anali of he fied poin i necear o deermine i characer. 5.8 Thermodnamic Man phical quaniie can be epreed in erm of parial derivaive, for eample The coefficien of hermal epanion α = 1 V ( V T P Molar hea capaci a conan volume c p = T ( T The reciproci relaion and he cclic relaion are ueful for manipulaing derivaive like hee. Calculaing he ae of a em can be done b minimiing i free energ. To find he abili of he em he naure of he aionar poin mu be found. The inabili of a phae of maer can poin o inereing phical effec, uch a phae eparaion which ou will learn more abou laer in our coure. P. 5.9 Problem 1A. Show ha 2 f = 2 f for he following funcion (a f(, = (b f(, = in( (c f(, = e ln( +

7 CHAPTER 5. PARTIAL DERIVATIVES 54 (d f(, = A. Evaluae + when (a f = 1 + (b f = ln (c f = f( 3B. Find which of he following differenial form Pd + Qd are eac. If he are eac, find a funcion f uch ha df = Pd + Qd, and he general oluion of he equaion Pd + Qd = 0. (i d + d (ii d + 2 d (iii ( + d + ( d (iv (coh co + coh cod (inhin inh in d (v (co in d + (in + cod (vi (d d/( A. An elecroaic poenial v i given b he funcion v(, = ep( /λ in(k i.e., i deca eponeniall along he ai and i periodic along he ai. (a Wrie down he oal differenial of v(,. (b Show ha 2 v and 2 v are equal remember ha for eample on he boom of a parial differenial mean ha we ake he derivaive wih repec o fir and hen ake he derivaive wih repec. 5B. The heigh h of each poin (, of an area of land i given b h(, = a( a 2 where a i a poiive conan. Find he locaion and heigh of he highe and lowe poin of he errain, and alo hoe along he and ae. Skech a map of he region b howing conour of conan h in he (, plane. 6B. Find du d in wo wa given ha (i u = n n and = coa, = in b, where a, b and n are conan, (ii u = and = ln

8 CHAPTER 5. PARTIAL DERIVATIVES 55 7B. The urface of a piece of ound-damping maerial undulae in boh he and direcion. The heigh of he urface of he maerial i decribed b an equaion which give he heigh h a a funcion of he and coordinae. Thi funcion i h(, = co(k co(k (a Calculae he wo parial derivaive of h. Now, if we move acro hi urface a a veloci v = (v, v hen a a ime our and coordinae are given b = v = v (b Uing he parial derivaive ou calculaed in a, find he derivaive dh/d which give he rae of change of he heigh a we move acro he urface a he veloci v. Wrie our anwer a a funcion onl of, no of and. 8C. For f(, = e, find /, and /. Check ha ( / (/ = ( / (/. Find (/ r θ and (/ θ r (i uing he chain rule, (ii b fir epreing f in erm of polar coordinae r, θ, and check ha he wo mehod give he ame reul. 9A. If z z 5 = 0 (an implici equaion for an of he variable,, z in erm of he oher wo, find, z, and how ha heir produc i 1. 10A. Van der Waal equaion (p + a/v 2 (V b = RT i an earl (and in man wa remarkabl ucceful aemp o repreen he relaion beween he preure p, volume V and emperaure T of a real ga (R, a, b are conan for a given ma of ga. Calculae epreion for p V, V T T p, T p, and verif heir produc i 1. V 11B. f(, i a calar funcion of poiion on he plane. Poiion ma alo be pecified b Careian coordinae u, v which are referred o ae roaed b an angle θ from he and aie. Show ha f 2 = 2 f u f v 2, i.e. he 2 dimenional 2 operaor i invarian under roaion of ae. 12A. Show ha V = (Ar n + Br n co(nθ + ǫ, where A, B, n and ǫ are arbirar conan, aifie he equaion 2 V r V r r V r 2 θ 2 = 0

9 CHAPTER 5. PARTIAL DERIVATIVES 56 13A. If V = f( c + g( + c, where f and g are arbirar funcion, and c i a conan, prove ha 2 V V c 2 2 = 0 14C. If u = +, v =, and f i a funcion of and, epre prove ha = 2 f u 2 + u 2 f u v + f v 2 c + v, in erm of u, v and 15C. The independen variable, are ranformed ino new variable X, Y given b he equaion X =, Y = 1/. If a funcion f(, i hu ranformed ino F(X, Y hen calculae F X and F Y in erm of and. 16B. If f i a funcion of and which can, uing he ubiuion = re θ, = re θ, be wrien a a funcion of r and θ how ha 2 = r r + θ, 2 = r r θ. 17B. Find a Talor epanion up o quadraic erm in 2 and 3 of f(, = e abou he poin = 2, = 3. 18B. Find he aionar value of he funcion (a ( (b inin (0 < < π,0 < < π (c ( e and deermine heir characer. 19B. (a Find he aionar poin of he funcion z = ( 2 2 e 2 2. (b Find he conour on which z = 0 and eamine he behaviour of z on he ae. Hence, or oherwie, deermine he characer of he aionar poin. Skech he conour. 20B. For he funcion f(, = ( find he componen of he vecor (/, /, known a he gradien vecor, a he poin ( 1, 0, (1, 0, ( 1, 1 and (1, 1. Make a kech howing he direcion of he gradien vecor a hee poin. 21B. Show ha f(, = ( ha aionar value a (0, 0 and (1/3, 1/3 and inveigae heir naure. 22C. A aionar poin of he funcion f(, i a minimum if he Heian mari (i.e. he mari of econd derivaive i poiive definie (i.e. i eigenvalue are boh poiive and a addle if he eigenvalue have oppoie ign. Inveigae he aionar poin of f(, =

10 CHAPTER 5. PARTIAL DERIVATIVES 57 Anwer o Chaper 5 Problem Noe: ou mu aemp he problem before referring o he anwer 1. (a (b (c (d = 1 = co( in( = e (+ 1 (+ 2 = ( (a + = (b + = 0 (c + = 2 dg(u du u= 3. Noe here f(, = K olve he equaion in general (i Eac, f(, =. (ii No eac. (iii Eac, f(, = (iv Eac, f(, = coh in + inhco. (v No eac. (vi Eac, f(, = arcan = arcan 4. (a dv = 1 λ e /λ in(kd + e /λ k co(kd (b = e /λ k co(k λ 5. Saionar poin a (, = (a/ 2, a/ 2 heigh 1/ 2, and ( a/ 2, a/ 2 heigh 1/ 2 6. (i du d = con 1 (ain m 1 (b(bm co(aco(b an in(ain(b (ii du d = ln ln 2 7. (a h h = k co(kin(k, = k co(kin(k (b dh d = h d d + h d d = k co(kv in(kv v k co(kv in(kv v 8. = e, = e, 2 f = e ( 1 (i r θ = r θ + r θ = e coθ e in θ = 2r coθ inθe r2 in θ co θ r = r + r = e ( r inθ e r coθ = r 2 e r2 in θ co θ (2 in 2 θ θ 1 θ θ (ii f = e r2 in θ co θ hen differeniae o obain anwer quoed above.

11 CHAPTER 5. PARTIAL DERIVATIVES z = z z = +5z4 z+4 3 = 32 +z +5z 4 p V T = p+2ab av V 3 (V b V p = RV 3 T T p p av +2ab V = V b R 11. The new coordinae are a roaed verion of he old, i.e. u = co θ + in θ and v = in θ + coθ. Uing hi coordinae ranformaion he reul in he queion follow. 12. V r = (Anrn 1 Bnr n 1 co(nθ + ǫ 2 V r 2 = (An(n 1r n 2 + Bn(n + 1r n 2 co(nθ + ǫ 2 V θ 2 = (Ar n + Br n co(nθ + ǫ Summing hee componen a inruced produce he required reul V 2 = f ( c + g ( + c 2 V 2 = c 2 f ( c + c 2 g ( + c Summing hee componen a inruced produce he required reul. 14. = u u + v v = u u + v v F 15. X = = 2 f u u 2 X + u + 2 f u v u v + 2 f u v u v + 2 f + v2 v = 2 f u f u v + 2 f = 2 f u f u v u + 2 f v 2 v + v X F Y = Y + Y 16. Changing variable produce he reul in he queion. 17. Denoe δ = 2 and δ = 3, hen f(2 + δ, 3 + δ 3e 6 + 9e 6 δ + 7e 6 δ u v + 2 f v v 2 v + v 2 v ( 27e 6 δ e 6 δδ + 16e 6 δ (a (0, 0 Maimum

12 CHAPTER 5. PARTIAL DERIVATIVES 59 (b (π/2, π/2 Maimum (c (1, 0 Saddle, (1 ± 1/ 2, ±1/ 2 Maimum if boh + ign, Minimum if mied ign. 19. (0, 0 Saddle, (±1, 0 Maima, (0, ±1 Minima 20. Repecivel ( 1, 2, ( 1, 2, (2, 4, (2, (0, 0 Maimum, (1/3, 1/3 Saddle. 22. (0, 0 Saddle, ( 1/12, 1/6 Minimum