Midterm exam 2, April 7, 2009 (solutions)

Transcription

1 Univrsiy of Pnnsylvania Dparmn of Mahmaics Mah 26 Honors Calculus II Spring Smsr 29 Prof Grassi, TA Ashr Aul Midrm xam 2, April 7, 29 (soluions) 1 Wri a basis for h spac of pairs (u, v) of smooh funcions u, v : R R ha saisfy h sysm of linar diffrnial quaions u 2u + 2v v 2u + v Soluion Th sysm can b xprssd as a marix quaion ] ] u 2 2 u v 2 1] v From class, w know a basis of soluions of his quaion if w can diagonaliz h marix To his nd, w find h ignvalus as h roos of h characrisic polynomial char() 2 + x 6 (x + 3)(x 2), which ar -3 and 2 Sinc h ignvalus ar disinc, w know ha h marix is diagonalizabl To find a diagonalizaion, w find (a basis of) ignvcors Of cours, w could now solv h ignvcor condiion quaion for ach ignvalu λ Hr is a gra rick for quickly ging a h ignvcors of a 2 2 marix (prov his as an xrcis, i s fun!): ] ] a b b if a 2 2 marix has an ignvalu λ hn is a λ-ignvcor c d λ a 2 2 Using his rick, w s ha and ar ignvcors for -3 and 2, rspcivly As always, 1] 4] 2 1 w can modify ignvcors by scalars, so w migh as wll ak and Now w can us 1] 2] h horm from class ha a basis of soluion is givn by ] ] 3 2, Drmin whhr h following ss in h plan ar opn closd or nihr Also dscrib h boundary s a) {(x, y) R 2 : 1 < x 2 + y 2 < 2} Soluion Opn This s is h inrscion of h opn ball of radius 2 wih h complmn of h closd ball of radius 1 So i s h inrscion of (wo) opn ss, so is opn Th boundary is h disjoin union of h circls of radius 1 and 2, {(x, y) R 2 : x 2 +y 2 1 or 2} b) {(x, y) R 2 : x 2 + y 2 1} Soluion Closd This s is h inrscion of h closd ball of radius 1 wih complmn of h opn ball of radius 1 So i s h inrscion of closd ss, so is closd I also can b sn o conain all of i s i poins I s is own boundary c) {(x, y) R 2 : x 2 + y 2 1} 1

2 Soluion Closd This s is h closd ball of radius 1 I s boundary is h circl of radius 1 d) {(x, y) R 2 : 1 x 2 + y 2 < 2} Soluion Nihr I s no opn bcaus whil (1, ) is in h s, no ball around (1, ) is conaind in h s I s no closd bcaus (2, ) is sm o b a i poin of h s, bu is no conaind in i Th boundary is h disjoin union of h circls of radius 1 and 2, {(x, y) R 2 : x 2 + y 2 1 or 2} ) R 2 { nonngaiv x axis } Soluion Opn Th nonngaiv x axis is closd (in R 2 ) bcaus i s sn o conain all of is i poins Thus i s complmn is opn Th boundary is h nonngaiv x axis f) R 2 { posiiv x axis } Soluion Nihr Th posiiv x axis is no opn (in R 2 ) bcaus no ball conaining a poin on h axis is sricly conaind in h axis Th posiiv x axis is no closd (in R 2 ) bcaus h origin is sn o b a boundary poin bu is no conaind in h posiiv x axis Sinc h posiiv x axis is nihr opn nor closd, hn nihr is is complmn Th boundary is h nonngaiv x axis g) R 2 { ingrs } Soluion Opn Th ingrs ar closd (in R 2 ) sinc hy conain all hir boundary poins Indd, any convrgn squnc of ingrs is vnually consan Thus hir complmn is opn Th boundary is h s of ingrs 3 Drmin if ach of h following is xiss If so, giv h i and an xplanaion, if no, prov i ( x 2 y 2 ) 2 a) (x,y) (,) x 2 + y 2 Soluion Dos no xis Along h lin y, w hav ha ( ) x (x,) (,) (x,) (,) x 2 whil along h lin x y, w hav ( x 2 x 2 (x,x) (,) x 2 + x 2 and so w s ha h i canno xis xy 2 b) (x,y) (1,1) x 2 + y 2 ) 2 (x,x) (,) 2 Soluion Th funcion xy2 is a quoin of coninuous funcions (polynomials) and h x 2 +y 2 dnominaor dos no vanish a (1, 1) Thus by ruls for coninuiy, h quoin is coninuous (1, 1), hnc xy 2 (x,y) (1,1) x 2 + y 2 xy2 x 2 + y 2 1 (1,1) 2 4 Find h quaion of h angn plan o h graph of f(x, y) 9 x 2 y 2 a h poin (1, 2)

3 Soluion Th graph of f in R 3 can b ralizd as h lvl surfac (or hyprsurfac) S {(x, y, z) R 3 : g(x, y, z) }, g(x, y, z) x 2 + y 2 + z 9 Our poin is hn (1, 2, 4) on h lvl surfac Th gnral formula for h angn plan T p S o a poin p on any lvl surfac S {x R n : g(x) } is givn by as long as D p g is no h zro map! T p S {v R n : D p g(v p) }, In our siuaion and in sandard coordinas, w hav g D g x, g y, z x Thus h rquird angn plan is ] 2, 4, 1] T S {(x, y, z) R 3 : 2(x 1) 4(y + 2) + (z 4) } {(x, y, z) R 3 : 2x 4y + z 14} 5 L f : R n R m and a R n For ach of h following, us xampls and samns from class and/or h homwork o drmin whhr such a funcion xiss a) f is diffrniabl a a bu no coninuous a a Soluion Dos no xis From a horm from class, if f is diffrniabl a a poin, hn i is coninuous hr b) All h parial drivaivs xis a a bu f is no diffrniabl a a Soluion On homwork 9, hr was an xampl of a funcion f : R 2 R dfind by { x xy if (x, y) (, ) f(x, y) 2 +y 2 if (x, y) (, ) whos parial drivaivs xis a (, ) bu is no diffrniabl hr c) Th qualiy 2 f x 1 x 2 2 f x 2 x 1 Soluion Th wording of his par is uninnionally ambiguous Sricly spaking, ys, any funcion whos scond parial drivaivs xis and ar coninuous vrywhr has an qualiy of mixd parials For xampl, h zro funcion (or any polynomial) saisfis his cririon Th qusion was supposd o b, Dos hr xis a funcion whr h mixd parials disagr? Ys, hr xis such funcions Exampls of such ar oulind in Aposol s book 6 L f : R 2 R 2 b dfind by f(x, y) (9 x 2 y 2, xy ) Explain why f is diffrniabl a vry poin (a, b) R 2 Compu h oal drivaiv D f Soluion A horm from class says ha f (f 1, f 2 ) : R 2 R 2 is diffrniabl if and only if h componn funcions f i of f ar diffrniabl Hr, 9 x 2 y 2 is a polynomial so is diffrniabl vrywhr, and also xy is diffrniabl vrywhr Finally, w compu, using h Jacobian marix xprssd in h sandard basis, f 1 f x 1 D f y ] 2a 2b for all (a, b) R 2 f 2 x f 2 y a+b a+b

4 7 L f : R 2 R and g : R 2 R 2 b dfind by f(x, y) sin 2 (x + y), Compu D (f g) for any (a, b) R 2 Soluion W ll us h chain rul g(u, v) (u 2 + uv 3, uv) D (f g) D g f D g To his nd, using h Jacobian marix wih rspc o h sandard bass, compu and Thn D (x,y) f 2 sin(x + y) cos(x + y), 2 sin(x + y) cos(x + y)] 2 sin(x + y) cos(x + y) 1, 1], 2u + v 3 3uv D (u,v) g 2 ] v u 2a + b 3 3ab D (f g) 2 sin(x + y) cos(x + y) (a 2 +ab 3,ab) 1, 1] 2 ] b a 2 sin(a 2 + ab 3 + ab) cos(a 2 + ab 3 + ab) 2a + b + b 2, a + 3ab 2 ] 8 L f : R 3 R b dfind by f(x, y, z) xy + xy + z 2 Drmin h criical poin(s) of f and a ach on, wri h quadraic approximaion Wha can you say abou h xrm valus (local/global min/max) of f? Soluion Firs no ha f is diffrniabl on R 3 Now, using h Jacobian marix wih rspc o h sandard basis, compu D (x,y,z) f y + y xy, x + x xy, 2z] Thn (x, y, z) R 3 is a criical poin of f if D (x,y,z) (,, ) This is quivaln o z, x(1 + xy ) and y(1 + xy ) Bu now sinc 1 + xy >, w finally g ha (x, y, z) is a criical poin for f if and only if (x, y, z) (,, ) For h quadraic approximaion, w ll nd h Hssian marix of f a h criical poin, y 2 xy 1 + xy + xy xy H (,,) f 1 + xy + xy xy x 2 xy (,,) 2 Th quadraic approximaion of f a (,, ) is now f(u, v, w) f(,, ) + D (,,) f(u, v, w) (u, v, w)h (,,)f(u, v, w) (4uv + 2w2 ) 1 + 2xy + z 2 No ha w could hav gussd his from h bginning, sinc h quadraic rm in h quadraic approximaion of xy is jus xy (coming from h Taylor xpansion), whil xy and z 2 ar hir own quadraic approximaions Finally, no ha d H (,,) f 8 and has a las on ignvalu of 2 (h vcor (1, 1, 1) works) Thus h produc of h rmaining wo ignvalus is 4, hnc on of hm is ngaiv Thus h Hssian is indfini a (,, ), indicaing ha f has a saddl poin hr As for global min or max, h xy rm in f guarans ha f gs arbirarily larg in boh posiiv and ngaiv dircions So hr ar no local and no global mins or maxs 9 Using h (i) dfiniion of h oal drivaiv, show ha 1, 1] is h oal drivaiv of h funcion f(x, y) x + y for any poin in R 2

5 Soluion For any (x, y) R 2, compu f(x + u, y + v) f(x, y) 1, 1](u, v) (u,v) (,) (u, v) x + u + y + v (x + y) (u + v) (u,v) (,) u 2 + v 2 (u,v) (,) u 2 + v 2 and hnc h linar funcion 1, 1] (i doing a vcor wih 1, 1]) is indd h drivaiv of f a (x, y) 1 EXTRA CREDIT Wri a basis for h spac of pairs (u, v) of smooh funcions u, v : R R ha saisfy h sysm of linar diffrnial quaions u 9u + 1v v 1u 11v Soluion Th sysm can b xprssd as a marix quaion ] ] u 9 1 u v 1 11] v From class, w know a basis of soluions of his quaion if w can diagonaliz h marix To his nd, w find h ignvalus as h roos of h characrisic polynomial char() 2 + 2x + 1 (x + 1) 2, and now w raliz ha afrall, h marix is no diagonalizabl A horm from class says ha h columns of h marix xponnial will b a basis of soluions of h sysm To his nd, no ha w can dcompos ] ] ] ino commuing marics (sinc scalar marics commu wih vryhing) Firs, rcall ha " # " 1 # 1 Also, no ha using h Taylor xpansion, w hav " # " # " # zro marics sinc all highr powrs of h marix vanish Now using h produc formula for h xponnial, w hav " # " # " # ] ] (1 + 1) 1 ] 1 (1 1) and so h columns of his marix form a basis for h sysm of diffrnial quaions On can furhr simplify his basis o ] ] (1 + 1), 1