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# sol5scan - MATH 237 Assignment 5 Winter 2007 Solutions 4.A3...

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Unformatted text preview: MATH 237 Assignment 5 Winter 2007 Solutions 4.A3 a) Given f(a:,y) = 6395—29 we have fa:(x, : 36301—2?! fy(\$,xy) = _263\$—2y fx\$(\$9y) : gen—2y fang/(33,29) : —6€32—2y fyy(m,y) = 4e3x—2y. so that Vf(2,3) = (fm(2,3),fy(2,3)) = (3, —2). and _ fmcc(2a3) fwy(2a3) _ 9 "'6 W“) ‘ ( mm) was) > ‘ l —6 4 b) The linear approximation at (2, 3) is Lemmy) = f(2, 3) + M2, 3) (Iv - 2) + M2, 3) (y — 3) = 1+3(:c—2)—2(y—3). The second—order Taylor polynomial is ' (\$_29y_3) + %(x_2ay_3)Hf(2i3) ( > 1 1+ 3(m — 2) — 2(y — 3) + §[9(x — 2)2 —12(:v — 2)(y — 3) +4(y - 3)2]. P2,(2,3) (13: y) C) We ﬁnd Vf(:z:,y) = (aw—29\$, —2), which is always parallel to (3 — 2). Thus the level curves of f, which are everywhere orthogonal to the gradient (and so to (3, —2)) are straight lines. This is clear from the form of f as the level curves are deﬁned by 3:73 — 2y = constant. 4.A5 1) With f(:1:,y) = (:c — y) sin(:1: + y), we ﬁnd fm(a:, y) : sin(a: + y) + (a: — y) cos(m + y) y) fylﬂo = — \$11017 + y) + (33 — y) COS(\$ + ’9) fax, .9) = 00805 + y) + 603(33 + y) - (a: — y) sinCI + y) fmytr, y) = 608(w + y) — 008(56 + y) — (9: - y) sinks + y) fyy(:1:,y)= — cos(a: + y) — cos(:c + y) — (:1: — y) sin(:r + Evaluating at (7T, 7r), we have mm) = 0, MW?) = 0, mm) = 0 fxx(7ra 7r) = 2: fxylnv”) : 0, Joya/(Wu?) : *2 Then the second order Taylor polynomial at (7r, 7r) is Iaaaew>= ﬂmwhthme—ﬂ+JMmﬂw-Wt+ g(tamwxm—wr+2tnmwxm—ee—wu+tnmwxy-ma = e—«V—e—«V ii) Since f and its partial and second partial derivatives are continuous (that is, f is a 02 function), Taylor’s Theorem tells us that P2,(,,‘,r)(:r, y) provides a good approximation to f (:12, 3/) near (7T, 7r). Thus the level curves of f near (7r, 7r) are well approxmiated by the level curves of P2, (7r, 7r) near (7r, 7r) These curves 0 = (:0 — 7r)2 — (y — 7r)2 are hyperbolae centred at (71', 7r), as shown below. We conclude that the function f looks like a saddle near (7r,7r). 4.B2 a) Let u(m,t) = Ae‘(\$”“)2. Then ux(a:, t) = —2A(:c — ewe—(“0”2 and we = We — ave-mew _ gig—W while “453, t) = 2Ac(x — Ct)e—(\$—ct)2 and WW”: 15) = 4A262(x — ct)2e_(1"c‘)2 _ 2AC2e—(m—ct)2 so cgum (m, t) 2 2mm, t) as required. b) me the graphs of it against :1: for various values of t, we see that the wave moves with speed 1% = c to the right. 0) Observing that the combination a: — ct lead to a rightward motion, we deduce that :1: + at will lead to a leftward motion. A leftward moving wave of the same shape is described by u(\$, t) = Ae‘(z+0‘)2. d) Provided f has continuous second partials (so that we may safely apply the chain rule in What follows), we let u(:1:, t) = f (:1: — ct) and ﬁnd I/ ux(a:,t) = fl(x— ct) and um(m,t) = f (:3 —ct) while // 21.L(\$,t) = —cfl(ar — ct) and utt(x,t) = 02f (at — ct), 2 which conﬁrms c 113:9;(1‘, t) = uu(m,t). 4B7 We suppose f : R2 —> R has continuous second partial derivatives which, for (32,31) in a neighbourhood of (a, b), are all bounded by some constant .M. Consider the error Rum) made by using the linear approximation to estimate f at a point (33, in this neighbourhood. Then, from Taylor’s Theorem (page 93 in course notes) there is some point C on the line segment joining (a, b) and (:r,y) such that R1.(a,b) = g _ (1)2 + 2f1y(x _ (IX?! _ b) + fyy(C(y _ (7)2] - Then mm s glMo—aero—axy—b)+M(y—b)2| = g[(\$—a)2+2(m—a)[y—b)+(y—b)2| s g|(m—a>2+<y—b)2|+MI<m—a)(y—b)l 3 s %||(m,y)—<a,biil2+ = M||(ﬂv,y)-(aab)l|2 M 31l(m,y)— (mm? as required. The second last line follows from — a)(y — b)| S — (1)2 + (y — b)2|, which follows from expanding — a) + (y — (1)]2 2 0. 5.A1(iii) a) With f(:r,y) = (a:2 + 3/2 — 1)y we ﬁnd fx(x,y) = 2:62; Many) = 9:2 + 3y2 — 1 The critical points are those (:13, y) for which f3; = fy = 0. The ﬁrst condition speciﬁes m = 0 or y = 0. If :r = 0, then fy = 0 implies 3y2 = 1 or y = if?“ while if y = 0 then fy = 0 implies 3:2 = 1 or m = :i:1. We conclude there are four critical points: (0, ii) and (i1, 0). To classify these critical points, we ﬁnd the second derivatives fmx(\$ay) : 2y fmy(\$ay) = 2m fyy<xay) : 631‘ Then ;_ a 0 whose determinant fmfW — fy is positive and whose ﬁrst term is positive. We conclude 1 that (0, W) is a local minimum. We next ﬁnd Hf(0 —i)— _% O 1 V3- _ O _\/i§ Whose determinant fmfm, — \$21, is positive and whose ﬁrst term is negative. We conclude that (0, is a local maximum. Treating the last two points together we have Hf(:i:1,0)= (i3 In both cases the determinant fmfyy — 12.3] is negative. These are saddle points. b) From the above analysis we have a classiﬁcation of the nature of the level curves near these critical points (i.e. ellipses and saddles). A more precise picture can be reached by considering the second order Taylor polynomials: while Putting these together, we have the following: c<<2.,- mil ll Extra 8. The function f (at, y) = 1 + 3:2 + y2 has derivatives 1: 1/ firm: = _ f 17,? =——-——-—-— ( y) ,————1+m2+y2 A J) ,—-——1+x2+y2 —a:y 1 + 51:2 fm(\$ay) = W fwy(\$,yl = W fyy(33,y) = W We note that f” is everywhere positive, and the determinant of the Hessian is ll (1_+3:21TyZ)3/_2 [(1 + y2)(1 + x2) — (_my)2] 1 (1+ on? + MW2 2 fmfyy _ my [1 + y2 + 3:2 + yzmg — y2m2] > 0. We conclude that the function is convex. ...
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## This note was uploaded on 01/17/2011 for the course MATH 237 taught by Professor Wolczuk during the Spring '08 term at Waterloo.

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sol5scan - MATH 237 Assignment 5 Winter 2007 Solutions 4.A3...

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