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Math2015_1011_T2_Solns

# Math2015_1011_T2_Solns - ZS’ Warlb‘s Math 2015 Applied...

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Unformatted text preview: ZS’ Warlb‘s Math 2015 Applied Multivariate and Vector Calculus: Test2 Friday October 22, 2010 10:30am to 11:20am gld‘l‘ I 0143 Student Number: Name: Instructions: Complete all 5 of the following problems in the space provided. Notes and calculators are not permitted. All cell phones and pagers are to be turned off. 1. Suppose a duck is swimming in the circle :1: = cos t, y = sint and that the water surface temperature is given by the formula A 4’ Nﬂrlﬂs _. Find dT/dt, the rate of change of temperature the duck might feel at time t = 7r/ 4. Row» bu, chain rule, cﬂ‘: + 21‘. cl’c BEL (it 33 (it T(:z:, y) = xzey — 371/3 c\_x;-sm is g} __ -" l ‘ Ak \lr-‘WH _ ' (3‘3 = 0090 “K” : +_l_ ‘ Lit - | At At tsm ‘8 ’2‘ I” 4i i: 1: 23L {3-— 3 :5 1T - BIL 33" l3 Bat/haw, Ea” ‘" ?D;[ 2 x7- ,_ ‘3 air \ t 3 *3le Z q, i . E! "I ‘- -E EB, L. L\[email protected]€,lu1’_éﬁ d L T‘— ‘2, L L} 2. Suppose that over a certain region of space the electric potential V is given by V(:c,y, z) = my + 2:102 — 23. S A (a) Find the rate of change of the potential at the point P(2, —3, —1) in the direction of the vector v = i — 2j + 2k. (b) Find a unit vector in the direction in which V is changing most rapidly at P. (c) Find the maximum rate of increase at P. 7_/\ W: (33.43;)? +04 -32 K =7 VV(1,—3,-13 a 5? +7.: — a? ’ Q1) M obftLlL‘h/uva AUtVAl “N? UJMQI’L A ‘ A A_ /\ A A A vaxrz ’1_:_.1 +744 a: 1(¢—1S+IL)~—® u vvufxrnn —. man; my 2 I373 5 Marks - 3. What horizontal plane is tangent to the surface z=m2—4my—2y2+12m—12y—1 and what is the point of tangency? (Here 2 is the vertical variable While a; and y are the horizontal variables). A Plant. lS hori Lox/{lag 0n kkg QSUa/f‘bvl Q) 27; = (AMglom‘E- Thm‘l is. The, énbukl L014 0? {Mr P\ “M i": luAfLPQMdLWl’ 0%: 3L Ml la . WL wws3r kwc. on M Plano (anal at Fob} «Av lansmcay 9?; ; Di _ O Nous Bi=Zx—Ll'\ﬁ'\-\'L =0 TNSQ. lus’r ‘lwo acct/Al CNS llAVL soagum (D For mu Mum a —'5\ so in rabmal MKS Q/cxuwl' tm It; -31 —® l“ Emwl a; lg k_q_) \ '-3\> ' :‘L‘ ’ 3 S— er= 4. Find a second degree Taylor approximation for the function f (x, y) = V1162 + 3/3 near the point (1, 2). §(N\3\:‘\§Llivﬁ3 3‘7 Sykhzgz ‘5 3;”- (BW) = L” =3 g U L3‘ ‘ l , _ __ {17% 3” 3 ‘—-® TwXur AWrowmu’thhn‘. \h “USMbJMOOCX J1) (,hl) ¥t\$ﬂ§3 £1 \$Uﬂ,)* ngﬂv UL—\> -\~ ¥3L\,L)[\3’13 +l¥x,pu.1)(3L-|)L Li¥gﬁt‘\1)(vj’1\l’ L + ksumtmww 4 = 3 +13Lx-q + M343 3r ggrbL—‘V’r £93437, _ 7___ (Mme-L7 ~———— (D ‘i 5. Suppose z = f (x, y) where f is a differentiable function. If and y = es sint (0 Marks .. '97; = 3% ‘ a} 31' B a: 5; as * 33 53: Q) Q} : '\~ 21% 23- I N: Db» 3‘: M 2t 0 (ngvibﬁ %: -€,\$Sm't ~ .— 7»; 3 M6) B1 €\$U>\$+a : DL '93 3‘3: {Sen/Ho : ‘3 ﬂ :3; (D 2‘5 _—_ es 0090 T- b“— Di: )7; ‘- (D'e 7' -: '3; 3} I .- 31: 3,2} L +L‘b_{) \xgmj—n g\ Jf\ \$51!} 93 .1 m, 7' I 31 ‘- 32. vi - x [5; \Mi «V 5x3 m I m a} 1.. m. m Jc y} Jr K k W) Zauﬁ 3 1. )1; "L = W ‘31) \igi * [535 \ —-—-® ...
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