W11_CalcII_test2_Purple_solution

W11_CalcII_test2_Purple_solution - Calculus II Midterm 2 W...

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Unformatted text preview: Calculus II Midterm 2 W 2011 PULeVz 2 1. (4 mark each; total 12 marks) (a) Find the probability of waiting between 7 and 10 minutes at the local bank branch, where average wait time is 6 minutes. lo :17 v (0 ~ _7 Phqw) = l 4:8 Aott = ~€%:l = -e %+€ /‘ m 0.\L 7 7 -‘7 —\°/ Answer: 8 (“6 A (b) Consider the initial value problem y’ = 1 — mg, with initial value condition y(2) = —1. Use Euler‘s Method with step-size h = 2/3 to approximate lab.)=—\ So xo:1l3°=-‘ goats}: ‘3 : \_x‘a’ ~ 3“: lam-\+L‘F(XV\-n3v\-l) ; xvx=xn-\_\'\A n 117 n yin. O X°=l ‘a,=—\ 2. xf‘Z tat: \a\+\,\$(x\,\3\3= \-\—-7§-~(\—%\)s =5;— 3 X?” ‘63“13‘J‘L‘C‘XW‘V'E‘" eo- TE?— (0) Find the slope of the tangent line to the graph of the curve given by parametric equations :1: = sin(3t) + 6—2 3; = tcost at the point Where t = 0. Smog M: MVLC 3 %+\Jc:o = L 30 AxAkkb‘i-O 2‘ Asia; ~. 3%: = 300% _e-t ’ %\£=o= l 453% = (90.95.. fimkt ’ Kt“: = 4 Answer: y). Calculus H Midterm 2 W 2011 3 2. (10 marks) Solve the following separable differential equation, isolating for y as an explicit function of 1': arm—1 = Big—2; 20(0) = 1- %/= fl = (xex)g—L 1. _ x Asqu ~. Sx exA av was $0 1‘3 -Sxe AM. x ujx’duvgolx &\1:Q’7‘Ax u=€x —\— 3 : x— x‘\’C ’ so {A} z 3xex’ 3€X+C = xe" —e" +C so (a =$\l;3>xe,x -3>e,x-\-C Dow (AC0) =2 & PLOQGNG m AbOVE eET “sf—375‘ =>1C=Hl So vagm Calculus II Midterm 2 W 2011 4 3. (8 marks) (a) (3 marks) Find fy, fa; for the function flay) : Bang—2% (b) (3 marks) Brine with a concentration of salt of 0.06 kg/ L is added at a rate of 10 L / min to a tank with 50 L of pure water. The solution is kept thoroughly mixed and exits the tank at the same rate. SET UP (DO NOT EVALUATE) the initial value problem (that is the differential equation AND the initial value condition) that models this process. L151 A=Am 66 Arkqu oc SALT As A Foucn'on at» Time i. 1?: \O-o.oc———~\o =o.e——>§— >1 A(o)=0- (c) (2 marks) Consider the differential equation for which the direction field is shown below. Sketch the solution to the initial value problem with ini- tial condition y(—1) = 0 \ \\‘\“-«.H \ \\NHH \l \\N‘H~«H \, \\NHH \ \\“H‘“~.H \\\N"~¢.M \! \\“M"‘~a“-q \\\NM-. I \EEZ'MN‘H‘} \HHH \R‘NNHH \\\\WH \u\\m‘-.-q \. \\‘-«r~«~u \. \\N*-r- \.\\‘w--- \ \\N“-w- Calculus H Midterm 2 W 2011 5 4. (10 marks) Find the area inside one 100p of the polar curve r = 2 8111(36) but outside the curve 7' = 1. his 1gTr/m 3% 5% = j 45(LkmiA‘be-OAIG =3 (amlse— five =Sp~fivm€e3~flae 1123 1178 V1 5%, 5% _ _ _\ - —j —UoQO)&9 — (1'9 *‘i—MQGSX 1‘7n, 171% _ E: _ J. a“; _ 1L: 4— ‘ 1E ‘ 3; I. ’M 3 36 + G M 3 = l t _\- Calculus II Midterm 2 W’ 2011 6 5. (10 marks total) Answer the following questions in the Space provided (a) (2 marks) Convert the point with polar coordinates (4, arr/4) to Carte- sian coordinates (exact values: not decimal representations). (:L‘l )GL'I: Lt x: (0’39: u'i—é‘t‘lfi) iatr-Meszfi Answer: (10139.5) (b) (2 marks) Convert the polar curve r = 30se6 to a Cartesian equa- tion. F: 30559 = M-az'é :50 Me=-%: 2< ‘3’ VMQ =3 mhe=%- $0 Answer: 13:3 (0) (1 marks) True / False The differential equation 2yy’ — 33695 @= 3 has order 2. Answer: 1 205 (d) (2 mark) Find the domain of the function NEED V0201. 36% 23061237“) 2m 2 2 + sin my 3; — 2 /§ ~ t No we‘STMcnon 113332;: 2’) = Answer: 17-- (e) (1 mark) True/Falsegy’ = sin(:t')sin(y) is a separable differential equation. ‘6: 7)» camgb Answer: €AL$E (f) (1 mark) True/False is a solution of the differential equation 3;” + Qy = 0. {F ‘6: 8‘5» 1 3/: ’38-‘536 ) 30:98.3”: Answer: FALSE (g (1 mark) Let f(u,v,w) = 2sin(7ru) +11) sec(v). Find f(1/2,7r/9,0). ) £(a,1g—’,o)= armlf+ omen; Answer: 9» ...
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This note was uploaded on 10/05/2011 for the course MATH 1020 taught by Professor Paulatu during the Spring '11 term at UOIT.

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W11_CalcII_test2_Purple_solution - Calculus II Midterm 2 W...

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