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Math20C Midterm2 - YE let/P Math 200 Calculus Midterm Exam...

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Unformatted text preview: YE let/P . Math 200 Calculus Midterm Exam II March 2, 2007 Naunc: Lecture I-Iour: ME— Sec-tion Hour: m Guidelines for the test: I No hooks. notes 01' calculators are allowed. 0 You may leave answers in symbolic form: like Jr}. unless they simplify further. like J?) = 3. e” = l. or eos(37r/J.) = —J§/‘2. «- Use the space provided. If necessary, write “see other side” and continue working 011 the heck of the same sheet. a Circle your final answers when relevant. 0 Show all steps in your solutions and make your reasoz'iing clear. Answers with no explanation will receive no credit. even if they are correct. 0 You have 50 minutes. ('1) (a) Match 1.110 follmving equatiuns (i)—(\-‘i) with the. graphs (I)-(VI). You need M explain ymlr choices. The first. our has been decided for you. (i) j‘(:r.y) = Hil12:1.'+ 591’ [1 (ii) f(:::.y) = (.-n:-:(::;in(3:r2 + '9'?» (iii) f(:r.y) = sin(:r) sin(3y) (iv) f(;:‘. y) = ('05(3J') msm) MPH (V) f(-?‘. y) = 5iII(I-T| + lyl) |H ‘ (vi) “Ir-y) = 3:17: Pugl: I (2) Suppose that , , _ if[a.n)#[010)? m" U) Q {0, + 1mm) = (U: 0.)- (a) Is f (one) continuous at (U. D) ? Clearly explain your answer. 1 M u‘s‘m'w 'a u-""_-—'_'—_ :: “Mt—7U). a) U14 V1 UL H-w (Qua-2 V‘) M1 2.- 0 V12 0 \‘tm (9'1 (11V :0 (X ,\ \fiwiul U7 *Viio l '1 LA 1V —0 d mm+vlzfi OLE-{IV lsgl‘n ‘1. 1"?) L U L 4:! IW SO OFU—LAVL _ l (3x1 SKI/WQ-Iu' Yet’: '11- “r, cav‘hflmOM-S (1)) Suppose further that 'u. = 110‘) and n = vhf) are continuous functions of f, for all t E R. (1) Write down the chain. rule that. would allow one to compute I‘ll: in terms of fm ft“ m and 'Ug. You need not compute any derivative ”EL-0E Ar explicitly. 4' OH1 _ “a; ‘llvl H+ +‘Qv‘lV-l' / \V l l "f ”l” (ii) Use the two tables of values given below to calculate fig i=1 Page 2 W 3 o: £L°r0\{ffij (3) Suppose f(3:,y) = 3:2 + yg, Where (3:,y) E R2. The surface 2 = f(3:._y) is a paraboloicl. (a) One of the two contour maps below is for a paraboloid and the other a. cone. Circle the one that corresponds to a paraboioid. (b) Let P = (2,1,5) is a point on the surface 2 = fist, 3}). Find a. unit vector 11' for which the directional derivative Dfif(P] is the largest possible. P=C2,1‘3\ 2:4(Yfl-1) :xqfl'l‘w-Z' 'F‘x:2,)( ”CH—'27 "3‘ v¥<¢fl31<4rz> De—HPPVHPE-M \ i P 4.; $7 TEL'z-rl7 (0) Consider all possible lines that are tangent. to the surface 2 = f [37,19] at P. One of these lines has the largest possible slope. What is the magnitude of this slope ".F' WIN—mm : id. two ,__"‘"=.. 11W: ”04.4 : ~20 (d)' Find the equation of the plane tangent to z = f(:z:, y) at 13(23 1, 5). F3 : W 1F”! :- 2‘1 FAM " O Fig : Zr Fug, ’ 2. Ziemzflo +¥y(1n7(x_;ji-pv(2n\(w'l) : S +‘iL><v7A+ 2044) 7:34 qx-g+2ul-Z 2 s-é‘xrufrlo **4x+2%'8 Page 3 (4) (a) For each of the following sets. detemliue whether it is closed and whether it is bounded. (i) {[:I:, y) : :32 + y2 >1} Closed? (Ye. i Bounded? ([email protected] (ii) Hay) = y 2 2} Closed? e“ Boundei‘l‘? (Yes. {ti-ac“ 4 Vii—NM + x142fi‘d Jr? (b) Suppose f(;::1y)=(:r, F 1)2 + (1!; — 1)2 + (:L'+ g)? 2%:qu +ZVI (i) Find all critical points of f[3:, y). ’Qx: 20 :1) + IMHO We»1 ._ 204 ~13 + “axed : 1 x-‘z-l-LX4'1 filx O 1 jx+l‘1"7‘ {L 1.1% saw +2 $xx:2+l'r"l pxq‘l- Qvifls’ZIT‘l/"L’l :2. '2 4x+7vl L“ +1” ”W: +1): Gw;O='2FW-"L+-LK"11W—f Hul'l17C72- 1X4h11'2'vl‘l-2 0 ”“4””? K11. (ii) Classifyr each of your critical points as local maximum. local mini- mum or saddle point. ”2.. mils-3 = more»: spam = quflb’r =W’qt'l>o \ ‘ 19‘4' \4. (ER?) \5 mmxmwm / J (iii) Find the shortest distance from the point (1.1.41) to the plane. a: + y — z = 1. P =mfi-u '1 VH7 + 01 P‘r—(x-nH-(vi—OIMHO‘ _2_ PX=2L.><—i‘) 'zx—2--o 1:2,; x4 ,5 ‘szCw—‘l 2‘4'2“) ‘1 vazcl-H) 114-51=O 97'? l1+ l,L 4' (' ll. rd; II Fayed (5) Use the method of Lagrange 111ultipliers to find the maximum and minimum values of the- function [(13 y) = r2 - '92 subject. to the constraint. 3:2+'y2=1. \—V\_/) _ 650‘!“ @(Vfi-‘JB = 3(1— ‘41—, Jazz; / 3)::2): , JEN—27 / =3 ./ 9w ‘1 V+=AV3 Page .5 [6) Note: You would have to have earned full credit on problem 2 to qualify for any portion of this extra credit. Consider again the function defined in problem '2 ml!— '1‘ 0.0-. m. w) = T (“ “é (. ’ (3. If (v.1?) --= (0.0). Is f(-u.. 1.!) (1il'l'm'clll.ia.hlo at. (l. l) '3' At ((1.0) '3’ Clearl}r and fully explain using the notion of dificrc”mobility. ‘ plan-"S Yes ‘l-‘C— Lu N3 ‘ MMMWU‘XA QT (‘9 I n . ‘3 cm «Sm/“d :0 0* 4”“ OIR‘(U‘-/(/1 rl K- Wu #91:?8‘ . mu‘Ll), WW ‘“ ' - how“ '1 0' 0‘ ‘5 C M-E IQ— PW Z/r’r IS ‘3 _ ‘fil [MUD -"’l'l.ll oxi'l UL Page 6‘ Formula. Sheet, {1} Some surfacme: f(.r. y) = 3‘2 # H2 is a saddle: f{..r y) : 3‘37": — .1":i is n lllOIlkfl}: Huddle; f(3.'. y) 2 Hi — Jr“! — “:3 has at [Jtmk m. ((1.0, 16). {1-3) 1” is continumm at. (0,3)) if gin: ”flag-y) = flan). (any - In! (3) A: :3 fz AJ? + 1",, Ag: d; =1} :33.- 4- f_,, dy. which is the 531110115 LU) = fln} + f“(n){.:r — a) (fur single. variable) w w L(:1:. y) -: f(n.h) + t—(r:.h}(3' — a] + .—{u.b)(y — b). ()3' 0y (-I) To find critical points. solve j". = I] and f." :2 0 simultaneously for 1:. y. (5) Suppose that. [H.b) is a critical [mint of f. Hllll 10!. D = in if" . Then 9;: .Im (:1) 11' D b (I am! f.,_.r{a,b) > (J. Llauu f(u. b) is a local minimum. (a) If D ) [l and fn.(u.b) < I]. then fin. h} is a local maximum. (:1) If D < I). then flab) is not a local maximum or millin'mm. ((5) Suppose z = f[.r. y}. .r 2 arm and y = y“). (1:: EU rim Bf dy __._+__ E‘mm mm' (7) Suppose z = f{J',y). ;r := 113.!) nml y = y(s.!]. If): _ ()2 i);- (5: By _. _ _ __ + _ _ {is fir (“b.- Uy as 62 _ r'): 3);? 02 fly a. - 77+Tym (8) Suppose F(J'.y. :} = f} and .. = {.r. 9). Then. r‘)‘: __ F; (Jr T F: (“)3 _ F” 0!; __ F: (9) Suppose I? = (Hub) is a unit vector. Duf[-£- y) = AER/(.1- +- "b.9111!” — f(.:.',;;) = Vflzlr. y} - u {1”} Solve Vfljgy] : A v g[.r,y} for .r. 3;. Compute f(.r,y} at. all solutions to find max and min. ...
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