Solution for MIdterm II 32b - Discussion Session A B C D...

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Unformatted text preview: Discussion Session: A B C D Math 32B Name(Print): Winter 2016 Midterm II UID 02/26/16 Time Limit: 50 mins(12:00—12:50pm) Signature This exam contains 6 page: (including this cover page) and 5 problems. Check to see if any pages are missing. Enter all requested information on the top of this page, and put your initials on the top of every page, in case the pages become separated. You may only use an non-graphic calculator on this exam. You are required to show your work on each problem on this exam. The following rules apply: 0 Organize your work, in a reasonably neat and coherent way, in the space provided. Work scat- tered all over the page without a clear ordering will receive very little credit. o Mysterious or unsupported answers will not receive full credit. A correct answer, unsup- ported by calculations, explanation, or algebraic work will receive no credit; an incorrect answer supported by substantially correct calculations and explanations might still receive partial credit. o If you need more space, use the back of the pages; clearly indicate when you have done this. Do not write in the table to the right. Math 32B Midterm II - Page 2 of 6 02/26/16 1. (a) (8 points) Let C is the curve of y = 1:2 from (0,0) to (3, 9), compute f0 3wds. 0.4% 3 L/j W x 9* w H L__7 0;) $11 6» g- ‘1 4. Q 3 ‘1 E: 0 ; 2g [Hay/Z (.1 : #‘372/2’3F ‘ (b) (8 points) Let C is the upper half of the circle of adius 2, with centre at (0, 0), compute fcyds. ' \ (6‘ 7— ~2$lu0 X : 2029 > i ‘ OS 9 5K 3 1 f/zzsxua 7“” “12039 Math 32B Midterm II — Page 3 of 6 02/26/16 2. (20 points) Calculate the integral of f (z, y, z) = a” over the portion of the plane z+2y+22 = 3, _ i — 7 _ x: 3-23'12 mauve. 1, 07 3:3 7') ‘ 2 :2 {-43.21 =<—2,o ,> 04_:<:</2V_20<Z<1 1 k l 1 —l< L, 27 0 D épk _ % 9 A _ XD\€)\° lf 3 (,1 1 _\ 1— ‘ir :sfiaé e-M-Ma -36 K-Bh 5(1 0) : 563% I-E-é-l' if .3. lg 136“”): Math 32B Midterm II - Page 4 of 6 02/26/16 3. (a) (6 points) Find constant a and b such that the two-dimensional vector field F(:L', y) =< 3x23; sin y — 2:1:2 cos y, azsy cosy + bz3 sin y > is a gradient of a function f (z,y). FM 3) U a Gamdlcui’ 9“: «Fuwi'im ff {10(6) {"Aev‘ '3'???" 93%— 3x231u3+3x156vsa+2x251ug : 955th +9335“? 36 = 50x23 675% i 3LX2$1M5L 5‘ \fm 2) 02'; b" 2 (b) (14 points) Suppose that a and b are as in part (a), compute the line integral / F- Tds, C whereC: r(t) =,<tt4>, 0<t<1. We fwd Cm EAR film VOJFBWHQQ {WWPM ‘Ei - 3X 3&m3 lY £0351» =7 {a a): Xz‘asmg 9*-Xm3. ? "la: Yga C033 '1';— 3;)(JwaL (,4) +8(3) (+7 => fauna) : ng‘wg + x’ama’réxgs‘wa +323) ;§xgs‘w3 ”v Xga m3 *3”) Btfiowpavisrm a“? be) 8231‘“) z) 8%; :KHR. PuCil 14:0 2) +th 3) X599 REX 053‘ fflo M :40 07 NH :4! 9° jg; 115qu n ~{to/0) 23mm -§—casm. Math 32B Midterm II - Page 5 of 6 02/26/16 4. (24 points) Let S be the part of the surface 2 = my where a:2+y2 S 1, with upward orientation. Compute the surface integral of the vector fields F(x, y, z) =< y, z, z >. X’- Yéosg a 2 wa9 2 1 x3 2 Y213M6a39 051’s] 05 5£2TL .— TV: (use, 5M , 2rsluaaase> Ye : <‘ rs‘mfi I sz9p Y2m529> i v 5 h erre ; C039 SW9 ngiupaze -Ys‘w9 Vane #50329 ‘ 1 (425L419 , ‘ngose, Y) uipwmj Med,“ ' (45MB, 42639. Vizwya’mé Math 32B Midterm II - Page 6 of 6 02/26/16 5. (20 points) (Multiple choice questions). Circle the right answer for each following statement; no justification is required for your answers. For each answer you will receive: 5 points if your answer is correct; 0 points if it is wrong or if you provide no answer. (1). Circle ONE statement that is TRUE (3) The Jacobian of a (II-transformation T : :1: = z(u,'u), y = y(u,'v) is always positive; _ (b) If f z D —> R is a scalar function and C is a smooth curve in D, then f0 fds = — LC fds; (c If F is a vector field in D and C is a smooth curve in D, then fCF - Tds = LCF - Tds. @- the above statements (a), (b) and (c) are incorrect. (2) Circle ONE statement that is TRUE. (a) Given a vector field F: < P, Q > on a domain D, if you find a simple closed curve C in D such that fC F - Tds = 0, then you may conclude that the vector field is conservative in D. (b) Any vector field F = < P,Q >, defined on D, satisfying 13% = 9% in D is conservative in D c Consider F x,y =< zlny — y, d > in R2. For any positively oriented simple closed curve 211 C C F - Tds = 0. d) or a conservative vector field F in R2 and a positively oriented simple closed curve C, we ve that fCF - Tds = 0. (3) Circle ONE statement that is FALSE. If a vector field, there are infinitely many potential functions corresponding to it. (b) If F(x, y)= < :12, —y > and C is a positively oriented circle of radius 1, then f0 F - Tds < 0. c If f (z) and g(y) are differentiable functions in IR, then F(x, y)= < f (1:), g(y) > is conser- vative in R2. (d) Consider the change of variables _ l u—-—§ln% v=,/a:y then the Jacobian a if” = 21). (4) Math the planar vector fields (a) — (d) wi the plots (A) — (D). (a) F(x,y)=< 2,2: > Plot: (b) F(x, y)= < y, cosa: > Plot: l; (c) F(x, y)=< 21: + 2,1; > Plot: (: (d) F(x,y) = < a: + y,:c — y > Plot: A , ...
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