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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 nongraphic 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
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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.
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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: 323'12 mauve. 1, 07
3:3 7') ‘
2 :2 {43.21 =<—2,o ,>
04_:<:</2V_20<Z<1 1 k
l
1
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136“”): Math 32B Midterm II  Page 4 of 6 02/26/16 3. (a) (6 points) Find constant a and b such that the twodimensional 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>
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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 (IItransformation 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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 Spring '08
 Rogawski
 Math, Multivariable Calculus, coherent, Manifold, Math 32B Midterm

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