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Unformatted text preview: Math 2263 Name (Print)
SPRING 2005
FINAL EXAM Signature Section___._. I.D.# Recitation Instructor READ AND FOLLOW THESE INSTRUCTIONS This booklet contains 16 pages including this cover page. Check to see if any are miss ing. PRINT on the upper righthand corner all the requested information, and sign your ' 't‘al n the to of every page, in case the pages become separated.
" E" ' . Do your work in the blank spaces and on the backs of pages I i
This exam consists of 12 multiple—choice questions, counting 15 points each, for a total of 180 points, and a handgraded part with 8 problems, for a total of 220. You have 3 hours
to work on this exam. INSTRUCTIONS FOR MACHINEGRADED PART (Questions 1—12):
You MUST use a soft pencil (No. 1 or No 2) to answer this part. Do not fold or tear the answer sheet, and carefully enter all the requested information according to the instruc
tions you receive. DO NOT MAKE ANY STRAY MARKS ON THE ANSWER
SHEET. When you have decided on a correct answer to a given question, circle the an
swer in this booklet and blacken completely the corresponding circle in the answer sheet.
If you erase something, do so completely. Each question has a unique correct answer. If
you give two different answers, the question will be marked wrong. There is no penalty
for guessing, but if you don’t answer a question, be sure to skip the corresponding line in the answer sheet. INSTRUCTIONS FOR THE HANDGRADED PART (Questions 13—20):
SHOW ALL WORK. Unsupported answers will receive little credit. Notice regarding the machine graded sections of this exam: Either the student
or the School of Mathematics may for any reason request a regrade of the machine graded
part. All regrades will be based on responses in the test booklet, and not on the machine
graded response sheet. Any problem for which the answer is not indicated in the test
booklet, or which has no relevant accompanying calculations will be marked wrong on the
regrade. Therefore work and answers must be clearly shown on the test booklet. AFTER YOU FINISH BOTH PARTS OF THE EXAM: Place the answer sheet
between two pages of this booklet (make a sandwich), with the side marked “GENERAL PURPOSE ANSWER SHEET” facing DOWN. Have your ID card in your hand when
turning in your exam. Multiple choice part __ Hand—graded part Total Letter Grade MATH 2263 NAME
FINAL EXAMINATION Page 2 MULTIPLE CHOICE PROBLEMS
Circle the letter of the answer on this sheet and ﬁll in the corresponding box on the answer sheet in #1 or #2 pencil. 1. Let the curve r(t) be given by 4
r(t) = (2t2 + 2)i + (2% + 3) j + (2t + 2)k. Then the curvature at the point corresponding to t = 1 is MATH 2263 NAME______________________.____. FINAL EXAMINATION
Page 3 3. An equation of the tangent plane to the surface
mzy + yzz + x22 = 19 at the point (1, —2,3) is
(a) 53: — 11y + 102 = 57 (b) 52: — 11y + 102 = 19 (C) 5x + 113/ + 102 = 13 (d) 5m — 11y+ 102 = 0 (e) (2553; + 22)(x  1) + (x2 + 2y2)(y + 2) + (112 + 21,2)(2 — 3) = 0 4. Let f(:c, y) = m3 + 2x2 + 2y2 — 2mg + 49: — 8y. Then at the point (0,2), a f does not have a critical point b (
(
(c) f has a local maximum
(
( )
) f has a local minimum d) f has a critical point but the second derivative test fails e) f has a saddle point MATH 2263 NAME
FINAL EXAMINATION Page 4 1/2 20/4
5. Changing the order of integration in / d1: / x2y dy leads to 1:3
23/4 1/2 0
(a) /3 dy/ x2yd$
z 0 1/8 9/37
(b) /0 dy/4y $2yd$ 1/8 4y
(c) /0 dy/szydm
2 4y
(d) / dy/ x2ydm
0 y
2 4y
2
(e) /0dy/y my d3: 6. Let B be the solid in R3 bounded from below by the elliptic paraboloid z = :02 + 2312 and from above
by the parabolic cylinder 2 = 10 — 332. Then the integral /// f (33,11, 2) dV can be evaluated as the
D following iterated integral: m 10 «10—2:i d d d
(a) /_m/O Wﬂw) z y x b m «542/2 lo—z2
( ) /—\/lﬁ/—‘/5—m2/2 :I:2+2y2 x/g v5—w§ 10
(c) /_¢5/_m2 0 f(x,y,z)dzdydx d x/5 JET—Tc? M d d d
()/_ﬁ/_W/Wf($,y,1) z y a: “5 J5? 10—m2 d d d
(e) "/_‘/§/\/5:£E/$2+2y2 f(a:,y,z) 2 y :13 my,» dzdydx MATH 2263 NAME________________________.______.__. FINAL EXAMINATION
Page 5 7. The volume of the solid E consisting of those points (x,y,z) in R3 which lie inside the cylinder
2:2 + y2 = 4 and outside the cone 2:2 + y2 = 22, i.e. E = {(55, y, z) z2 S x2 + y2 S 4}, equals ( A a)
b)
C) 32
3
256%
3
167r
3 (
(d) 871'
(e) 47r 8. Let S be the portion of the surface 2 = 21/3: , which lies over the triangle in the x,yplane with
vertices at (0,0), (2,0) and (0,1). The area of S is then (a) /:/:2ﬁ§dydx 7» /:/:f «3733
/:/:“'” «377
(d) /:/7f7f?dydw
/:/;«3—7m MATH 2263 NAME
FINAL EXAMINATION Page 6 9. Let C be the closed curve consisting of the2 segment on the x—axis joining the points (—1,0) and (1,0),
and of the upper semicircle {(33, y) law:2 + y2 =,1 y_ > 0}, oriented counterclockwise. Then /(12x + 4y) dx + (7x — 2y) dy equals
C 0" 301—?
) _.3_7"
2 OJ ) 7”
d) 771'
6) AA?AA 10. Let the vector ﬁeld F(a:, y, z) = (6x + 33/22  yzz, 3mg —— 3y22, yz2 — 4207:) be deﬁned on all of R3.
Then curlF equals ( ( (c) (35132 + 3x312 2 — 173,122 , 533312—3132, §yz3 — 22:22)
( (3y2 +22, 3y2 — 2yz+4z, 3y—6yz+z2) ( ( MATH 2263 NAME
FINAL EXAMINATION Page 7 11. The vector ﬁeld (wry + ﬂyz + 3x2, 6332 + 2533;  yz) involving parameters oz and ﬂ, is conservative when
a) 01:2 and ﬂz—l (
(b) a=6 and ﬁz—l
(c) a=6 and 3:1 (d) 01212 and ﬁz—l
(e) a=12 and ﬂzl 12. Let S be the portion of the surface 2 = my which lies above the square, in the x,y—plane, with
vertices (0,0, 0,), (1,0,0,), (0,1,0,) and (1,1,0). Let the normal to S be chosen to point in the general downward direction. Let F(x, y, z) = (y, —x,a:y). Then //curlF  dS equals (c) 2t times the area of S (d) (—2 )times the area of S (e) /;/; (2xy+2) dydz
END OF MULTIPLE CHOICE PROBLEMS MATH 2263 NAME FINAL EXAMINATION
Page 8 THE REMAINING PROBLEMS ARE HANDGRADED. SHOW YOUR WORK. 13. (25 points). Find the length of the curve MATH 2263 NAME FINAL EXAMINATION
Page 9 14. {25 points). If z=ln(x+\/a:2+y2), Where xzrcost9 and y=rsin0, 82 82 ﬁnd the ﬁrst partlal derlvatlves 5—7“ and E. MATH 2263 NAME/I FINAL EXAMINATION
Page 10 15. (30 points ) Using Lagrange multipliers, ﬁnd the maximal value of the function f(a:, y, z) = my2z3 subject to the constraint g(:c, y, z) = a: + y + z = 12, in the domain D = {at > 0,y > 0,25 > 0}. MATH 2263 NAME
FINAL EXAMINATION
Page 11 16. (25 points). Evaluate the double integral _————1 2 2
I/J/ngdA, where D={<z,y>lx +y $1}. MATH 2263 NAME FINAL EXAMINATION
Page 12 17. (30 points). Find the volume of the solid bounded by the plane 2 = 2m and the paraboloid z = x2+y2. MATH 2263 NAME
FINAL EXAMINATION Page 13 18. (30 points). Change the variables a: = u + v, y = u — v, to evaluate the double integral // eﬁdA,
D where D is the region in the ﬁrst quadrant {(x, y) I a: 2 0, y 2 0}, between the lines at + y = 2 and
$+y=4 MATH 2263 NAME FINAL EXAMINATION
Page 14 19. (25 points). Evaluate the surface integral //zzdS, Where S={(m,y,z)x2+y2=z2, 0323 1}.
s MATH 2263 NAME FINAL EXAMINATION Page 15
paraboloid z = 1 — $2 — y2 which lies above the 20. (30 points). Let S be the portion of the elliptic
general upward direction, i.e. to the outside of the plane 2: = 0. Let the normal to S be chosen in the
paraboloid. Evaluate the surface integral f/F  dS, where F = ($3,313, 1 +23).
5 FIIVHI: r‘./\l—\l\/Ill\ll—\IHH\J Formulas for Final Exam, Math 2263, Spring 2005
Length of the curve r(t) : a S t g b is : / [r'(t) dt, its curvature at the point r(t) is lr'(t)l3 Tangent plane to the graph of z = f(x, y) at the point (do, go, 20) is given by the equation z—zoz fx(a‘0,y0) (x—mo)+fy(93o,y0) ' (VJ—yo) Tangent plane to the surface F(:I:7 y, z) = C' : const at the point (do, go, 20) is given by the equation
VF(1‘0)' (1‘ ”‘ 1‘0) Z 0; i.e. Fz(1‘0) ' (5v — $0) + Fy(r0) ' (3/ — 310) + Fz(1‘0) ‘ (Z _ 20) = 0, Where r = (x,y,z), r0 = (960790720)
Chain rule: if z = f(ac, y), and at = :L'(U, v), y = y(u, v), then zuzzz'$u+zy'yua szzx'wv+zy'yv Second derivatives test. Let fx(a,b) : fy(a, b) = 0, and D :2 (fmfyy — 3y)(a, b). (a) If D > 0
and fm(a,b) > 07 then f(a,b) is a local minimum; (b) If D > 0 and fm(a, b) < 0, then f(a, b) is a
local maximum; (c) If D < 0, then f(a, b) is a not a local maximum or minimum. Method of Lagrange multipliers. The maximum and minimum values of f(x, 3/, 2) subject to the constraint g(a:, y) = k satisfy Vf(:r, a; Z) = W907, 21,2), 906,11) = k Double integral in polar coordinates. if( (r, 0) )E R <=> (:6, y): (r 0050, rsin 0) E D, then //f( (z y )dxdyz/ f( (,rcosH rsind) rdrdd A similar formula holds true in cylindrical coordinates.
Change of variables 1n double integrals: if( (u ,v) E R <2> (a: y): (a: (u, v), y(u,v)) E D, then /Df(x,y)dxdy = //f(x(: My v))J(u,v)ldudv, mu Km 21!: ll Where .J(u, v) det — —$uyv — xvyu. VolumeofasolidE:{(a:, y,z() my) )6D f1(ZL" y)£z Sf2($ 3/)13 V=/ [(f2 a: v) —f1(rr v)ld$dy Surface area of the graph of function 5' : 2 = 2(x, y), (at, y) E D is =//,/1+2§+2§d$dy.
D Surface mtegral of f( :2: ,2y, 2) over S: 2(51: 7,y) (my) 6 D is //f(,2)xy, )dS=/:( f(,2:z:y, 2(a:,y) )1/1+22+22dxdy. Green’s Theorem. If C is the boundary of D with positive (counterclockwise) orientation, then /Pda: + Qdy = é/(Qz — Pg) dxdy. C Stokes’ Theorem. If If C is the boundary of S with positive orientation, then /Fdr : // carlFdS,
s C
i j k
where Fdr = Pda: + Qdy + Rd2, curlF = V X F 2 8/855 8/8y 8/82
P Q R Divergence Theorem. If S is the boundary of E with positive (outward) orientation, then //FdS=///dz'deV, where diszVF=P$+Qy+RZ.
S E The vector ﬁeld F :(P, Q, R) on R3 is conservative, i.e. F : Vf for some function f(:c, y, 2) if
and only if curl F 2 0. ...
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