Fall 2001 Calc 3- Exam 1 Solutions

Fall 2001 Calc 3- Exam 1 Solutions - MA 261 EXAM I Fall...

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 4
Background image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: MA 261 EXAM I Fall 2001 Name: ______ Page 2/6 1. Determine a so that the line is parallel to the plane 2:10 + 3y — 52 = 14. DiYéClltWA VQL'HV 075(4) Z14> A. a:—4 Manama fizzle/35mm fi:(zl3/ -5> B_a:3 T146 art/*8 aufll /.0€&M€ fluff Fa/aflzez a:::4 whem w-T’l-‘2a+6-ZO =@. F11 g H \‘l 2. The line through (3, 2, 1) and (5, 1, 2) intersects the plane x + y + z = 14 at the point Diredmm Ve07L0V: J = (5,1,2) "43,2.0 __ (2’4) 1) B. (8,4,2) C. 15, ,— Parametfl“ 67‘4‘371‘0‘6’ D. ho 0—1 %:3+Ztl ySZ’t, Z;/+i E. (9,—3,8) III-ferfiec‘tlc‘an mef ,’ x+a+z=[3+2t)+/Z'f)+{/+f)‘/‘/z {:9 x://) ?:-Z / 2:; MA 261 EXAM I Fall 2001 Name: —_______ Page 3/6 3. The vector—valued function F(t) 2 5+ (t cos 03+ (t sin 01;, 0 g t < 00 describes a : l S 5: . x / g f {2 a 2‘51“? A. circle in a horizontal plane The C ‘4 ryé a? S {a e B. circle in a vertical plane C. spiral in a horizontal plane 1“ Ea X‘/- W “rug/fl “8 [[4 F0 gar C9 ordj E. ellipse in a vertical plane 4. In spherical coordinates the two equations p = 2, ¢ = 7r / 6 describe I“ Cfiugrl‘Cbg warp/5 A. acone Z :/0 3 B @ V‘ r 10 9;k% : I C. aplane D. a sphere 711i; (3 a unfit: Clifié I.“ E. acylinder Lort'zoanZ [£4995 5 ‘E ' ‘ aw i WSeofr‘om W, 1““ ‘5 “ u mg (“he S/ahere /0=Z 01.4491 Me. ’0an haKJQ’C/o”£ ff: 7/6 MA 261 EXAM I Fall 2001 Name: *h— Page 4/6 5. The curves F1 (t) = (t, t2, 1) and F2(t) = (sin t, sin 2t, 1) intersect at (0,0,1) at an angle ‘0, Where cos (9 = .Z-M'IZCVSec7’1‘0n flatly?! fl/f W: <1, 240» E’zo):<w> B. % 72W) :(oost/zmszz; 0)) 570/442,» 9/ a / ' 5 5059: Klara/[0) :: —')/;“ 3% lE/M/ {pi/0)] 5 6. The level surfaces of the function f (ac, y, z) = a; — 3/2 — z2 are me give! 5‘4"" 72“ 623 : A. ellipsoids 7", Z7" 37 B cones C. cylinders -~ L (X, k) — y + 2 TLI‘C (C 6'V‘CMZQI") flaI/fibfl E. hyperbolic paraboloids MA 261 EXAM I Fall 2001 Name: —___—.__ Page 5/6 9? B —2fi C. 0 9273 '37 9/— 94 D- W “ " f “ E. m 9175 (57/0): e” smhr) + flea-Jiu/fifizzieotqfld 9X3? :_ 277 l . 9 foo, fine com sv‘ourf W‘”‘ (5;; 2 8. The area of the triangle With vertices (1,0,1), (1,1,0) and (0,1,1) is [61‘ P: {420/ MW), I9: 0, w M 52¢ 40, 4—1) / PI? : NW C.\/§ //-7—ZQXPI€/ D1 E.\/3 MA 261 EXAM I Fall 2001 Name: _____—__ Page 6/6 9. A particle has acceleration 5(t) = 6tj+ 21;. The initial position is 77(0) 2 and the ' initial velocity is 17(0) = i — j. The distance from the position of the particle at time t = 1 to the point (2, 2,3) is J=53Jf+a=<®3f§2t> +64 E" _. firm): (1)4) 0> B. f7 - 1 F’SJJi ‘I'EIZ=(.t/1f3”f/f >+CZ D4 ’ E2 5;: f/v):<0/ I) fix): (I, I, /> [(2,2,3>- {MD}: /(l) I, 9/43“ 10. The curvature of the curve defined by the intersection of the cylinder x2 +3;2 2 1 with the plane y + z = 2 at (0,1,1) is (you may use the formula H = Pa/‘QMQILVTC e%¢%l‘ous ; ‘ f z: '5 : 5' 'M f 2 :Z‘ 5‘“ X 603 / a ( / ‘ , ~ ' Vector 27w;me . F‘Zwa/ Sm I D g FIJZ’Sc‘Mf) (’0st -CDS{> E_2 f" .. 4‘00515/ “Sit/1'6) 5“” 7" > / .— s )7 -7 I I corms/001MB 1‘0 t Z V ' <0) ) > m, ‘-" .r 29/ 7' -' I I ) a! a” :- SV 5 (37/) -F”(Iz:;)r0, /F? ’ffi/a)]‘/r/%)IIV igll E {MC ; scmzr’zfl/z)1:t, 96 r 13/,3 :4; ...
View Full Document

Page1 / 5

Fall 2001 Calc 3- Exam 1 Solutions - MA 261 EXAM I Fall...

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online