Fall 2003 Prelim 2 Solutions

# Fall 2003 Prelim 2 Solutions - MATH 192 FALL 2003...

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Unformatted text preview: MATH 192, FALL 2003 ‘PRE-LIM 2, SOLUTIONS 10 points on each problem. 1,1?Q=2i+2j+4k L: x=2t, y=1+2t, z=4t. 2. a) Intersection point: (2,2,0), using s = —2 or t = 0 . b) Direction vectors : V1 : i+3j +4k , v2 = 2i+j. Normal vector: 11 2 VI xv2 = —4i+8j—5k. Plane: —4x+8y—52 =8. 3.21) 2(2+3t)+5(1+t)+2(4+2t)=2:>t=—l. Intersection point: (-l,0,2). b)Vectorﬁ‘om (2,1,4) to (-l,0,2)is v=3i+1j+2k. n=2i+5j+2k. Distance=lvﬂ- l6+5+4l -15 lnl _W_ﬁ' 4. LinealongPR: x=2+t, y=—3+t, z=ti LineaiongQR: x=5+4s, y=1+83, z=2. Coordinate of R: (4,4,2), using I = 2 . 1;}?de FR=2i+2j—2k, §R='—i—2j. Area: /2=|—4i+2j—2k|/2=JE. 5. Volume: u.(vxw)| = [(2i —j—k)-(—15i+5j—k)l = 34. 6- 8) r(t) = %(e” +1)i+(cost+t)j—~1-12—t‘k. b) a(t) = 2e2‘i — (cost)j—t2k. 7 _ 1' (5+ y+2)(J——Jm) ' Wanl—Tm J}- W sin£lj s I . lirn ysin[~l-) = 0. x (Ln—40.0) x 9. Domain: y— x2 —2 > 0 . Geometrically, domain is the inside of the parabola y = J:2 + 2 . =4. 8. Range: (0,00). Domain is open and unbounded. 10. ﬂ = e1” + erzx” +ysin (xy) . 62f = (1+2x)e2“" + sin (xy) + xycos(xy). ax ayax af 62f —- = erw +xsin (xy) 2 2 = xez‘” +2:2 cos(xy). ...
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