spprelim1 - “l{a The plane should contain the points(1.0...

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Unformatted text preview: “l. {a} The plane should contain the points (1.0? i) and (O.—~l.0} and be parallel to the normal vector of a: —— y + z = {3, ie. imj ~%— k. So; a normal vector is —21 ~%— 2k, the cross product of i - j + k and i + j + k, the vector between the two. given points. So, an equation for the plane is ~52: + 2 = O. The given lines are parallel to the vectors '2j ~— k and 251+ 3k) Whose cross gorodoct is 6i - ‘2j -~ 4k, a vector perpendicular to both lines. {o} The plane must contain the points {1, —2, and (270., 2) and be parallel to the vectors ——i + 33} ~ 2k and Si — ‘93 + 6k. Notice that these vectors are parallel. As in part (a): we take the cross product of —-i—%—33——2k and i+2j+2k, the vector between the points. to get the normal vector 11 = lOi —» 5k. Then an equation for the plane is 10x - 52 r: 10. 2. {a} like compute fit) 2 ~etj+v€cos{\/§t)k and 842‘) : «elj—f’asinh/gtlk. So, at t 2 O, the velocity is Vile} :— ~j + x/‘gk? the speed is x/l + 2: 2, and the acceleration is 21(0) :: ~3. (is) The angle 9 between a and V satisfies cos 69 2 Egg: : so {9 2 1. , 3 {c} A normal vector to this plane is given lay the cross product of viii} and a{t}§ which is (3et sin{\,/§r’} + fie: cosh/gt» i. So, an equation for the plane is x z 8. . . . . i . . ,. ‘ f“ - {)er. . U3 3. The. eneral antiderivative of Eegiiwsinr mt‘ik is e““1+cost ~%—k+C, Where s i I a. C any constant vector The initial condition gives il-j + C 2: -~i +j *— 4k. . gr 7 x i a I! g r \ So, C = Misti—w ek and : {er: w 231 + cost} w +4} k. 4. {a} Points in the. don sin must satisfy +y2~9 > 0. so D 2 such that $24,— > 9} which is all points; outside the circle of radius 3 about the origin. Conversely, on the liee 3;; z: 3 and .r > 0, we can achieve any positive real value for 332 + y? w filliesP the range is all real numbers greater than zero, 9, {la} The domain is open since any point is a positive distance from the circle of radius and so is the center of a disc contained in the domain. lt is not closed since it does not eontai'o points on the circle boundary. it is unbounded since it contains points outside any large circle. (c) The level curve x is ———2-\/~:L1—~2_——.—E : ~31 and so 3:2 + y? —— 9 :2 16, ‘ x 3; —J which is the circle of radius 5 about the origin. The other level curve satisfies $2 + y? — 9 2 7. which is the circle of radius 4 about the origin. Ac ‘ Q W1: v n\ Av \ . w» a A gnaw :x M» 3&3 MEG 9% fist “Egg 835$: QVRS firm 5.3K Yfififivwayng‘ ~ ~ w . Q 3 x V Em: All“ ‘ x r37 £3 3% A$gmflfi 9» ,N a x ES 2% N ; fimm v immixl q. s a QQCL» CAKEM m3; ...
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