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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
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- Fall '06
- PANTANO
- Multivariable Calculus, normal vector