exam_1_version_b_solutions

exam_1_version_b_solutions - Ma 231 Exam I Version B...

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Unformatted text preview: Ma 231 Exam I Version B Solutions (l) (20 points) (a) Let a=(—2,—1,—2) and b=(—2,—6,3). (i) —2a+ 3b = Answer: —2a + 3b = —2(—2, —1,—2) + 3 (—2,—6,3) = (—2,-16,13) (ii) What is the unit vector in the same direction as a? a (—2,-1,-2)=<2 1 2> Answer: — - |a|_ J4+1+4 (iii) What is the scalar projection of b on a, comp_(b)? a-b_4+6—6_i Answer: comp_ (b) = W - 3 3 (iv) What is the vector projection of b on a, proj_(b)? - 4 Answer: pm], (b) = (comp, (b))ﬁ = —3, —%> (b) Draw the vector projection of v on w : Solution: w (2) (20 points) (a) For the line through the points (2,—1,3) and (—2, 2, 5), ﬁnd the vector, parametric and symmetric equations. Solution: Direction: D = (—2,2,5) — (2,—1,3)=(-4,3,2) Point: (2,-1,3) Vector parametric: r(t) = (2,4,3) +t(-4, 3,2) = (2 — 4t,—l +3t,3 + 2t) Parametric: x=2——4t, y=-1+3t, z =3+21 x-2 _y+l __ 2—3 —-4 3 2 Symmetric: (b) Check that the lines intersect, ﬁnd the point of intersection and the cosine of the angle between the lines: x+l v—7 2—5 L" 7— —1 _ 3 La: x+8=ﬂ=z—2' ‘ 1 5 —1 Solution: Parametric Equations: L,: x=2t-—l, y=—t+7,z=3t+5; L2: x=s-8,y=Ss-6, z=-s+2 2t—1=s—8 Solve: —t+7=5s—6 3t+5=-—s+2 Adding twice second to ﬁrst, 13 =11s —20, s = 3, t = -2; Check third: 3(—2) + 5 = -1, — (3) + 2 = —1 Point of intersection: (—5, 9, -—1) Directions: I)l =(2,—1,3), D2 = (1, 5,—1), Cosine of angle of intersection: cos(9)= Dl-D2 = 2-5-3 z 6 =_2_ ID.IID2I Jl—4J2_7 Til/2‘7 m (3) (20 points) Let R :(1,2,2), P2 :(3,—l,1), I; :(4, 3, 4) be three points in space. (:1) Find the equation of the plane through R , P2 and 13,. Solution: Let P1 :(1, 2, 2), P2 :(3,——1,1), P3 :(4, 3,4) be three points in space. E172=(2,—3,—1) and EE=(3,1,2) i j k N=ﬁxﬁ=2 —3 —1 3 l 2 ,—3 -—l .2 —l 2 -3 , , =1 —] +k =—51-7]+llk l 2 3 2 3 l Check: (2, —3,—1) - {-5, —7, 1 1)=o, (3,1,2)-(-5,—7,11)= 0 Equation of plane: —5(x—l)—7(y-—2)+11(z—2)= O or-Sx—7y+llz= 3 (b) What is the area of the parallelogram with sides Ple and PIP3 ? Solution: Amalia—mm =|—5i—7j+11k|=J25+49+12 =J195 (4) (15 points) For the quadratic surface y = 4x2 + z2 : (a) Sketch the x = 0 and z = 0 slices: 1:0 Z 2-!) y J’ I (b) Sketch the y = constant slices: 1-}: 2J5 (c) Sketch the surface : «*0 (5) (10 points) (it) Give a vector parametric representation for the curve formed by the intersection of the surfacexzy2 —l andtheplane 2x+y+z =3. Solution: y=t, x=t2 ~1, x=3——2x—y=4—212 —t r(r)=<t2 —1,t,4—2t2 —t) (b) Write the arclength of the curve x = 2cos(31), y = sin(3t), 0 s t S III 2 as an integral. (Do M evaluate the integral). Solution: m) = (2 cos(3t), sin(3t)), r'(t) = (—6sin(3t), 3cos(3t)) Arclength= I‘lr'(t)|dt= JI-J36sinz(3t)+9cos2(3t)dt 0 0 (6) (15 points) (a) An object in space has a velocity of v(t) = -2 cos(t)i + 2sin(t) j + 312k and the initial position r(O) = -2i + 3j + 5k, ﬁnd the position, r(t), at time t. Solution: r(t) = [ v(t) dt = (—25in(t) + C, )i + (—2 cos(t) + CZ) j + (t3 + Ca»; —2i+ 3j+ 5k = r(O) = Cli+ (—2+Cz)j+ Cak :9 C1: —2,C2 = 5,C3 = 5 r(t) = (—2sin(t) — 2)i+ (—2 cos(t) + 5)j + (t3 + 5)k (b) Find the tangential and normal components of the acceleration of the object in (a). Solution: v(t) = —2 cos(t)i + Zsin(t) j + 3:21: , v(t) = |v(t)l = J4 + 9:4 a(t) = 2sin(t)i + 2cos(t)j + 6th 1 a, = v'(t) =;(4+ s»‘)‘1 2 (361‘3) = W ’4 2 2 2 324t6 16 +180!4 “Hal "’1 = (“36‘ "4m 47:57?— 18:3 ...
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exam_1_version_b_solutions - Ma 231 Exam I Version B...

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