23s08-Exam1-OLD_1

23s08-Exam1-OLD_1 - 2 k , 0 t 1. Your numerical answer...

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OLD MATHEMATICS 23 EXAMINATION 1 Reminder: The actual examination may be very different. (1) Let u = i - j +2 k and let v = 2 i - j + k . Are the vectors u and v orthogonal, parallel, or neither? Explain carefully. (2) Find an equation of the plane P containing the points (3 , - 1 , 2), (8 , 2 , 4) and ( - 1 , - 2 , - 3). (3) Let P be the plane in the above problem. Find a parametric equation for the line through ( - 2 , 2 , 4) and perpendicular to the plane P . (4) Carefully sketch the surface 4 y = x 2 + z 2 . (5) Describe the surface S given by the spherical coordinate equation ρ sin ϕ = 2 in terms of an equation involving rectangular coordinates x, y, z . Identify the surface S . (6) let r ( t ) = h 2 sin t, 5 t, cos t i . Find the curvature of the curve r ( t ) at the point (0 , 0 , 1). Simplify your answer. Note . The point was incorrectly listed as (0 , 0 , 2) in the first version. (7) Consider the helix r ( t ) = h 2 cos t, sin t, t i where t ( -∞ , ). Fine a parametric equation of the tangent line to the helix at the point ( 0 , 1 , π 2 ) . (8) Find the length of the space curve f ( t ) = 12 t i + 8 t 3 / 2 j + 3 t
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Unformatted text preview: 2 k , 0 t 1. Your numerical answer should be in simplest possible form. (9) Find the space curve r ( t ), t [ a, b ], which represents the curve of intersection of the cylinder x 2 + y 2 = 4 and the plane y + z = 3. You will get some partial credit for correctly sketching the surfaces. You need to nd both r ( t ) and the interval [ a, b ]. Answers on page 2. 1 Answers (not solutions) (1) u is neither orthogonal nor parallel to v . (2)-13 x + 17 y + 7 z + 42 = 0. (3) x =-2-13 t y = 2 + 17 t z = 4 + 7 t (4) (5) S is given by x 2 + y 2 = 4; a (right circular) cylinder. (6) = 1 / 29 (the point was incorrectly listed as (0 , , 2) in the rst version). (7) x =-2 t y = 1 z = 2 + t (8) L = 15 (9) x = 2 cos t y = 2 sin t z = 3-2 sin t for 0 t 2 . 2...
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23s08-Exam1-OLD_1 - 2 k , 0 t 1. Your numerical answer...

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