Week 10 - Solution

# 3 parameter equations of straight lines 1 a b x t

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Unformatted text preview: ) = y 2 + xz 1 =⇒gy (y, z ) = y 2 =⇒ g (y, z ) = y 3 + h (z ) 3 13 =⇒f (x, y, z ) = xyz + y + h (z ) 3 fz (x, y, z ) = xy + h′ (z ) =⇒ h′ (z ) = 0 =⇒ h (z ) = C Therefore, 1 f (x, y, z ) = xyz + y 3 + C . 3 Parameter Equations of Straight Lines 1. (a) (b) x (t) = (6 − 2) t + 2 = 2 + 4t y (t) = (0 − 1) t + 1 = 1 − t , where 0 ≤ t ≤ 1 z (t) = (3 − 8) t + 8 = 8 − 5t x (t) = −1 + 5t, y (t) = −3t, z (t) = 5 − 2t, where 0 ≤ t ≤ 1 (c) x (t) = 3, y (t) = 1 + t, z (t) = −1 − 5t, where 0 ≤ t ≤ 1 (d) x (t) = 2 − 6t, y (t) = −7 + 9t, z (t) = 5, where 0 ≤ t ≤ 1 Line Integral 2 1. (a) C = {x (t) = −1 + 3t, y (t) = 1 + 2t, 0 ≤ t ≤ 1} 1 xy ds = 0 C 2 dx dt dy dx = 3, = 2, ds = dt dt + 2 dy dt √ √ (−1 + 3t) (1 + 2t) 13 dt = 13 √ dt = 1 13 dt √ 3 13 6t + t − 1 dt = 2 2 0 (b) C = {x (t) = 1 − t, y (t) = 3t, z (t) = 1 + 5t, 0 ≤ t ≤ 1} dx dt dy dz dx = −1, = 3, = 5, ds = dt dt dt 1 xy 2 z ds = 0 C 2 + 2 dy dt + √ (1 − t) (3t)2 (1 + 5t) dz dt 2 dt = √ 35 dt √ 35 dt = 3 35 3 (c) C = {x (t) = t , y (t) = t, 0 ≤ t ≤ 1} 1 x ds = 0 C √ dx dy = 3t2 , = 1, ds = 9t4 + 1 dt dt dt √ 31 2 9t4 + 1 2 9t4 + 1 dt = t3 3 36 Note: use substitution, let u = 9t4 + 1. (d) C = x (t) = 4 cos t, y (t) = 4 sin t, − π ≤ t ≤ 2 1 0 3 10 2 − 1 = 54 π 2 dx dy = −4 sin t, = 4 cos t, ds = 4 dt dt dt π 2 4 4 6 (4 cos t) (4 sin t) (4 dt) = 4 xy ds = −π 2 C 15 sin t 5 π 2 = −π 2 8192 5 Note: use substitution, let u = sin t. (e) C = {x (t) = t, y (t) = t2 , − 2 ≤ t ≤ 1} dy = 2t dt 1 C x − 2y 2 dy = −2 (t) − 2 t2 2 (2t dt) = 48 (f) Let Γ1 be the line segment from (0, 0) to (2, 0) and Γ2 be the segment from (2, 0) to (3, 2), then Γ1 = {x (t) = 2t, y (t) = 0, 0 ≤ t ≤ 1}, Γ2 = {x (t) = 2 + t, y (t) = 2t, 0 ≤ t ≤ 1} and C = Γ1 + Γ2. For Γ1 , dx = 2 dt and dy = 0 1 Γ1 xy dx + (x − y ) dy = [(2t) (0) (2 dt) + (0)] = 0. 0 For Γ...
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## This document was uploaded on 01/23/2014.

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