Week 10 - Solution

A t3 f r t r t c f dr 2 t4 t3 d3d4 t t dt dt

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Unformatted text preview: 1 , dx = dt and dy = 2 dt 1 Γ2 xy dx + (x − y ) dy = 0 {(2 + t) (2t) (dt) + [(2 + t) − (2t)] (2 dt)} = 17 . 3 Therefore, C xy dx + (x − y ) dy = Γ1 xy dx + (x − y ) dy + 3 Γ2 xy dx + (x − y ) dy = 17 . 3 (g) C = x (t) = √ t, y (t) = t, z (t) = t2 , 0 ≤ t ≤ 1 dy = dt and dz = 2t dt √ 1 (t) t2 (dt) + yz dy + xy dz = t (t) (2t dt) = 0 C 23 28 2. (a) t3 F (r (t)) = r′ (t) = C F • dr = 2 t4 , − t3 d3d4 t, t dt dt = t10 , −t7 t4 = 3t2 , 4t3 F (r (t)) • r′ (t) dt C 1 1 t10 , −t7 • 3t2 , 4t3 dt = = 0 3t12 − 4t10 dt = − 0 19 143 (b) 2 F (r (t)) = et , t5 1 C F • dr = Note: for nt and r′ (t) = 2t, 3t2 1 2 et , t5 • 2t, 3t2 dt = 0 n−1 tn 2 2tet + 3t7 dt = e − 0 5 8 n e , use substitution, let u = t . (c) F (r (t)) = 2t, t2 , 3t and r′ (t) = [2, 3, −2t] 1 1 C F • dr = −1 2t, t2 , 3t • [2, 3, −2t] dt = −1 4t − 3t2 dt = −2 (d) F (r (t)) = t2 + t4 , −t2 , −4t4 and r′ (t) = 1, 2t, 4t3 1 1 C F • dr = 0 t2 + t4 , −t2 , −4t4 • 1, 2t, 4t3 dt = 0 t2 − 2t3 + t4 − 16t7 dt 59 =− 30 3. The mass of the wire with density function ρ (x, y ) is m= ρ (x, y ) ds. C The center of mass of the wire is (¯, y ), where x¯ x= ¯ My = Mx My and y = ¯ with m m xρ (x, y ) ds and Mx = C yρ (x, y ) ds. C 4 Since the density is a constant k , then ρ (x, y ) = k . C = x (t) = 2 cos t, y (t) = 2 sin t, − π 2 π 2 ≤t≤ π 2 m= ρ (x, y ) ds = (k ) (2 dt) = 2kπ −π 2 C π 2 My = (2 cos t) (k ) (2 dt) = 8k xρ (x, y ) ds = −π 2 C π 2 Mx = yρ (x, y ) ds = (2 sin t) (k ) (2 dt) = 0 −π 2 C 8k ,0 2kπ 2kπ Therefore, the mass of the wire is 2kπ and the center of mass is m= ρ (x, y ) ds = r (cos t + sin t) (r dt) = 2r2 π 2 xρ (x, y ) ds = r3 (π + 2) 4 r3 (π + 2) (r sin t) [r (cos t + sin t)] (r dt) = 4 (r cos t) [r (cos t + sin t)] (r dt) = 0 C π 2 Mx = yρ (x, y ) ds = 0 C r (π +2) r (π +2) ,8 8 Therefore, the...
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This document was uploaded on 01/23/2014.

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