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Mathematic Methods HW Solutions 40

# Mathematic Methods HW Solutions 40 - Chapter 8 1.4 1.5 1.6...

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Chapter 8 1.4 x = k - 1 gt + k - 2 g ( e - kt - 1) 1.5 x = - - 2 sin ωt + v 0 t + x 0 1.6 (a) 15 months (b) t = 30(1 - 2 - 1 / 3 ) = 6 . 19 months 1.7 x = ( c/F )[( m 2 c 2 + F 2 t 2 ) 1 / 2 - mc ] 2.1 y = mx , m = 3 / 2 2.2 (1 - x 2 ) 1 / 2 + (1 - y 2 ) 1 / 2 = C , C = 3 2.3 ln y = A (csc x - cot x ), A = 3 2.4 x 2 (1 + y 2 ) = K , K = 25 2.5 y = axe x , a = 1 /e 2.6 2 y 2 + 1 = A ( x 2 - 1) 2 , A = 1 2.7 y 2 = 8 + e K - x 2 , K = 1 2.8 y ( x 2 + C ) = 1, C = - 3 2.9 ye y = ae x , a = 1 2.10 y + 1 = ke x 2 / 2 , k = 2 2.11 ( y - 2) 2 = ( x + C ) 3 , C = 0 2.12 xye y = K , K = e 2.13 y 1, y ≡ - 1, x 1, x ≡ - 1 2.14 y 0 2.15 y 2 2.16 4 y = ( x + C ) 2 , C = 0 2.17 x = ( t - t 0 ) 2 / 4 2.19 (a) I/I 0 = e - 0 . 5 = 0 . 6 for s = 50ft Half value thickness = (ln 2) = 69 . 3ft (b) Half life T = (ln 2) 2.20 (a) q = q 0 e - t/ ( RC ) (b) I = I 0 e - ( R/L ) t (c) τ = RC , τ = L/R Corresponding quantities are a , λ = (ln 2) /T , μ , 1 . 2.21 N = N 0 e Kt 2.22 N = N 0 e Kt - ( R/K )( e Kt - 1) where N 0 = number of bacteria at t = 0, KN = rate of increase, R = removal rate. 2.23 T = 100[1 - (ln r ) / (ln 2)] 2.24 T = 100(2 r - 1 - 1) 2.26 (a) k = weight divided by terminal speed. (b) t = g - 1 · (terminal speed) · (ln 100); typical terminal speeds are 0 . 02 to 0 . 1 cm / sec, so t is of the order of 10 - 4 sec. 2.27
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