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A5F08Sol-138

# A5F08Sol-138 - Math 138 Assignment 5 Solutions Fall 2008 1...

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Math 138 Assignment 5 Solutions Fall 2008 1. Biologists stocked a lake with 400 fish and estimated the carrying capacity (the maximum population for the fish of that species in the lake) to be 10 000. The number of fish tripled in the first year. a) Assuming that the size of the fish population satisfies the logistic equation (see page 592 of the text), find an expression for the size of the population after t years. Solution Let P ( t ) represent the population of fish in the lake t years after the stocking of the lake. Then P (0) = P 0 = 400, P (1) = 1200, and K = 10 000. From the solution to the logistic differential equation P ( t ) = P 0 K P 0 + ( K - P 0 ) e - kt we get P = 400(10 000) 400 + 9600 e - kt = 10 000 1 + 24 e - kt . Since P (1) = 1200, we have 1200 = 10 000 1 + 24 e - k 1 + 24 e - k = 100 12 24 e - k = 88 12 e k = 288 88 k = ln 36 11 . So P = 10 000 1 + 24 e - t ln 36 11 P = 10 000 1 + 24 ( 11 36 ) t . b) How long will it take for the population to increase to 5000? Solution Solve for t in: 5000 = 10 000 1 + 24 ( 11 36 ) t 1 + 24 11 36 t = 2 11 36 t = 1 24 t ln 11 36 = ln 1 24 t 2 . 68 . The fish population will reach 5000 in approximately 2.68 years.

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Math 138 Assignment 5 Solutions Fall 2008 2. a) Use Euler’s method with step-size 0.2 to estimate y (0 . 4) where y ( x ) is the solution to the initial- value problem: y = 2 xy 2 y (0) = 1 Solution Euler’s method with h = 0 . 2, x 0 = 0, y 0 = 1, and F ( x, y ) = 2 xy 2 , we need to find y 2 .
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A5F08Sol-138 - Math 138 Assignment 5 Solutions Fall 2008 1...

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