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Unformatted text preview: h thousands per day. b. We have y = 1 10 y (1y 9 )1 5 = 1 90 y (9y )1 5 = 1 90 (y 2 +9 y18) =1 90 ( y3)( y6). The zeroes of1 90 ( y3)( y6) are at y = 3 and y = 6, so these are the stable equilibria. The graph of1 90 ( y3)( y6) is a parabola which is negative for y < 3, positive for 3 < y < 6 and negative for y > 6. Hence y = 3 is an unstable equilibrium, while y = 6 is a stable equlibrium. c. If the initial population of the lake is 7500 sh, then y (0) = 7 . 5 since y measures thousands of sh. The sh population will decrease and eventually approach 6000 (the stable equilibrium point y = 6.) If the initial population of the lake is 2500 sh, then y (0) = 2 . 5. Since1 90 ( y3)( y6) is negative for y < 3, y ( t ) will continue to decrease, and eventually the lake will be completely farmed out. 1...
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This note was uploaded on 03/05/2008 for the course MAE 33B taught by Professor Lee during the Spring '07 term at UCLA.
 Spring '07
 lee

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