Unformatted text preview: h thousands per day. b. We have y = 1 10 y (1y 8 )3 16 = 1 80 y (8y )3 16 = 1 80 (y 2 +8 y15) =1 80 ( y3)( y5). The zeroes of1 80 ( y3)( y5) are at y = 3 and y = 5, so these are the equilibrium points. The graph of1 80 ( y3)( y5) is a parabola which is negative for y < 3, positive for 3 < y < 5 and negative for y > 5. Hence y = 3 is an unstable equilibrium, while y = 5 is a stable equlibrium. c. If the initial population of the lake is 6000 ﬁsh, then y (0) = 6 since y measures thousands of ﬁsh. The ﬁsh population will decrease and eventually approach 5000 (the stable equilibrium point 5). If the initial population of the lake is 2500 ﬁsh, then y (0) = 2 . 5. Since1 80 ( y3)( y5) 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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 Spring '07
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 Math, 2 pts, Stability theory, Logistic function, initial population, 1  k

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