Biologists stocked a lake with 400 fish and estimated the carrying capacity (the

Question

Biologists stocked a lake with 400 fish and estimated the carrying capacity (the maximal population for the fish of that species in that lake) to be 10000. The number of fish tripled in the first year. (a) Assuming that the size of the fish population satisfies the logistic equation, find an expression for the size of the population after t years. P(t) = 400*3^(t) (b) How long will it take for the population to increase to 5000? ____________________ years I keep getting for part B- 2.299 years but it won't accept my answer or 3 or 2.3 years.....help please!

Explanation / Answer

Start with: y = 10000 / [1 + e^(-t)] Find where 400 would lie, then find where 1200 would lie 1+ e^(-t) = 10000/y e^-t = (10000-y)/y ln e^-t = ln [(10000-y)/y] -t =ln [(10000-y)/y] t = - ln [(10000-y)/y] For 400: t = - ln [(10000-400)/400] t = -ln(9600/400) t = - ln (24) t = - 3.178 For 1200: t = -ln[(10000-1200)/1200] t = -ln(8800/1200) t = -ln (7.33) t = - 1.992 The difference is ~ -1.186 So, we shift the time by 3.178, and multiply it by 1.186, giving us: P = 10000 /[ 1 + e^(-((1.186 n) - 3.178)] P = 10000 /[ 1 + e^(3.178 - 1.186n)] Sub back in to test: P(0) = 10000 / [1 + e^(3.178)] = 10000/(1 +24) = 10000/ 25 = 400 P(1) = 10000/[1 + e^(3.178-1.186)] = 10000/[1 + e^(1.992)] = 10000/(1 + 7.330) = 10000/8.330 = 1200 P = 10000 /[ 1 + e^(3.178 - 1.186n)]

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