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 4400. The number of fish tripled in the first year. Assuming that the size of the fish population satisfies the logistic equation dP/dt = kP(1 - P/K), determine the constant k, and then solve the equation to find an expression for the size of the population after t years. How long will it take for the population to increase to 2200 (half of the carrying capacity)? It will take years.

Explanation / Answer

K = 4400; DSolve[P'[t] == k*P[t]*(1 - P[t]/K), P, t] p(t) = (4400 e^(k t) ) / ( e^(k t) - E^(4400 C1) ) p(0)= 400 ==> p(t) = (4400 e^(k t))/(10 + e^(k t)) p(0) = 400 The number of fish tripled in the firt year. Means if t=1 year fish population = 3* 400 =1200 p(t) = (4400 e^(k t))/(10 + e^(k t)) p(1) = 1200= (4400 e^(k ))/(10 + e^(k )) 1200= (4400 )/( 10 e^(-k) + 1 ) 1200 10 e^(-k) + 1200 = 4400 12 000 e^(-k) = 3200 e^(-k) = 3200/1200 e^(-k) = 4/15 e^(k) = 15/4 k= ln(15/4) k= 1.32176 p(t) = (4400 e^(k t))/(10 + e^(k t)) p(t) = (4400 )/(10 e^( -k t) + 1) p(t) = (4400 )/(10 e^( -1.32176 t) + 1) how long will it take 2200 fish 2200 = (4400 )/(10 e^( -1.32176 t) + 1) t=1.74206/ ln(e) = 1.74206 years

Related Questions

Navigate

• Browse (All)
• Browse B
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.