The Pacific halibut fishery has been modeled by the differential equation

Question

The Pacific halibut fishery has been modeled by the differential equation

dy/dt = ky(1-y/M)

where y(t) is the biomass (the total mass of the members of the population) in kilograms at time t (measured in years), the carrying capacity is estimated to be M = 6 × 10⁷ kg, and k = 0.75 per year.

(a) If y(0) = 2 × 10⁷ kg, find the biomass a year later. (Round your answer to two decimal places.)

(b) How long will it take for the biomass to reach 4 × 10⁷ kg? (Round your answer to two decimal places.)

Answer 1

Given The differential equation is given as,

dydt=ky(1−yM) .

Rearrange it as follows.

dydt−ky=−kMy2

To solve this equation, let y−1=u

Therefore,

−1y2dydt=dudt

Replace these values in the original equation.

Therefore, the differential equation becomes,

−1y2dydt+ky=kMdudt+ku=kM

The integrating factor of this equation is given by,

I.F.=e∫kdt=ekt

Therefore, the solution of the given equation is,

u⋅ekt=∫kMektdt=kM×ektk+c=1Mekt+c

Replace the value of u.

Therefore, the solution of the original differential equation is given by,

y−1=1M+ce−kt

Rearrange the equation.

1y=1+Mce−ktMy=M1+Mce−kt

According to the question,

M=6×10^7 kg, k=0.75 per year and y(0)=2×10^7 kg.

Therefore,

2×10^7=(6×10^7)/(1+6×10^7c)

12×10^14c=4×10^7

c=0.34×10−7

Thus,

y=(6×10^7)/(1+2e−0.75t)

Thus, the weight of biomass after one year,

y(1)=(6×10^7)/(1+2e−0.75)

=(6×10^7)/1.944

=3.08×10^7kg

Now, y(t)=4×10^7

Therefore,

4×10^7=(6×10^7)/(1+2e−0.75t)

1+2e−0.75t=1.5

e−0.75t=0.25

Take log on both sides.

−0.75t=−1.386

t=1.84years

Thus, it will take approximately 1.84 years for the biomass to reach 4×10^7 kg.

Answer 2

f(t) = M/(1+e^-k(t-t0))

Substituting at t=0, we get 2*10^7 = 6 * 10^7/(1+e^.75 t0)

cross-multiplying and dividing, 3 = 1 + e^ .75 t0

t0 = 4/3 ln(2)

Then, a year later, f(t) = 6 * 10^7/(1+e^-.75 (1 - 4/3 ln(2))) =

To get the time until we get to 4 * 10^7, we can either solve or notice that the middle of the curve is at 3*10^7, and 4*10^7 and 2^10^7 will be equal times away

Thus, the time is 8/3 ln 2 = = 1.85.