Answer:
a. The biomass weighs 2.30 * 10^7 kg after a year
b. It'll take 2.56 years to get to 4*10^7kg
Step-by-step explanation:
a.
k = 0.78,K = 6E7 kg
Given
dy/dt = ky(1- y/K)
Make ky dt the subject of formula
ky dt = dy/(1-y/K) --- make k dt the subject of formula
k dt = dy/(y(1-y/K))
k dt = K dy / y(K-y)
k dt = ((1/y) + (1/(K-y)))dy ---- integrate both sides
kt + c = ln(y/(K-y))
Ce^(kt) = y/(K-y)
Substitute the values of k and K
Ce^(0.78t) = y/(6*10^7 - y) ----- (1)
Given that y(0) = 2 * 10^7kg
(1) becomes
Ce^(0.78*0) = (2 * 10^7)/(6*10^7 - 2*10^7)
Ce° = (2*10^7)/(4*10^7
C = 2/7
Substitute 2/7 for C in (1)
2/7e^0.78t = y/(6*10^7 - y) ---(2)
We're to find the biomass a year later
So, t = 1
2/7e^0.78 = y/(6*10^7 - y)
0.62 = y/(6*10^7 - y)
y = 0.62(6*10^7 - y)
y = 0.62*6*10^7 - 0.62y
y + 0.62y = 0.62*6*10^7
1.62y = 0.62*6*10^7
1.62y = 3.72 * 10^7
y = 2.30 * 10^7kg.
Hence, the biomass weighs 2.30 * 10^7 kg after a year
b.
Here, we're to calculate the time it'll take the biomass to get to 4*10^7 kg
Substitute 4*10^7 for y in (2)
2/7e^0.78t = 4*10^7/(6*10^7 - 4*10^7)
2/7e^0.78t = 4*10^7/2*10^7
2/7e^0.78t = 2
e^0.78t = (2*7)/2
e^0.78t = 2
t = 2 * 1/0.78
t = 2.56 years
Hence, it'll take 2.56 years to get to 4*10^7kg
⇒2.4