#1

can anyone help with the following:

a fishery uses the following model for the absolute growth rate:

dP/dt = rP(1-[P/K]) - h

where h is the constant absolute harvest rate

a) if r = 1.04 and K = 100, calculate the harvest rate h which gives an equilibrium at P(eq) = 55

b) using the above values of r, K and h, show that the model can be re-written in the factorised form

dP/dt = (r/K)(P-a)(b-P)

and find the values of a and b

c) thus, use separation of variables and partial fractions to show that

ln|P-45| - ln|P-55| = K(A)t + C

where C is an arbitrary constant. find the value of K(A) -------> note! i dno how to get subscripts. so K(A) reads as "K sub A"

d) next, calculate the value of the arbitrary constant if P = 46 when t = 0. then, find the time taken for the population to increasefrom 46 to 54

cheers in advance

a fishery uses the following model for the absolute growth rate:

dP/dt = rP(1-[P/K]) - h

where h is the constant absolute harvest rate

a) if r = 1.04 and K = 100, calculate the harvest rate h which gives an equilibrium at P(eq) = 55

b) using the above values of r, K and h, show that the model can be re-written in the factorised form

dP/dt = (r/K)(P-a)(b-P)

and find the values of a and b

c) thus, use separation of variables and partial fractions to show that

ln|P-45| - ln|P-55| = K(A)t + C

where C is an arbitrary constant. find the value of K(A) -------> note! i dno how to get subscripts. so K(A) reads as "K sub A"

d) next, calculate the value of the arbitrary constant if P = 46 when t = 0. then, find the time taken for the population to increasefrom 46 to 54

cheers in advance

#2

Fascinating, quite intriguing.

#3

part a,c and d are relatively easy...cant be bothered to go through b... here's a:

0=rP(1-[P/K])-h

h=25.74

0=rP(1-[P/K])-h

h=25.74

#4

Fascinating, quite intriguing.

Anyway, how about doing your homework yourself?

#5

Give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime.

#6

Give a man a fish and you feed him for a day; teach a man to fish and you feed him for a lifetime.

this thread is now about fish:

(and creepy old men with measuring sticks).

have a bump, threadstarter!

#7

*reported*

put it in the maths/science thread - i'll answer it there

put it in the maths/science thread - i'll answer it there

#8

a) Stick your values in, dP/dt=0 since the population isn't changing. Rearrange to find h, easy enough aye?

b) Expand everything out, factorise it again to get your answer.

c) From b):

dP/dt = (r/K)(P-a)(b-P)

So (integral) 1/[(P-a)(b-P)]=(integral) r/K dt.

Fix up the LHS with partial fractions, integrate and use laws of logs to get the 2 lns in the answer, hopefully divided by something, which you can name K(A). Put your constant on the RHS, name it K so you can say K(A)K=C, so the answers is found.

d) is just sticking numbers into c). First part, just throw your numbers in, rearrange for C. Second part, sub P=45 and P=54 and find the difference between the corresponding values of t this gives.

(PS, the reasons I've not actually done a questions are that a)I haven't slept in about 30 hours, and b) I haven't done this kinda stuff since last June, so I'm bound to screw up)

b) Expand everything out, factorise it again to get your answer.

c) From b):

dP/dt = (r/K)(P-a)(b-P)

So (integral) 1/[(P-a)(b-P)]=(integral) r/K dt.

Fix up the LHS with partial fractions, integrate and use laws of logs to get the 2 lns in the answer, hopefully divided by something, which you can name K(A). Put your constant on the RHS, name it K so you can say K(A)K=C, so the answers is found.

d) is just sticking numbers into c). First part, just throw your numbers in, rearrange for C. Second part, sub P=45 and P=54 and find the difference between the corresponding values of t this gives.

(PS, the reasons I've not actually done a questions are that a)I haven't slept in about 30 hours, and b) I haven't done this kinda stuff since last June, so I'm bound to screw up)

#9

Mind = blown.

#10

we aren't that smart on the west coast. sorry.

#11

Mind = blown.

just look at the calming picture of the huge fish. it'll assuage/massage your mind