Further Growth and Decay (1 Viewer)

[SunnyScience]

Can someone please help me with these two questions. I tried proving these the as per normal way, but the added variable at the front is putting me off. So, how do you do these questions, please:

1. In actual population studies the rate of growth is given by dP/dt = kP(1-RP) where k and R are constants. This reflects limitation on growth, such as by lack of food and existence of predators. This is called the logistic growth equation.

Show by differentiation that P = I/[RI + (1-RI)e^-kt] where I is the initial population (a constant), is a solution of the logistic equation.

Also,

2. An isolate insect population is attached to a single plant which has a carrying capacity of 100 insects. Its logistic law of growth is given by dP/dt = 0.001P(100 - P).

Show by substation that P = 100/[1+(1/k).e^(-0.1t)]

Thank you sooo much!


Just one more thing. With questions dealing with applications of calculus to the physical world, my teacher says that we have to derive the formulas every time we need to use them. Is this necessary - will you lose marks in the HSC?

Thanks again.

 

[Sanjeet]

Very nice question but it's actually got me stumped.

 

[Timske]

Show by different?

 

[SunnyScience]

*differentiation, sorry

nobody can get it? lol....?

 

[Carrotsticks]

(1) and (2): Just differentiate the given expression P(t) with respect to t...

I, R and K are constants so treat them as such. I'll do them if nobody else gets it but I'm pretty sure you'e just getting intimidated by all the constants and whatnot =)

 

[Aesytic]

P = I/[RI + (1-RI)e^(-kt)]
dP/dt = [0 - I(-k(1-RI)e^(-kt)]/[RI + (1-RI)e^(-kt)]^2 using quotient rule
= [kI(1-RI)e^(-kt)]/[RI + (1-RI)e^(-kt)]^2

now,
P[RI + (1-RI)e^(-kt)] = I
[RI + (1-RI)e^(-kt)] = I/P
[RI + (1-RI)e^(-kt)]^2 = I^2/P^2 --- (1)

[RI + (1-RI)e^(-kt)] = I/P
(1-RI)e^(-kt) = I/P - RI
kI(1-RI)e^(-kt) = kI(I/P - RI) --- (2)

.'. dP/dt = (2)/(1)
= [kI(I/P - RI)]/[I^2/P^2]
= kI{[I/P-RI]/[I^2/P^2]}
= kI{[IP-RIP^2]/I^2]} (multiply top and bottom by P^2)
= kI{[P-RP^2]/I} (divide top and bottom by I)
= k{[P-RP^2]} (I on the outside and the I in the denominator cancel out)
= kP(1-RP)

 

[Sanjeet]

Also,

2. An isolate insect population is attached to a single plant which has a carrying capacity of 100 insects. Its logistic law of growth is given by dP/dt = 0.001P(100 - P).

Show by substation that P = 100/[1+(1/k).e^(-0.1t)]

Substitution* ?