Fish population model calc problem

Question

The fish population is represented by dN/dt= r*ln(N)(K-p^2)

p= number of predator

r= intrinsic growth rate of the fish population

K= carrying capacity of the population

1. Show that the rate of change in the fish population decreases with respect to the predator population is N>1.
2. Show that the rate of change in the fish population increases with respect to increasing density N is only the number of predators p is less than sqrtK.

Answer 1

I am assuming that there are two typos in the questions: "is" should be "if".

I am assuming that N, p, r, K are positive.

I am assuming that "rate of change in the fish population" means

"rate of change in the fish population with respect to time".

In that case I denote "rate of change in the fish population" by

F(N, p, r, K) = r×ln(N)(K-p²)

1.

The "rate of change in the fish population" decreases with respect to predator population if

∂F/∂p < 0

That evaluates to

r×ln(N)(-2p) < 0

Under the assumption that r > 0 and p > 0, this is equivalent to

ln(N) > 0

That is equivalent to

N > 1.

2.

This question refers to N as "density", while above it refers to it as "fish population".

The word "density" is normally a ratio between some quantity and volume.

The problem does not mention anything about the volume of water available to the fish.

Therefore I assume that the volume is constant.

In that case the density of fish has the same behavior as their number N,

so "density N" means the same thing as "population N".

Then

"rate of change in the fish population increases with respect to increasing density N"

means the same thing as

"rate of change in the fish population increases with respect to N".

And that means

∂F/∂N > 0

That evaluates to

r(K-p²)/N > 0

Under the above assumption that all the quantities are positive, this is equivalent to

K-p² > 0

That is equivalent to

p < √K