Population Modelling, with a model that decreases linearly

Mathsgla

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I'm having problems constructing a suitable expression that could be used to model the growth/decay of a population.

For example,

Write down expressions for the growth rate that could be used to model the following situations:

1) A growth rate with value a0 > 0 at t=0 and -a0 at t=T and decreases linearly.

It's the linearity bit that is confusing me. I know it must be some form of the straight line, perhaps a(t)=-a0t+a0. But I don't know if this implies that a0 > 0 for t = 0. But then with t=1, I would get a(1)=0, a(2)=-a0 and a(3)=-2a0. Hence for all t=T a(t) does not equal -a0 so this surely cannot be the answer.



2) A growth rate that is zero except for 0 ≤ t ≤ τ where it is constantand positive. The value of the non-zero growth rate should besuch that the relative increase (increase in N divided by N(0)) inpopulation size from t = 0 to t = τ is r, assuming dN/dt = α(t)N.

I'm a bit stumped by this one. I know that I am working with the exponential model due to the differential equation given. Solving for N=eat. Not really sure where to go from here. I guessing that by the growth rate being a constant it must be of the form a(t)=ta0 but this doesn't take into account the upper limit of time.

Any tips/help/solutions will be more than welcomed!
 
I'm having problems constructing a suitable expression that could be used to model the growth/decay of a population.

For example,

Write down expressions for the growth rate that could be used to model the following situations:

1) A growth rate with value a0 > 0 at t=0 and -a0 at t=T and decreases linearly.

It's the linearity bit that is confusing me. I know it must be some form of the straight line, perhaps a(t)=-a0t+a0. But I don't know if this implies that a0 > 0 for t = 0. But then with t=1, I would get a(1)=0, a(2)=-a0 and a(3)=-2a0. Hence for all t=T a(t) does not equal -a0 so this surely cannot be the answer.



2) A growth rate that is zero except for 0 ≤ t ≤ τ where it is constantand positive. The value of the non-zero growth rate should besuch that the relative increase (increase in N divided by N(0)) inpopulation size from t = 0 to t = τ is r, assuming dN/dt = α(t)N.

I'm a bit stumped by this one. I know that I am working with the exponential model due to the differential equation given. Solving for N=eat. Not really sure where to go from here. I guessing that by the growth rate being a constant it must be of the form a(t)=ta0 but this doesn't take into account the upper limit of time.

Any tips/help/solutions will be more than welcomed!
(1) You are correct in that linearly means in a straight line. So, use the point-slope form of a line to start with point (0,a0) and you have
a(t) = m (t - 0) + a0 = m t + a0
Now put the point (T, -a0) to get
-a0 = m T + a0
What is m?

(2) First, solving for N would give
N(t)=N0e0ta(t)dt\displaystyle N(t)\, =\, N_0\, e^{\int_0^t\, a(t)\, dt}
Since N/N0 is the constant r, what does that make a(t) and N(t)?
 
(1) You are correct in that linearly means in a straight line. So, use the point-slope form of a line to start with point (0,a0) and you have
a(t) = m (t - 0) + a0 = m t + a0
Now put the point (T, -a0) to get
-a0 = m T + a0
What is m?

(2) First, solving for N would give
N(t)=N0e0ta(t)dt\displaystyle N(t)\, =\, N_0\, e^{\int_0^t\, a(t)\, dt}
Since N/N0 is the constant r, what does that make a(t) and N(t)?

1) so by the definition of a gradient, m=(y2-y1)/(x2-X1) so, m= -(2a0)/T so m is the rate of decay of the population with respect to time T?
 
1) so by the definition of a gradient, m=(y2-y1)/(x2-X1) so, m= -(2a0)/T so m is the rate of decay of the population with respect to time T?
Correct. In math speak, rate means derivative and the derivative of a(t)
a(t) = m t + a0
is m.
 
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