Solving Logistic Equation Homework Problem

  • Thread starter roam
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In summary, the conversation discusses finding the equilibrium points of a given equation using the quadratic formula. The attempt at a solution uses the formula but obtains a different answer than the one in the textbook. The discrepancy is resolved by realizing that the denominator in the solution should be multiplied by 2k.
  • #1
roam
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Homework Statement



I'm having some difficulty understanding the last step of the following solved problem from my textbook:

We will compute the equilibrium points of [itex]kP \left( 1- \frac{P}{N} \right) - C[/itex].

[itex]kP \left( 1- \frac{P}{N} \right) - C =0[/itex]

[itex]-kP^2+kNP - CN =0[/itex]

The quadratic equations has solutions

[itex]P= \frac{N}{2} \pm \sqrt{\frac{N^2}{4}-\frac{CN}{k}}[/itex]

The Attempt at a Solution



So if we start off by

[itex]-kP^2+kNP - CN =0[/itex]

and using the quadratic equation:

[itex]x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}[/itex]

I get:

[itex]P= \frac{-kN \pm \sqrt{(kN)^2-4kC}}{2k}[/itex]

[itex]= \frac{N}{2} \pm \frac{\sqrt{k^2N^2-4kC}}{2k}[/itex]

So, why is my answer different from the one in the textbook? And how can I end up with the solution in the book (that I've posted above)? :confused:
 
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  • #2
roam said:
[itex]P= \frac{-kN \pm \sqrt{(kN)^2-4kC}}{2k}[/itex]

[itex]= \frac{N}{2} \pm \frac{\sqrt{k^2N^2-4kC}}{2k}[/itex]

[itex]2k = \sqrt{\text{(what?)}}[/itex]
 
  • #3
scurty said:
[itex]2k = \sqrt{\text{(what?)}}[/itex]

Oh, I see. I never thought of it. Thank you very much. :smile:
 

1. How do I solve a logistic equation homework problem?

To solve a logistic equation homework problem, you first need to understand the basic form of a logistic equation: P(t) = K / (1 + A * e-rt). You will need to have values for the carrying capacity (K), the growth rate (r), and the initial population (P0). Then, you can use algebraic techniques or a graphing calculator to solve for the population at a specific time (t).

2. What is the difference between a logistic equation and an exponential equation?

A logistic equation takes into account a limiting factor, or carrying capacity, which is the maximum population that an environment can support. This results in a sigmoid (S-shaped) curve, as opposed to an exponential equation which produces a J-shaped curve. Additionally, a logistic equation approaches the carrying capacity as time goes to infinity, while an exponential equation continues to increase without limit.

3. How do I determine the carrying capacity from a logistic equation?

The carrying capacity (K) is the maximum population that an environment can support. In a logistic equation, it is represented by the coefficient in the numerator. To determine the carrying capacity, you can look at the behavior of the population as time goes to infinity. At this point, the population will level off and approach the carrying capacity.

4. Can a logistic equation be used to model real-life situations?

Yes, logistic equations can be used to model real-life situations such as population growth, spread of diseases, and adoption of new technologies. They are particularly useful when a population is limited by resources or other factors, and can accurately predict the behavior of the population over time.

5. What is the significance of the growth rate (r) in a logistic equation?

The growth rate (r) in a logistic equation represents the rate at which the population is growing. It is a measure of the birth rate minus the death rate, and is usually expressed as a decimal or percentage. A higher growth rate will result in a faster increase in population, while a lower growth rate will result in a slower increase. The value of r also affects the shape of the sigmoid curve, with higher values producing steeper curves and lower values producing flatter curves.

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