# Differential Equations + Iteration

• Aug 23rd 2012, 07:14 AM
SkyCapri
Differential Equations + Iteration
I need some help with a maths problem here.

A biologist is researching the population of a species. She tries a number of different models for the rate of growth of the population and solves them to compare with observed data. Her first model is dp/dt=kp(1-(p/m)) where p is the population at time t years, k is a constant and m is the maximum population sustainable by the environment. Find the general solution of the differential equation.

I've separated the variables and integrated both sides of the equation. The general solution is p= mAe^kt/(1+Ae^kt).

Her observations suggest that k= 0.2 and m= 100 000. If the initial population is 30 000, estimate the population after 5 years to 2 sig. fig.
I found the value of A to be 3/7 . Then I substituted it into the general solution to obtain p = 54 000 after 5 years.

Now comes the difficult part. (Thinking)
She decides that the model needs to be refined. She proposes a new model dp/dt= kp(1-(p/m)^α) and investigates suitable values of α. Her observations lead her to the conclusion that the maximum growth rate occurs when the population is 70% of its maximum. Show that (α+1)0.7^α=1 and that an approximate solution of this equation is α=5. How do I prove this?

Express the time that it will take the population to reach 54 000 according to this model as a definite integral, and use the trapezium rule to find this time approximately.
I know how to obtain the definite integral, which is http://upload.wikimedia.org/wikipedi...eb485cd0f6.png5/(p(1-((1/100 000)p)^5) dp with the upper and lower limits being 54 000 and 30 000 respectively. The time taken is 3.0 years. My question is how many intervals should I use when I apply the trapezium rule?
• Aug 23rd 2012, 07:41 AM
Prove It
Re: Differential Equations + Iteration
Quote:

Originally Posted by SkyCapri
I need some help with a maths problem here.

A biologist is researching the population of a species. She tries a number of different models for the rate of growth of the population and solves them to compare with observed data. Her first model is dp/dt=kp(1-(p/m)) where p is the population at time t years, k is a constant and m is the maximum population sustainable by the environment. Find the general solution of the differential equation.

I've separated the variables and integrated both sides of the equation. The general solution is p= mAe^kt/(1+Ae^kt).

Her observations suggest that k= 0.2 and m= 100 000. If the initial population is 30 000, estimate the population after 5 years to 2 sig. fig.
I found the value of A to be 3/7 . Then I substituted it into the general solution to obtain p = 54 000 after 5 years.

Now comes the difficult part. (Thinking)
She decides that the model needs to be refined. She proposes a new model dp/dt= kp(1-(p/m)^α) and investigates suitable values of α. Her observations lead her to the conclusion that the maximum growth rate occurs when the population is 70% of its maximum. Show that (α+1)0.7^α=1 and that an approximate solution of this equation is α=5. How do I prove this?

Express the time that it will take the population to reach 54 000 according to this model as a definite integral, and use the trapezium rule to find this time approximately.
I know how to obtain the definite integral, which is http://upload.wikimedia.org/wikipedi...eb485cd0f6.png5/(p(1-((1/100 000)p)^5) dp with the upper and lower limits being 54 000 and 30 000 respectively. The time taken is 3.0 years. My question is how many intervals should I use when I apply the trapezium rule?

What values can \displaystyle \displaystyle \begin{align*} \alpha \end{align*} take?
• Aug 23rd 2012, 07:45 AM
HallsofIvy
Re: Differential Equations + Iteration
Quote:

Originally Posted by SkyCapri
I need some help with a maths problem here.

A biologist is researching the population of a species. She tries a number of different models for the rate of growth of the population and solves them to compare with observed data. Her first model is dp/dt=kp(1-(p/m)) where p is the population at time t years, k is a constant and m is the maximum population sustainable by the environment. Find the general solution of the differential equation.

I've separated the variables and integrated both sides of the equation. The general solution is p= mAe^kt/(1+Ae^kt).

Her observations suggest that k= 0.2 and m= 100 000. If the initial population is 30 000, estimate the population after 5 years to 2 sig. fig.
I found the value of A to be 3/7 . Then I substituted it into the general solution to obtain p = 54 000 after 5 years.

Now comes the difficult part. (Thinking)
She decides that the model needs to be refined. She proposes a new model dp/dt= kp(1-(p/m)^α) and investigates suitable values of α. Her observations lead her to the conclusion that the maximum growth rate occurs when the population is 70% of its maximum. Show that (α+1)0.7^α=1 and that an approximate solution of this equation is α=5. How do I prove this?

What do you get when you calculate (5+1)0.7^5?

Quote:

Express the time that it will take the population to reach 54 000 according to this model as a definite integral, and use the trapezium rule to find this time approximately.
I know how to obtain the definite integral, which is http://upload.wikimedia.org/wikipedi...eb485cd0f6.png5/(p(1-((1/100 000)p)^5) dp with the upper and lower limits being 54 000 and 30 000 respectively. The time taken is 3.0 years. My question is how many intervals should I use when I apply the trapezium rule?
That's for you to decide. How accurate do you want your integration to be?
• Aug 23rd 2012, 07:06 PM
SkyCapri
Re: Differential Equations + Iteration
(5+1)0.7^5=1.00842. But how do I show that (α+1)0.7^α=1?
• Aug 25th 2012, 09:42 AM
HallsofIvy
Re: Differential Equations + Iteration
You have the equation dp/dt= kp(1-(p/m)^α) where dp/dt is the rate of growth. To find the maximum rate of growth, do exactly what you learned to do in Calculus to find max and min: differentiate with respect to t and set the derivative equal to 0:
Using the chain rule, the derivative with respect to p is
$\displaystyle k(1- (p/m)^a)+ kp[-a(p/m)^{a-1}]$ so the derivative with respect to t is
$\displaystyle (k(1- (p/m)^a)+ kp[-a(p/m)^{a-1}])(dp/dt)$
Set that equal to 0.
• Sep 1st 2012, 03:35 AM
SkyCapri
Re: Differential Equations + Iteration
Quote:

Originally Posted by HallsofIvy
You have the equation dp/dt= kp(1-(p/m)^α) where dp/dt is the rate of growth. To find the maximum rate of growth, do exactly what you learned to do in Calculus to find max and min: differentiate with respect to t and set the derivative equal to 0:
Using the chain rule, the derivative with respect to p is
$\displaystyle k(1- (p/m)^a)+ kp[-a(p/m)^{a-1}]$ so the derivative with respect to t is
$\displaystyle (k(1- (p/m)^a)+ kp[-a(p/m)^{a-1}])(dp/dt)$
Set that equal to 0.

Thanks a lot.