|Aug22-12, 06:11 AM||#1|
Would someone be able to help me with the quesiton
Life cycle of butterfly is egg -> caterpillar -> butterfly.
Each week 5% of the eggs will hatch and 2% of the caterpillars will turn to butterflies.
You start with 100 eggs and 60 caterpillars.
Using calculus to obtain an equation which will refine your model for the amount of caterpillars at any time.
How many caterpillars are there after 40 weeks?
|Aug22-12, 07:37 AM||#2|
How is this not homework? And what attempt have you made to do it yourself?
Let C(t) be the number of caterpillars at time t (in weeks) and E(t) the number of eggs. Then dE/dt is the rate of change of number of eggs. What is it equal to each week? ("Each week 5% of the eggs will hatch".) dC/dt is the rate at which the number of caterpillars changes. What is that equal to? ("2% of the caterpillars will turn to butterflies". And, of course, the hatched eggs become caterpillars.)
|Aug22-12, 08:08 AM||#3|
You will get the most out of these forums if you attempt the problem yourself and show us the attempt. That way we can help you with the particular area where you get stuck.
|Aug23-12, 03:35 AM||#4|
Dw, i figured it. I just found a function of e with regards to time and subbed that into the equation dc/ct=0.05e-0.02c. After that i integrated it, solved for a homogenous equation and than a partial solution
|Aug23-12, 08:14 AM||#5|
Any old function e(t)?
Don't butterflies lay eggs?
|Aug27-12, 07:08 AM||#6|
Ok, so i got the equation C(t)= 226.667e^(-0.02(t)) - 166.667e^(-0.05(t))
and at t=40 the equation gives the answer 79.2921
however, if the relationship stated in the initial post about the question is model in excel it gives a value of 78.55. What assumptions are made by the differential equations at proves such a small difference?
|Aug28-12, 08:35 PM||#7|
It looks like you used a first order forward integration formula with time steps of 1 week in excel to approximate the solution to the differential equations. The problem formulation implies "continuous compounding" of the production and destruction rates, rather than first order forward difference approximation. This accounts for most of the difference in the results. If you had used a higher order integration formula, the difference in results would have been less.
|Aug30-12, 08:58 PM||#8|
Is this relationship like compound interest?
|Aug31-12, 11:51 AM||#9|
It's not exactly the same as compound interest, but the comparison between the exact solution of the differential equations and the (approximate) solution you got in your spreadsheet algorithm is analogous to the comparison between "continuous compounding" of interest, and compounding only at discrete time intervals (like say, monthly).
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