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ganeshie8
 one year ago
A certain kind of migratory birds start reproducing at the age of \(2\) years.
\(2.5\%\) of the baby birds die each year.
\(10\%\) of the baby birds that survive become adult each year.
The adult birds reproduce at a rate of \(4\%\) per year and die at a rate of \(3\%\) per year.
Estimate the number of baby birds and adult birds at the end of \(30\) th year if the starting population of baby birds and adult birds are \(10, 20\) respectively.
ganeshie8
 one year ago
A certain kind of migratory birds start reproducing at the age of \(2\) years. \(2.5\%\) of the baby birds die each year. \(10\%\) of the baby birds that survive become adult each year. The adult birds reproduce at a rate of \(4\%\) per year and die at a rate of \(3\%\) per year. Estimate the number of baby birds and adult birds at the end of \(30\) th year if the starting population of baby birds and adult birds are \(10, 20\) respectively.

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dan815
 one year ago
Best ResponseYou've already chosen the best response.1system of equations this time, everything else same

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1can these be solved individually by decoupling or something

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1anyways lets see if i can setup the recurrence relations

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1baby birds \(b_n = b_{n1}(10.0250.1)\) adult birds \(a_n = a_{n1}(1+0.040.03) + 0.1*b_n\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$\begin{align*}B(n)&=B(n1)0.025\cdot B(n1)+0.04\cdot A(n1)\\&=0.975\cdot B(n1)+0.04\cdot A(n1)\\A(n)&=A(n1)0.03\cdot A(n1)+0.1\cdot B(n1)\\&=0.97\cdot A(n1)+0.1\cdot B(n1)\end{align*}$$ so: $$\begin{bmatrix}B(n)\\A(n)\end{bmatrix}=\begin{bmatrix}0.975&0.04\\0.97&0.1\end{bmatrix}\begin{bmatrix}B(n1)\\A(n1)\end{bmatrix}$$ so it follows $$\begin{bmatrix}B(n)\\A(n)\end{bmatrix}=\begin{bmatrix}0.975&0.04\\0.97&0.1\end{bmatrix}^n\begin{bmatrix}B(0)\\A(0)\end{bmatrix}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, the bottom row hsould be \(0.1\quad0.97\)

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i think il need to use matlab for this

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm, well what we have is not diagonalizable unfortunately so there's no way to decouple this cleanly

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Oh it cannot have independent eigenvectors is it

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1but since all the numbers are less than 1, is it okay to assume the matrix approaches 0 for large n ?

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2are they golden plovers.. ..?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[ b_n = (10.025)(0.9)b_{n1}+(0.04)a_{n1}\\ a_n = (10.025)(0.1)b_{n1}+(10.03)a_{n1} \\ A_n = \begin{bmatrix} 0.8775&0.04\\ 0.0975&0.97 \end{bmatrix}^{n1}A_{1} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1Haha similar kind id guess but the saddest thing is that they gonna extinct sooner based on our matrix!

campbell_st
 one year ago
Best ResponseYou've already chosen the best response.2ok... blame global warming

Empty
 one year ago
Best ResponseYou've already chosen the best response.2Plug it into matlab or octave ;P

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm, interesting premultiplication of the 0.1 maturation rate @wio. is that matrix diagonalizable

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://www.wolframalpha.com/input/?i=Diagonalize+%7B%7B0.8775%2C+0.04%7D%2C%7B0.0975%2C+0.97%7D%7D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmm, well that's no fun

Empty
 one year ago
Best ResponseYou've already chosen the best response.2I think there's something elegant about being able to use computers :P

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.1i think wolframalpha is lying, two independent eigenvectors means we can diagonalize http://www.wolframalpha.com/input/?i=eigenvectors+%7B%7B0.8775%2C+0.04%7D%2C%7B0.0975%2C+0.97%7D%7D

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0well, the point of asking whether we can diagonalize is because it'll make the matrix power very, very easy :) and yes, the fact the eigenvalues are below 0 tells us that in the long run the behavior is for the populations to die out

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0but if they're not both below 0, then that doesn't necessarily hold
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