ganeshie8
  • ganeshie8
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.
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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dan815
  • dan815
system of equations this time, everything else same
dan815
  • dan815
2* system
ganeshie8
  • ganeshie8
can these be solved individually by decoupling or something

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More answers

ganeshie8
  • ganeshie8
anyways lets see if i can setup the recurrence relations
ganeshie8
  • ganeshie8
baby birds \(b_n = b_{n-1}(1-0.025-0.1)\) adult birds \(a_n = a_{n-1}(1+0.04-0.03) + 0.1*b_n\)
anonymous
  • anonymous
$$\begin{align*}B(n)&=B(n-1)-0.025\cdot B(n-1)+0.04\cdot A(n-1)\\&=0.975\cdot B(n-1)+0.04\cdot A(n-1)\\A(n)&=A(n-1)-0.03\cdot A(n-1)+0.1\cdot B(n-1)\\&=0.97\cdot A(n-1)+0.1\cdot B(n-1)\end{align*}$$ so: $$\begin{bmatrix}B(n)\\A(n)\end{bmatrix}=\begin{bmatrix}0.975&0.04\\0.97&0.1\end{bmatrix}\begin{bmatrix}B(n-1)\\A(n-1)\end{bmatrix}$$ 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
  • anonymous
oops, the bottom row hsould be \(0.1\quad0.97\)
ganeshie8
  • ganeshie8
i think il need to use matlab for this
anonymous
  • anonymous
hmm, well what we have is not diagonalizable unfortunately so there's no way to decouple this cleanly
ganeshie8
  • ganeshie8
Oh it cannot have independent eigenvectors is it
ganeshie8
  • ganeshie8
but since all the numbers are less than 1, is it okay to assume the matrix approaches 0 for large n ?
campbell_st
  • campbell_st
are they golden plovers.. ..?
anonymous
  • anonymous
\[ b_n = (1-0.025)(0.9)b_{n-1}+(0.04)a_{n-1}\\ a_n = (1-0.025)(0.1)b_{n-1}+(1-0.03)a_{n-1} \\ A_n = \begin{bmatrix} 0.8775&0.04\\ 0.0975&0.97 \end{bmatrix}^{n-1}A_{1} \]
ganeshie8
  • ganeshie8
Haha similar kind id guess but the saddest thing is that they gonna extinct sooner based on our matrix!
campbell_st
  • campbell_st
ok... blame global warming
Empty
  • Empty
Plug it into matlab or octave ;P
anonymous
  • anonymous
hmm, interesting premultiplication of the 0.1 maturation rate @wio. is that matrix diagonalizable
anonymous
  • anonymous
http://www.wolframalpha.com/input/?i=Diagonalize+%7B%7B0.8775%2C+0.04%7D%2C%7B0.0975%2C+0.97%7D%7D
anonymous
  • anonymous
hmm, well that's no fun
Empty
  • Empty
I think there's something elegant about being able to use computers :P
ganeshie8
  • ganeshie8
i 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
  • anonymous
well, 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
  • anonymous
but if they're not both below 0, then that doesn't necessarily hold

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