A community for students.

Here's the question you clicked on:

55 members online
  • 0 replying
  • 0 viewing

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.

  • This Question is Closed
  1. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    system of equations this time, everything else same

  2. dan815
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    2* system

  3. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    can these be solved individually by decoupling or something

  4. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    anyways lets see if i can setup the recurrence relations

  5. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    baby birds \(b_n = b_{n-1}(1-0.025-0.1)\) adult birds \(a_n = a_{n-1}(1+0.04-0.03) + 0.1*b_n\)

  6. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    $$\begin{align*}B(n)&=B(n-1)-0.025\cdot B(n-1)+0.04\cdot A(n-1)\\&=0.975\cdot B(n-1)+0.04\cdot A(n-1)\\A(n)&=A(n-1)-0.03\cdot A(n-1)+0.1\cdot B(n-1)\\&=0.97\cdot A(n-1)+0.1\cdot B(n-1)\end{align*}$$ 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 $$\begin{bmatrix}B(n)\\A(n)\end{bmatrix}=\begin{bmatrix}0.975&0.04\\0.97&0.1\end{bmatrix}^n\begin{bmatrix}B(0)\\A(0)\end{bmatrix}$$

  7. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    oops, the bottom row hsould be \(0.1\quad0.97\)

  8. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    i think il need to use matlab for this

  9. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm, well what we have is not diagonalizable unfortunately so there's no way to decouple this cleanly

  10. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Oh it cannot have independent eigenvectors is it

  11. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    but since all the numbers are less than 1, is it okay to assume the matrix approaches 0 for large n ?

  12. campbell_st
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    are they golden plovers.. ..?

  13. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    \[ b_n = (1-0.025)(0.9)b_{n-1}+(0.04)a_{n-1}\\ a_n = (1-0.025)(0.1)b_{n-1}+(1-0.03)a_{n-1} \\ A_n = \begin{bmatrix} 0.8775&0.04\\ 0.0975&0.97 \end{bmatrix}^{n-1}A_{1} \]

  14. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    Haha similar kind id guess but the saddest thing is that they gonna extinct sooner based on our matrix!

  15. campbell_st
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    ok... blame global warming

  16. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    Plug it into matlab or octave ;P

  17. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm, interesting premultiplication of the 0.1 maturation rate @wio. is that matrix diagonalizable

  18. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    hmm, well that's no fun

  19. Empty
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 2

    I think there's something elegant about being able to use computers :P

  20. ganeshie8
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 1

    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

  21. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    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

  22. anonymous
    • one year ago
    Best Response
    You've already chosen the best response.
    Medals 0

    but if they're not both below 0, then that doesn't necessarily hold

  23. Not the answer you are looking for?
    Search for more explanations.

    • Attachments:

Ask your own question

Sign Up
Find more explanations on OpenStudy
Privacy Policy

Your question is ready. Sign up for free to start getting answers.

spraguer (Moderator)
5 → View Detailed Profile

is replying to Can someone tell me what button the professor is hitting...

23

  • Teamwork 19 Teammate
  • Problem Solving 19 Hero
  • You have blocked this person.
  • ✔ You're a fan Checking fan status...

Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.

This is the testimonial you wrote.
You haven't written a testimonial for Owlfred.