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Applications of Differentiation II

  1. Mar 7, 2009 #1
    1. The problem statement, all variables and given/known data

    A research studied the commercial fishing situation in a certain fishing zone. Denoting the total catch of corel fish in that zone in t years time from January 1, 1992 by N(t) (in thousand tonnes), he obtained the following data:

    t=2, N(t)=55
    t=4, N(t)=98

    The research modelled N(t) by ln N(t) = a - e^(1-kt) where a and k are constants.

    (a) Show that e^(-4k) - e^(-2k) + ln (98/55) / e = 0

    Hence find, to 2 decimal places, two sets of values of a and k.

    (b) The research later found out that N(7) = 170. Determine which set of values of a and k obtained in (a) will make the model fit for the known data.

    Hence estimate, to the nearest thousand tonnes, the total possible catch of coral fish in that zone since January 1, 1992.

    (c) Skipped (involved differentiation).

    Answers
    (a): k = 0.59, a = 4.84 or k = 0.18, a = 5.89
    (b): k = 0.18, a = 5.89; 361 thousand tonnes
    (c): skipped

    2. Relevant equations

    Quadratic Equations and Differentiation Rules

    3. The attempt at a solution

    I can solve (a) and first part of (b) successfully.

    However, I don't understand why second part of (b) should be 361 thousand tonnes.

    The formula I used is ln N(t) = 5.90 - e^(1-0.18t),

    when I performed backward calculation, t is 30.55 years.

    It is quite strange and I can't get it.

    Can anyone tell me how to solve it?

    Thank you very much!
     
    Last edited: Mar 8, 2009
  2. jcsd
  3. Mar 8, 2009 #2
    Should I need to set t -> infinity to get the result?

    However, N(t) -> 365 (but not 361) when t -> infinity.
     
  4. Mar 9, 2009 #3
    It is due to round-off error. If you use the rounded a=5.89 to calculate the total catch, you will get 361.
     
  5. Mar 10, 2009 #4
    I got it!

    Thank you very much!
     
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