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Math Help - Derivative to find rate of change/population.

  1. #1
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    Derivative to find rate of change/population.

    I'm just looking for someone to point me in the right direction for this question. It's easy in theory but I'm stuck on it for some reason!

    8. The fish population in a lake can be modeled by the function p(t) = 15(t^2 + 30)(t + 8), where t is time in years from now.
    a) Determine the rate of change of the fish population when there are 5000 fish in the lake.
    b) When will the fish population double from its current level? What is the rate of change in the fish population at this time?

    So p'(t)= 45t^2 + 240t + 450

    a) 0= 15t^3 + 120t^2 + 450t - 1400
    b) 0= 15t^3 + 120t^2 + 450t - 3600

    From there I just can't seem to factor and work out the answer. Thanks.
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  2. #2
    MHF Contributor Unknown008's Avatar
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    a) look for:

    5000 = 15t^3 + 120t^2 + 450t + 3600
    Solve for t. Then use this value in p'(t)

    b) Find the number of fish at the current time, that is when t = 0. Let that amount by Y

    Double it, and solve for Y = 15t^3 + 120t^2 + 450t + 3600

    Then, use that time in p'(t).
    Last edited by Unknown008; October 18th 2010 at 08:41 AM. Reason: Typo
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  3. #3
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    Thanks for the reply.
    "a) 0= 15t^3 + 120t^2 + 450t - 1400
    b) 0= 15t^3 + 120t^2 + 450t - 3600

    From there I just can't seem to factor and work out the answer. "

    I've set up my equations to solve but I just can't seem to factor them.
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  4. #4
    MHF Contributor Unknown008's Avatar
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    Factor 5.

    0= 3t^3 + 24t^2 + 90t - 280

    Ok... Hm... Maybe use an iteration?

    t_{n+1} = \sqrt{\dfrac{280-90t}{3t +24}}

    I get:

    t = 1.91

    Starting from 1.9

    (I used wolframalpha to get an idea of the first point to use)

    b) 0 = 15t^3 + 120t^2 + 450t - 3600

    Well, I use nearly the same iteration.

    t_{n+1} = \sqrt{\dfrac{240-30t}{t +8}}

    I get t = 3.45

    You can now find the rate of growth with the two values of t now.
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  5. #5
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    So far we've learned to use the factor theorem for higher degree polynomials, which is apparently why I couldn't work this out! Thanks for your help!
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  6. #6
    MHF Contributor Unknown008's Avatar
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    I can give an example to iteration if you like

    It's a way to find roots that are otherwise not easy to get.

    If you have 0 = x^3 -27

    The usualy way is to make it like this: x^3 = 27

    Hence, x = ^3\sqrt{27} = 3

    The same principle is used, but with x.

    0 = 15t^3 + 120t^2 + 450t - 3600

    0 = t^3 + 8t^2 + 30t - 240

    By preference, I find this way better:

    t^3 + 8t^2 = 240 - 30t

    Factorise t^2;

    t^2(t + 8) = 240 - 30t

    Divide by (t+8) then take the square root.

    t = \sqrt{\dfrac{240 - 30t}{t+8}}

    Now, put a random number not too far fetched. The previous number had 5000 and t was 1.9. I take 3 for 7200.

    t_{n+1} = \sqrt{\dfrac{240 - 30t_n}{t_n+8}}

    Take t_n = 3

    t_{1} = \sqrt{\dfrac{240 - 30(3)}{(3)+8}} = 3.69

    It's quite close to 3 itself... take this number into the equation now.

    t_{2} = \sqrt{\dfrac{240 - 30(3.69)}{(3.69)+8}} = 3.32

    Again, do the same thing.

    t_{3} = \sqrt{\dfrac{240 - 30(3.32)}{(3.32)+8}} = 3.51

    As you can see, the values oscillate about a number. This number is the solution we are looking for. Continue this process until the desired accuracy is obtained and we get t = 3.45 at 3 significant figures
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