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Math Help - Differential equations help , rate of change

  1. #1
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    Differential equations help , rate of change

    A mathematician is selling goods at a car boot sale. She believes that the rate at
    which she makes sales depends on the length of time since the start of the sale,
    t hours, and the total value of sales she has made up to that time, x.
    She uses the model

     \frac{dx}{dt} = \frac{k(5-t)}{x}
    where k is a constant.

    Given that after two hours she has made sales of 96 in total,
    (a) solve the differential equation and show that she made 72 in the first hour of
    the sale.

    The mathematician believes that is it not worth staying at the sale once she is making sales at a rate of less than 10 per hour.
    (b) Verify that at 3 hours and 5 minutes after the start of the sale, she should have
    already left.


     \int x dx = \int k(5-t)

     \frac{1}{2}x^{2} = 5kt - \frac{1}{2}t^{2} + c

    t= 0, x =0 , c = 0

    t = 2 , x = 96 k = 576

    I really suck on part'b', how would I go about it?

    thank you
    Last edited by mash; March 6th 2012 at 05:11 PM.
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  2. #2
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    There's an error in your t integration. I would revisit that.
    Last edited by mash; March 6th 2012 at 05:12 PM.
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  3. #3
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    Although, your solution of k comes about as the result of a correct integration.
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  4. #4
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    You haven't quite solved part a) yet. What's left?
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  5. #5
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    But I get x = 72, with my equation? Also, I cant seem to spot what is wrong with my integration, could you please tell me?

    Thanks.
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  6. #6
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    Plugging in the value of t and solving for x at the appropriate time was what was required. A simple step that you've already done in your head, no doubt; however, in order to show it, you just need to write it down.

    As to the integral, it's a situation with an unclear notation. The correct t integration is k(5t-\frac{1}{2}\,t^{2})+C, not
    5kt - \frac{1}{2}t^{2} + C. See the difference?

    As for part b), try finding out what her sales rate is at the given time.
    Last edited by mash; March 6th 2012 at 05:12 PM.
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  7. #7
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    Quote Originally Posted by Ackbeet View Post
    Plugging in the value of t and solving for x at the appropriate time was what was required. A simple step that you've already done in your head, no doubt; however, in order to show it, you just need to write it down.

    As to the integral, it's a situation with an unclear notation. The correct t integration is k(5t-\frac{1}{2}\,t^{2})+C, not
    5kt - \frac{1}{2}t^{2} + C. See the difference?

    As for part b), try finding out what her sales rate is at the given time.
    Oh I see what you mean, I forgot to distribute the 'k' properly.

    Also would I be using my formula and input t = 3.05?
    Last edited by mash; March 6th 2012 at 05:12 PM.
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  8. #8
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    Not sure if you should use 3.05 or 3.08\overline{3}; I'm inclined to the latter, since you're probably using decimals in the DE, and so you'd have to convert minutes to percent of an hour.
    Last edited by mash; March 6th 2012 at 05:12 PM.
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  9. #9
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    Thank you very much, I have it worked it out now, using t= 3.0833
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