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Thread: 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 04:11 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 04:12 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 04:12 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 04:12 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 04:12 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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