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Math Help - Rates of change

  1. #1
    Newbie
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    Feb 2009
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    Rates of change

    I'm having trouble fully understanding the following question. Can anyone help please?


    p = e^(β0 + β1x) where β0 = 10.17 and β1 = -0.01581

    dp/dx = β1e^(β0 + β1x)


    The Question

    A small change in mileage (x) will lead to a small change Δp in price.
    They are related by the approximation:

    Δp ≈ (dp/dx) Δx

    Using this, derive an expression for the ratio of fractional change in price associated to a small change in mileage:

    (Δp/p)/Δx

    Hence find the percentage fall in price for every 1000 miles (x) travelled.
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  2. #2
    MHF Contributor
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    Hi

    \frac{\frac{\Delta p}{p}}{\Delta x} = \frac{1}{p}\:\frac{\Delta p}{\Delta x} = \frac{1}{p}\:\frac{dp}{dx} = \beta_1
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