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Thread: Using Differentials

  1. #1
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    Using Differentials

    The problem is to estimate $\displaystyle y + \Delta{y}$ given that $\displaystyle y = x^2 - 5x$, x = 4, and $\displaystyle \Delta{x} = 0.01$

    I have tried this and found that the approximate change in y is dy = 0.03 and the actual change in y is 0.0301, but I am not sure what to do with this information.
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  2. #2
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    Re: Using Differentials

    $\displaystyle y = x^2 - 5x$

    $\displaystyle dy = (2x-5)dx$

    at $\displaystyle x = 4$ , $\displaystyle y = -4$ and $\displaystyle \Delta y \approx dy = 0.03$

    $\displaystyle y + \Delta y \approx y + dy = -4 + 0.03 = -3.97$

    (note that y(4.01) = -3.9699 for reference)
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  3. #3
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    Re: Using Differentials

    Oh, I see. I wonder why they didn't write $\displaystyle y(4) + \Delta{x}$ or something to that effect; it would have been a little easier for a simpleton like me.
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  4. #4
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    Re: Using Differentials

    Telling you that "x= 4" does tell you that "y" refers to "y(4)".
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