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Math Help - Approximating Squareroot

  1. #1
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    Approximating Squareroot

    Can someone please show me how to approximate the square root of 172, by using calculus. thank you
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  2. #2
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    You can use Newton's method. Alternatively, write \sqrt{\frac{172}{14^2}} = \frac1{14} \sqrt{172} and approximate \sqrt{\frac{172}{14^2}} by using the Taylor series \sqrt{x} = 1 + \frac12 (x-1) - \frac18(x-1)^2+\cdots which is valid for |x-1|<1. You would get \approx13.1166181 by this method.
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  3. #3
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    Hello, mzkmikey!

    Approximate \sqrt{172} using calculus.
    I believe they want us to use differentials.


    We have the function: . f(x) \;=\;\sqrt{x} \;=\; x^{\frac{1}{2}}

    The differential is: . df \;=\;\tfrac{1}{2}x^{-\frac{1}{2}}dx \;=\;\frac{dx}{2\sqrt{x}}


    Let x = 169,\;dx = 3


    We know that: . f(172) \;=\;f(169 + 3) \;=\;f(169) + \Delta x

    . . where \Delta x is actual difference between \sqrt{172} and \sqrt{169}


    And we know:. . f(172) \;=\;f(169+3) \;\approx\; f(169) + dx .[1]

    . . where dx is an approximation of \Delta x.



    We have: . \begin{Bmatrix}f(169) &=& \sqrt{169} &=& 13 \\ \\[-3mm]<br />
\dfrac{dx}{2\sqrt{x}} &=& \dfrac{3}{2\sqrt{169}} &=& \dfrac{3}{26} \end{Bmatrix}


    Therefore: . f(172) \;\approx\;13 + \frac{3}{26} \;=\;\boxed{13.11538462}

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  4. #4
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    thank you
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