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Thread: 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 $\displaystyle \sqrt{\frac{172}{14^2}} = \frac1{14} \sqrt{172}$ and approximate $\displaystyle \sqrt{\frac{172}{14^2}}$ by using the Taylor series $\displaystyle \sqrt{x} = 1 + \frac12 (x-1) - \frac18(x-1)^2+\cdots$ which is valid for $\displaystyle |x-1|<1$. You would get $\displaystyle \approx13.1166181$ by this method.
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  3. #3
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    Hello, mzkmikey!

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


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

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


    Let $\displaystyle x = 169,\;dx = 3$


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

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


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

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



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


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

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  4. #4
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    thank you
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