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Thread: calculating error via linear approximation

  1. #1
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    calculating error via linear approximation

    Trying to understand this. Problem states "The radius of a circular disk is given as 24cm with a maximum error of 0.2cm. An estimate, via linearisation, of the maximum error in the calculated area of the disk is ?

    This is what I have done and think it's right.
    $\displaystyle
    \begin{array}{l}
    A = \pi r^2 \\
    A'(r) = 2\pi r \\
    \Delta r = 0.2 \times r = 0.2r \\
    \Delta A = A'(r) \times \Delta r = 2\pi r \times 0.2r = 0.4\pi r^2 = 0.4A \\
    \end{array}
    $

    The answer is given as being $\displaystyle 9.6\pi {\rm cm}^2 $

    But I would have thought it to be $\displaystyle r^2 \pi \times 0.4 = 0.4 \times (24)^2 \pi = 576 \times 0.4\pi = 230.4\pi$

    Am I completely out of my mind? What am I not getting here?
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Craka View Post
    Trying to understand this. Problem states "The radius of a circular disk is given as 24cm with a maximum error of 0.2cm. An estimate, via linearisation, of the maximum error in the calculated area of the disk is ?

    This is what I have done and think it's right.
    $\displaystyle
    \begin{array}{l}
    A = \pi r^2 \\
    A'(r) = 2\pi r \\
    \Delta r = 0.2 \times r = 0.2r \\
    \Delta A = A'(r) \times \Delta r = 2\pi r \times 0.2r = 0.4\pi r^2 = 0.4A \\
    \end{array}
    $

    The answer is given as being $\displaystyle 9.6\pi {\rm cm}^2 $

    But I would have thought it to be $\displaystyle r^2 \pi \times 0.4 = 0.4 \times (24)^2 \pi = 576 \times 0.4\pi = 230.4\pi$

    Am I completely out of my mind? What am I not getting here?
    $\displaystyle A(r+\delta )=A(r)+\delta A'(r)$

    The maximum error in the area of the disc is therefore approximatly $\displaystyle \delta A'(r)$ , where $\displaystyle r=24$ and $\displaystyle \delta=0.2$ cm

    RonL
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