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Math Help - 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.
    <br />
\begin{array}{l}<br />
 A = \pi r^2  \\ <br />
 A'(r) = 2\pi r \\ <br />
 \Delta r = 0.2 \times r = 0.2r \\ <br />
 \Delta A = A'(r) \times \Delta r = 2\pi r \times 0.2r = 0.4\pi r^2  = 0.4A \\ <br />
 \end{array}<br />

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

    But I would have thought it to be 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.
    <br />
\begin{array}{l}<br />
A = \pi r^2 \\ <br />
A'(r) = 2\pi r \\ <br />
\Delta r = 0.2 \times r = 0.2r \\ <br />
\Delta A = A'(r) \times \Delta r = 2\pi r \times 0.2r = 0.4\pi r^2 = 0.4A \\ <br />
\end{array}<br />

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

    But I would have thought it to be 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?
    A(r+\delta )=A(r)+\delta A'(r)

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

    RonL
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