# calculating error via linear approximation

• Sep 10th 2008, 05:22 AM
Craka
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.
$
\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 $9.6\pi {\rm cm}^2$

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?(Headbang)
• Sep 10th 2008, 06:30 AM
CaptainBlack
Quote:

Originally Posted by Craka
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.
$
\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 $9.6\pi {\rm cm}^2$

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?(Headbang)

$A(r+\delta )=A(r)+\delta A'(r)$

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

RonL