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If the radius of a sphere is measured as \(9 \: m\) with an error of \(0.03 \: m\), then find the approximate error in calculating its surface area.

This question has appeared in model paper 2012

1 Answer

  • Let $y=f(x)$
  • $\Delta x$ denote a small increment in $x$
  • $\Delta y=f(x+\Delta x)-f(x)$
  • $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
  • Surface area=S=$4\pi r^2$
Step 1:
Radius of the sphere =$9m$
Error in measurement =$\Delta r$
Surface area=S=$4\pi r^2$
$S=4\pi r^2$
$\large\frac{dS}{dr}$$=8\pi r$[Differentiating with respect to r]
Step 2:
$\Delta S=\large\frac{ds}{dr}$$\times \Delta r$
$\quad\;\;=8\pi r\times 0.03$
$\quad\;\;=8\pi \times 9\times 0.03$
$\quad\;\;=72\pi \times 0.03$
$\quad\;\;=2.16\pi m^2$


answered Aug 6, 2013 by sreemathi.v
edited Aug 6, 2013 by sharmaaparna1

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