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If the radius of a sphere is measured as $70$ cm with an error of $0.03$cm, then find the approximate error in calculating its surface area

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)\ $$\Delta x$
  • Surface area=S=$4\pi r^2$
Step 1:
Radius of the sphere =$70$ cm
Error in measurement =$\Delta r = 0.03$ cm
Surface area=S=$4\pi r^2$
$\Rightarrow \large\frac{dS}{dr}$$=8\pi r$ [Differentiating with respect to r]
Step 2:
$\Delta S=\large\frac{ds}{dr}$$\times \Delta r$
$\quad\;\;=8\pi r\times 0.03$
$\quad\;\;=8\pi \times 70 \times 0.03$
$\quad\;\;= 52.8\; cm^2$


answered Dec 22, 2013 by balaji.thirumalai

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