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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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If the radius of a sphere is measured as \(7 \: m\) with an error of \(0.02 \: m\), then find the approximate error in calculating its volume.

$\begin{array}{1 1} (A)\;3.90 \pi m^3 \\ (B)\;3.92\;\pi m^3 \\(C)\;3.94\; \pi\; m^3 \\ (D)\;3.9\;\pi m^3 \end{array} $

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1 Answer

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Toolbox:
  • Let $y=f(x)$
  • $\Delta y=f(x+\Delta x)-f(x)$
  • $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
  • Volume of the sphere $V=\large\frac{4}{3}$$\pi r^3$
Step 1:
Radius of the sphere=$7m$
Error in measurement of radius =$\Delta r$
$\qquad\qquad\qquad\qquad\quad\;\;\;\;\;\;=0.02m$
Volume of the sphere $V=\large\frac{4}{3}$$\pi r^3$
$\Delta V=\large\frac{dV}{dr}\times \Delta r$------(1)
$V=\large\frac{4}{3}\pi r^3$
$\large\frac{dv}{dx}=\frac{4}{3}$$\times 3r^2\times \pi$[Differentiating with respect to x]
Step 2:
Substitute the value of $\large\frac{dv}{dr}$ and $\Delta r$ in equation (1)
$\Delta V=\large\frac{4}{3}$$\times 3r^2\times \pi\times 0.02$
$\quad\quad=4\pi r^2\times 0.02$
$\quad\quad=4\pi 7^2\times 0.02$
$\quad\quad=4\pi\times 49\times 0.02$
$\quad\quad=392\pi \times 0.01$
$\quad\quad=3.92\pi m^3$
answered Aug 6, 2013 by sharmaaparna1
 

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