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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The approximate change in the volume of a cube of side \(x\) metres caused by increasing the side by $3\%$ is

\[(A)\; 0.06 x^3 \: m^3 \quad (B)\; 0.6 x^3 \: m^3 \quad (C)\; 0.09 x^3 \: m^3 \quad (D)\; 0.9 x^3 \: m^3\]

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  • Let $y=f(x)$
  • $\Delta y=f(x+\Delta x)-f(x)$
  • $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
  • Volume of the cube=$x^3m^3$
Step 1:
Side of a cube =$x$
Increase in the side=3%
Volume of the cube=$x^3m^3$
$\Delta x$=3% of $x=0.03x$
$\Delta V$ be the change in volume.
Step 2:
$\Delta V=\large\frac{dV}{dx}$$\times \Delta x$----(1)
$\large\frac{dV}{dx}=$$3x^2$[Differentiating with respect to x]
Step 3:
Substitute the value of $\large\frac{dV}{dx}$ in equation (1) we get,
$\Delta V=3x^2\times 0.03x$
$\quad\;\; =0.09x^3m^3$
$\Delta V=0.09x^3m^3$
Part (C) is the correct answer.
answered Aug 6, 2013 by sreemathi.v
edited Aug 7, 2013 by sharmaaparna1

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