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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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Find the approximate change in the volume V of a cube of side \(x\) metres caused by increasing the side by $1\%$.

$\begin{array}{1 1} 0.03\; x^3m^3 \\ 0.06\; x^3m^3 \\ 0.05\; x^3m^3 \\ 0.04\; x^3m^3 \end{array} $

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1 Answer

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Toolbox:
  • Let $y=f(x)$
  • $\Delta x$ denote a small increment in $x$
  • $\Delta y=f(x+\Delta x)-f(x)$
  • $dy=\big(\large\frac{dy}{dx}\big)\Delta x$
  • Volume of a cube =$l\times b\times h$
Step 1:
Volume =V
Side of the cube=$x$ metres
Increase in side=1%
$\qquad\qquad\;\;\;\;=0.01\times x$
$\qquad\qquad\;\;\;\;=0.01 x$
Volume of a cube =$l\times b\times h$
$\qquad\qquad\qquad=x\times x\times x$
$\qquad\qquad\qquad=x^3$
Step 2:
Approximate change in volume =$\Delta V$
$\qquad\qquad\qquad\qquad\qquad\;\;=\large\frac{dv}{dx}\times \Delta x$
$V=x^3$
$\large\frac{dv}{dx}$$=3x^2$[Differentiating with respect to x]
$\Delta V=\large\frac{dv}{dx}$$\times \Delta x$
$\qquad=3x^2\times 0.01x$
$\qquad=0.03x^3m^3$
answered Aug 6, 2013 by sharmaaparna1
 

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