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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The volume of a cube is increasing at the rate of \( 8cm^ 3/s\) . How fast is the volume of a cube is increasing at the rate of \(8cm^ 3 /s\) . How fast is the surface area increasing when the length of an edge is $12 cm$?

$\begin{array}{1 1} (A)\;\frac{8}{3}cm^2/s \\ (B)\;\frac{-8}{3}cm^2/s \\(C)\;\large\frac{8}{3} \normalsize m^2/s \\ (D)\;\frac{-8}{3}m^2/s\end{array} $

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1 Answer

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Toolbox:
  • If $y=f(x)$,then $\large\frac{dy}{dx}$ measures the rate of change of $y$ w.r.t $x$.
  • $\big(\large\frac{dy}{dx}\big)_{x=x_0}$ represents the rate of change of $y$ w.r.t $x$ at $x=x_0$
Step 1:
Given : $\large\frac{dv}{dt}=$$8cm^3/s$
Area of a cube is $6a^2\;cm^2$
Length of the edge of the cube =12cm.
Rate of change in volume of the cube $\large\frac{dv}{dt}$$=8cm^3/s$
Volume of the cube =$a^3$
$v=a^3$
$\large\frac{dv}{dt}$$=3a^2.\large\frac{da}{dt}$
Substituting for $\large\frac{dv}{dt}$ and $a$ we get,
$8=3\times 12\times 12\times \large\frac{da}{dt}$
$\Rightarrow \large\frac{da}{dt}=\frac{8}{3\times 12\times 12}\;$$ cm/s$
$\qquad\quad=\large\frac{1}{54}$$\;cm/s$
Step 2:
Surface area of a cube =$6a^3$
Therefore rate of change in surface area is $\large\frac{dA}{dt}$
$A=6a^2$
$\large\frac{dA}{dt}=$$12a.\large\frac{da}{dt}$
Substituting for $a$ and $\large\frac{da}{dt}$ we get,
$\qquad=12\times 12\times \large\frac{1}{54}$
$\large\frac{dA}{dt}=\frac{8}{3}$$cm^2/s$
answered Jul 5, 2013 by sreemathi.v
edited Jul 5, 2013 by sreemathi.v
 

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