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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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The radius of an air bubble is increasing at the rate of \( \large\frac{1}{2}\)cm/s. At what rate is the volume of the bubble increasing when the radius is $1\; cm$?

$\begin{array}{1 1} (A)\;\large\frac{1}{2\pi}cm^3/s \\ (B)\;\large\frac{-1}{2\pi}cm^3/s \\ (C)\;-2 \; cm^3/s \\ (D)\;2\pi \; cm^3/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 : The radius of the air bubble is increasing at the rate of $\large\frac{1}{2}$$cm/s$
(i.e)$\large\frac{dr}{dt}=\frac{1}{2}$$cm/s$
Radius $r=1cm$
Volume of the sphere $v=\large\frac{4}{3}$$\pi r^3$
Differentiating w.r.t $t$ on both sides we get,
$\large\frac{dv}{dt}=\frac{4}{3}$$\times \pi\times 3r^2.\large\frac{dr}{dt}$
Step 2:
Substituting for $r$ and $\large\frac{dr}{dt}$ we get,
$\large\frac{dv}{dt}=\frac{4}{3}$$\times \pi\times 3\times (1)^2\times \large\frac{1}{2}$
$\Rightarrow \large\frac{dv}{dt}$$=2\pi cm^3/s$
The rate at which the volume of the air bubble increases is $2\pi cm^3/s$
answered Jul 8, 2013 by sreemathi.v
 

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