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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A balloon, which always remains spherical on inflation, is being inflated by pumping in $900$ cubic centimetres of gas per second. Find the rate at which the radius of the balloon increases when the radius is $15\; cm$.

$\begin{array}{1 1} (A)\;\frac{-1}{\pi}cm/s \\ (B)\;\frac{-1}{\pi}m^2/s \\ (C)\;\frac{-1}{\pi}m/s \\(D)\;\frac{1}{\pi}cm/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}=$$900cm^3/sec$
Radius $r=15cm$
Volume of the sphere is $v=\large\frac{4}{3}$$\pi r^3$
Differentiating w.r.t $t$ on both sides we get,
$\large\frac{dv}{dt}$$=\large\frac{4}{3}$$\pi3r^2\large\frac{dr}{dt}$
Step 2:
Substituting the values of $r$ and $\large\frac{dr}{dt}$ we get,
$900=\large\frac{4}{3}$$\times \pi\times 3\times(15)^2\times \large\frac{dr}{dt}$
On simplifying we get,
$\large\frac{dr}{dt}$$=\large\frac{900\times 3}{4\times \pi\times 3\times 15\times 15}$
$\quad\;=\large\frac{1}{\pi}$$cm/s$
The rate at which the radius of the balloon increases is $\large\frac{1}{\pi}$$cm/s$
answered Jul 8, 2013 by sreemathi.v
 

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