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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A balloon, which always remains spherical has a variable radius. Find the rate at which its volume is increasing with the radius when the later is $10\; cm$.

$\begin{array}{1 1} (A)\;400\pi\;cm^3/cm \\ (B)\;200\pi\;cm^3/cm \\(C)\;100\pi\;cm^3/cm \\ (D)\;450\pi\;cm^3/cm \end{array} $

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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:
Volume of the sphere is $v=\large\frac{4}{3}$$\pi r^3$
Radius $r=10cm$
Hence differentiating w.r.t $r$ on both sides,
$\large\frac{dv}{dr}=\large\frac{4}{3}$$\pi 3r^2$
Step 2:
Substituting the values for $r$ we get,
$\large\frac{dv}{dr}$$=\large\frac{4}{3}$$\pi\times 3(10)^2$
$\qquad=400\pi cm^3/cm$
Hence the rate at which the volume is increasing is $400\pi cm^3/cm$
answered Jul 8, 2013 by sreemathi.v
 

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