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Home  >>  CBSE XII  >>  Math  >>  Application of Derivatives
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A balloon, which always remains spherical, has a variable diameter $\large\frac{3}{2}$$(2x+1)$. Find the rate of change of its volume with respect to $x.$

$\begin{array}{1 1} (A)\;\frac{27}{8}\pi(2x+1)^2 \\ (B)\;\frac{-27}{8}\pi(2x+1)^2 \\ (C)\;\frac{-27}{4}\pi(2x+1)^2 \\ (D)\;\frac{27}{4}\pi(2x+1)^2 \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:
Diameter of the sphere =$\large\frac{3}{2}$$(2x+1)$
Therefore radius $r=\large\frac{3}{4}$$(2x+1)$
Volume of the sphere $v=\large\frac{4}{3}$$\pi r^3$
Substituting for $r$ we get,
$v=\large\frac{4}{3}$$\pi\big[\large\frac{3}{4}$$(2x+1)\big]^3$
Step 2:
Differentiating w.r.t $x$ on both sides,
$\large\frac{dv}{dx}=\frac{4}{3}$$x\times \big(\large\frac{3}{4}\big)^3$$\times 3(2x+1)^2(2)$
$\quad\;\;=\large\frac{27}{8}$$\pi(2x+1)^2$
answered Jul 8, 2013 by sreemathi.v
 

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