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Home  >>  CBSE XII  >>  Math  >>  Differential Equations
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The volume of spherical balloon being inflated changes at a constant rate.If initially its radius is 3units and after 3seconds it is 6units.Find the radius of balloon after t seconds.

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  • Volume of the sphere=$\large\frac{ 4}{3}$$\pi r^3$
Step 1:
let the rate of change of the volume of balloon be $k$
Hence $\large\frac{dv}{dt}$$= k$
$\large\frac{d}{dt} (\frac{4}{3}$$ \pi r^3) = k$
$\large\frac{4}{3}$$\pi(3r^2).(\large\frac{dr}{dt}) $$= k$
Now seperating the variables,
we get $4 \pi r^2.dr = k.dt$
Step 2:
Integrating on both sides we get
$4\pi \int r^2dr=k \int dt $
$4\pi r^3 = 3.(kt+C)$
Step 3:
Given $t=0,r=3$
$C= 36 \pi$
Step 4:
when $t=3, r=6$
$4\pi . 6^3 = 3.(k.3+C)$
Now Dividing throughout by 3 we get
Hence $k=84\pi$
Step 5:
Now substituting the values of $k$ and $c$
we get $4\pi r^3=3(84\pi.t+36\pi)$
Taking $4\pi$ as a common factor
Dividing throughout by $4\pi$
we get $r^3=63t+27$
Thus the radius of the balloon after t seconds is $(63t+27)^{\Large\frac{1}{3}}$
answered Aug 14, 2013 by sreemathi.v

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