Question

The surface area of a spherical balloon,being inflated changes at a rate proportional to time the if initially its radius is 3 units and after 2 seconds it is 5 units, find the radius after t seconds.

Answer 1

Surface area of sphere,$A=4\pi r^2$
Given $\frac{dA}{dt}=Kt$
$\Rightarrow =4\pi 2r.\frac{dr}{dt}=kt$
$\Rightarrow 8\pi \int rdr=k\int tdt$
$\Rightarrow 4\pi r^2=k{t}^2+c$
$t=0\Rightarrow r=3$
$\therefore 4\pi \times 9=c$
$\Rightarrow c=36\pi$
$\therefore r^2=k^2+36\pi$
$t=2\Rightarrow r=5$
$\Rightarrow 4k=64\pi$
$\Rightarrow k=16\pi$
$\therefore 4\pi r^2=16\pi t^2+36\pi$
$\Rightarrow r^2=4t^2+9$
$\Rightarrow r=\sqrt{4t^2+9}$