The surface area of a balloon of spherical shape being inflated, increases at a constant rate. If initially the radius of balloon is 3 units and after 5 seconds, it becomes 7 units, then its radius after 9 seconds is__

The surface area of the spherical baloon is $A=4{\mathrm{\pi r}}^2$

Differentiate both sides with respect to time $t$

$\frac{dA}{dt}=4\mathrm{\pi }\left(2\mathrm{r}\right)\frac{\mathrm{dr}}{\mathrm{dt}}$

This is given as constant, it is equal to $k$

$\frac{dA}{dt}=k,\Rightarrow \frac{dr}{dt}=\frac{k}{8\mathrm{\pi r}}$

Hence, $A=kt +C$

Initially $A=36\mathrm{\pi }$, so that $C=36\mathrm{\pi }$

Hence, $A=kt + 36\mathrm{\pi }$

Get the value of $k$ using $t=5, r=7$

Therefore, $A=32\mathrm{\pi t}+36\mathrm{\pi }$

it implies that $r^2=8t+9$

Susbtitute $t=9$

$r^2=81\Rightarrow r=9$