The surface area of a spherical balloon being | vedclass

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(A) Let $r$ be the radius of the spherical balloon.The surface area $S$ is given by $S = 4 \pi r^2$.Given that the surface area increases at a constan

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$9$

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(A) Let $r$ be the radius of the spherical balloon.The surface area $S$ is given by $S = 4 \pi r^2$.Given that the surface area increases at a constant rate,we have $\frac{dS}{dt} = k$,where $k$ is a constant.Integrating with respect to $t$,we get $S = kt + C$,where $C$ is the constant of integration.Substituting $S = 4 \pi r^2$,we have $4 \pi r^2 = kt + C$.At $t = 0$,$r = 3$,so $4 \pi (3)^2 = k(0) + C \Rightarrow C = 36 \pi$.At $t = 5$,$r = 7$,so $4 \pi (7)^2 = k(5) + 36 \pi \Rightarrow 196 \pi = 5k + 36 \pi \Rightarrow 5k = 160 \pi \Rightarrow k = 32 \pi$.Thus,the equation becomes $4 \pi r^2 = 32 \pi t + 36 \pi$.Dividing by $4 \pi$,we get $r^2 = 8t + 9$.For $t = 9$,$r^2 = 8(9) + 9 = 72 + 9 = 81$.Therefore,$r = \sqrt{81} = 9$ units.

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

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