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(A) Let the volume of the spherical balloon be $V = \frac{4}{3} \pi r^3$. Given that the rate of change of volume is constant,let $\frac{dV}{dt} = k$,

Vedclass · Accepted Answer

$(63t + 27)^{\frac{1}{3}}$

Vedclass · Answer

(A) Let the volume of the spherical balloon be $V = \frac{4}{3} \pi r^3$. Given that the rate of change of volume is constant,let $\frac{dV}{dt} = k$,where $k$ is a constant. Integrating with respect to $t$,we get $V = kt + C$. Substituting $V = \frac{4}{3} \pi r^3$,we have $\frac{4}{3} \pi r^3 = kt + C$. At $t = 0$,$r = 3$,so $\frac{4}{3} \pi (3)^3 = k(0) + C \Rightarrow C = 36 \pi$. At $t = 3$,$r = 6$,so $\frac{4}{3} \pi (6)^3 = k(3) + 36 \pi$. $288 \pi = 3k + 36 \pi \Rightarrow 3k = 252 \pi \Rightarrow k = 84 \pi$. Substituting $k$ and $C$ back into the equation: $\frac{4}{3} \pi r^3 = 84 \pi t + 36 \pi$. Dividing by $\frac{4}{3} \pi$,we get $r^3 = 63t + 27$. Therefore,$r = (63t + 27)^{\frac{1}{3}}$.

The volume of a spherical balloon being inflated changes at a constant rate. If initially its radius is $3$ units and after $3$ seconds it is $6$ units,find the radius of the balloon after $t$ seconds.

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