The surface area of a balloon of spherical shape being inflated,...

Let surface area of the spherical balloon A = 4\pi{r}^{2}$</p><p>\dfrac{dA}{dt} =$8\pir$\dfrac{dr}{dt} =k$ (let) $. . . \left(1\right)</p><p>On integrating on both sides w.r.t$t$, we get</p><p>$4\pi{r}^{2} = kt + C$. </p><p>Given that, at$t = 0, r = 3$.</p><p>$⇒ 36\pi = C$</p><p>Also given that, at$t = 5, r = 7$</p><p>$⇒ 4\pi \times 49 = 5k + 36\pi$</p><p>$⇒ 5k = 4\pi $\left(49 - 9\right)$ ⇒ 5k = 4\pi \times 40$</p><p>$⇒ k = 32\pi$</p><p>On substituting$kvalue in $equation \left(1\right)we get,$4\pi{r}^{2} = 32\pit + 36\pi$</p><p>$⇒ {r}^{2} = 8t + 9$</p><p>Given$t = 9$.</p><p>$⇒ {r}^{2} = 81 ⇒ r = 9$.</p>$