The surface area of a spherical balloon,being inflated changes at a rate proportional to time

From the question it is given that,

Initial radius of balloon = 1 unit

After 3 seconds radius of balloon = 2 units

So, let us assume A be the surface area of balloon,

(\( \dfrac{dA}{dt}\)) ∝ t

Then,

\( \dfrac{dA}{dt}\) = λt

\( \dfrac{d(4\pi r^2)}{dt}= λt \)

\( 8πr (\dfrac{dr}{dt}) = λt\)

Integrating on both side we get,

\( 8π ∫rdr = λ ∫t dt\)

\( 8π (\dfrac{r^2}{2}) = (\dfrac{\lambda t ^2}{2}) + c\)

4πr² = (\( \dfrac{\lambda t ^2}{2}\)) + c …. … [equation (i)]

From question, r = 1 unit when t = 0,

4π (1)² = 0 + c

4π = c

From equation (i) c = 4πr² – (\( \dfrac{\lambda t ^2}{2}\))

Then, 4πr² = (\( \dfrac{\lambda t ^2}{2}\)) + 4π … [equation (ii)]

And also form question, given r = 2 units when t = 3 sec

4π(2)² =\( \dfrac{\lambda(3)2}{2} + 4π\)

16π = \(( \dfrac{9}{2})λ + 4π\)

\( ( \dfrac{9}{2})λ\) = 12π

By cross multiplication we get,

λ = (\( \dfrac{24}{9}\)) π

λ = (\( \dfrac{8}{3}\)) π

So, now equation (ii) becomes,

4πr² = (\( \dfrac{8\pi}{6}\)) t² + 4π

4π (r² – 1) = (\( \dfrac{4\pi}{3}\)) πt²

By cross multiplication,

r² – 1 =\( (\dfrac{1}{3}) t^2\)

r² = 1 +\( (\dfrac{1}{3}) t^2\)

r = \( \sqrt{(1 + (\dfrac{1}{3})t^2)}\)