55. The volume of the 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 balloon after t seconds.

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Vishal Baghel | Contributor-Level 10

8 months ago

Let 'r' and U be the radius and volume of the spherical balloon.

Then, $\frac{d U}{d t} = k,$ k = constant

$\begin{array}{l} \frac{d}{d t} (\frac{4}{3} π r^{3}) = k \\ \Rightarrow 4 π r^{2} \frac{d r}{d t} = k \\ \Rightarrow 4 π r^{2} d r = k d t \end{array}$

Integrating both sides,

$\begin{array}{l} \int 4 π r^{2} d r = \int k d t \\ \Rightarrow \frac{4}{3} π r^{3} = k t + c \end{array}$

Given at t = 0, r = 3

So, 4π(3)³ = c

C = 36π

And, at t=3, r=6

So, $\frac{4}{3} π {(6)}^{3} = 3 k + 36 π (c = 36 π)$

$\begin{array}{l} \Rightarrow 288 π - 36 π = 3 k \\ \Rightarrow k = \frac{252 π}{3} = 84 π \end{array}$

Hence, putting value of c and k in,

$\frac{4}{3} π r^{3} = k t + c$ , we get,

$\begin{array}{l} \frac{4}{3} π r^{3} = 84 π . t + 36 π \\ \Rightarrow r^{3} = \frac{3}{4 π} (84 π . t + 36 π) \\ \Rightarrow r^{3} = 63 t + 27 \\ \Rightarrow r = {[63 t + 27]}^{\frac{1}{3}} \end{array}$

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