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19. The volume of 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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Volume of a sphere,  V = \frac{4}{3}\pi r ^3

Given that the rate of change is constant,.

\\ \therefore \frac{dV}{dt} = c \\ \implies \frac{d}{dt} (\frac{4}{3}\pi r ^3) = c \\ \implies \int d(\frac{4}{3}\pi r ^3) = c\int dt \\ \implies \frac{4}{3}\pi r ^3 = ct + k

Now, at t = 0, r = 3, and at t = 3, r = 6

Putting these values:

\frac{4}{3}\pi (3) ^3 = c(0) + k \\ \implies k = 36\pi

Also,

 \frac{4}{3}\pi (6) ^3 = c(3) + 36\pi \\ \implies 3c = 252\pi \\ \implies c = 84\pi

Putting the values of c and k:

\\ \frac{4}{3}\pi r ^3 = 84\pi t + 36\pi \\ \implies r ^3 = (21 t + 9)(3) = 62t + 27 \\ \implies r = \sqrt[3]{62t + 27}

Posted by

HARSH KANKARIA

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