Question

The radius of a spherical balloon a is increasing at a constant rate of 0.1 cm/s. another balloon, b, is being continuously deflated in such a way that the total volume of a and b remains constant. suppose that at the beginning the radius of a is 10 cm and that of b is 9 cm. find the rate of change of the radius of b when the radius of a is 12 cm

Answer 1

Let $r_A$ and $r_B$ be the respective radii of balloons A and B. If the fixed total volume is V, then

$V = \dfrac{4\pi}3\left({r_A}^3 + {r_B}^3\right)$

and knowing $r_A=10\,\rm cm$ and $r_B=9\,\rm cm$ at the start, we have V = 6916π/3 cm³. Then when $r_A=12\,\rm cm$, the radius of the other sphere is $r_B=1\,\rm cm$.

Differentiating both sides with respect to time t gives a relation between the rates of change of the radii:

$0 = 4\pi \left({r_A}^2 \dfrac{dr_A}{dt} + {r_B}^2 \dfrac{dr_B}{dt}\right) \implies \dfrac{dr_B}{dt} = -\left(\dfrac{r_A}{r_B}\right)^2 \dfrac{dr_A}{dt}$

We're given $\frac{dr_A}{dt} = 0.1\frac{\rm cm}{\rm s}$ the whole time. At the moment $r_A=12\,\rm cm$, the radius of balloon B is changing at a rate of

$\dfrac{dr_B}{dt} = -\left(\dfrac{12\,\rm cm}{1\,\rm cm}\right)^2 \left(0.1\dfrac{\rm cm}{\rm s}\right) = \boxed{-14.4 \dfrac{\rm cm}{\rm s}}$