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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

Sagot :

LammettHashidn

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

[tex]V = \dfrac{4\pi}3\left({r_A}^3 + {r_B}^3\right)[/tex]

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

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

[tex]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}[/tex]

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

[tex]\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}}[/tex]

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