Course:MATH102/Question Challenge/2008 December Q4

Question

Two spherical balloons are connected so that one inflates as the other deflates, the sum of their volumes remaining constant. When the first balloon has radius 10 cm and its radius is increasing at 3 cm/sec, the second balloon has radius 20 cm. What is the rate of change of the radius of the second balloon? [The volume of a sphere of radius r is $V={\frac {4}{3}}\pi r^{3}$ ].

Hints

Hint
Consider writing down an equation for the total volume (which is fixed)$V_{1}+V_{2}=V=$ constant.

Solutions

Solution

$V=4/3\pi x^{3}+4/3\pi y^{3}$ where $V=$ the total volume of the two ballons, $x=$ the radius of the first balloon, and $y=$ the radius of the second balloon $dv/dt=0$ The first step is to take the first derivative: $V=4/3\pi x^{3}+4/3\pi y^{3}$ $V'=4\pi x^{2}dx/dt+4\pi y^{2}dy/dt$ $4\pi x^{2}dx/dt=-4\pi y^{2}dy/dt$ Substituting in the given radii and rate of change of the radii: $4\pi (10)^{2}(3)=-4\pi (20)^{2}dy/dt$ Solving for $dy/dt$: $1200\pi =-1600\pi dy/dt$ $dy/dt=-3/4$