Question

Suppose a balloon of volume V and radius r is being inflated, so that V and r are both functions of time. If dv/dt is constant, what can be said about dr/dt (without calculation) as r increases.

Answer 1

If I used the chain rule I would come to the conclusion that dr/dt decreases as r increases, but in this question no calculations are allowed. I understand that if dr/dt increases as r increases dv/dt would also increase thus contradicting the assumption.

Answer 2

It depends on the shape of the thing i would imagine

Answer 3

Say it is a sphere - If a constant rate is pumped in dV/dt=+c , the radius expansion rate would have to slow down as time goes on

Answer 4

\[V = \frac{ 4 }{ 3 }\pi*r^3\]
\[\frac{ dV }{ dt }=\frac{ 4 }{ 3 }*\pi*\frac{ d }{ dr }*[r^3] * \frac{ dr }{ dt } = 4\pi*r^2*\frac{ dr }{ dt }\]

Answer 5

If dV/dt is a constant on the Left... then for larger and larger 'r' , dr/dt would have to become smaller and smaller.

4*pi*r^2 * dr/dt = constant value

Answer 6

You could say maybe, for a set time interval say 1sec, a new "Shell" of a Volume is added to the sphere.
As the balloon becomes larger, the "shell" thickness required to contain the Volume of 1sec of inflation gets thinner and thinner. So the radius must slow down expansion