Question

A spherical balloon is being inflated at a rate of 3 cm/sec. If the volume of the balloon is 0 at time 0,at what rate is the volume increase after 5 min?

Plz help, i really don't know how to start...

Answer 1

Volume of a sphere is given by \(\large\rm V=\frac43\pi r^3\)

And we're given... the change in volume.
Ya I think that's what they mean by "inflated", air is going in at this rate.
\(\large\rm V'=3\)

Answer 2

So ummm.. hmm what can we do >.< Thinking...

Answer 3

This is being inflated so that the diameter of the balloon is increasing at a steady rate.
The rate of inflation is given in a change of the diameter 3 cm/s. or also the radius increases at half that,
dr/dt = 1.5 cm/s

Answer 4

differentiate both sides of the volume of a sphere with respect to time t

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

use the given dr/dt, and you have an equation for volume at some time t