Question

A sperical balloon is being blown up in such a way that it's volume increases at the constant rate of 2 cubic metres per minute. Find the rate at which the radius is increasing at the instant when it is 3 metres. (HINT- the formula for the volume of a sphere of radius r is V= 3/4 πr^3)
Thanks in advance!

Answer 1

You seem to have the volum of a sphere written incorrectly its

V = 4/3 pi r^3

isn't this is case of related rates.


\[\frac{dV}{dt} = \frac{dV}{dr} \times \frac{dr}{dt} \]

so

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

so

\[2 = 4 \pi r^2 \times \frac{dr}{dt}\]

or

\[\frac{dr}{dt} = \frac{2}{2 \pi r^2} \]

now just substitute r = 3 to find dr/dt

Answer 2

oops it should read

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

substitute r = 3 into the right hand side of the equation to find dr/dt