Answer to Question #195615 in Calculus for Kobz

Question

A balloon is being inflated by pumped air at the rate of 2 cubic inches per second. How fast is the diameter of the balloon increasing when the radius is ½ inch?

Answer 1

Volume of spherical balloon is given by, V = $\dfrac{4{\pi}r^{3}}{3}$ ,

Now differentiating with respect to r we get ,

$\dfrac{dV}{dr}=4{\pi}r^{2}$

$dV=4{\pi}r^{2}dr$ .....1)

$\dfrac{dV}{dt}=2$ cubic inches per second

$\dfrac{dr}{dt}=$

Radius r = $\dfrac{1}{2}$ inch

now differentiating equation 1 with respect to time , we get $\dfrac{dV}{dt}=4{\pi}r^{2}\dfrac{dr}{dt}$

now puting all the value in the given equation , we get -

= $2=$ $4{\pi}\dfrac{1}{4} \dfrac{dr}{dt}$

=$\dfrac{dr}{dt}=\dfrac{2}{\pi}$ inches ......2) , relation between diameter and radius is given by r = $\dfrac{D}{2}$

=$\dfrac{d(D)}{dt}=\dfrac{4}{\pi}\ inches$