A spherical balloon is inflated so that its volume is increasing at the rate of 2.7 ft3/min. How rapidly is the diameter of the balloon increasing when the diameter is 1.3 feet
The formula for volume of a sphere is [tex]V=\frac{4}{3}\pi r^{3}[/tex] [tex]\rightarrow\frac{dV}{dt}=4\pi r^{2}\frac{dr}{dt}[/tex]
If the diameter is 1.3, then [tex]r=\frac{1.3}{2}=0.65[/tex]
We are given [tex]\frac{dV}{dt}=2.7[/tex] and [tex]r=0.65[/tex] [tex]\rightarrow2.7=4\pi(0.65)^{2}\frac{dr}{dt}[/tex] [tex]\rightarrow\frac{dr}{dt}=\frac{2.7}{4\pi(0.65)^{2}}\approx0.5085\ ft/min[/tex]
Change in diameter is twice change in radius, so we get 2*0.5085=1.017 ft/min