Answer:
3ft/min
Explanation:
Complete question:
A spherical balloon is inflating with helium at a rate of 192pi ft^3/min. How fast is the balloon's radius increasing at the instant the radius is 4ft?
Volume of a sphere = [tex]V = \frac{4}3}\pi r^3[/tex] where:
r is the radius of the sphere:
Using the chain rule to find dV/dt we have:
[tex]\frac{dV}{dt} = \frac{dV}{dr} * \frac{dr}{dt}\\\frac{dr}{dt} = \frac{(\frac{dV}{dt})}{(\frac{dV}{dr})}[/tex]
[tex]\frac{dr}{dt}[/tex] is the rate at which the balloon's radius is increasing
[tex]\frac{dV}{dt}[/tex]is the rate at which the volume is increasing
Given
[tex]\frac{dV}{dt} = 192\pi ft^3/min\\[/tex]
radius = 4ft
Required
[tex]\frac{dr}{dt}[/tex]
From the formula:
[tex]\frac{dV}{dr} = 3(\frac{4}{3})\pi r^2 \\\frac{dV}{dr} = 4\pi r^2\\\frac{dV}{dr} = 4\pi (4)^2\\\frac{dV}{dr} = 64 \pi cm^3/ft[/tex]
Get [tex]\frac{dr}{dt}[/tex]
[tex]\frac{dr}{dt} = \frac{(\frac{dV}{dt})}{(\frac{dV}{dr})}\\\frac{dr}{dt} = \frac{192\pi}{64\pi}\\\frac{dr}{dt} = 3ft/min[/tex]
Hence the balloon radius is increasing at the rate of 3ft/min.
