Question

A spherical balloon is inflating with helium at a rate of 8080piπ StartFraction ft cubed Over min EndFraction ft3 min. How fast is the​ balloon's radius increasing at the instant the radius is 22 ​ft? Write an equation relating the volume of a​ sphere, V, and the radius of the​ sphere, r.

Answer 1

Answer:

$\frac{dr}{dt}$=0.04132 ft/min

Step-by-step explanation:

V = volume of the sphere

r = radius of the sphere

$\frac{dV}{dt}$ = 80π ft³/min

now,

For sphere, V = $\frac{4}{3}\pi r^3$

for rate of change of volume with respect to time (t)

$\frac{dV}{dt}=\frac{d(\frac{4}{3}\pi r^3)}{dt}$

or

$\frac{dV}{dt}=3\times\frac{4}{3}\pi r^2\times\frac{dr}{dt}$

or

$\frac{dV}{dr}=4\pi r^2\times\frac{dr}{dt}$

at r = 22ft, $\frac{dV}{dt}$ = 80π ft³/min

80π = $4\pi(22)^2\times\frac{dr}{dt}$

or

$\frac{dr}{dt}$=0.04132 ft/min