Question 151280
A high-altitude spherical weather balloon expands as it rises due to the drop in atmospheric pressure. Suppose that the radius r increases at the rate of 0.03in./sec and that r = 48in. at time t = 0; Determine an equation that models the volume v of the balloon at time t, and find the volume v of the balloon at time t, and find the volume when t = 300 sec
:
Volume of a sphere: V = {{{(4/3)pi*r^3}}}
:
Volume when the balloon is launched (t=0):
V = {{{(4/3)pi*48^3}}}
V = 463,246.7 cu inches
:
The equation for this problem: t = time in seconds
V = {{{(4/3)pi*(48+.03t)^3}}}
:
For t = 300
V = {{{(4/3)pi*(48+.03*300)^3}}}
V = {{{(4/3)pi*(48+9)^3}}}
V = {{{(4/3)pi*57^3}}}
V = 775,734.6 cu inches after 300 sec