Helium is pumped into a spherical balloon at a constant rate of 2 cubic feet per

Question

Helium is pumped into a spherical balloon at a constant rate of 2 cubic feet per second. How fast is the radius increasing after 3 minutes? At what time (if any) is the radius increasing at a rate of 120 feet per second? The radius is increasing at a rate of ft/sec. (Do not round until the final answer. Then round to four decimal places as needed.) Select the correct choice below and, if necessary, fill in the answer box to complete your choice. The time at which the radius increases at a rate of 120 feet per second is sec. (Do not round until the final answer. Then round to five decimal places as needed.) There is no solution.

Explanation / Answer

V=4/3*pi*r^3

dV/dt=4pi*r^2*dr/dt

dV/dt=2

2=4*pi*r^2*dr/dt

dt=2pir^2*dr

integrasting we get

t=2pi*r^3/3

at t=0 r=0

at t=3 mins=180 s

r^3=3*180/(2*pi)

=85.94

r=4.41

dr/dt=2/(4*pi*r^2)

=2/(4*pi*4.41^2)

=0.0082 ft/s

b)

dr/dt=120

r^2=2/(4*pi*120)

=0.0013

r=0.0364

t=2*pi*(0.0364)^3/3

=0.0001 s

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