Asked by Artis
Can you give me step by step instruction on how to solve this problem
Helium is pumped into a spherical balloon at a constant rate of 4cubic 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?
Helium is pumped into a spherical balloon at a constant rate of 4cubic 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?
Answers
Answered by
Reiny
given: dV/dt = 4 ft^3/sec
V= (4/3)πr^3
dV/dt = 4πr^2 dr/dt
3 min = 180 sec
So after 3 min, V = 180(4) or 720 ft^3
720 = (4/3)πr^3
solve for r, then sub into
dV/dt = 4πr^2 dr/dt
solve for dr/dt
For the second part, replace dr/dt by 120
That way you will be able to solve for r
Using that r, find the Volume,
divide that volume by 4 to get the number of seconds it took to fill it that much.
V= (4/3)πr^3
dV/dt = 4πr^2 dr/dt
3 min = 180 sec
So after 3 min, V = 180(4) or 720 ft^3
720 = (4/3)πr^3
solve for r, then sub into
dV/dt = 4πr^2 dr/dt
solve for dr/dt
For the second part, replace dr/dt by 120
That way you will be able to solve for r
Using that r, find the Volume,
divide that volume by 4 to get the number of seconds it took to fill it that much.
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