Helium is pumped into a spherical balloon at a constant rate of 2 cubic feet per second. How fast is the radius increasing after 4 minutes? At what time (if any) is the radius increasing at a rate of 140 feet per second?
2 Answers
Where did this question come from? Why isn't it simply rejected?
Explanation:
What is the temperature of the ambience. Are we at sea level or in outer hard vacuum space. What is the resistance of the balloon to being inflated? Why shouldn't the equation PV = nRT be highly relevant? And about 80 other parameters. Yeah it's fun to help people with their homework, but I don't like being a sucker.
After 4 minutes the radius is increasing at
The radius is increasing at
Explanation:
Let us set up the following variables:
# {(r, "Radius of sphere at time t","(feet)"), (V, "Volume of sphere at time t", "(feet"^3")"), (t, "time", "(sec)") :} #
Our aim is to find
The volume of a sphere is
# (dV)/(dr) = 4pir^2 #
Now we need to find a relationship between
# (dV)/(dt) = 2 #
If we treat this as a separable DE, we get:
# V = 2t + C #
And assuming that
# V=2t => 4/3pir^3=2t #
# " "=> r^3=(3t)/(2pi) #
# " "=> r=((3t)/(2pi))^(1/3) #
And Applying the chain rule we have:
# \ \ \ \(dV)/(dt) = (dV)/(dr) * (dr)/(dt) #
# :. \ \ \ 2 = 4pi((3t)/(2pi))^(2/3)* (dr)/(dt) #
# :. \ \ \ 1 = 2pi((3t)/(2pi))^(2/3)* (dr)/(dt) #
And so when
# 1 = 2pi(360/pi)^(2/3)* (dr)/(dt) #
# :. (dr)/(dt) = 0.00674608 ... = 0.00675 " feet s"^-1# (3sf)
And for the second part, when
# 1 = 2pi((3t)/(2pi))^(2/3)* 140 #
# :. t = 0.000080278 = 0.0000803 s # (3sf)
We can make sense of this if we look at the radius/time graph, and we see that initially the radius increases incredibly rapidly wrt time
graph{((3x)/(2pi))^(1/3) [-2.162, 2.165, -1.08, 1.082]}
