A spherical balloon , initially inflated to a volume of over 30000 cubic meters, springs a leak which causes the radius to decrease at a rate of 2 meters per minute. At the instant when the radius is 3 meters , how fast ( in cubic meters per minute ) is the hor air escaping from the balloon?
VOLUME=4/3 pi R^3
dV/dt=4/3 pi 3r^2 dr/dt=4PIr^2 dr/dt
Put in dr/dt, r, and calculate dv/dt
To find the rate at which the air is escaping from the balloon, we need to calculate the rate of change of the volume with respect to time.
We are given that the radius is decreasing at a rate of 2 meters per minute. We can use this information to find the rate at which the volume is changing, as the volume of a sphere is given by the formula V = (4/3) * π * r^3, where V is the volume and r is the radius.
Firstly, let's set up the given information:
Initial volume (V0) = 30000 cubic meters
Rate of change of radius (dr/dt) = -2 meters per minute (negative sign because the radius is decreasing)
Radius when we want to find the rate of air escaping (r) = 3 meters
To find the rate of change of volume (dV/dt), we can differentiate the volume equation with respect to time:
dV/dt = d/dt [(4/3) * π * r^3]
Applying the chain rule, we get:
dV/dt = (4/3) * π * 3 * (dr/dt) * r^2
Substituting the given values:
dV/dt = (4/3) * π * 3 * (-2) * 3^2
= -72π cubic meters per minute
Therefore, the rate at which the hot air is escaping from the balloon is -72π cubic meters per minute. The negative sign indicates that the volume is decreasing.