When the volume of a spherical balloon is increasing at a constant rate of 10 cubic meters per hour, radius is increasing can be determined using the derivative of the volume with respect to time.
To find the rate at which the radius is increasing, we need to relate the volume and the radius of the spherical balloon. The volume of a sphere is given by the formula V = (4/3)πr^3, where V is the volume and r is the radius.
Taking the derivative of the volume with respect to time will give us the rate of change of the volume, which is 10 cubic meters per hour in this case. Let's denote the rate of change of the radius as dr/dt.
Differentiating the volume equation with respect to time, we have dV/dt = 4πr^2 (dr/dt). Since the volume is increasing at a constant rate of 10 cubic meters per hour, we can substitute dV/dt with 10.
10 = 4πr^2 (dr/dt)
Now, we can solve for dr/dt, which represents the rate at which the radius is increasing. Plugging in the given radius of 5 meters, we have:
10 = 4π(5^2)(dr/dt)
10 = 100π(dr/dt)
Simplifying the equation, we find:
dr/dt = 10/(100π)
dr/dt = 1/(10π) meters per hour
Therefore, when the radius is 5 meters, the rate at which it is increasing is approximately 1/(10π) meters per hour.