A spherical balloon is inflected with gas at the rate of 20cm^3 per mint. How fast the radius of the balloon changing at the when the radius is 2cm?
We can use the formula for the volume of a sphere: V = (4/3)πr^3, where V is the volume and r is the radius.
Differentiating both sides with respect to time (t), we get:
dV/dt = 4πr^2 (dr/dt)
where dV/dt is the rate of change of volume (which is 20 cm^3/min), and dr/dt is the rate of change of radius (which we need to find).
At the given radius of 2 cm, we can plug in the values and solve for dr/dt:
20 = 4π(2)^2 (dr/dt)
dr/dt = 20 / (16π)
dr/dt ≈ 0.397 cm/min
Therefore, the radius of the balloon is changing at a rate of approximately 0.397 cm/min when the radius is 2 cm.
To solve this problem, we can use the formula for the volume of a sphere:
V = (4/3) * π * r^3
Where V is the volume of the sphere and r is the radius. We are given the rate of change of the volume, so we can differentiate the above equation with respect to time (t):
dV/dt = 4πr^2 * dr/dt
Here, dV/dt represents the rate of change of the volume, and dr/dt represents the rate of change of the radius.
Given that dV/dt = 20 cm^3/min and r = 2 cm, we can substitute these values into the equation and solve for dr/dt:
20 = 4π(2^2) * dr/dt
20 = 16π * dr/dt
Dividing both sides of the equation by 16π:
dr/dt = 20 / (16π)
Simplifying the right side:
dr/dt ≈ 0.397 cm/min
Therefore, the radius of the balloon is changing at a rate of approximately 0.397 cm/min when the radius is 2 cm.