Final answer:
The rate at which the radius of a spherical balloon increases can be determined using related rates and the formula for the volume of a sphere. When the balloon's radius is 4 cm, and it is being filled at a rate of 2 cm³/s, the radius increases at a rate of 1 / (16π) cm/s.
Step-by-step explanation:
Rate of Change of a Balloon's Radius
The student is asking about how fast the radius of a spherical balloon is increasing while it is being filled with air at a constant rate. To solve this problem, we need to use the formula for the volume of a sphere, which is V = (4/3)πr^3, where V is volume and r is the radius in centimeters. Given that the balloon is being filled at a rate of 2 cm³/s, we can use related rates to find the rate at which the radius is increasing when the radius is precisely 4 cm.
We start by taking the derivative of the volume with respect to time to get dV/dt = 4πr^2(dr/dt). Plugging in the known values, dV/dt = 2 and r = 4, we can solve for dr/dt, which will give us the rate at which the radius is increasing.
Substituting the values, we get:
2 = 4π(4)^2(dr/dt)
=> dr/dt = 2 / (4π(4)^2)
=> dr/dt = 1 / (16π)
Therefore, when the radius of the balloon is 4 cm, the radius is increasing at a rate of 1 / (16π) cm/s.
To find the rate of change of the radius as the balloon is being filled with air, we can use the formula for the volume of a sphere, which is given by V = (4/3)πr³, where V is the volume and r is the radius. Taking the derivative with respect to time, we get dV/dt = 4πr²(dr/dt). Since the rate of change of the volume is constant at 2 cm³/s, we have dV/dt = 2. Substituting the given radius (r = 4 cm) into the equation allows us to solve for the rate at which the radius is increasing, dr/dt.
dV/dt = 4πr²(dr/dt)
2 = 4π(4)²(dr/dt)
dr/dt = 2/(4π(4)²)
dr/dt ≈ 0.004 cm/s