To find the rate of increase of the radius of the spherical ball when its volume is increasing, we start with the formula for the volume of a sphere:
V = (4/3)πr³
Here, V is the volume and r is the radius of the sphere. We know that the volume is increasing at a rate of dV/dt = 4π cc/sec.
We need to find dr/dt, the rate of change of the radius with respect to time.
To do this, we differentiate the volume formula with respect to time t:
dV/dt = 4πr² (dr/dt)
Now, we can express dr/dt in terms of dV/dt and r:
dr/dt = (dV/dt) / (4πr²)
Next, we need to find the value of r when V = 288π cc. We can solve for r:
288π = (4/3)πr³
Canceling π from both sides, we get:
288 = (4/3)r³
Multiplying both sides by (3/4):
r³ = 288 * (3/4) = 216
Now, taking the cube root:
r = 6 cm
Now we can substitute r and dV/dt into our equation for dr/dt:
dr/dt = (4π) / (4π(6)²)
Simply the equation:
dr/dt = 4π / (144π) = 4 / 144 = 1 / 36
Thus, the rate of increase of the radius when the volume is 288π cc is:
dr/dt = 1/36 cm/sec