Final answer:
To find the rate at which the balloon's radius is increasing, differentiate the volume of a sphere with respect to radius and use the given volume rate of change. By calculating dr/dt using the chain rule, the radius is found to be increasing at a rate of ⅔/(25π) cm/sec when the radius is 5 cm.
Step-by-step explanation:
To solve for the rate at which the balloon's radius is increasing when the radius is 5 cm, we first need to recognize that we are dealing with a volume rate of change which involves calculus. The volume V of a sphere (which we can model our balloon as) is given by the formula V = ⅔πr^3, where r is the radius. The rate of change of the volume with respect to time, dV/dt, is given as 4 cm³/sec.
The rate of change of the radius with respect to time, dr/dt, can be found by differentiating the volume with respect to the radius and then using the chain rule to relate dV/dt to dr/dt. Differentiating both sides of the formula for the volume of a sphere with respect to r gives dV/dr = 4πr^2. Then, by the chain rule, dV/dt = (dV/dr)(dr/dt), and substituting the known rate of 4 cm³/sec for dV/dt gives us 4 = 4π(5)^2(dr/dt).
Solving for dr/dt, we find that dr/dt = ⅔/(25π) cm/sec. Thus, the radius of the water balloon is increasing at a rate of ⅔/(25π) cm/sec when the radius is 5 cm.