Final answer:
To determine the rate at which the surface area of the balloon is increasing, we first use the formula S = 4πr^2, take the derivative with respect to time, and then plug in the given values. The surface area is increasing at a rate of 16π cm^2/sec when the radius is 4 cm.
Step-by-step explanation:
The question asks at what rate the surface area of a spherical balloon increases given that the radius of the balloon is expanding at a constant rate. We will apply the concept of related rates in calculus to solve this problem.
Step-by-Step Solution:
- First, we recognize the formula for the surface area of a sphere, which is S = 4πr^2, where S is the surface area and r is the radius of the sphere.
- To determine the rate at which the surface area changes with respect to time, we take the derivative of the surface area with respect to time t: dS/dt = 8πr(dr/dt), where (dr/dt) is the rate of change of the radius over time, which is given as 0.5 cm/sec.
- Now we substitute the given radius of 4 cm into the derivative equation and the rate of change of the radius. This gives us dS/dt = 8π(4 cm)(0.5 cm/sec) = 16π cm/sec.
- The rate at which the surface area of the balloon increases when the radius is 4 cm is 16π cm^2/sec.
Thus, the surface area of the spherical balloon is increasing at a rate of 16π cm^2/sec when its radius is 4 cm.