Related Rates And Surface Area

To solve this problem, let us denote the radius of the spherical chocolate ball as cm. The ice-cream layer on the chocolate ball has a uniform thickness of 1 cm, making the total radius of the chocolate ball with the ice-cream layer cm.

We are given that the ice-cream melts at a rate of 81 cm³/min, which is the rate of change of its volume. Also, the thickness of the ice-cream layer decreases at the rate of cm/min. Using these values, we can find the surface area of the chocolate ball:

1. First, write the formula for the volume of the sphere (which represents the chocolate ball with ice-cream): .
2. The rate of change of the volume, when the thickness changes, is given by: .
3. We are given that cm/min. Substituting this into the formula, we have: .
4. Since the ice-cream layer alone melts (i.e., cm³/min), we equate: .
5. Solving gives , so .
6. The surface area of the chocolate ball (without the ice-cream layer) is : cm².

Thus, the surface area of the chocolate ball without the ice-cream layer is cm².