Key Concepts and Formulas

• The volume of a sphere is given by $V = \frac{4}{3}\pi r^3$, where $r$ is the radius.
• The surface area of a sphere is given by $A = 4\pi r^2$, where $r$ is the radius.
• The chain rule of differentiation: $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt}$

Step-by-Step Solution

Step 1: Define Variables and Given Information

Let $r$ be the radius of the chocolate ball and $x$ be the thickness of the ice-cream layer. Then the radius of the outer sphere (chocolate ball + ice-cream) is $r+x$. We are given:

• $\frac{dV}{dt} = -81 \mathrm{~cm}^3 / \mathrm{min}$ (The volume of ice-cream is decreasing, hence the negative sign)
• $\frac{dx}{dt} = -\frac{1}{4\pi} \mathrm{~cm} / \mathrm{min}$ (The thickness of the ice-cream layer is decreasing, hence the negative sign)
• $x = 1 \mathrm{~cm}$

Step 2: Express the Volume of the Ice-cream Layer

The volume $V$ of the ice-cream layer is the difference between the volume of the outer sphere and the volume of the chocolate ball: $V = \frac{4}{3}\pi (r+x)^3 - \frac{4}{3}\pi r^3$

Step 3: Differentiate the Volume with Respect to Time

Differentiating both sides of the volume equation with respect to time $t$, we get: $\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi (r+x)^3 - \frac{4}{3}\pi r^3 \right)$ Applying the chain rule: $\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3(r+x)^2 \cdot \frac{d(r+x)}{dt} - \frac{4}{3}\pi \cdot 3r^2 \cdot \frac{dr}{dt}$ $\frac{dV}{dt} = 4\pi (r+x)^2 \left( \frac{dr}{dt} + \frac{dx}{dt} \right) - 4\pi r^2 \frac{dr}{dt}$ $\frac{dV}{dt} = 4\pi (r+x)^2 \frac{dr}{dt} + 4\pi (r+x)^2 \frac{dx}{dt} - 4\pi r^2 \frac{dr}{dt}$ $\frac{dV}{dt} = 4\pi \frac{dr}{dt} \left[ (r+x)^2 - r^2 \right] + 4\pi (r+x)^2 \frac{dx}{dt}$

Step 4: Simplify and Substitute Given Values

We are given $\frac{dV}{dt} = -81$ and $\frac{dx}{dt} = -\frac{1}{4\pi}$, and $x = 1$. Substituting these values into the equation: $-81 = 4\pi \frac{dr}{dt} \left[ (r+1)^2 - r^2 \right] + 4\pi (r+1)^2 \left( -\frac{1}{4\pi} \right)$ $-81 = 4\pi \frac{dr}{dt} (r^2 + 2r + 1 - r^2) - (r+1)^2$ $-81 = 4\pi \frac{dr}{dt} (2r+1) - (r^2 + 2r + 1)$ $-81 + r^2 + 2r + 1 = 4\pi (2r+1) \frac{dr}{dt}$ $r^2 + 2r - 80 = 4\pi (2r+1) \frac{dr}{dt}$

Step 5: Realize that the radius of the chocolate ball is constant

Since the chocolate ball is not melting, its radius $r$ is constant with respect to time, which means $\frac{dr}{dt} = 0$.

Step 6: Revisit the volume equation

The mistake was in differentiating the volume of ice cream as the difference between two volumes. It is correct, but it introduces $\frac{dr}{dt}$ which is zero. The correct approach is to consider the volume of the ice cream layer as the volume that is changing. So, $V = \frac{4}{3}\pi (r+x)^3 - \frac{4}{3}\pi r^3$. Now, only $x$ is changing with time (since $r$ is constant). $\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi (r+x)^3 - \frac{4}{3}\pi r^3 \right)$ $\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3(r+x)^2 \cdot \frac{dx}{dt} - 0$ $\frac{dV}{dt} = 4\pi (r+x)^2 \frac{dx}{dt}$

Step 7: Substitute given values

We are given $\frac{dV}{dt} = -81$, $\frac{dx}{dt} = -\frac{1}{4\pi}$, and $x = 1$. Substituting these values into the equation: $-81 = 4\pi (r+1)^2 \left( -\frac{1}{4\pi} \right)$ $-81 = -(r+1)^2$ $(r+1)^2 = 81$ $r+1 = \pm 9$ Since $r$ must be positive, we have $r+1 = 9$, which gives $r = 8$.

Step 8: Calculate the Surface Area of the Chocolate Ball

The surface area of the chocolate ball is $A = 4\pi r^2$. Substituting $r = 8$, we get: $A = 4\pi (8^2) = 4\pi (64) = 256\pi$

Common Mistakes & Tips

• Be careful with signs when dealing with decreasing rates. Make sure to use negative signs appropriately.
• Remember that the radius of the inner sphere (chocolate ball) is constant.
• Distinguish between the rate of change of the total volume and the rate of change of the ice-cream volume.

Summary

We first defined the variables and then expressed the volume of the ice-cream layer in terms of the radius of the chocolate ball and the thickness of the ice-cream. Differentiating the volume with respect to time, and using the fact that the radius of the chocolate ball is constant, we solved for the radius of the chocolate ball. Finally, we calculated the surface area of the chocolate ball using the formula $A = 4\pi r^2$.

The final answer is $\boxed{256 \pi}$, which corresponds to option (D).