Key Concepts and Formulas

• Volume of a Sphere: $V = \frac{4}{3}\pi r^3$, where $r$ is the radius.
• Related Rates: Using implicit differentiation to find the relationship between the rates of change of different variables.
• Chain Rule: $\frac{d}{dt}f(g(t)) = f'(g(t)) \cdot g'(t)$

Step-by-Step Solution

Step 1: Define variables and state given information

Let $R$ be the radius of the iron ball, $h$ be the thickness of the ice, and $V$ be the volume of the ice. We are given that $R = 10$ cm and $\frac{dV}{dt} = -50$ cm$^3$/min (negative because the volume is decreasing). We want to find $\frac{dh}{dt}$ when $h = 5$ cm.

Step 2: Express the volume of the ice as a function of its thickness

The ice forms a spherical shell around the iron ball. The radius of the outer sphere (iron ball + ice) is $R + h = 10 + h$. The volume of the ice is the difference between the volume of the outer sphere and the volume of the iron ball: $V = \frac{4}{3}\pi (R+h)^3 - \frac{4}{3}\pi R^3$ Substituting $R=10$ cm: $V = \frac{4}{3}\pi (10+h)^3 - \frac{4}{3}\pi (10)^3$ $V = \frac{4}{3}\pi [(10+h)^3 - 1000]$ This equation expresses the volume of the ice, $V$, as a function of its thickness, $h$.

Step 3: Differentiate the volume equation with respect to time

Differentiate both sides of the equation with respect to $t$: $\frac{dV}{dt} = \frac{d}{dt} \left( \frac{4}{3}\pi [(10+h)^3 - 1000] \right)$ $\frac{dV}{dt} = \frac{4}{3}\pi \frac{d}{dt} [(10+h)^3 - 1000]$ Using the chain rule, we get: $\frac{dV}{dt} = \frac{4}{3}\pi \left[ 3(10+h)^2 \frac{dh}{dt} - 0 \right]$ $\frac{dV}{dt} = 4\pi (10+h)^2 \frac{dh}{dt}$ This equation relates the rate of change of the volume of the ice to the rate of change of its thickness.

Step 4: Substitute the given values and solve for $\frac{dh}{dt}$

We are given $\frac{dV}{dt} = -50$ cm$^3$/min and $h = 5$ cm. Substituting these values into the equation: $-50 = 4\pi (10+5)^2 \frac{dh}{dt}$ $-50 = 4\pi (15)^2 \frac{dh}{dt}$ $-50 = 4\pi (225) \frac{dh}{dt}$ $-50 = 900\pi \frac{dh}{dt}$ Solving for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{-50}{900\pi}$ $\frac{dh}{dt} = \frac{-1}{18\pi}$

Since the question asks for the rate at which the thickness decreases, we take the absolute value of the negative rate of change: $\left| \frac{dh}{dt} \right| = \frac{1}{18\pi} \text{ cm/min}$

Common Mistakes & Tips

• Sign Convention: Remember to include the negative sign for decreasing quantities. Failing to do so will result in an incorrect sign for $\frac{dh}{dt}$.
• Chain Rule Application: Ensure the chain rule is applied correctly when differentiating $(10+h)^3$ with respect to time.
• Units: Always keep track of the units to ensure consistency and to verify the final answer's unit is correct.

Summary

We used the concept of related rates and implicit differentiation to find the rate at which the thickness of the ice decreases. We related the volume of the ice to its thickness, differentiated with respect to time, and then plugged in the given values to solve for the unknown rate. The rate at which the thickness of the ice decreases is $\frac{1}{18\pi}$ cm/min.

Final Answer: The final answer is $\boxed{\frac{1}{18\pi}}$, which corresponds to option (A).