Key Concepts and Formulas

• Volume of a Sphere: The volume $V$ of a sphere with radius $r$ is given by $V = \frac{4}{3}\pi r^3$.
• Related Rates: If two or more variables are related and each is a function of time, then their rates of change are also related. We can find these relationships by differentiating the equation relating the variables with respect to time.
• Chain Rule: If $y$ is a function of $x$, and $x$ is a function of $t$, then $\frac{dy}{dt} = \frac{dy}{dx} \cdot \frac{dx}{dt}$.

Step-by-Step Solution

1. Define Variables and Establish the Relationship

Let $R$ be the radius of the iron ball, which is constant at $R = 10$ cm. Let $x$ be the thickness of the ice layer at time $t$. The radius of the ice-covered ball is then $r = R + x = 10 + x$. The volume $V$ of the ice is the volume of the ice-covered ball minus the volume of the iron ball: $V = \frac{4}{3}\pi (10+x)^3 - \frac{4}{3}\pi (10)^3$ We want to find $\frac{dx}{dt}$ when $x = 5$ cm, given that $\frac{dV}{dt} = -50$ cm$^3$/min (since the ice is melting, the volume is decreasing).

2. Differentiate with Respect to Time

Differentiate both sides of the volume equation with respect to time $t$: $\frac{dV}{dt} = \frac{d}{dt} \left[ \frac{4}{3}\pi (10+x)^3 - \frac{4}{3}\pi (10)^3 \right]$ Since $\frac{4}{3}\pi (10)^3$ is a constant, its derivative is zero. Using the chain rule, we get: $\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3(10+x)^2 \cdot \frac{dx}{dt} = 4\pi (10+x)^2 \frac{dx}{dt}$

3. Substitute Given Values

We are given $\frac{dV}{dt} = -50$ cm$^3$/min and we want to find $\frac{dx}{dt}$ when $x = 5$ cm. Substitute these values into the equation: $-50 = 4\pi (10+5)^2 \frac{dx}{dt}$ $-50 = 4\pi (15)^2 \frac{dx}{dt}$ $-50 = 4\pi (225) \frac{dx}{dt}$

4. Solve for $\frac{dx}{dt}$

Solve for $\frac{dx}{dt}$: $\frac{dx}{dt} = \frac{-50}{4\pi (225)} = \frac{-50}{900\pi} = \frac{-1}{18\pi}$ The rate at which the thickness of the ice decreases is the absolute value of this, which is $\frac{1}{18\pi}$ cm/min.

Common Mistakes & Tips

• Sign Convention: Be careful with the sign of $\frac{dV}{dt}$. Since the volume is decreasing, $\frac{dV}{dt}$ is negative.
• Units: Always include units in your calculations to ensure consistency and avoid errors.
• Constant Radius: Remember that the radius of the iron ball itself is constant and does not change with time.

Summary

We established a relationship between the volume of the ice and its thickness, differentiated it with respect to time using the chain rule, and then substituted the given values to solve for the rate at which the thickness of the ice decreases. The rate at which the thickness of the ice decreases when the thickness is 5 cm is $\frac{1}{18\pi}$ cm/min.

Final Answer

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