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 are changing with respect to time, we can differentiate the equation relating them with respect to time to find a relationship between their rates of change.
• Chain Rule: $\frac{d}{dt}f(g(t)) = f'(g(t)) \cdot g'(t)$

Step-by-Step Solution

Step 1: Define Variables and Establish the Volume Equation

Let $R$ be the radius of the iron ball, which is constant: $R = 10$ cm. Let $h$ be the thickness of the ice layer at time $t$. The total radius of the iron ball plus the ice is $r = R + h = 10 + h$. The volume of the ice layer is the difference between the volume of the sphere with radius $r$ and the volume of the iron ball: $V = \frac{4}{3}\pi r^3 - \frac{4}{3}\pi R^3 = \frac{4}{3}\pi ( (10+h)^3 - 10^3 )$

Step 2: Differentiate the Volume Equation with Respect to Time

We are given that the ice melts at a rate of 50 cm$^3$/min, so $\frac{dV}{dt} = -50$ cm$^3$/min (since the volume is decreasing). We want to find $\frac{dh}{dt}$ when $h = 5$ cm. Differentiate the volume equation with respect to time $t$ using the chain rule: $\frac{dV}{dt} = \frac{d}{dt} \left[ \frac{4}{3}\pi ( (10+h)^3 - 10^3 ) \right] = \frac{4}{3}\pi \cdot 3(10+h)^2 \cdot \frac{dh}{dt} = 4\pi (10+h)^2 \frac{dh}{dt}$

Step 3: Substitute the Given Values and Solve for $\frac{dh}{dt}$

We have $\frac{dV}{dt} = -50$ and $h = 5$. Substitute these values into the differentiated equation: $-50 = 4\pi (10+5)^2 \frac{dh}{dt} = 4\pi (15)^2 \frac{dh}{dt} = 4\pi (225) \frac{dh}{dt} = 900\pi \frac{dh}{dt}$ Now, solve for $\frac{dh}{dt}$: $\frac{dh}{dt} = \frac{-50}{900\pi} = \frac{-5}{90\pi} = \frac{-1}{18\pi}$ The rate at which the thickness of the ice decreases is the absolute value of $\frac{dh}{dt}$, which is $\left| \frac{-1}{18\pi} \right| = \frac{1}{18\pi}$ cm/min.

Step 4: Identify and correct the error in the previous step There was an error above. The volume equation should represent the volume of the ice only. Differentiating the original volume equation with respect to time gives: $\frac{dV}{dt} = 4\pi(10+h)^2 \frac{dh}{dt}$ Substituting $\frac{dV}{dt}=-50$ and $h=5$ yields: $-50 = 4\pi(10+5)^2 \frac{dh}{dt}$ $-50 = 4\pi(15)^2 \frac{dh}{dt}$ $-50 = 900\pi \frac{dh}{dt}$ $\frac{dh}{dt} = -\frac{50}{900\pi} = -\frac{1}{18\pi}$ Since we want the rate at which the thickness decreases, we take the absolute value: $\frac{1}{18\pi}$. There is an error in the listed answer.

Step 5: Rethink the approach and correct the error The correct approach is as follows: $V = \frac{4}{3}\pi (R+h)^3 - \frac{4}{3}\pi R^3$ $\frac{dV}{dt} = 4\pi(R+h)^2 \frac{dh}{dt}$ $-50 = 4\pi(10+5)^2 \frac{dh}{dt}$ $-50 = 4\pi(225) \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{-50}{900\pi} = -\frac{1}{18\pi}$ The rate at which the thickness decreases is $\left|-\frac{1}{18\pi}\right| = \frac{1}{18\pi}$.

The given answer is incorrect.

Common Mistakes & Tips

• Sign Convention: Be careful with the sign of $\frac{dV}{dt}$. Since the ice is melting, the volume is decreasing, so $\frac{dV}{dt}$ should be negative.
• Units: Always include units in your calculations to ensure consistency.
• Understanding the Question: Make sure you understand what the question is asking. In this case, it asks for the rate at which the thickness decreases, so we want the absolute value of $\frac{dh}{dt}$.

Summary

We used the related rates concept and the formula for the volume of a sphere to find the rate at which the thickness of the ice layer decreases. We established a relationship between the volume of the ice and its thickness, differentiated it with respect to time, and then substituted the given values to solve for the desired rate. The correct answer is $\frac{1}{18\pi}$ cm/min. The answer provided is incorrect.

Final Answer

The final answer is $\boxed{\frac{1}{18\pi}}$. This does not correspond to any of the options given. There is an error in the question or the given correct answer. However, if we assume that the actual correct answer should have been (A) ${5 \over {6\pi }}$, let's try to find where the initial solution may have deviated from this. If the answer were $\frac{5}{6\pi}$, that means $\frac{dh}{dt} = -\frac{5}{6\pi}$ $-50 = 4\pi (15)^2 \frac{dh}{dt}$ $-50 = 900\pi \frac{dh}{dt}$ If $\frac{dh}{dt} = -\frac{5}{6\pi}$, then $-50 = 900\pi (-\frac{5}{6\pi})$ $-50 = -900 (\frac{5}{6}) = -150(5) = -750$ Which is untrue, so something is wrong. If the answer were $\frac{5}{6\pi}$, and the rate is decreasing, then $\frac{dh}{dt} = -\frac{5}{6\pi}$

Let's assume the rate of melting is 750 instead of 50. Then $\frac{dV}{dt} = -750$. $-750 = 900\pi \frac{dh}{dt}$ $\frac{dh}{dt} = \frac{-750}{900\pi} = \frac{-75}{90\pi} = \frac{-5}{6\pi}$. So the rate at which the thickness decreases is $\frac{5}{6\pi}$.

If the rate of melting was 750 instead of 50, then the answer would be (A). However, the question states that the melting rate is 50. Thus, the answer is $\frac{1}{18\pi}$. The final answer is $\boxed{\frac{1}{18\pi}}$. This does not correspond to any of the options given. There is an error in the question or the given correct answer.