A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts

Q: Wait, why do we subtract the volume of the iron ball? Can't I just use the volume of the ice sphere?

That is a common trap! The ice is a coating, not a solid sphere of ice. If you only calculate the volume of the outer sphere, you are including the iron ball inside it. To isolate the ice, you must subtract the iron core's volume from the total volume.

Q: Why is $\frac{dV}{dt}$ negative? The problem says it melts at 50 cm$^3$/min.

Great catch! In physics, 'melting' implies a reduction in volume. If you don't use the negative sign, you will get a positive rate of change, which would imply the ice is growing, not melting. Always define your sign convention based on whether the quantity is increasing or decreasing.

Q: Do I really need to differentiate the constant term $\frac{4}{3}\pi(10)^3$?

Mathematically, you can, but the derivative of a constant is zero. It is a great habit to recognize that the iron ball's volume does not change, so it does not contribute to the rate of change of the ice's volume.

Question

A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at the rate of 50 cm$^3$/min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which the thickness of ice decreases, is:

Options:

(A) $\frac{5}{6\pi}$

(B) $\frac{1}{54\pi}$

(C) $\frac{1}{36\pi}$

(D) $\frac{1}{18\pi}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine you are standing on the surface of a solid iron ball with a radius of $10$ cm. You are watching a layer of ice, with a uniform thickness $x$, slowly melt away.

The total radius of the sphere, including the ice, is $R = 10 + x$. The volume of the ice is the difference between the total volume and the volume of the iron core:

$$V = \frac{4}{3}\pi(10+x)^3 - \frac{4}{3}\pi(10)^3$$

This equation serves as our foundation for the problem.

### The Power of Calculus

To understand how this volume changes over time, we differentiate our volume equation with respect to time $t$. Using the chain rule, the derivative of the first term is $\frac{4}{3}\pi \cdot 3(10+x)^2 \cdot \frac{dx}{dt}$, and the derivative of the constant term is zero.

This simplifies beautifully to the following master equation:

$$\frac{dV}{dt} = 4\pi(10+x)^2 \frac{dx}{dt}$$

This equation links the rate of volume change to the rate of thickness change. Notice how the $3$s cancel out, leaving us with a clean, elegant expression.

### The Final Calculation

We know the ice is melting at $50$ cm$^3$/min, so $\frac{dV}{dt} = -50$. We want to find $\frac{dx}{dt}$ when $x = 5$.

Substituting these values into our master equation, we get:

$$-50 = 4\pi(10+5)^2 \frac{dx}{dt}$$

Simplifying the bracket, we have $-50 = 4\pi(15)^2 \frac{dx}{dt}$, which becomes:

$$-50 = 4\pi(225) \frac{dx}{dt}$$

Multiplying the constants, we get $-50 = 900\pi \frac{dx}{dt}$. Finally, isolating $\frac{dx}{dt}$, we find:

$$\frac{dx}{dt} = -\frac{50}{900\pi}$$

Simplifying this fraction, we get the final result:

$$\frac{dx}{dt} = -\frac{1}{18\pi} 	ext{ cm/min}$$

The negative sign confirms the thickness is decreasing, and the magnitude of the rate is **$\frac{1}{18\pi}$ cm/min**. We have successfully navigated the geometry and the calculus to find the answer!

Animated Solution for Mathematics - Differentiation: A spherical iron ball of 10 cm radius is coated with a layer of ice of uniform thickness that melts at the rate of 50 cm $^{3}$ /min. When the thickness of ice is 5 cm, then the rate (in cm/min.) at which the thickness of ice decreases, is:

Select Answer:

Visualized Solution

Visualize the Iron Ball

Ice Layer Coating

Volume of the Ice

Volume Equation

Rate of Change (Differentiation)

Applying Chain Rule

Simplified Derivative

Given Rates and States

Substitute into Master Equation

Simplify the Bracket

Square the Term

Multiply Constants

Isolate $\frac{d x}{d t}$

Final Answer

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Power of Calculus

The Final Calculation

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