A spherical iron ball 10 cm in 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? Isn't the ice just a sphere?

The iron ball is the core, and it does not melt. The ice layer is only the shell around it. To find the volume of the ice, we must remove the volume of the core from the total volume of the sphere. The ice is a spherical shell, not a solid sphere.

Q: Why is the derivative of the iron ball volume zero?

The iron ball has a fixed radius of $10$ cm. Since it does not change with time, its volume is a constant. The derivative of any constant with respect to time is zero. This is a crucial step in simplifying your differential equation!

Q: Why do we use $10+x$ for the radius instead of just $x$?

The thickness $x$ is added to the radius of the iron ball. The total radius of the sphere is the distance from the center to the outer edge of the ice, which is the iron radius ($10$ cm) plus the thickness $x$. Always be careful to define your variables relative to the center of the sphere.

Question

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

Options:

(A) $\frac{1}{36\pi} 	ext{ cm/min}$

(B) $\frac{1}{18\pi} 	ext{ cm/min}$

(C) $\frac{1}{54\pi} 	ext{ cm/min}$

(D) $\frac{5}{6\pi} 	ext{ cm/min}$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

To solve this problem, we must first define our geometry. The total radius of our object, let us call it $R$, is the sum of the iron core's radius and the thickness of the ice, $x$.

Thus, we have the relationship:
$$R = 10 + x$$

The volume of the ice is the volume of the entire outer sphere minus the volume of the inner iron core. Mathematically, this is expressed as:
$$V = \frac{4}{3}\pi(10+x)^3 - \frac{4}{3}\pi(10)^3$$

### The Calculus of Change

We are given that the ice melts at a rate of $50$ cm$^3$/min. This represents the rate of change of volume, or $\frac{dV}{dt}$.

To find the rate at which the thickness $x$ changes, we differentiate our volume equation with respect to time $t$. Using the chain rule, the derivative of the total volume is:
$$\frac{dV}{dt} = \frac{4}{3}\pi \cdot 3(10+x)^2 \cdot \frac{dx}{dt}$$

The constant term, representing the volume of the iron ball, vanishes because its derivative is zero. This leaves us with the elegant relationship:
$$\frac{dV}{dt} = 4\pi(10+x)^2 \frac{dx}{dt}$$

### The Final Calculation

We are interested in the moment when the thickness $x$ is exactly $5$ cm. Substituting $\frac{dV}{dt} = 50$ and $x = 5$ into our equation, we get:
$$50 = 4\pi(10+5)^2 \frac{dx}{dt}$$

This simplifies to:
$$50 = 4\pi(15)^2 \frac{dx}{dt}$$
$$50 = 4\pi(225) \frac{dx}{dt}$$

Multiplying the constants, we find:
$$50 = 900\pi \frac{dx}{dt}$$

Finally, isolating $\frac{dx}{dt}$, we obtain:
$$\frac{dx}{dt} = \frac{50}{900\pi}$$

Simplifying the fraction, the rate at which the thickness is decreasing is:
$$\frac{dx}{dt} = \frac{1}{18\pi} 	ext{ cm/min}$$

You have successfully modeled a dynamic physical system using the chain rule. Keep this intuition, and you will master any problem.

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

Select Answer:

Visualized Solution

The Iron Core

The Ice Coating

Total Radius of the Sphere

Volume of the Ice Layer

Equation for Ice Volume

The Melting Rate

Differentiating with Respect to Time

Applying the Chain Rule

Simplifying the Equation

Substituting the Given Values

Evaluating the Radius Term

Squaring the Radius

Simplifying the Constants

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

Final Conclusion

The Sigma Insight: Tangents, Normals and Rate Measure

Analyzing the Setup

The Calculus of Change

The Final Calculation

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