Related Rates: Melting Snowball Volume Decrease

Problem Statement

  • A snowball is melting such that its diameter is decreasing at a constant rate of 0.3\text{ cm/min}.
  • Task: Find the rate at which the volume of the snowball is decreasing when the diameter is 9\text{ cm}.

Key Geometric & Calculus Facts

  • Volume of a sphere: V = \frac{4}{3}\pi r^{3}.
  • Diameter–radius relationship: d = 2r \;\Longrightarrow\; r = \frac{d}{2}.
  • A rate “decreasing” implies the derivative is negative.
  • Chain Rule (implicit differentiation): If V=\frac{4}{3}\pi r^{3} and r=r(t), then \frac{dV}{dt} = \frac{dV}{dr}\cdot\frac{dr}{dt}.

Converting Diameter Rate to Radius Rate

  • Given: \frac{dd}{dt} = -0.3\;\text{cm/min}.
  • Since r = \frac{d}{2},
    \frac{dr}{dt} = \frac{1}{2}\frac{dd}{dt} = \frac{1}{2}(-0.3) = -0.15\;\text{cm/min}.

Value of the Radius at the Specified Moment

  • When d = 9\text{ cm},
    r = \frac{9}{2} = 4.5\text{ cm}.

Implicit Differentiation of the Volume Formula

  • Differentiate V = \frac{4}{3}\pi r^{3} with respect to time t:
    • Treat \pi and \frac{4}{3} as constants.
    • Apply the power rule to r^{3}, then multiply by \frac{dr}{dt} (chain rule).
      \frac{dV}{dt} = 4\pi r^{2}\frac{dr}{dt}.

Numerical Substitution & Computation

  • Plug r = 4.5 and \frac{dr}{dt} = -0.15 into the differentiated formula:
    \frac{dV}{dt} = 4\pi (4.5)^{2}(-0.15)
    = 4\pi (20.25)(-0.15)
    = 4\pi (-3.0375)
    \approx -38.1704\;\text{cm}^{3}!\big/!\text{min} (to four decimal places).

Interpretation of the Result

  • The negative sign confirms volume is decreasing.
  • Verbal statement (omit the sign in wording):
    "When the radius is 4.5\text{ cm} and shrinking at 0.15\text{ cm/min}, the snowball’s volume is decreasing at approximately 38.17\text{ cm}^{3}\text{/min}."
  • Always pair the numeric magnitude with the phrase “decreasing” to communicate directional change.

Conceptual Takeaways

  • Chain Rule is essential whenever a geometric variable (radius) is itself a function of time.
  • Translating between diameter and radius ensures you plug compatible quantities into formulas.
  • Rates of change carry units and signs that must be preserved for correct physical interpretation.

Common Mistakes to Avoid

  • Forgetting to halve the diameter rate before using it as the radius rate.
  • Dropping the \frac{dr}{dt} factor during differentiation.
  • Ignoring the negative sign when interpreting “decreasing” in a numerical answer.

Review of Numerical Data & Units

  • \frac{dd}{dt} = -0.3\text{ cm/min}.
  • \frac{dr}{dt} = -0.15\text{ cm/min}.
  • At d = 9\text{ cm}, r = 4.5\text{ cm}.
  • \frac{dV}{dt} \approx -38.1704\text{ cm}^{3}/\text{min}$$.

Broader Connections & Real-World Relevance

  • Similar procedures apply to any changing-shape problem in physics, chemistry (e.g.
    dissolving pellets), and engineering (e.g.
    shrinking bubbles).
  • Illustrates how local rate data (surface measure) link to global quantities (volume) via calculus.
  • Demonstrates clear usage of implicit differentiation—a cornerstone skill for related-rates word problems.