Differential Equations
Applications — evaporation model
Grade Class 12
Question:
<p>Spherical rain drops evaporate at a rate proportional to their surface area at any instant \(t\). The differential equation for radius \(r\) is:</p>
<span>\(\frac{dr}{dt} = -kr^2\)</span>
<span>\(\frac{dr}{dt} = -k\)</span>
<span>\(\frac{dr}{dt} = \frac{k}{r}\)</span>
<span>\(\frac{dr}{dt} = -\frac{k}{r^2}\)</span>
Step-by-Step Solution
Key Concept: dV/dt = -k \cdot S, then use V = 4/3\pir^3 and S = 4\pir^2.
<div class='solution'><p><strong>Step 1:</strong> Volume \(V = \tfrac{4}{3}\pi r^3\), Surface area \(S = 4\pi r^2\).</p>
<p><strong>Step 2:</strong> Evaporation rate: \(\dfrac{dV}{dt} = -k \cdot S = -4k\pi r^2\).</p>
<p><strong>Step 3:</strong> Also \(\dfrac{dV}{dt} = 4\pi r^2 \cdot \dfrac{dr}{dt}\).</p>
<p><strong>Step 4:</strong> Equating: \(4\pi r^2 \cdot \dfrac{dr}{dt} = -4k\pi r^2 \implies \boxed{\dfrac{dr}{dt} = -k}\).</p>
<p><strong>Answer: (B)</strong> The radius decreases at a constant rate.</p>
<p class='key-concept'>🔑 Key Concept: dV/dt = S·(dr/dt) for a sphere. When rate ∝ S, the S cancels and radius decreases at constant rate — a beautiful result.</p>
<p class='trap-warning'>⚠️ Trap: Forgetting to compute dV/dt via chain rule and writing dV/dt = -kr (instead of -k·4πr²), leading to wrong answer (A).</p></div>
Correct Answer: 2