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Exercise 10.2 · Q2

Q.Assume that a spherical rain drop evaporates at a rate proportional to its surface area. Form a differential equation involving the rate of change of the radius of the rain drop.

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Translate "evaporates at a rate proportional to surface area" into dVdt=−kS\dfrac{dV}{dt}=-kS, then use V=43πr3, S=4πr2V=\frac43\pi r^3,\ S=4\pi r^2 to convert this into a differential equation for the radius rr alone.

Step 1. Set up the volume rate law. Let V(t)V(t) be the raindrop's volume at time tt. "Evaporates at a rate proportional to its surface area SS" gives dVdt=−kS\dfrac{dV}{dt}=-kS (k>0k>0, since VV decreases as the drop evaporates).

Step 2. Express VV and SS in terms of rr. For a sphere, V=43πr3V=\dfrac43\pi r^3 and S=4πr2S=4\pi r^2.

Step 3. Differentiate VV with respect to tt via the chain rule. dVdt=dVdr⋅drdt=4πr2drdt\dfrac{dV}{dt}=\dfrac{dV}{dr}\cdot\dfrac{dr}{dt}=4\pi r^2\dfrac{dr}{dt}. …

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