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Exercise 6.6 · Q93

Q.Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3mm and 1 hour later has been reduced to 2mm, find an expression for the radius of the raindrop at any time tt.

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Let V=43πr3V=\dfrac43\pi r^3 be the drop's volume; evaporation proportional to surface area means dVdt=−k⋅4πr2\dfrac{dV}{dt}=-k\cdot4\pi r^2. But by the chain rule dVdt=4πr2drdt\dfrac{dV}{dt}=4\pi r^2\dfrac{dr}{dt}, so 4πr2drdt=−4πkr2⇒drdt=−k4\pi r^2\dfrac{dr}{dt}=-4\pi kr^2\Rightarrow\dfrac{dr}{dt}=-k: the radius shrinks at a CONSTANT …

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