Question

Growing Raindrop Suppose that a droplet of mist is a perfect sphere and that, through condensation, the droplet picks up moisture at a rate proportional to its surface area. Show that under these circumstances the droplet's radius increases at a constant rate.

Short answer

Under the given conditions, the radius of the droplet increases at a constant rate.

Step-by-step solution

Establish relations

We start by understanding the relationship between the volume \(V\) of the droplet and its radius \(r\). The volume of a sphere is given by \(V = \frac{4}{3}\pi r^3\). The surface area \(A\) is \(A = 4\pi r^2\). It's mentioned in the problem that the rate of moisture accumulation is proportional to the surface area, which gives us \(\frac{dV}{dt} = k A = k 4\pi r^2\). Here, \(k\) is the proportionality constant.

Differentiate the Volume formula

Differentiating the volume of a sphere with respect to time \(t\), we get \(\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}\).

Equate and Simplify

Equate the two values of \(\frac{dV}{dt}\) from Step 1 and Step 2. This results in \(4\pi r^2 \frac{dr}{dt}= k 4\pi r^2\). Dividing both sides by \(4\pi r^2\) gives us \(\frac{dr}{dt} = k\). This implies that the rate of increase of radius is a constant, irrespective of its current size.