Calculus Problem of the Week June 27, 2011

Sometimes easy and sometimes hard, our calculus problem of the week could come from any calculus topic. If you really want to get better at calculus, following these problems is a great way to make yourself practice! Past calculus problems of the week. Note: these are no longer updated, but you should find the old problems helpful!

This week’s problem (taken from Calculus – Larson 4th ed.)

As a spherical raindrop falls, it reaches a layer of dry air and begins to evaporate at a rate that is proportional to its surface area. Show that the radius decreases at a constant rate.


To start out, let’s figure out exactly what this problem is asking for. We want to show that the radius of this raindrop decreases at a constant rate. The first thing you should notice, is that this is a way of describing a rate of change, i.e. a derivative. We want to show that \(\dfrac{dr}{dt}\) is a constant where \(r\) is the radius and \(t\) is some unit of time.

So, we are being asked to show something about a rate of change, and if you read carefully, you will see we are also given a rate of change. We are told the rate at which the raindrop evaporates is proportional to its surface area. This would mean the raindrop is getting smaller at a rate that’s proportional to its surface area – i.e., it is decreasing in volume. The surface area of a sphere is:

\(\text{surface area} = 4\pi r^2\)

Mathematically, we can translate the fact that the rate is proportional to this into:

\(\dfrac{dV}{dt}=c(4\pi r^2)\)
where \(V\) is volume and \(c\) is a constant.

But, we could also find \(\dfrac{dV}{dt}\) using implicit differentiation on the volume formula for a sphere. The formula for the volume of a sphere is:

\(V=\dfrac{4}{3}\pi r^3\)

Taking \(\dfrac{d}{dt}\) of both sides:

\(\begin{align} \dfrac{dV}{dt} &= \dfrac{4}{3}\pi (3r^2)\dfrac{dr}{dt}\\ &= 4\pi r^2\dfrac{dr}{dt}\end{align}\)

Now we have two formulas for \(\dfrac{dV}{dt}\), so they must equal:

\(c(4\pi r^2)=4\pi r^2\dfrac{dr}{dt}\)

Solving for \(\dfrac{dr}{dt}\) now gives \(c\), a constant, which is what we were trying to show.

\(\begin{align} c(4\pi r^2)& = 4\pi r^2\dfrac{dr}{dt} \\ \dfrac{c(4\pi r^2)}{4\pi r^2} & = \dfrac{dr}{dt}\\ \dfrac{dr}{dt} &= c\end{align}\)

Further study

If you are taking a calculus course, you should take a look at the free calculus lessons on MathBootCamps!