*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 Note: these are no longer updated, but you should find the old problems helpful!*

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## 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.

### Solution

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!