# Flore-D's question at Yahoo! Answers involving related rates

#### MarkFL

Staff member
Here is the question:

Assume that a spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is 3mm, and one-half hour later been reduced to 2mm, find an expression for the radius of the raindrop at any time.
Here is a link to the question:

I have posted a link there to this topic so the OP may find my response.

#### MarkFL

Staff member
Hello Flore-D,

Let $V$ be the volume of the raindrop in $\text{mm}^3$ at time $t$ in $\text{hr}$. Let $r$ be the radius of the drop at time $t$.

We are told:

$\displaystyle \frac{dV}{dt}=-k\left(4\pi r^2 \right)$ where $0<k\in\mathbb{R}$

Since the drop is spherical, we may state:

$\displaystyle V=\frac{4}{3}\pi r^3$

Differentiate with respect to time $t$:

$\displaystyle \frac{dV}{dt}=4\pi r^2\frac{dr}{dt}$

Equate the two expressions for $\displaystyle \frac{dV}{dt}$:

$\displaystyle 4\pi r^2\frac{dr}{dt}=-k\left(4\pi r^2 \right)$

$\displaystyle \frac{dr}{dt}=-k$

We find that the radius will decrease at a constant rate. This stems from the fact that the derivative of a sphere with respect to its radius is its surface area. Think of a sphere being decomposed into spherical shells. So, we have the initial value problem:

$\displaystyle \frac{dr}{dt}=-k$ where $\displaystyle r(0)=3,\,r\left(\frac{1}{2} \right)=2$

Integrating, we find:

$\displaystyle r(t)=-kt+C$

Use initial value to determine the parameter $C$:

$\displaystyle r(0)=-k(0)+C=3\,\therefore\,C=3$

and so we have:

$\displaystyle r(t)=-kt+3$

Now, use the other known point to determine the constant of proportionality $k$:

$\displaystyle r\left(\frac{1}{2} \right)=-k\left(\frac{1}{2} \right)+3=2\,\therefore\,k=2$

Hence, we find the radius at any time $0\le t$ is given by:

$\displaystyle r(t)=-2t+3$