Differential Equations to solve the changing radius of a drop of liquid This is the question:     
"Your lab partner leaves a drop of bleach on the lab bench, which takes the shape of a hemisphere. The drop initially has a radius of 1.6mm, and evaporates at a rate proportional to its surface area. After 10 minutes, the radius is 1.5mm. How long until the drop is gone?"    
What I currently have is $\frac{dv}{dr}=A$ where $v$ is volume and $A$ is the surface area of the drop. I was thinking of using the equation $\frac{dv}{dr} \frac{dr}{dt}$ and trying to solve the equation using $\frac{dr}{dt}=\frac{1.6-1.5}{10}$, is my approach to this correct? 

Your suggestion is correct, since it turns out that $\frac{dr}{dt}$ is constant. This is not completely obvious, so we do the calculation.
The (curved) surface area $A$ is $2\pi r^2$, where $r$ is the radius, and the volume $V$ is $\frac{2}{3}\pi r^3$.
We are told that $\frac{dV}{dt}=-kA=-2k\pi r^2$, where $k$ is a positive constant. Note that
$$\frac{dV}{dt}=\frac{dV}{dr}\frac{dr}{dt}=2\pi r^2\frac{dr}{dt}.$$
Thus 
$$2\pi r^2 \frac{dr}{dt}=-2k\pi r^2.$$
This simplifies to the very nice
$$\frac{dr}{dt}=-k.\tag{1}$$
We leave the rest to you. The differential equation (1) is very easy to solve.
