The area of a circle is decreasing at the rate of 5m^2/s when its radius is 3m.
a) at what rate is the radius decreasing at that moment?
b) at what rate is the diameter decreasing at that moment?
a) Since the area of a circle is $\displaystyle A=\pi r^2$, we see that $\displaystyle \frac{\,dA}{\,dt}=2\pi r\frac{\,dr}{\,dt}$. We know the rate of change of the area, and we know the radius at that instant in time. Do you think that you can find $\displaystyle \frac{\,dr}{\,dt}$?
b) From a), we said that $\displaystyle A=\pi r^2$. Since we're looking for the rate of change of the diameter, we need to introduce this value into this equation. Since $\displaystyle r=\frac{D}{2}$, we see that $\displaystyle A=\pi\left(\frac{D}{2}\right)^2=\tfrac{1}{4}\pi D^2$.
Now, $\displaystyle \frac{\,dA}{\,dt}=\tfrac{1}{2}\pi D\frac{\,dD}{\,dt}$
You know the rate of change of the area and you can figure out the diameter at that instant in time (since $\displaystyle D=2r$). Can you continue and find $\displaystyle \frac{\,dD}{\,dt}$?