rates of change with area and radius I am doing Cambridge AS level maths past papers and came across a question who's answer I don't understand.
The question is:
An oil pipeline under the sea is leaking oil and a circular patch of oil has formed on the surface of the sea. At midday the radius of the patch of oil is $50~\text{m}$ and is increasing at a rate of $3~\text{m}$ per hour. Find the rate at which the area of the oil is increasing at midday.
The answer is:
$A= \pi r^2$ leads to $\frac{dA}{dr} = 2\pi r$
it then continues by using the chain rule and the fact that $\frac{dr}{dt}= 3$ which I all understand.
But why is $\frac{dA}{dr} =2\pi r$?
 A: Does the the following statement make sense to you?
If $y=x^2$ then ${dy}/{dx}=2x$
Multiply both side by $pi$:
If $y=\pi x^2$ then ${dy}/{dx}=2\pi x$
Substitute $A$ for $y$ and $r$ for $x$...
If $A=\pi r^2$ then ${dA}/{dr}=2\pi r$
or, in other words,
$A=\pi r^2$ would lead to ${dA}/{dr}=2\pi r$
A: For a Connected Rate of change problem such as this one, first define your variables: 
Let $r$ = Radius of Oil Patch (in meters).
Let $t$ = Time taken (in hours) for radius/area of Oil Patch to increment.
Let $A$ = Area of Oil Patch (in meters squared).
Let $\dfrac{dA}{dt}=\,$Rate at which area of Oil Patch is increasing (in meters squared per hour). 
Which can be written via the chain rule as $\dfrac{dA}{dt}=\color{red}{\dfrac{dA}{dr}}\cdot\color{blue}{\dfrac{dr}{dt}}\tag{1}$
We already know the term marked $\color{blue}{\rm{blue}}$; it is equal to $3\,\text{meters per hour}$ as you already mentioned.
Since you know that the Oil Patch area $A$ is approximately given by a circle of area $A=\pi r^2\tag{2}$

Getting closer to answering your question:
If we differentiate both sides of $(2)$ with respect to $r$ we get $$\color{red}{\frac{dA}{dr}}=\frac{d (\pi r^2)}{dr}=2\pi r \quad\text{(The Circumference of the Oil Patch)}$$
As $r=50\,\text{meters}$ we know the value of the $\color{red}{\rm{red}}$ term.
All that remains to be done is to substitute the $\color{blue}{\rm{blue}}$ and $\color{red}{\rm{red}}$ terms into equation $(1)$:
$$\dfrac{dA}{dt}=\color{red}{\dfrac{dA}{dr}}\cdot\color{blue}{\dfrac{dr}{dt}}=2\pi\cdot 50\cdot 3=300\pi \,\,\text{meters per hour}$$ 
