[Math] rates of change with area and radius

calculus

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 $\frac{dA}{dr} =2\pi r$?

Best Answer

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 ${dA}/{dr}=2\pi r$