# 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.

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

• What do you mean? $\pi$ is just a constant so $\frac {dA}{dr}=\frac {d(\pi r^2)}{dr}=\pi\frac {d(r^2)}{dr}=2\pi r$ – lulu Apr 22 '16 at 10:42
• Use the power rule. If $f(r) = r^n$, then $f'(r) = nr^{n - 1}$. – N. F. Taussig Apr 22 '16 at 10:43
• Presumably the answer also says $\frac{dA}{dt}=\frac{dA}{dr}\frac{dr}{dt}$. – almagest Apr 22 '16 at 10:44
• @Mirte The answer is $300\pi\,$ meters per hour – BLAZE Apr 22 '16 at 11:20

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

• As well as working with the algebra, it's also nice to think geometrically - a small change in the area of a circle $\pi r^2$ can be approximated by a narrow strip which winds around the circle's circumference $2\pi r$. – Jean Flower Apr 22 '16 at 10:51

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 $(1)$: $$\dfrac{dA}{dt}=\color{red}{\dfrac{dA}{dr}}\cdot\color{blue}{\dfrac{dr}{dt}}=2\pi\cdot 50\cdot 3=300\pi \,\,\text{meters per hour}$$