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

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

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.