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So here is the question:

The fuel from a ship leaks into the sea forming a circular oil slick. The area of this circle is increasing at the rate of $20$ $m^2$ per minute.

They asked me to prove that when the radius of the circle = $3$, it is increasing at $\frac{10}{3\pi}$ metres per minute.

What I know so far:

$\frac{dA}{dt}$=20 --> $A = 20t + C$

From the question, it seems that they are asking for $\frac{dr}{dt}$.

So, the expression I found is as follows: $\frac{dr}{dt}$=$\frac{dA}{dt} * \frac{dr}{dA}$

I have no idea what $\frac{dr}{dA}$ is, so could I get some help?

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    $\begingroup$ It is forming in a circular oil slick. $A = \pi r^2, dA/dt = 2 \pi r (dr/dt)$ $\endgroup$
    – Math Lover
    Commented Oct 8, 2021 at 12:58

1 Answer 1

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In questions like this what they are usually wanting you to do is to use an equation for the geometry as the input to the derivative to get differentials. You can then plug in the derivatives that you know to find the derivatives that you don't.

Here, it is a circle. The equation relating the area to the radius of the circus is $A = \pi r^2$. The derivative with respect to time is $\frac{dA}{dt} = 2\pi r\,\frac{dr}{dt}$. As you noted, the problem gives you $\frac{dA}{dt} = 20$. This allows you to solve for $\frac{dr}{dt}$, which is what is being asked for.

$$ \frac{dr}{dt} = 20\cdot\frac{1}{2\pi r} = \frac{10}{\pi r}$$

This gives you a formula for the rate of increase of the radius for any given radius length.

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