calculus - Why isn't $dy/dx$ (in this case, $dr/dt) $ simply the rate or slope?

I came late to the Maths party & have no undergraduate Mathematics qualification (am doing a Mathsy postgraduate qualification so I've tried my best to bring myself up to speed but the foundations are dreadfully shaky) so please explain like I'm five, if at all possible.

I have read the other "isn't $dy/dx$ a ratio" / "when can $dy/dx$ be a ratio for all intents and purposes" posts but the answers are either overly complex for my purposes or too removed from my particular problem for me to apply the answers sensibly.

I'm helping my lovely neighbour with her high-school level Maths at the moment and she asked me a question I simply can't provide a satisfactory and simple answer to.

We're dealing with rudimentary calculus, an oil slick to be precise. We're using cm as units and we have already calculated dr/dt:
$$\frac{dr}{dt}=\frac{2\times10^7}{\pi r}\text{ cm}$$
The next part of the question asks us how many hours until the radius is 1km so 1x10^5 cm for our particular question.

Would someone please be so kind as to explain simply why we can't divide the radius by the "rate"? Why can't this simply be treated as a rate/slope?

I do know that I instead must set the integral of $r.dr =\int (2x10^7/\pi).dt$ to find the correct answer. And I do of course notice that this elicits a value that is half that which is obtained from simple division of the radius by $dr/dt.$ Where/why/how in the simpler approach am I neglecting to divide by 2?

I do apologize for the low-level of this question. I did scour a series of resources and did find semblances of answers but none that have truly clicked with me / none that I can relay v simply to my tutee. I'm aware I'm likely being incredibly dense so I do apologize for this also and understand if it's not worth the time to answer! As I said, I tried in vain to solve the problem myself before deciding to hassle you here at MathExchange.

Many many thanks!

Answer

$\frac {dy}{dx}$ is the instantaneous rate. You can't just divide the radius by the rate because the rate is changing with time. If you do your calculation when the radius is $1$ cm, you get a rate of $\frac {2 \cdot 10^7}{\pi}$ and it takes only $\frac \pi{200}\cdot \frac {10^5-1}{10^5}$ seconds to get to $r=10^5$. If you start when $r=100$ the rate is $\frac {2 \cdot 10^5}{\pi}$ and it takes $\frac \pi{2}\cdot \frac {10^5-100}{10^5}$. It can't take longer to get from $100$ to $10^5$ than it takes to get from $1$ to $10^5$ because from $1$ you have to go through $100$.