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!

  • $\begingroup$ The rate isn’t constant. Compare this to accelerating a car. If the acceleration is constant, you can just divide the difference in the starting and ending speeds by this rate to get the needed time, but if the acceleration is changing, you can’t do that. In this problem, the rate decreases with increasing radius, so it takes a lot longer to grow by that last cm than it does the first one. $\endgroup$ – amd May 26 '17 at 1:04
  • $\begingroup$ Hi there, thanks so much for taking the time to reply! Ah so intuitively this is what I said, I know to use integrals for non-constant rates but I couldn't explain simply why her answer was twice mine? Was that a red herring? $\endgroup$ – EJK May 26 '17 at 1:09

$\frac {dy}{dx}$ 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 $100$.

  • $\begingroup$ Hi there, thanks so much for taking the time to reply! That's great and that is what I said, although in a much less clear way and with little conviction, "oooh so the rate's not constant as it's a function of r... and as it's not linear, it's not a y = mx ting"! Did I read too much into the fact that her answer was exactly twice mine (and the correct answer)? If so, wow, I've been on something of a fool's errand for a good while! $\endgroup$ – EJK May 26 '17 at 1:15
  • $\begingroup$ When you integrate the differential equation, you get $r=k\sqrt t$ I don't think the factor $2$ means anything. It depends on where you start. $\endgroup$ – Ross Millikan May 26 '17 at 1:26
  • $\begingroup$ Thank you for replying again! Ah okay, I fixated upon the multiple of two and couldn't see the wood for the trees as they say! I shall repeat your answer verbatim to my tutee; I'm sure she'll be most grateful. Thanks so much! $\endgroup$ – EJK May 26 '17 at 1:29

(This is to address the factor of two. It ended up being too long for a comment.)

The reason that you can’t simply divide by the rate is, as Ross Millikan explains, because the growth rate is constantly changing. You could, however, divide by the average growth rate over the given range, and that’s where the factor of two comes in. You can see a factor of two sneak in when you integrate $r\,dr$, but there’s a better way to see that it’s not coincidental.

So far, you’ve been looking at the radius of the slick as a function $f$ of time, but to solve this problem you really want the inverse $f^{-1}$ of this function, that is, you’d like to know how much time it takes to get to a certain radius. The Inverse Function theorem tells us that, under some easily-met conditions, this inverse function exists and, moreover, that $(f^{-1})'=1/f'$. Loosely speaking, under suitable conditions differentiation and inversion “commute.” (This is one of the theorems that lets us get away with treating ${dy\over dx}$ as an ordinary fraction.) This theorem is also one of the ones that you’re using when you separate variables and integrate, as you do in your question.

So, invert ${dr\over dt}$ to get the instantaneous hours-per-cm rate, which has the form $kr$, and then integrate to get $\frac k2(b^2-a^2)$ as the time that it takes the slick to grow from a radius of $a$ to a radius of $b$. Now, however, the rate you’re dealing with is a linear function, so that the average growth rate over this interval is simply ${b+a\over2}k$. If you multiply this average rate by the length $b-a$ of the interval, you get the same answer as you did by integrating. (This is a special case of an important theorem about definite integrals, but if you think about them as areas under a curve, it should be intuitively obvious.) In short, you can view the “stray” factor of $2$ as coming from having to average the growth rate over the range of slick sizes.


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