# Simple derivative task, no function, only values given, how the graph might look?

I was sure I had this subject nailed by now, but I'm curious. The task below.

The amount of lava coming out every hour during an eruption can be described with $f(t)$. The output is in metric tons and t is hours after the eruption began. You know the following
$f(0) = 300$
$f'(10) = 0$
$f''(10) = -10$
What can you say about the eruption based on this?

What I think so far is that it starts at 300 tons. After 10 hours there is either a top, bottom or shoulder(?) point. Because the second derivative is negative this implies that the graph is "sad". With a top. The probable thing is just that it starts at 300 and increases at some rate until the rate of change is 0 at 10 hours. Constant flow. But the rate of change (lave flow per hour) will decrease after 10 hours.

How do I know that it's not a shoulder point at t=10 when I can't check the sign "around" the derivative like I normally would to rule that out? Maybe the negative second derivative is enough to rule it out?

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Second derivative negative shows you don't have an inflection point at $10$. You had already observed this. – André Nicolas Oct 19 '12 at 18:09
@AndréNicolas what if the second derivative was positive, could that imply it? But you say so because it's negative and it will cause decrease after 10? – Algific Oct 19 '12 at 18:15
Second derivative positive, plus first derivative $0$, would imply you have a local minimum at $10$. – André Nicolas Oct 19 '12 at 18:19
@AndréNicolas the second derivative must be 0 for it to be an inflection/shoulder point? – Algific Oct 19 '12 at 18:21

It should be noted that in reality you can not derive anything meaningful at all from the provided information. While it's true that mathematically speaking, $f$ must have a local maximum at $t=10$, the information is pretty useless because you cannot distinguish between a small bump and a major peak.

The values of $f$ at $t=0,1,2,\ldots$ hours could be

$300,100,30,10,3,1,0,0,0,0,5,0,\ldots$,

i.e. the local maximum at $t=10$ could be a small additional burp. Or, of course, the values of $f$ could be

$300,600,1200,2400,4800,9600,19200,38400,76800,153600,153650,153600,307200$,

i.e. the flow rate rises exponentially with a small irregularity around $t=10$. If it's your house that is potentially in the way of this lava flow, then believe me, there's a huge difference between those two cases...

Plus, there's no information about the time resolution of $f$. Since $f$ is a flow rate, it's basically meaningless without that. Say you wanted to know how much lava has errupted in total 5 hours after the erruption started? Could you simply compute $f(0)+\ldots+f(5)$ to get an approximate result? You can't! For that to work, you'd need to know that $f(t)$ is the average flow over one-hour intervals. As it stands, $f$ could be the average flow-rate over one-minute intervals. In which case, of course, $f(0)+\ldots+f(5)$ is probably more than a magnitude from the actual total errupted mass between $t=0$ and $t=5$.

In conclusion, this is an example of an absolutely horrible piece of mathematical exercise. It takes mathematicaly concepts, extremal points and derivatives, and applies them to a practical problem with total disregard of whether they are applicable or not.

Mathematics has tons of tools to deal with partial information about functions, measurement errors and extremal point estimation, which allow you to derive actual and dependable results. Trying to find extremal points from sampled values by methods developed for differentiable function is not one of them!

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Since f'(10)=0 and f"(10) is negative, there is a local maxima at t=10. If it was a shoulder point, f"(10) would also be equal to 0.

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Say for the graph x^3, is x=0 both an inflection point and a shoulder point? – Algific Oct 19 '12 at 18:17
I assumed that a 'shoulder point' IS the same as an 'inflection point'. Inflection point: A point where the function changes from a concave function to a convex function. – RahulHP Oct 21 '12 at 11:07