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It's been hard to come up with a question title that makes sense so please bear with me.

On an exponential curve there's a point on the x axis where the y axis starts increasing significantly. The exact location/ calculation of the point isn't important for my question.

Here's an example:

Exponential curve showing point of interest

When people are talking about every day life that involve exponential curves we sometimes use terms like "hit a steep learning curve" to describe our experience of reaching this point.

In calculus we use the term "inflection point" for when a curve changes between positive and negative.

Is there also a term for the above point on the exponential curve?

Hans Lundmark
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BaronGrivet
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    The point you describe sounds like the point where the tangent line is parallel to $y=x$. For a function like $y=e^x$ where the derivative is $y'=e^x$, this means $y'=1$ at $x=0$... – abiessu Feb 26 '14 at 03:38
  • In a common application such as loan or credit-debt payment, the function is more like $ \ e^{kx} \ $ with $ \ k \ $ having a value on the order of 0.1 to 0.2 . – colormegone Feb 26 '14 at 04:18

2 Answers2

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The English idiom is "the knee in the curve."

This doesn't have much, if anything, to do with mathematics, however.

For a good explanation of why it's a subjective issue and not a mathematical one, you can look at this article: http://www.growth-dynamics.com/articles/Kurzweil.htm (archived) About 1/3 down the page there are a few graphs one over the other with the title "where is the knee?" They have different $y$ axes, but show the same function. You can see that your $y$ axis determines where you think the "knee" should fall.

nealmcb
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Jeff Snider
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  • In dealing with compound-interest situations, such as credit card debt, it is helpful to people to advise them as to about where the "knee" lies for their interest rate. Those who has let their debt "travel beyond the knee" are generally the ones who find themselves in something of a trap... While it is reasonable to say that this is something of a colloquial expression, people who work with applications would probably put it "around" the point where the slope reaches one. – colormegone Feb 26 '14 at 04:13
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    The link has died, but it probably is saying that $e^{x+\ln k}=ke^{x}$, which is to say shifting an exponential horizontally corresponds to scaling it vertically, so the "knee" moves depending on how you plot the graph. @colormegone What does slope 1 mean when each axis has different units? (I'd think for advising about compound interest, it'd be better to give doubling time or a similar measure.) – Kyle Miller Aug 16 '17 at 18:39
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    @KyleMiller Agreed. Note that the article is available at the internet archive: https://web.archive.org/web/20160306052314/http://www.growth-dynamics.com/articles/Kurzweil.htm – nealmcb Dec 03 '19 at 18:25
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This is an old question, but a very good rule of thumb for such a point would be the minimum of the radius of curvature function.

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The exponential function and its radius of curvature:

It takes a minimum value at $x=-\frac{\log (2)}{2}=-0.346574$

Another example, arctan function:

enter image description here

The radius is smallest at $x=\pm 0.831576$

Gappy Hilmore
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    Just to point out a potential pitfall for the unwary: for an exponential curve relating two different measurements (so likely different units), the exact point of minimal curvature depends on the relative scalings of the axes. – Kyle Miller Aug 16 '17 at 18:30