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In introductory calculus classes we learn the utility of calculating $f'(x)$ and $f''(x)$ for sketching the curve $f(x)$. My question is given $f'''(a)$, how does this value affect the shape a curve $f(x)$ for $x$ near $a$?

Consider an example where $f'(x)>0$ and $f''(x)<0$ on the whole domain. How does the sign of $f'''(x)$ change the shape of the curve?

Applied Squared Mathematician
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    @TylerHG I think the OP means just by looking at the graph of $f$ –  Nov 02 '14 at 00:33
  • @TedShifrin I think you are referring to either the example or my deleted comment. Both were clearly talking about an example and not a general $f$. I suppose you are stating that constructing such an example would be difficult? Or are you saying that constructing such an example is not possible? If so, can you explain why there is no such function $f(x,\theta)$ such that one can vary $\theta$ along some smooth trajectory in parameter space such that only the third derivative changes? – Applied Squared Mathematician Nov 02 '14 at 04:42
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    Note that if $\dfrac{\partial^2 f(x,\theta)}{\partial x^2}$ is independent of $\theta$, then so is $\dfrac{\partial^3 f(x,\theta)}{\partial x^3}$. If you don't believe this, one way to see it is this: $\dfrac{\partial}{\partial\theta}\dfrac{\partial^2 f(x,\theta)}{\partial x^2}=0 \implies \dfrac{\partial}{\partial\theta}\dfrac{\partial^3 f(x,\theta)}{\partial x^3}=0$, by interchanging partial derivatives. – Ted Shifrin Nov 02 '14 at 12:42
  • OK, yes that makes sense, my comment about fixing one derivative and varying another make no sense but I think the original question is still interesting. – Applied Squared Mathematician Nov 02 '14 at 23:09

2 Answers2

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Let's change the notation a little, to make things easier to read. Instead of a function $x \mapsto f(x)$, let's consider a function $t \mapsto x(t)$, and let's use dots to denote differentiation with respect to $t$.

Then the curvature of $x$ is given by $$ \kappa(t) = \frac{\ddot x}{\sqrt{1 + {\dot x}^2}} $$ and the derivative of curvature is $$ \dot\kappa = \frac{d\kappa}{dt} = \frac{\dddot x(1+ {\dot x}^2 ) - \dot x \ddot x^2 } { ({1 + {\dot x}^2})^{3/2} } $$ So, we can say the following:

  • If $\dddot x > 0$ and $\dot x < 0$, then $\dot\kappa > 0$, so $\kappa$ is increasing, so radius of curvature is decreasing
  • If $\dddot x < 0$ and $\dot x > 0$, then $\dot\kappa < 0$, so $\kappa$ is decreasing, so radius of curvature is increasing

So, the sign of $\dddot x$ gives us some information about growth of curvature in some situations, at least.

But, in the unfortunate regions where $\dddot x > 0$ and $\dot x > 0$, or where $\dddot x < 0$ and $\dot x < 0$, this line of reasoning doesn't tell us anything about curvature, as far as I can see.

So, it's not quite as simple as I indicated in my original (lazy) answer. Thanks to @TedShifrin for pointing out the error in my reasoning.

bubba
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    Bubba, this is not quite so clear to me. Since $\kappa = f''/\sqrt{1+f'^2}$, we get something not so pretty for $$\kappa' = \frac{f'''(1+f'^2)-f'f''^2}{(1+f'^2)^{3/2}}.$$ I guess it is clear that if $f'''<0$ and $f'>0$, then $\kappa$ must decrease and that if $f'''>0$ and $f'<0$, then $\kappa$ must increase. – Ted Shifrin Nov 02 '14 at 12:36
  • @TedShifrin -- Yes, you're right, of course. I didn't think the mess in the denominator would have much impact, and I was too lazy to work through the details. I was wrong. If you write an answer, I will delete mine. Or, if you don't want to write an answer, I will correct mine, using info from your comment. I think it's a good question, so it deserves a correct answer. – bubba Nov 02 '14 at 13:09
  • Go ahead and correct :) Actually, I still think it's informative when $f'$ has constant sign. I agree it's a nice question. – Ted Shifrin Nov 02 '14 at 13:25
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You can also say that the third derivative $f′′′(x)$ gives you information about convexity (or concavity) of the first derivative (and geometric interpretation of the first derivative is obvious).

Maybe you have heard about the famous quotation of Richard Nixon: "The rate of increase of inflation is decreasing." That is a use of the third derivation in practice.

Joseph
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