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What's the intuition for understanding that the second derivative does correspond to the curvature of the function?

For first derivatives one can think of the tangent lines, but what to think for second derivatives?

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3 Answers 3

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The second derivative is the instantaneous rate of change of the first derivative.

So if the second derivative is large and positive, then the slope of the tangent line is increasing quickly, which means the graph is curving sharply.

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How about the definition of curvature? Let $y=f(x)$ be a function. Then we have:

enter image description here

$sgn(\kappa) = sgn(y'')$

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The curvature of a function can be approximated by the second derivative when the first derivative at that point is close to zero.

If the curvature is given by $$\frac{y''}{\sqrt{(1+y'^2)^3}}$$

and $y' \approx 0$, then the curvature can be approximated by

$$\frac{y''}{\sqrt{(1+0^2)^3}} = \frac{y''}{1^{3/2}} =y''$$

I know essentially nothing about the curvature of surfaces but I suspect a similar relation will apply to the partial derivatives when the gradient is close to zero.

Intuitively, it makes sense that the second derivative approximates curvature when the line itself is almost flat. Consider $x^4$ between x=10 and x=12.

enter image description here

This slope appears to be hardly curving, but if we take its second derivative at x=11, we get $y''=12(11)^2 = 1452$ which is huge. It just happens to be almost completely swamped by the slope (at x=11 it is 5324), which when plugged into the curvature equation results in a tiny curvature of 0.0037.


If you want the second derivative to be within 99% of the curvature, the first derivative needs to be at maximum $\sqrt{\sqrt[3]{(\frac{1}{0.99})^2}-1} = 0.081$ the magnitude of the second derivative.

When the second derivative comes up in differential equations, whether it's the deflection of a beam or vibrations in a string or membrane, one of the implicit assumptions is that the magnitude of deflection/vibration is tiny relative to the curvature. If you were to use the curvature itself, these equations would be much harder!

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