What does the square of a differential mean?

Leo Liu
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Homework Statement
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Relevant Equations
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When I was following the calculations of finding the potential energy of a spring standing on a table under gravity, I encountered the integral shown below, where ##d\xi## is the compression of a tiny segment of the spring and ##k'## is the effective spring constant of that segment. The integral sums up the elastic potential energy of each segment along the spring. However, it is not a run of the mill integral in that its differential term is squared. From my experience, the square of a differential is 0, but in this case it obviously possesses some unique properties. Could someone explain why this makes sense? And how do you compute this integral?
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Origin: https://www.zhihu.com/question/324405110/answer/1860254582
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From your text
k'=A(dx)^{-1} though I am not familiar at all with such a treatment of infinitesimal. Anyway
d\xi=B(dx)
k'(d\xi^2)=C dx
So
\int Dk'(d\xi^2)= \int E dx
as we expect.
 
I don't think it's helpful to think of it as a square differential. Let's call u(x) the compression of a segment of length δx. Then
u(x) = λg(L0-x)δx/kL0
k' = kL0/δx
Evaluate 1/2 k'u(x)2, and THEN (only then) let δx → 0. That's when it becomes a differential.
 
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