Solving the Mystery of dl in Integrals

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SUMMARY

The discussion centers on the interpretation of the differential element dl in the context of line integrals, specifically in the formula {\cal{E}} = \int_{a}^{b} \vec{E} \cdot \vec{dl}. Participants clarify that dl represents a vector differential along a curve parameterized by l(t) = (x(t), y(t)), where dl is expressed as (x'(t)*dt, y'(t)*dt). This formulation indicates that dl is analogous to dx in traditional integrals, with the lowercase 'l' denoting the curve's path rather than a variable of integration.

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Goldenwind
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I'm trying to deal a problem using this formula, but I'm unclear as to what dl represents (Or if it is the same as dx in most integrals, then in that case I don't know what the lowercase L is)

[tex]{\cal{E}} = \int_{a}^{b} \vec{E} \cdot \vec{dl}[/tex]
 
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Along a general curve l(t)=(x(t),y(t)), dl=(x'(t)*dt,y'(t)*dt). So if E=(Ex,Ey) that becomes Ex*x'(t)*dt+Ey*y'(t)*dt.
 

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