Re-arranging Pythagorian theorem for centroid of a line

[Starblazer]

I am trying to understand what is going on algebraically.

For centroid of a line, the Length of the differential element is given by Pythagorian theorem
\(\displaystyle dL = \sqrt {{{\left( {dx} \right)}^2} + {{\left( {dy} \right)}^2}} \)

The text book states it can be written in the form

\(\displaystyle \begin{array}{l}
dL = \sqrt {{{\left( {\frac{{dx}}{{dx}}} \right)}^2}d{x^2} + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}d{x^2}} \\
dL = \left( {\sqrt {1 + {{\left( {\frac{{dy}}{{dx}}} \right)}^2}} } \right)dx
\end{array}\)
or
\(\displaystyle \begin{array}{l}
dL = \sqrt {{{\left( {\frac{{dx}}{{dy}}} \right)}^2}d{y^2} + {{\left( {\frac{{dy}}{{dy}}} \right)}^2}d{y^2}} \\
dL = \left( {\sqrt {{{\left( {\frac{{dx}}{{dy}}} \right)}^2} + 1} } \right)dy
\end{array}\)
What processes have been performed on this formula to get it into this form?

 

[stapel]

I think they're using "dx" and "dy" as variables, rather than functional-type processes (in the form "dy/dx", etc).

So, for instance, they've taken "(dx)^2" and (dy)^2" and "multiplied" them by "(dx/dx)^2", where "dx/dx" is regarded as a fraction equal to 1 (so multiplying by it doesn't change anything) or just using the fact that the derivative of x, with respect to x, is 1: (d/dx)(x) = 1.

Then they rearrange: (dx/dx)^2 * (dy)^2 = (dy/dx)^2 * (dx)^2, etc. Then they "factor out" the (dx)^2, and remove it from the square root, converting it to a plain "dx". I think.

I wouldn't swear to the mathematical purity of what they're doing....