Change in length of a right triangle

Question

And so my question is how do I prove the ??

Answer 1

$\newcommand{\pder}[2]{\frac{\partial #1}{\partial #2}}$ This can be done through a straightforward application of related-rates to the Pythagorean theorem: $$ x^2 + a^2 = L^2 $$ $$ \pder{}{x} (x^2 + a^2) = \pder{}{x} L^2 $$ $$ 2x = 2L \pder{L}{x} $$ $$ \pder{L}{x} = \frac{x}{L} = \frac{x}{\sqrt{x^2 + a^2}} = \frac{\sqrt{L^2 - a^2}}{L} = \cos \varphi $$

All of the expressions in the last line are equivalent, use whichever one works best for you.