Ladder problem

[sallycats]

A ladder leans against a frictionless wall as pictured. The top of the ladder is y units from the ground and the bottom of the ladder is x units from the wall. The ladder moves to the right with a speed of v.

According to the Pythagorean Theorem, x[sup:gahbd3v2]2[/sup:gahbd3v2]+y[sup:gahbd3v2]2[/sup:gahbd3v2]=L[sup:gahbd3v2]2[/sup:gahbd3v2], so y= sqrt(L[sup:gahbd3v2]2[/sup:gahbd3v2]-x[sup:gahbd3v2]2[/sup:gahbd3v2]).

Differentiate both sides with respect to time and dy/dt = -x/sqrt(L[sup:gahbd3v2]2[/sup:gahbd3v2]-x[sup:gahbd3v2]2[/sup:gahbd3v2]) dx/dt.

dx/dt = v, so dy/dt = -xv/sqrt(L[sup:gahbd3v2]2[/sup:gahbd3v2]-x[sup:gahbd3v2]2[/sup:gahbd3v2])

As x approaches L, the nominator, -xv, approaches -Lv, which is a negative number.
Also, as x approaches L, the denominator, sqrt(L[sup:gahbd3v2]2[/sup:gahbd3v2]-x[sup:gahbd3v2]2[/sup:gahbd3v2]), approaches 0.
This indicates that dy/dt goes to negative infinity, but how is it possible for a ladder sliding along the ground to approach negative infinity?

 

[BigGlenntheHeavy]

\(\displaystyle x^{2}+y^{2} \ = \ L^{2}, \ x \ and \ y \ are \ > \ 0.\)

\(\displaystyle 2x\frac{dx}{dt}+2y\frac{dy}{dt} \ = \ 0\)

\(\displaystyle y\frac{dy}{dt} \ = \ -x\frac{dx}{dt}\)

\(\displaystyle \frac{dy}{dt} \ = \ \frac{-x}{y}\frac{dx}{dt}, is \ decreasing, \ as \ \frac{dx}{dt} \ is \ positive.\)

\(\displaystyle \frac{dx}{dt} \ = \ \frac{-y}{x}\frac{dy}{dt}, is \ increasing, \ as \ \frac{dy}{dt} \ is \ negative.\)