Answer
$\frac{dy}{dt} = -\frac{x}{y}~\frac{dx}{dt}$
If the top of the ladder approaches the ground, then the value of $y$ gets smaller and the value of $x$ gets closer to $L$. As $y$ gets smaller and smaller, the speed at which the top of the ladder approaches the ground, that is $\vert \frac{dy}{dt} \vert$, gets faster and faster. In fact, as $y \to 0$, then $\vert \frac{dy}{dt} \vert \to \infty$
Clearly, this is not physically possible. Therefore, the model is not appropriate for very small values of $y$.
Work Step by Step
Let $L$ be the length of the ladder. Then:
$x^2 + y^2 = L^2$
We can differentiate this equation on both sides with respect to $t$:
$x^2 + y^2 = L^2$
$2x\frac{dx}{dt} + 2y~\frac{dy}{dt} = 0$
$\frac{dy}{dt} = -\frac{x}{y}~\frac{dx}{dt}$
If the top of the ladder approaches the ground, then the value of $y$ gets smaller and the value of $x$ gets closer to $L$. As $y$ gets smaller and smaller, the speed at which the top of the ladder approaches the ground, that is $\vert \frac{dy}{dt} \vert$, gets faster and faster. In fact, as $y \to 0$, then $\vert \frac{dy}{dt} \vert \to \infty$
Clearly, this is not physically possible. Therefore, the model is not appropriate for very small values of $y$.
