Calculus question (explain the errors)?

Question

Suppose a ladder of length L is leaning against a frictionless wall. The top of the ladder reaches a point that is y units from the ground. The bottom of the ladder is x units from the wall. The bottom of the ladder is moving to the right with speed v. Therefore x and y are quantities that change with time. By the Pythagorean Theorem x^2+y^2=L^2. Solving for y gives √L^2-x^2 . Differentiating both sides with respect to time gives dy/dt= -x( √L^2-x^2) (dx/dt). But dx/dt=v , so dy/dt=-xv/√L^2-x^2 . As x approaches L, the numerator approaches -Lv (a nonzero negative number) while the denominator approaches 0. Hence, the quotient is going to negative infinity. In other words, the top of the ladder is falling infinitely fast when the bottom of the ladder has been pulled a distance of L from the wall.

Answer 1

x^2+y^2=L^2
Derentiating both sides with respect to time gives,
2xx'+2yy' = 0
Solve for y',
y' = -xx'/y
When y approaches zero, y' is undefined.

However, in real situation, y' has to satisfy physics law, such as rotational moment cannot be infinity,

Answer 2

The error is because you assume that the wall and the floor are frictionless, so the change in x over time (dx/dt) keeps increasing. Therefore the ladder will continue to move to the right even after x approaches L

Answer 3

I can't see any error.

However, this is something that would never happen in the real world. Try to describe the forces acting on the ladder that would cause it to behave that way. I think you'll fail.

In the simplest case, namely a force pulling the bottom of the ladder along the ground, the top of the ladder will move away from the wall. Gravity doesn't cause it to accelerate downward quickly enough to remain in contact with the wall.

Answer 4

what then is the question?