Calculus Problem: Dy/dt When Ladder Hits Ground?

[Gimp]

Calculus Problem??

Suppose that we have a ladder 20feet long, leaning against a wall. Let x be the distance from the wall to the bottom of the ladder and y be the distance from the ground to the top of the ladder.

a. What will be the value of dy/dt when the top of the ladder hits the ground? What is going on here?

So far I found the value of dy/dt to be negative infinity, but i cannot figure out what is actually going on here. Does someone have an idea?? Thanks in advance.

 

[Galileo]

That's correct under certain assumptions (which are nonphysical).

Investigate the assumptions you made when making the problem. You probably assumed the left side of ladder sticks to the wall.

 

[Gimp]

I do understand that it's not the height that is going to infinity, but instead it's the velocity that's going to infinity. However, I don't understand how it can to infinity because the velocity should slow down at some point, but before the ladder hits the ground. However it's not the friction against the wall and the ladder. It's right before it hits the ground that something happens, but I'm not sure exactly it is.

 

[Galileo]

The problem is with the assumption that $L^2=y^2+x^2$, where L is the length of the ladder. This will hold in the beginning, but not during the entire fall. In fact, the ladder will at a certain point 'detach' itself from the wall.

 

[Gimp]

Ahhh...! I think I got it, thank you very much!

 

[Galileo]

I found a very nice website listing several mathematical fallacies.
http://www.math.toronto.edu/mathnet/plain/falseProofs/fallacies.html

The ladder example is there too. Check it out.