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In Calculus there is a "classic" related rates problem involving a falling ladder. Say the ladder is $25$ ft tall and is leaning against a wall. The bottom edge of the ladder is pulled away from the wall at a constant rate of $2$ ft/sec; as it moves, the top of the ladder slides down the wall. The student is asked to express the downward velocity of the top of the ladder in terms of the position $x$ of the bottom of the ladder, and finds that $$\frac{dy}{dt}=-\frac{2x}{\sqrt{625-x^2}}$$ Of course it makes sense that the velocity should only be defined up to $x=25$, because beyond that point the ladder comes away from the wall. But it seems strange (even to me, who has taught this stuff) that the downward velocity approaches $\infty$ as $x \to 25$. Why is that a "reasonable" result? If I imagine a speedometer attached to the top of the ladder, it's hard for me to believe that in the moments before the ladder hits the ground the speedometer readout increases without bound.

Is there an intuitive explanation of why the downward velocity of the top of the ladder ought to diverge to infinity as the ladder hits the ground?

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    $\begingroup$ The ladder will detach from the wall well before that happens. The velocity only approaches infinity if the end of the ladder is forced to remain in contact with the wall all the way down. $\endgroup$
    – user856
    Feb 17, 2016 at 3:57
  • $\begingroup$ In other words, mathematically, there is nothing wrong with the result you stated, but in the real world, such a situation cannot reasonably be simulated... $\endgroup$
    – imranfat
    Feb 17, 2016 at 3:59
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    $\begingroup$ One could imagine that the end of the ladder is on frictionless wheels that run on rails attached to the wall, and the rails enclose the wheels on both sides. Then one can see that when the ladder is nearly horizontal, it becomes extremely difficult to pull the other end at a constant velocity, because the rails are pulling back equally hard... $\endgroup$
    – user856
    Feb 17, 2016 at 4:04
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    $\begingroup$ Adding to @Rahul's idea: we could imagine the whole contraption lying, flipped on its side, on some frictionless surface, so that one could do away with gravity. And, cough, I too have taught this, and have felt the same unease. $\endgroup$
    – peter a g
    Feb 17, 2016 at 4:07
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    $\begingroup$ This may be an example of a physical model attached to a mathematical model where no physical model belongs. Sometimes we should just do pure math. As $dx/dy$ tends to zero, $dy/dx$ tends to infinity. $\endgroup$
    – David K
    Feb 17, 2016 at 4:34

1 Answer 1

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In terms of mechanical systems, this is what we would call a singular configuration which is a position where the system looses a degree of freedom. In this case when $x=25$ the system cannot move by pushing in the $x$ direction (note that I did not say that the system cannot move, only that it cannot move by pushing in the $x$ direction). The reason that $\frac{dy}{dt}=\infty$ is unreasonable is because that it is equally unreasonable that $\frac{dx}{dt}=2$ at $x=25$. The same thing happens in other systems. An example would be a piston in an engine when it is at top dead center... the system looses a degree of freedom.. i.e. no matter how hard you push on the piston, it cannot (in theory) move the crank shaft. Furthermore if we computed the crankshaft speed based on the piston speed, we would get the same infinite result. The case of Gimbal Lock is another example of a situation where this occurs... the system looses a degree of freedom, in this case from three to two.

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