Consider a ladder of length L leaning against a frictionless wall which is at right angles to the ground. You pull the bottom of the ladder horizontally away from the wall, at constant speed v. The claim is that this causes the top of the ladder to fall infinitely fast.

Common sense tells us this can't possibly be true, but can you find the flaw in the following supposed "proof" of this claim?

The Fallacious Proof:

* Step 1: As shown, let x denote the horizontal distance from the bottom of the ladder to the wall, at time t.

* Step 2: As shown, let y denote the height of the top of the ladder from the ground, at time t.

* Step 3: Since the ladder, the ground, and the wall form a right triangle, $\displaystyle x^2+y^2=L^2$ .

* Step 4: Therefore, $\displaystyle y=\sqrt{L^2-x^2}$ .

* Step 5: Differentiating, and letting x' and y' (respectively) denote the derivatives of x and y with respect to t, we get that

$\displaystyle y'=\frac{xx'}{\sqrt{L^2-x^2}}$

* Step 6: Since the bottom of the ladder is being pulled with constant speed v, we have x' = v, and therefore

$\displaystyle y'=\frac{xv}{\sqrt{L^2-x^2}}$

* Step 7: As x approaches L, the numerator in this expression for y' approaches -Lv which is nonzero, while the denominator approaches zero.

* Step 8: Therefore, y' approaches $\displaystyle -\infty$ as x approaches L. In other words, the top of the ladder is falling infinitely fast by the time the bottom has been pulled a distance L away from the wall.