Related Rates Ladder When a ladder is falling, why doesn't the top of the ladder fall at the same speed as the bottom of the ladder? I understand how one can solve the derivative using $x^2+y^2=z^2$, but I don't really understand how to intuitively think about this.
 A: Consider a ladder leaning against a wall such that $l-b>h$, where


*

*$h$ is the height at which the ladder meets the wall, 

*$b$ is the distance from the base of the ladder to the wall, and

*$l$ is the actual length of the ladder.
Note that this is equivalent to $b+h<l$, and this will hold whenever the ladder is neither fully vertical or fully horizontal. (Why? If $x, y>0$, then $(x+y)^2=x^2+y^2+2xy>x^2+y^2$; now use the Pythagorean theorem ...)
Now let's think about what happens as the ladder slides down the wall. From the time it starts sliding to the time it stops (= lies fully horizontal), the base has to travel a distance of $l-b$ along the ground, but the top of the ladder will only travel a distance of $h$ along the wall. So intuitively, we should expect the top of the ladder to move down the wall more slowly than the base of the ladder moves along the ground because the base has further to go in the same amount of time.
A: What makes you think that the top of a ladder doesn't fall at the same rate as the bottom?  If you carry a 20 ft ladder to the top of a 120 ft high building and drop the ladder off, the "top" and "bottom" of the ladder will fall at the same speed.  Or are you thinking of the case where the ladder is leaned up against a wall, the top starts sliding down the wall while the bottom moves away from the wall?  Of course, in that case,  the bottom isn't "falling" at all!
Are you actually asking "when the top of a ladder slides down a wall and the bottom slides away from the wall, does the top of the ladder slide down at the same rate at which the botton slides away from the wall?
If that is the case, suppose a ladder, of length L, is leaning up against a wall with its top at height h.  In that case, by the "Pythagorean theorem", the base of the ladder is at distance $x= \sqrt{L^2- h^2}$.  Squaring both sides, $x^2= L^2- h^2$.  Differentiating with respect to time, t, we have $2x\frac{dx}{dt}= -2h\frac{dh}{dt}$ or $x\frac{dx}{dt}= -h\frac{dh}{dt}$.  $-\frac{dh}{dt}$ is the rate at which the top is sliding down the wall (so negative) and $\frac{dx}{dt}$ is the rate at which the bottom is sliding away from the wall.  We can write $x\frac{dx}{dt}= -h\frac{dh}{dt}$ as $\frac{\frac{dx}{dt}}{\frac{dh}{dt}}= -\frac{h}{x}$.  The two rates (ignoring the negative sign) are the same only when x= h- that is, when the ladder is at a 45 degree angle.
