Why Are the Rates of the Sliding Ladder Problem Different?
Date: 12/17/2003 at 12:38:38 From: Tom Subject: related rates classic "ladder sliding down a wall" problem As the top of the ladder slides down the wall, the bottom of the ladder slides away from the wall. Why are those rates not the same? I know how to calculate the different rates in the problem, but I am unsure why they are different.
Date: 12/18/2003 at 12:01:40 From: Doctor Peterson Subject: Re: related rates classic "ladder sliding down a wall" problem Hi, Tom. You are talking about essentially the problem discussed here: Related Rates http://mathforum.org/library/drmath/view/53412.html I think you are saying that you can solve the problem mathematically, but you don't believe your answer because it contradicts your expectations. You need to retrain your intuition so that it makes sense to you that the two ends would move at different rates. You might start by just doing the experiment. Take a book and stand it up against a wall, or the side of your computer. Pull the bottom out at a constant rate (which I think is easier than making the top slide down at a constant rate). Watch what happens to the top: it will barely move at first, then move faster until it plops down on the table at the end as fast as it can go. The speed of the top is not the same as the speed of the bottom. Now look at it from the side. When it is standing up steeply, making a change in the width of the base of the triangle hardly changes the height; but when it lies nearly flat, a change in the height hardly changes the base. Why? Consider a circle centered at A, the top of the ladder, with radius AC, the length of the ladder (hypotenuse of the right triangle). When the ladder is at a steep slope, the arc through C is nearly parallel to the base BC; if you swing AC upward to decrease the slope, C rises very little, so A will drop little if we keep C on the floor: A * /| * /| * / | * / | * / | ** / | * / | ** / | ** / | *** / | -------------*****C*----B ***** When the ladder is nearly horizontal, the situation is reversed; the arc at C is nearly vertical, and C moves out very little while A has to move down considerably to keep C on the floor: --A * ------ | * ------ | * ------ | -------C----------------------B * | ** | So as the angle changes, the relationship between a movement of A and the corresponding movement of C changes; at first C moves more, then A moves more. Now look at the calculus. If BC and AB are x and y respectively, and AC is L, we have x^2 + y^2 = L^2 2x dx/dt + 2y dy/dt = 0 y dy/dt = -x dx/dt dy/dt ----- = -x/y dx/dt So the ratio of a change in y to a change in x is proportional to x and inversely proportional to y; the steeper the ladder, the smaller the change in y for a given change in x. That is just what we have observed. If you have any further questions, feel free to write back. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/
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