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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/ 
Associated Topics:
College Calculus
High School Calculus

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