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What is the Longest Pole?


Date: 04/04/99 at 21:31:09
From: Marya marinsek
Subject: Pole Length Problem

A metal pole L ft long is pushed on the floor from a corridor 5 ft 
wide into another corridor 4 ft wide and at a right angle. What is the 
longest pipe that can be pushed into the 4-ft corridor?


Date: 04/30/99 at 11:33:05
From: Doctor Fwg
Subject: Re: Pole Length Problem

Dear Marya,

Let's restate the problem: 

   Find the length of the longest rigid (and slender) pole that can be 
   carried around the intersection of two hallways (one of width "X" 
   and the other of width "Y") if the pole axis is kept parallel to 
   the plane of the floor when it turns around the corner:

  

If one lets the entire pole length = L, then:

   L = L1 + L2,

but:  L1 = Y/[Sin(a)] and L2 = X/[Cos(a)], so:

   L =  Y/[Sin(a)] + X/[Cos(a)].

Now: L is expressed as a function of the hall widths (which are 
constants) and the angle "a" (which is a variable) that occurs when 
the pole is just touching the corner and the intersecting walls of the 
corridors. Substituting varying arbitrary values for the angle "a" 
(ranging between, but not including, 0 and 90 degrees), an extensive 
range of "L" values can be calculated. If one plots these values of L 
as a function of the angle "a", a bowl-shaped curve (or line) will be 
produced (concave upward). The slope of this line is zero at the 
minimum value of L located on this curved line. The minimum value of 
L, at this angle "a", is the length of the longest pole that can just 
turn around the corner that exists due to the intersecting corridors.

The classical technique for finding a minimum (or maximum) along any 
curved line is to take the first derivative of the line function, set 
that result equal to zero, and solve for the minimum (or maximum) 
value of interest. In this case, one can take the first derivative of 
L with respect to the angle "a", set this result equal to zero, and 
solve for the minimum value of L. This minimum in L is, in fact, the 
longest value that L can have which will permit the pole to turn 
completely around the corner. Any pole length longer than this will 
not allow the pole to turn completely and thus pass from one hallway 
to the other. Note also that this procedure should produce a result 
that is independent of travel direction.

Using the equation last written above and following this outline, one 
obtains:

dL/da = -Y[Cos(a)]/[Sin(a)]^2 + X[Sin(a)]/[Cos(a)]^2 = 0.

Simplifying this equation, one will find that:

Tan(a) = [Y^(1/3)]/[X^(1/3)], therefore:

A right triangle having an abscissa side length of X^(1/3) and an 
ordinate side length of Y^(1/3), as well as angles of "a", "(90-a)", 
and "90" degrees, exists here. Using the Pythagorean Theorem, one can 
show that the hypotenuse (C) of this triangle is:

C = [X^(2/3) + Y^(2/3)]^(1/2), so:

Sin(a) = [Y^(1/3)]/C and Cos(a) = [X^(1/3)]/C.

Now, re-writing L using these expressions:

L = Y/[Sin(a)] + X/[Cos(a)] = Y/[C*Y^(1/3)] +  X/[C*X^(1/3)].

(Ed. note: This equation contains an error, which another student and 
doctor discuss below.)

Simplifying and factoring the expression immediately above yields:

L = [X^(2/3) +  Y^(2/3)]^(3/2).

I hope I haven't made any errors - make sure you check this over in 
detail. After checking this solution, see if you can do this problem 
by yourself without looking at the illustration.

With best regards,

- Doctor Fwg, The Math Forum
http://mathforum.org/dr.math/   
    

Date: 02/12/2016 at 18:51:25
From: Nathaniel Riley
Subject: Maximum length of a ladder including width around a corner

I am trying to find the maximum length ladder that can go around a corner, 
but am having difficulty incorporating the width of the ladder into 
my work.

I can use trigonometry, or even just coordinate geometry, to approach 
this question; but neither method takes into account the real-world 
constraint of the ladder's width (or for that matter, its thickness). How 
could this be completed?

My own two solutions, which assume infinitely thin ladders, rely on 
these approaches:
   http://mathforum.org/library/drmath/view/53712.html
   http://archives.math.utk.edu/visual.calculus/3/applications.2/

Date: 02/13/2016 at 09:53:41
From: Doctor Anthony
Subject: Re: Maximum length of a ladder including width around a corner

Hi, Nathaniel.

Let L = length of the ladder.

             A
    |-------/|--------------------------
    |      / |
    |     /  | a    width of passage = a
    |    /   |
    |  B|@---|---------------------------
    |  /|
    | / |
    |/  |
   C|---| 
    | b |  width of passage = b
    |   |

If   L = AB + BC
       =  a/sin@ + b/cos@

Then L = a.cosec@ + b.sec@

To find @ for max L, we set dL/d@ = 0  
                            dL/d@ = -a.cosec@.cot@ + b.sec@.tan@
                    a.cosec@.cot@ = b.sec@.tan@
                  a.cos@/sin^2(@) = b.sin@/cos^2(@)
                              a/b = sin^3(@)/cos^3(@)
                              a/b = tan^3(@)

As a check, if a = b, then tan@ = 1, and @ = 45 degrees, which is correct.

For example, suppose a = 6, b = 3.

Then a/b = 2.

So tan@ = 2^(1/3) 
        = 1.256
      @ = 51.56 degrees

      L = a/sin@ + b/cos@
        = 6/sin(51.56) + 3/cos(51.56)
        =  12.485 metres  <-----

- Doctor Anthony, The Math Forum
  http://mathforum.org/dr.math/

Date: 02/13/2016 at 18:15:29
From: Nathaniel Riley
Subject: Re: Maximum length of a ladder including width around a corner

This solution does not take the width of the ladder into account. That is 
precisely what I asked for help with.

Date: 02/13/2016 at 21:56:47
From: Doctor Peterson
Subject: Re: Maximum length of a ladder including width around a corner

Hi, Nathaniel.

I can vaguely recall trying to solve this extended problem in the past, 
and getting stuck.

   +-----------------------+--------
   |                    /   W\
   |                 /         +
   |         L    /         /
   |           /         /         A
   |        +         /
   |     /         /
   |  /         +-------------------
   +         /  |
   | \    /     |
   |   +        |
   |            |
   |            |
   |      B     |

I have made an attempt using this picture, re-labeling the old angle theta 
as t:

   +-----------------------+------
   |                 t  /  :
   |                 /     :
   |         L    /        :
   |           / t         :     A
   |        +..............:
   |     /  :W\t:
   |  / t   :...+-----------------
   +........:   |
   |            |
   |            |
   |      B     |

This yielded the following equation to maximize L:

   A cos^3(t) - B sin^3(t) + W(sin^2(t) - cos^2(t)) = 0

This looks a little hard, so I took a break and searched for references to 
this natural extension to the familiar "ladder problem." The extension can 
be called the "couch problem" for obvious reasons. The following PDF 
discusses it on page 12:

http://www1.american.edu/academic.depts/cas/mathstat/People/
kalman/pdffiles/ladderprob.pdf Thus, in the idealized geometry of the problem statement, perhaps we should try to maneuver a rectangle rather than a line segment around the corner. If the width of the rectangle is fixed at w, what is the greatest length L that permits the rectangle to go around the corner? This version of the problem also provides a reasonable model for moving bulkier objects than ladders. For example, trying to push a desk or a couch around a corner in a corridor is naturally idealized to the problem of moving a rectangle around the corner in figure 11. This is the motivation given by Moretti in his analysis of the rectangle version of the ladder problem. In honor of his work, we refer to the rectangular version hereafter as the couch problem. It should not be confused with the sofa problem, which concerns the area of a figure to be moved around a corner. Kalman eventually comes to the same equation I got. His equation (8) on page 15 says At this point, finding alpha appears to depend on solving a sixth degree polynomial equation.... It is easy to solve this equation numerically (given values for a, b, and w), and very likely impossible to solve it symbolically. You can probably find many other discussions of the problem. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/

Date: 02/14/2016 at 00:13:58
From: Nathaniel Riley
Subject: Re: Maximum length of a ladder including width around a corner

Hello,

Thanks for your help. I do have another question.

This Prezi gets to the same point, but doesn't require trigonometry:
   https://prezi.com/xjuffaugg1_d/moving-a-couch-around-a-corner/

Would you be able to explain how Giesler arrives at b = -1 and a = m on 
slide 9?

Thanks.

Date: 02/14/2016 at 14:03:58
From: Doctor Peterson
Subject: Re: Maximum length of a ladder including width around a corner

Hi, Nathaniel.

Giesler is using "b" in a couple different ways, which is confusing. But 
the graph of the problem shows the line as (slope-intercept form)

   y = mx + c

The (general) form for the line in the distance formula is

   ax + by + c = 0

Putting the given equation into this form, it becomes

   mx - y + c = 0

Here, the "a" in the general form (the coefficient of x) is m, and the "b" 
(the coefficient of y) is -1. These are then put into the formula for 
the distance:

       |ax_0 + by_0 + c|   |mx_0 - y_0 + c|
   d = ----------------- = ---------------- = w
        sqrt(a^2 + b^2)     sqrt(m^2 + 1)

Solving for c in terms of w, with x_0 = y_0 = 0, and knowing that c is 
negative since it is the y-intercept of the line in the graph, we find

  |mx_0 - y_0 + c| = w sqrt(m^2 + 1)
               |c| = w sqrt(m^2 + 1)
                c  = -w sqrt(m^2 + 1)

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/


Date: 02/14/2016 at 15:31:13
From: Nathaniel Riley
Subject: Maximum length of a ladder including width around a corner

Thank you so much for this help!

I do have one more question; sorry to keep asking.

On slide 10, Giesler derives the coordinates of the points. I am a little 
unsure how she gets them. I know that there is a mistake in point A -- the 
y value should be -a -- a correction that appears on the next slide.

Would you be able to help me in understanding this?

Thanks again,
Nathaniel

Date: 02/14/2016 at 16:20:36
From: Doctor Peterson
Subject: Re: Maximum length of a ladder including width around a corner

We have the equation of the line

   y = mx - w sqrt(m^2 + 1)

Point B is where x = b, which we plug in to get y:

   y = mb - w sqrt(m^2 + 1)

Point A is where y = -a, so we have to solve for x:

   mx - w sqrt(m^2 + 1) = -a
                     mx = -a + w sqrt(m^2 + 1)
                      x = [-a + w sqrt(m^2 + 1)]/m

Does that help?

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/

Date: 02/15/2016 at 00:09:43
From: Nathaniel Riley
Subject: Thank you (Maximum length of a ladder including width around ...)

Thank you so much. I don't know why I didn't think of that.

I think I should be fine from here.

Date: 02/16/2016 at 03:20:50
From: Nathaniel Riley
Subject: Maximum length of a ladder including width around a corner

I have yet another question. Sorry to ask!

It is based off this conversation again:
  http://mathforum.org/library/drmath/view/53712.html

How do you go from this ...

   Y/[C*Y^(1/3)] +  X/[C*X^(1/3)]

... to this?

   L = [X^(2/3) +  Y^(2/3)]^(3/2)

Thanks for any help.

Nathaniel

Date: 02/16/2016 at 09:32:24
From: Doctor Peterson
Subject: Re: Maximum length of a ladder including width around a corner

Hi, Nathaniel.

Let's try it. Doctor Fwg says the latter expression comes from 
"simplifying and factoring the expression immediately above." We have

           Y          X
   L = --------- + ---------
       C*Y^(1/3)   C*X^(1/3)

We can carry out the divisions to get

       Y^(2/3)   X^(2/3)   Y^(2/3) + X^(2/3)
   L = ------- + ------- = -----------------
          C         C              C

Looking back further, we find

   C = [X^(2/3) + Y^(2/3)]^(1/2)

So we can put this in:

       [X^(2/3) + Y^(2/3)]
   L = -------------------------
       [X^(2/3) + Y^(2/3)]^(1/2)

But this would just become

   L = [X^(2/3) + Y^(2/3)]^(1/2)

So we need to find his mistake (or our error in copying). Looking further 
back, I see that Doctor Fwg did this:

   C = [X^(2/3) + Y^(2/3)]^(1/2), so:

   Sin(a) = [Y^(1/3)]/C and Cos(a) = [X^(1/3)]/C.

   Now, re-writing L using these expressions:

   L = Y/[Sin(a)] + X/[Cos(a)] = Y/[C*Y^(1/3)] +  X/[C*X^(1/3)].

This last line should have been:

   L = Y/[Sin(a)] + X/[Cos(a)] = Y/[Y^(1/3)/C] +  X/[X^(1/3)/C]

So let's do the work over again from here:

         CY        CX
   L = ------- + ------- 
       Y^(1/3)   X^(1/3)

     = CY^(2/3) + CX^(2/3) 
     = C[Y^(2/3) + X^(2/3)]
     = [X^(2/3) + Y^(2/3)]^(1/2) [Y^(2/3) + X^(2/3)]
     = [Y^(2/3) + X^(2/3)]^(3/2)

So Doctor Fwg's final result is correct, but your confusion understandably 
comes from this one line, which was indeed wrong:

   L = Y/[Sin(a)] + X/[Cos(a)]
     = Y/[C*Y^(1/3)] +  X/[C*X^(1/3)]

- Doctor Peterson, The Math Forum
  http://mathforum.org/dr.math/

Date: 02/16/2016 at 16:58:26
From: Nathaniel Riley
Subject: Thank you (Maximum length of a ladder including width around ...)

Thank you so much. 

I didn't know the C should be on the top; but thinking about it, now I 
see why. 

Thanks again,

Nathaniel
Associated Topics:
High School Calculus
High School Trigonometry

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