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/ 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 |
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