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)]. 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/ |
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