Leaning Ladders
Consider two ladders leaning in opposite directions between two
parallel walls. One ladder is 40 feet long and the other is 60
feet long. Each ladder is touching the base of the wall opposite
the wall on which it is leaning. The two ladders cross at a point
10 feet above the level ground. How far apart are the walls?
This type of problem often leads to questions about what constitutes
a "closed-form" solution, and about the relative utility of implicit
vs explicit definitions. If the tops of the ladders contact the
walls at elevations a,b, then the height h of the crossing point is
given by
1 1 1
--- = --- + --- (1)
h a b
regardless of the distance between the two walls. Of course, if w
is the distance between the walls and A,B are the lengths of the
two ladders, then
a = sqrt( A^2 - w^2 ) b = sqrt( B^2 - w^2 )
Thus, given the value of w we can directly compute the corresponding
value of h. Notice, however, that this computation involves the
evaluation of square roots, which is an algorithmic process that
generally yields only approximate answers.
Now suppose, instead, that we're given some value of h and asked to
compute the value of w. Our first impulse is to seek an formula
involving only standard "calculator key" operations giving w as an
explicit function of h. To do this we would probably try to "invert"
equation (1) and isolate w. If we define D = A^2 - B^2 and
x = B^2 - w^2, this leads to the quartic
x^4 - 2(2h^2-D) x^3 + D(D-6h^2) x^2 - 2(h^2 D^2) x + (h^4 D^2) = 0
For example, given the values A = 40, B = 60, h = 10, and making
the substitution x = 200y, we arrive at
y^4 + 18y^3 + 70y^2 - 100y + 25 = 0 (2)
Given the appropriate root y = 0.81149... of this equation, we have
w = sqrt(1600-200y) = 37.917 feet. This would typically be presented
as "the solution" of the problem. However, notice that (2) is no
more explicit than (1). Rather than deriving the polynomial (2) and
then solving for the root, we could just as easily have immediately
solved for the root of the equation
1 1 1
-------------- + -------------- - --- = 0 (3)
sqrt(3600-w^2) sqrt(1600-w^2) h
since most root-finding algorithms will work just as well with a
general function. The only real advantage I can see in converting
to a pure polynomial is that polynomials are sort of a "canonical form"
and we can check easily for possible factorizations, etc. If, for
example, we find that the polynomial form is irreducible, then we feel
confident we're not missing some trivial simplification, which may not
be obvious from equation (3).
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