Computer Algebra Recipes

198 CHAPTER 4. NONLINEAR ODE MODELS
determined, the curve that minimizes the time of descent will be known.
At the starting point x = 0, we shall choose to set the parameter μ equal
to zero. Some care must be taken in evaluating the arctan term at μ =0. The
limit must be taken from the positive side (i.e., from the \right") of μ =0.
> eq:=limit(x,theta=0,right)=0;
eq := B ¡ A¼
4 =0
The resulting eq is easily solved for the constant B.
> sol2:=B=solve(eq,B);
sol2 := B = A¼
4
The solution sol2 is assigned and x displayed.
> assign(sol2); x:=x;
x := ¡ 1
μ
1
cos(2 μ)
A sin(2 μ) ¡ A arctan
+
2 2 sin(2 μ)
A¼
4
We still have to determine the constant A and the range of μ. Now it gets a bit
tricky. Wehavenoideayetwhatthecoordinates(x1 ; y1 ) of the endpoint on
the museum roof should be.
Hand me that copy of Mathematical Methods in Physics [MW71]. Ah, here
we go. The case in which one endpoint of the sought-after curve is ¯xed and
the other endpoint is allowed to vary along a line g(x; y) = 0 is considered.
It is shown that if the Euler{Lagrange function F is of the structure8 F =
f(x; y) p 1+(y0 ) 2 ; which it is in our case, then the condition for determining
the unknown endpoint is that the slope of y(x) must satisfy the condition y 0 =
(@g=@y)=(@g=@x) atthatpoint. Butthisisjustamathematicalstatement
that the curve y(x) of quickest descent must intersect the destination curve
g(x; y) = 0 at right angles. Here the equation for the slanting museum roof is
the straight line g(x; y) =y + x ¡ 200 = 0: But both @g=@y and @g=@x are
equal to 1, so we have y 0 = 1 at the museum roof. Since the parameter μ has
been introduced, the slope of y(x) mustbecalculatedintermsofμ.
> slope:=simplify(diff(rhs(Y),theta)/diff(x,theta));
slope := cos(μ)
sin(μ)
Setting the slope to 1, the value £ that the parameter μ must have at the
museum roof is determined, assuming that μ is positive.
> Theta:=solve(slope=1,theta) assuming theta>0;
£:= ¼
4
8If F is not of this structure, the endpoint condition is more complicated, taking the form
³
0 @F
F ¡ y
@y 0
´
@g @F
¡
@y @y 0
@g
@x =0: