Computer Algebra Recipes

4.2. SECOND-ORDER MODELS 165
> restart:
> eq:=diff(y(x),x)=(h-y(x))/(0-x);
eq := d
¡ y(x)
y(x) =¡h
dx x
We solve eq for Heather's vertical coordinate h.
> h:=solve(eq,h);
μ
d
h := ¡
dx y(x)
x + y(x)
Heather's speed is r times that of Patches, so dh=dt = r (ds=dt), where t is time
and ds = p 1+(dy=dx) 2 dx is an element of arclength along Patches' path.
But then dh = rds,ordh=dx = r (ds=dx) =r p 1+(dy=dx) 2 . Entering this
last relation yields the relevant second-order nonlinear ODE for y(x),
> ode:=diff(h,x)=r*sqrt(1+diff(y(x),x)^2);
μ
s
μ 2 d d
ode := ¡ y(x) x = r 1+
dx2 dx y(x)
2
the expression for h having been automatically substituted. Although ode is
nonlinear, it can be analytically solved. Patches is initially at x =1;y=0and
the tangent line there has zero slope. These starting conditions are used in the
dsolve command.
> sol:=dsolve(fode,y(1)=0,D(y)(1)=0g,y(x));
x xxr
sol := y(x) =¡
¡
2(¡1+r) xr 2(1+r) +
r
;
¡1+r2 x xxr
y(x) =
+
2(¡1+r) xr 2(1+r) ¡
r
¡1+r2 Two forms of the solution are generated, the ¯rst corresponding to Heather
moving downward (in the negative y direction) along the x = 0 line, the second
corresponding to moving upward. The upward (second one here) solution is
selected and simpli¯ed with respect to powers of x.
> y:=simplify(rhs(sol[2]),power);
y := x(1¡r) x(1+r)
+
2(¡1+r) 2(1+r) ¡
r
¡1+r2 Evaluating y at x = 0 yield's Heather's vertical coordinate when Patches reaches
her. Notice that as r approaches 1 from below, Y goes to in¯nity, i.e., Patches
cannot catch Heather in a ¯nite distance. Obviously, Patches must run faster
than Heather can walk to catch her.
> Y:=eval(y,x=0);
r
Y := ¡
The given ratio r = 3
8
> r:=3/8: y:=y; Y:=evalf(Y);
¡1+r 2
is entered, and y and Y are automatically evaluated.