along the curve with equation
(1.1) y = y(x).
Note that it will frequently be convenient to use the same symbol—
here y—to denote both a univariate function and the ordinate of it
graph, because the correct interpretation will be obvious from contex
Needless to say, the endpoints must lie on the curve, and so
(1.2) y(0) = 1, y(1) = 0.
Let the bead have velocity
(1.3) v =
ds
dt
τ =
dx
dt
i +
dy
dt
j =
dx
dt
i +
dy
dx
j ,
where s denotes arc length, t denotes time and i, j and τ are uni
vectors in the (rightward) horizontal, (upward) vertical and tangentia
directions, respectively, so that the particle’s speed is
(1.4) v = |v| =
ds
dt
=
dx
dt
2
+
dy
dt
2
= 1 +
dy
dx
2
dx
dt
implying
(1.5) ds = 1 + (y )2 dx,
where y denotes
dy
dx
.2
Let the particle start at (0, 1) at time 0 an
reach (1, 0) at time tf after travelling distance sf along the curve
Then its transit time is
(1.6)
tf
0
dt =
sf
0
ds
v
=
1
0
1 + (y )2
v
dx.
If g is the acceleration due to gravity and the bead has mass m, the
its kinetic energy is
1
2
mv2,
its potential energy is mgy and—becaus
there is no friction—the sum of the kinetic and potential energies mus
be a constant. Because the sum was
1
2
m02 + mg · 1 = mg initially, w
have
1
2
mv2 + mgy = mg or
(1.7) v = 2g(1 y),
2As
remarked above, the symbol y may denote either a derivative, i.e., a functio
or the value, say y (x), that this function assigns to an arbitrary element—here x—
its domain. The correct interpretation is obvious from context; e.g., y in (1.8)-(1.9
denotes the assigned value, for otherwise the integral would not be well defined.
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