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.

(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.