SME
Calculus II · Problem type
Arc length of a cycloid
Use this for “Find the length of one arch of x = 2(t − sin t), y = 2(1 − cos t).”
Diagram
At t = 2π the point is (12.57, 0).
The arrows show the way t runs.
Length
from
t
=
0
to
2π:
L
=
∫
√
x′² + y′²
dt
=
16
This is an example. Type your own values to start a new problem.
x
=
r
(t
−
sin
t)
y
=
r
(1
−
cos
t)
L
=
∫
0
(dx/dt)²
+
(dy/dt)²
dt
=
x
x at T
worked out (not typed)
y
y at T
worked out (not typed)
Formulas
x
=
r
×
(
T
−
sin(
T
))
12.5664 = 2 × (2π − sin(2π))
y
=
r
×
(1
−
cos(
T
))
0 = 2 × (1 − cos(2π))
L
=
4
×
r
×
(1
−
cos(
T
2
))
16
=
4
×
2
×
(1
−
cos(
2π
2
))
Assumptions
•
A circle of radius r rolls along the x-axis without slipping; t is the angle it has turned, in radians.
•
The speed is √((dx/dt)² + (dy/dt)²) = r√(2 − 2 cos t) = 2r sin(t/2), which is never negative for 0 ≤ t ≤ 2π.
•
One arch runs from t = 0 to T = 2π, so T stays in that range.
Step-by-step
WE KNOW
r = 2, T = 2π
FIND
x at T (x), y at T (y), arc length (L)
Step 1 · Find x at T (x)
Use x = r × (T − sin(T))
The
center
has
rolled
r
×
T
along
the
axis;
the
point
sits
r
sin
T
behind
it.
x
=
2
×
(2π
−
sin(2π))
x
=
2
×
(2π
−
0)
x
=
12.5664
Step 2 · Find y at T (y)
Use y = r × (1 − cos(T))
The
center
is
r
up;
the
point
sits
r
cos
T
below
it.
y
=
2
×
(1
−
cos(2π))
y
=
0
Step 3 · Find arc length (L)
Use L = 4 × r × (1 − cos(T ÷ 2))
Integrate
the
speed
2
r
sin(
t
2
)
from
0
to
T
:
the
antiderivative
is
−
4
r
cos(
t
2
).
L
=
4
×
2
×
(1
−
cos(
2π
2
))
L
=
16
Check
Put the numbers back in. Both sides should match.
12.5664 = 2 × (2π − sin(2π)) ✓
0 = 2 × (1 − cos(2π)) ✓
16
=
4
×
2
×
(1
−
cos(
2π
2
))
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.