SME
Calculus III (Multivariable) · Problem type
A helix: speed, length and curvature
Use this for “For r(t) = ⟨3 cos t, 3 sin t, 4t⟩, find the velocity at t = 0, the speed, the length of one turn and the curvature.”
Diagram
r(t) = ⟨3 cos t, 3 sin t, 4t⟩.
At t = 0: r = (3, 0, 0) and v = r′(t) = ⟨0, 3, 4⟩.
Speed
√
3
2
+
4
2
=
5
at
every
t,
so
from
0
to
6.283
the
length
is
31.42.
κ
=
3
3
2
+
4
2
=
0.12.
This is an example. Type your own values to start a new problem.
r(t)
=
⟨
a
cos
t,
a
sin
t,
c
t⟩
r′(
t
)
=
⟨
v₁
,
v₂
,
v₃
⟩,
|r′|
=
|r′|
L
=
|r′|
×
T
=
L
κ
Curvature
worked out (not typed)
Formulas
v₁
=
−
a
×
sin(
t
)
0 = −3 × sin(0)
v₂
=
a
×
cos(
t
)
3 = 3 × cos(0)
v₃
=
c
4 = 4
|r′|
=
√
a
2
+
c
2
5
=
√
3
2
+
4
2
L
=
|r′|
×
T
10π = 5 × 2π
κ
=
a
a
2
+
c
2
0.12
=
3
3
2
+
4
2
Assumptions
•
r(t) = ⟨a cos t, a sin t, ct⟩ winds round the z-axis at radius a and rises 2πc each turn.
•
t is in radians; one full turn is a span of T = 2π ≈ 6.283.
•
The speed |r′(t)| is the same at every t, so the length is the speed times the span.
Step-by-step
WE KNOW
a = 3, c = 4, t = 0, T = 2π
FIND
velocity, x part (v₁), velocity, y part (v₂), velocity, z part (v₃), speed (|r′|), length over the span (L), curvature (κ)
Step 1 · Find velocity, x part (v₁)
Use v₁ = −a × sin(t)
r′(
t
)
part
by
part:
the
derivative
of
a
cos
t
is
−a
sin
t
.
v₁
=
−3
×
sin(0)
v₁
=
−3
×
0
v₁
=
0
Step 2 · Find velocity, y part (v₂)
Use v₂ = a × cos(t)
The
derivative
of
a
sin
t
is
a
cos
t
.
v₂
=
3
×
cos(0)
v₂
=
3
Step 3 · Find velocity, z part (v₃)
Use v₃ = c
The
height
ct
rises
at
the
steady
rate
c
.
v₃
=
4
Step 4 · Find speed (|r′|)
Use |r′| = √(a² + c²)
|r′(
t
)|²
=
a
2
sin²
t
+
a
2
cos²
t
+
c
2
=
a
2
+
c
2
,
since
sin²
t
+
cos²
t
=
1.
|r′|
=
√
3
2
+
4
2
|r′|
=
√
9 + 16
|r′|
=
√
25
|r′| = 5
Step 5 · Find length over the span (L)
Use L = |r′| × T
Arc
length
is
the
integral
of
the
speed
over
t
;
a
steady
speed
makes
it
speed
times
span.
L
=
5
×
2π
L
=
10π
(≈
31.42)
Step 6 · Find curvature (κ)
Use κ = a ÷ (a² + c²)
κ
=
|r′
×
r″|
÷
|r′|³;
here
|r′
×
r″|
=
a
√
a
2
+
c
2
and
|r′|³
=
(
a
2
+
c
2
)
√
a
2
+
c
2
.
κ
=
3
3
2
+
4
2
κ
=
3
9 + 16
κ
=
3
25
κ
=
0.12
Check
Put the numbers back in. Both sides should match.
0 = −3 × sin(0) ✓
3 = 3 × cos(0) ✓
4 = 4 ✓
5
=
√
3
2
+
4
2
✓
10π = 5 × 2π ✓
0.12
=
3
3
2
+
4
2
✓
Topic check
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