SME
Calculus III (Multivariable)
Vector-valued functions, arc length and curvature
Use this for “A particle moves on r(t) = ⟨3 cos t, 2 sin t⟩. At t = π/4, find its speed, the tangential and normal acceleration, and the curvature.”
Diagram
At t = π/4 the point is (2.12, 1.41).
The arrows show the way t runs.
This is an example. Type your own values to start a new problem.
r(t)
=
⟨
a
cos
t,
b
sin
t⟩,
t
=
t
κ
=
|r′
×
r″|
÷
|r′|³
=
a
×
b
÷
v
3
=
κ
Here r′ × r″ = ab sin² t + ab cos² t = ab at every t, and |r′| is the speed v.
aₜ
Tangential acceleration
worked out (not typed)
aₙ
Normal acceleration
worked out (not typed)
ρ
Radius of curvature
worked out (not typed)
Formulas
v
=
√(
a
2
×
sin
(
t
)
2
+
b
2
×
cos
(
t
)
2
)
2.5495
=
√(
3
2
×
sin
(
π
4
)
2
+
2
2
×
cos
(
π
4
)
2
)
aₜ
=
(
a
2
−
b
2
)
×
sin(
t
)
×
cos(
t
)
v
0.9806
=
(
3
2
−
2
2
)
×
sin(
π
4
)
×
cos(
π
4
)
2.5495
aₙ
=
ab
v
2.3534
=
3
×
2
2.5495
κ
=
ab
v
3
0.3621
=
3
×
2
2.5495
3
ρ
=
1
κ
2.762
=
1
0.3621
Assumptions
•
t is in radians, typed as a fraction of π (π/4) or a decimal; one lap of the ellipse is 0 ≤ t ≤ 2π.
•
r′(t) = ⟨−a sin t, b cos t⟩ and r″(t) = ⟨−a cos t, −b sin t⟩; aₜ = (r′ · r″) ÷ v changes the speed, aₙ = |r′ × r″| ÷ v turns the path.
•
κ = aₙ ÷ v², and ρ = 1 ÷ κ is the radius of the circle that fits the path best at that point.
Step-by-step
WE KNOW
a = 3, b = 2, t = π/4
FIND
speed (v), tangential acceleration (aₜ), normal acceleration (aₙ), curvature (κ), radius of curvature (ρ)
Step 1 · Find speed (v)
Use v = √(a² × sin(t)² + b² × cos(t)²)
The
speed
is
the
length
of
r′(
t
)
=
⟨−a
sin
t
,
b
cos
t
⟩.
v
=
√(
3
2
×
sin
(
π
4
)
2
+
2
2
×
cos
(
π
4
)
2
)
v
=
√
9 × 0.5 + 4 × 0.5
v
=
√
6.5
v
=
2.5495
Step 2 · Find tangential acceleration (aₜ)
Use aₜ = (a² − b²) × sin(t) × cos(t) ÷ v
r′
·
r″
=
a
2
sin
t
cos
t
−
b
2
sin
t
cos
t
;
divide
by
the
speed
to
get
the
part
along
the
path.
aₜ
=
(
3
2
−
2
2
)
×
sin(
π
4
)
×
cos(
π
4
)
2.5495
aₜ
=
5
×
0.5
2.5495
aₜ = 0.9806
Step 3 · Find normal acceleration (aₙ)
Use aₙ = ab ÷ v
r′
×
r″
=
ab
cos²
t
+
ab
sin²
t
=
ab,
since
sin²
t
+
cos²
t
=
1;
divide
by
the
speed.
aₙ
=
3
×
2
2.5495
aₙ
=
6
2.5495
aₙ = 2.3534
Step 4 · Find curvature (κ)
Use κ = ab ÷ v³
κ
=
|r′
×
r″|
÷
|r′|³,
which
is
aₙ
v
2
:
the
turning
per
unit
of
speed
squared.
κ
=
3
×
2
2.5495
3
κ
=
3
×
2
16.57
κ
=
6
16.57
κ
=
0.3621
Step 5 · Find radius of curvature (ρ)
Use ρ = 1 ÷ κ
The
best-fitting
circle
bends
as
much
as
the
path,
so
its
radius
is
1
κ
.
ρ
=
1
0.3621
ρ
=
2.762
Check
Put the numbers back in. Both sides should match.
2.5495
=
√(
3
2
×
sin
(
π
4
)
2
+
2
2
×
cos
(
π
4
)
2
)
✓
0.9806
=
(
3
2
−
2
2
)
×
sin(
π
4
)
×
cos(
π
4
)
2.5495
✓
2.3534
=
3
×
2
2.5495
✓
0.3621
=
3
×
2
2.5495
3
✓
2.762
=
1
0.3621
✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.