Topic 21 Notes
Jeremy Orloff
21
Parametric equations and vector derivatives
Old, compressed version of topic 21 notes.
Parametric Equations
General parametric equations:
−
Notation: →
r (t) = hx(t), y(t)i = x(t) bi + y(t) bj = position vector
We can view (x(t), y(t)) as the coordinates of a particle moving in the plane or we
−
can view →
r (t) = hx(t), y(t)i as its position vector.
Example: A rocket takes off from the origin with initial xvelocity v0,x and initial y-velocity v0,y . Find the parametric
equations for its path.
1
Physics ⇒ x(t) = v0,x t, y(t) = − gt2 + v0,y t.
2
At time t the rocket is at point P = (x(t), y(t)). We call the
−−→
vector r(t) = OP = x(t) i + y(t) j the position vector.
Velocity, speed and arclength
Now we see the benefits of using the position vector.
change in position
dr
=
= velocity = v(t) = x0 (t) i + y 0 (t) j = hx0 , y 0 i
dt
change in time
Geometically: the picture shows a parametric curve for a moving
point. Over a small time dt the point moves a (vector) dr =
dx i + dy j and its (instantaneous) velocity
displacement
dr
dx
dy
=
=
i+
j = hx0 (t), y 0 (t)i
time
dt
dt
dt
The picture for arclength, s, shows
p
ds = |dr| = (dx)2 + (dy)2
s 2 2
ds
distance dr dx
dy
⇒ speed =
=
= =
+
.
dt
time
dt
dt
dt
y
P•4B
v(t) =
r(t)
x
JT9H
−→
∆r
•
∆x i
∆s
•
∆x
Acceleration (change in velocity/change in time)
dv
d2 r
a(t) =
= 2 = x00 (t) i + y 00 (t) j = hx00 , y 00 i
dt
dt
Example: Find velocity, speed, acceleration and arclength for the rocket example.
r(t) = x(t) i + y(t) j = v0,x t i + (− g2 t2 + v0,y t)j.
dr
dv
⇒ v(t) =
= v0,x i + (−gt + v0,y )j,
⇒ a(t) =
= −gj.
dt
dt
1
+
∆y j
*4
∆y
dr
dt
21 PARAMETRIC EQUATIONS AND VECTOR DERIVATIVES
2
q
ds
2
+ (−gt + v0,y )2 .
= |v(t)| = v0,x
Speed =
dt
To avoid too many symbols. Let v0,x = 10, v0,y = 10 g = 10 and find the arclength
of the path from t = 0 to t = 1.
Z 1
Z 1p
ds
Arclength L =
1 + (−t + 1)2 dt
dt = 10
dt
0
0
Make the change of variables u = −t + 1
Z 1√
1 + u2 du.
⇒ du = −dt, t = 0 → u = 1, t = 1 → u = 0. ⇒ L = 10
0
Use trig. subst. u = tan θ
Z π/4
√
√
⇒ L = 10
sec3 θ dθ = 5 [sec θ tan θ + ln(sec θ + tan θ)|π/4
=
5(
2
+
ln(
2 + 1)).
0
0
Tangent vector: (same thing as velocity) In the picture above, we see that as ∆t
becomes tangent to the curve
shrinks to 0 the vector ∆r
∆t
When the parameter is t we can refer to r0 (t) as the velocity. In general, the derivative
is given its geometric name: the tangent vector.
Unit tangent vector
The unit vector in the direction of the tangent vector is denoted T =
r0 (t)
v
=
.
0
|r (t)|
|v|
It’s called the unit tangent vector.
Note ds
T = r0 (t).
dt
_
Curvature: How sharply curved is the trajectory?
That is, how fast does the tangent vector turn per unit arclenth?
This is tricky so pay attention.
is the rate T is turning per
Curvature
dT unit arclength. That is, κ = .
ds
(Smaller circle = faster turning = greater curvature.)
o
T
O
•
ds
Note: the book doesn’t use the absolute value in its definition of κ, but it’s more standard to include it.
∆sO
T
T
∆s
Note well, curvature is a geometric idea– we measure the rate with respect to arclength. The speed the point moves over the trajectory is irrelevant.
T is a unit vector ⇒ T = hcos φ, sin φi where φ is the tangent angle.
y
dT
d
dφ
⇒
= hcos φ, sin φi =
h− sin φ, cos φi.
ds
ds
ds
_
?
Both magnitude and direction of dT
are useful:
ds
N
T
dT dφ φ
Curvature = κ = = .
ds
ds
x
dT
Direction of
= N = unit normal ⊥ T.
Radius of curvature = κ1 .
T
•
•
•
21 PARAMETRIC EQUATIONS AND VECTOR DERIVATIVES
The center of curvature and the osculating circle:
The osculating (kissing) circle is the best fitting circle to
the curve.
Radius = radius of curvature.
Center along normal direction.
3
•
o
radius of curvature
Nomenclature summary:
Here are a list of names and formulas. We will motivate and derive them below.
r(t) = position.
.
s = arclength, speed = v = ds
dt
ds
0
v(t) = r (t) = dt T = tangent vector, velocity.
2
= ddt2r = acceleration.
a(t) = dv
dt
T = unit tangent vector, N = unit normal vector.
κ = curvature, R = 1/κ = radius of curvature.
φ = tangent angle.
C = Center of curvature = center of best fitting circle (has radius = radius of curvature).
Formulas:
1.
2.
3.
4.
(explained in the following pages)
p
ds
= |v(t)| = (x0 )2 + (y 0 )2 .
Speed =
dt
v
ds
v = T, T =
dt
ds/dt
2
2
ds
ds
v2
d2 s
a(t) = 2 T + κ
N = 2T+ N
dt
dt
dt
R
dφ dT |a × v|
dT
.,
= κN.
κ = = =
3
ds
ds
|v|
ds
4a. For plane curves r(t) = x(t) bi + y(t) bj : κ =
|x00 y 0 − x0 y 00 |
.
((x0 )2 + (y 0 )2 )3/2
5. v × (a × v) = κv 4 N.
1
6. C = r + R N = r + N.
κ
Example: For the parabola r(t) = t i + t2 j find v, a, T, ds/dt, κ, R, N and C for arbitrary t.
v = i + 2t j, a = 2j.
√
v
1
√ 2t
⇒ ds/dt = |v| = 1 + 4t2 , T = |v|
j.
= √1+4t
2 i +
1+4t2
Formula 4:
a × v = −2k.
⇒ κ=
2
.
(1+4t2 )3/2
(Maximum curvature at t = 0 as expected.)
•j
2 3/2
R = 1/κ = (1+4t2 ) .
Formula 5: v × (a × v) = −4t i + 2 j. ⇒ N =
Formula 6:
C = r + RN = (t i + t2 j) +
Proofs of formulas 3-5:
√ 1
(−4t i
2 1+4t2
1+4t2
(−4t i
4
+ 2 j).
+ 2 j).
t=0
21 PARAMETRIC EQUATIONS AND VECTOR DERIVATIVES
4
Note: we will repeatedly use that v = ds
.
dt
Formula 3 is an application of the product and chain rules:
ds
Start with v = T.
dt
dv
d2 s
d2 s
ds dT
ds dT ds
⇒ a =
= 2T+
= 2T+
dt
dt
dt
dt ds dt
2 dt dt
d2 s
ds
=
κN.
T+
dt2
dt
In physics this is the decomposition of acceleration into tangential and radial components.
Formula 4 now follows from formula 3 since T and N are orthogonal unit vectors:
!
3
2
ds
ds
ds
d2 s
T
=
T
+
κN
×
κ(N × T).
a×v =
dt2
dt
dt
dt
Since N and T are orthogonal unit vectors N × T is a unit vector
)3 κ = v 3 κ.
⇒ |a × v| = ( ds
dt
The second part of formula 4 is just the first in coordinates:
v = x0 bi + y 0 bj and a = x00 bi + y 00 bj
p
b and v = (x0 )2 + (y 0 )2
⇒ a × v = (x00 y 0 − x0 y 00 )k
⇒ what we want.
Formula 5 now follows from what we just did. We found
a × v = v 3 κ(N × T). ⇒ v × (a × v) = v 4 κ T × (N × T) = v 4 κ N.
The last equality is easy using your right hand (since T and N are orthogonal unit
vectors).
Next time we’ll do a number of examples of all these concepts.