Curvature and Acceleration Summary
De…nition For the path r (t) de…ne the unit tangent vector T (t) by r0 (t) = kr0 (t)k
De…nition Let r (s) be a unit speed parametrization of a curve. Then de…ne the curvature
(s) by
dT
=
ds
Theorems 1 & 2 & more: Curvature Formulae: For a path r (t) with arbitrary speed:
(t) = kT0 (t)k =v (t)
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(t) = kr0
(x) =
jf 00 (x)j
1 + (f 0 (x))2
(t) =
jx0 y 00
r00 k = kr0 k (space curve)
3=2
(function y = f (x) so plane curve)
y 0 x00 j
x0 (t)2 + y 0 (t)2
3=2
(plane curve hx (t) ; y (t)i )
Important geometric interpretation: to visualize the curvature at a point on the curve,
imagine a circle touching the curve with the same amount of "bending". Then =reciprocal
radius of that circle.
De…nition Let r (t) parameterize a curve. Then de…ne the normal vector N (t) by
N (t) = T0 (t) = kT0 (t)k
Relationship of the parallel vectors N and T0 :
T0 (t) =
(t) v (t) N (t)
The Binormal is
B=T
N
the third basis vector in the Frenet Frame fT; N; Bg (another right handed system - for
project use only - will not assigned on homework or tests).
Osculating Plane: The vectors T and N generate the osculating plane.
Osculating Circle: The osculating circle lies in the osculating plane, and is de…ned at the
point r (t0 ) as follows:
– Radius: 1=
– Center: r (t0 ) + 1 N
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Motion in Space
De…nition Let r (t) parameterize a curve. Then de…ne
velocity as r0 (t)
acceleration r00 (t)
As before we have speed as the scalar v (t) = kr0 (t)k
Acceleration vector and normal vs tangential components
a = aT T + aN N
where aT is the tangential component of acceleration and aN is the normal component of
acceleration
Scalar Components:
aT = v 0 and aN = v 2
q
a v
and aN = a N = kak2
aT = a T =
kvk
a2T
Vector components:
a = aT T + aN N
a v
v and aN N by subtraction
aT T =
v v
Geometric Interpretation:
If a lags behind the normal vector (with respect to the tangent), then the path is decelerating.
If a points ahead of the normal vector, then the path is accelerating.
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