Chapter 2:curves in R2 and curvature Flashcards
Vector valued function
Function with values in R²
Eg vector(x) = (x(t),y(t))
Components are functions of t
Derivative of vector(x)
d/dt of vector(x) = ( dx/dt, dy/dt)
Differentiate components
This equals zero exactly when components are constant: when differentiating a constant vector function
Dot product of two vector function
vector(x ₁)•vector(x₂)(t) =
x ₁•y ₁ +x ₂•y₂
Given two vector-valued functions vector(x₁)= (x1, y1): R → R^2 and vector(x₂) = (x2, y2): R → R2 , their dot product
vector(x ₁)•vector(x₂) = : R → R is defined by
vector(x ₁)•vector(x₂)(t)
= x1(t) · x2(t) = x1(t)x2(t) + y1(t)y2(t)
Theorem 2.1.4: Derivative of dot
Derivative of dot product satisfies product rule:
derivative( vector(x ₁)•vector(x₂)(t)) =
(dvector(x ₁)/dt)•vector(x₂)(t)
+
vector(x ₁)•(dvector(x₂)(t)/dt)
=
Where their derivatives are defined
ie uses product rule
Derivatives and orthogonality
by considering
derivative of x dot x (itself)
when is x orthogonal to xdot
||x|| constant IMPLIES x˙(t) ORTHOGONAL to x(t) for all t
ie if x has constant magnitude then always orthogonal to its derivative
BECAUSE
d/dx ( ||x||² ) = d/dx ( x • x) = 2x•x˙ ( 2x•xdot)
hence if ||x||² constant derivative =0 ..
eg circle
DEF 2.2.1 LEVEL CURVE
trajectory traced by a point
A level curve is a set of points C = {(x, y) ∈ R² | f(x, y) =
c}.
not every eq defines a level curve eg no sols x²+y²=-1
DEF 2.2.2 A Parametrized curve
A parametrized curve in R² is a CONTINUOUS map
γ: ]α, β[ to R²
for some α, β in R with -∞ ≤ α less than t less than β ≤ ∞
(STATE THE SINGLE CONTINUOUS INTERVAL)
Parametrization for line and line segment
each component: through a point + t gradient component
eg through many points and parallel to many multiples of gradient
para: γ_1: R to R² , γ_1(t)=(t, t + 1).
* line segment from one point to the other: interval affected, eg (2-cost,4-2cost) for interval 0 to 4pi (larger than__) is segment (1,2) to (3,6)?
parametrization for unit circle
- level curve { (x,y) in R² : x² +y²=1} any satisfying this and open interval ]α,β[ ‘s LENGTH β-α must be larger than 2pi
square root of 1-x² gives two semicircles (no good)
γ: ]α,β[ to R² , γ(t)=(cost,sint).
with INTERVAL length bigger than 2pi eg R
or
γ: ]α,β[ to R² , γ(t)=(cos2t,sin2t). with INTERVAL ]α,β[ length β- αbigger than pi eg R
Parametrization for Parabola
and parabolic arc
- Parabola x+y² = 1 or y² = 1-x or x= 1- y² =|)
*parabolic arc is st |y| is less than 1 (segment of parabola) - PARABOLA : γ: ]-∞
,∞[ to R² , γ(t)=(1-t²,t). - PARABOLIC ARC γ: ]-1,1[ to R^2 , γ(t)=(1-t^2,t).
γ: ]0,pi[ to R^2 , γ(t)=(1-cos^2t, cost).
note that changing the last domain to the whole of R doesnt give the whole parabola like the first one does
general graph parametrisation
proving
*For a function f : R to R, the graph Γf of f is the curve on R² defined
by
Γf := {(x, y) | y - f(x) = 0}.
A parametrization is
γf : R to R² , t to (t, f(t)).
- to prove this we show it gives the image,
every point on the level curve must be in the image and equation must be satisfied
show they are both contained in each other than they must be equal (image and level set}
*or disprove find a point given not in the image
stereographic projection
unit circle has parameterization
γ:R to R^2
t to
( 2t/(t^2+1), (t^2-1)/(t^2+1))
formed by taking t to Q on arc that lies on the line PA ( O is point (0,1) and A a point on the x-axis. PA has gradient (0-`)/(t-0) = -1/t and passes through (0,1) giving y = 1 -(1/t)x. Q is not equal to A so y is less than 1 and t^2(1-y)=(1+y) rearranges
TRIG IDENTITIES
tan^2 + 1 = sec^2
1+cot^2 =cosec^2
cosx = sin(pi/2 - x) etc
DEF 2.3.1 Re-parametrization
A parametrized curve γ˜: ]α ˜, β ˜ [ to R² is a re-parametrization
of a parametrized curve γ: ] α , β [ to R² if there is a bijective continuous
map
Ф: ]α ˜, β ˜ [ to : ]α , β[ (the re-parametrization map)
such that the inverse Ф -1
is also continuous
and γ˜ = γ○Ф = γ(Ф(t˜)) st t˜ ∈ ]α ˜, β ˜ [
“sub in new “t” as Ф(t˜) “
REPARA properties
*if repara funct Ф is bijective
then
γ and γ○Ф have the same image in R²
* γ○Ф and γ are both re-parametrizations of each other
DEF 2.4.1 TANGENT VECTOR
and SPEED
The tangent vector to the parametrized curve γ at t is its
derivative γ˙(t); (dot)
and the speed of γ at t is the magnitude ||γ˙(t)||
(from limit of derivative becoming instantaneous as time interval tends to 0)
Example checking a repara:
line l = {(x,y) in R^2 : y-x=1}
γ: R to R^2 γ(t) = (t,t+1)
and
γ˜: R to R^2 γ˜(t) = (2t˜+1,2t˜+2)
Repara function is Ф(t) = 2t+1. This is continuous and bijective.
Ф-1(t ˜)= (t˜-1)/2.
γ○Ф (Ф(t), Ф(t)+1) = γ ˜(t˜) as required
DEF 2.5.1 SMOOTH
A parametrized curve γ: ] α , β [ to R² is said to be
smooth if it is differentiable infinitely many times, i.e. if the two components
x, y of γ = (x, y) are differentiable infinitely many times
differentiate once, twice, pattern to see infinitely?
DEF 2.5.1 REGULAR
The parametrized curve is regular if γ˙(t) ≠ 0 for all t in ]α, β[ .
ie derivative not equal to 0
ie SUB in t
If smooth and regular then parameterized curve has no sharp corners
EXAMPLE check if smooth and regular
γ :R to R^2 γ(t) =(t^2, t^3)
γ :R to R² γ(t) =(t^2, t^3)
can be differentiated infinitely many (give values )
but at t=0 γ˙(t) =0 so it is smooth but not regular sketch shows sharp corner at t=0 (never be parameterized so that regular) |/ -- |\
ASSUMPTION
our parametrized curves will always be smooth and regular, unless
stated otherwise.
THEOREM 2.5.3
SMOOTH AND REGULAR
(relationship between parameterizations being so)
proof: γ is smooth and regular
Suppose that Ф is a smooth reparametrization function of γ: ] α , β [ to R^2 so that Ф and Ф-1
are infinitely differentiable.
Then γ is smooth and regular if
and only if
γ˜ = γ○Ф is smooth and regular
DERIVATIVE OFγ˜
= γ’(Ф(t˜)) Ф’(t˜) is also
Relationship between tangent vectors to the curve and parametrisations
CHAIN RULE
the tangent vectors to the curve relative to its
two parametrizations are related by
γ˜DOT(˜t) = Ф ‘ ( t~) • γ˜(Ф(t~))
for all t~ in ]α ˜, β ˜ [
ie dγ˜/dt = dФ/dt * dγ/dt
REMEMBER t~ not t
DEF 2.6.2 ARC LENGTH
Fixing t_0, the arc length, s(t), of the parametrized curve γ from t0 to t is the integral s(t) = INTEGRAL from( t_0, t) of || γ DOT(τ)|| .dτ = s(t) of para curve from t_0 to t ie magnitude of derivative of parametrisation
ARC LENGTH derivation
approximating arc length γ: ] α , β [ to R^2
by choosing
α less than t_0 less than t less than β
then length γ(t_0) to γ(t).
Better by choosing
t_0 less than t_1………. less than t_k =t
and considering Straight Lines length
=
Σ from i=1 to k
√ (̅x̅(̅t̅_i)̅-̅x̅(̅t̅_i-1)̅)² ̅+̅ ̅(̅y̅(̅t̅_i)̅-̅y̅(̅t̅_i-1)̅)̅²
LEMMA 2.6.1
Taking an appropriate limit and asum tendsto
taking an appropriate limit, sum tends to
∫ from t_0 to t √[(x(τ)² + y(τ)²] . dτ = ∫ from t_0 to t ||γDOT(τ) || dτ
ARC LENGTH PROPERTIES
- length should not depend on the chosen parametrization, but only on
the segment of curve between =(a) and γ(b). - arc length is positive if t is bigger than initial t_0 and negative if t is less than initial t_0
- s(t_0)=0
- different starting point arc length differs by constant ∫ from t_0 to t_0~ of ||γ’(u)||. du:
t_0 ~t_0 t
∫ from t_0~ to t of ||γ’(u)||. du = ∫ from t_0~ to t_0 of ||γ’(u)||. du
+ ∫ from t_0 to t of ||γ’(u)||. du
*for simplicity assume that parameterization function is increasing
PROP 2.6.3 arc length and. Reparametrization
The arc length of γ between parameter values t0 and t can be
computed using any re-parametrization of γ.
proof:
Suppose we have γ~: ] α~ , β~ [ to R^2 and re para funct Ф st γ~ = γ○Ф,
Ф is bijective and continuous.
Arc length for γ:
∫ from t_0 to t of ||γDOT(τ) || dτ.
∫ from t~_0 to t~ ||γ~DOT(τ~) || dτ~ = ∫ from t~_0 to t~ ||γDOT(Ф(τ~))Ф'(τ~) || dτ~
(BY CHANGE OF VARS: τ to τ~ (applying Ф-1, τ=Ф(τ~) dτ/dτ~ = Ф’(τ~ ),
Ф’(τ~ ) is scalar)
=
∫ from t_0 to t
||γDOT(τ)Ф’(τ~) || (Ф’(τ~ )-1dτ~)
=∫ from t_0 to t of ||γDOT(τ) || dτ.
ARC LENGTH PROOF
PROOF:
by the mean value theorem
suppose f: [R,S] to R is smooth.
Then there exists p ∈[R,S] such that f(S) -f(R) = (S-R)f’(p).
For interval [t_i-1, t_i] i=1,..,k function x We have a p_i within interval st x(t_i) -x(t_i-1) = (t_i-t_i-1)x'(p_i)
function y We have a q_i within interval st y(t_i) -y(t_i-1) = (t_i-t_i-1)y'(q_i)
Δt = t_i - t_i-1 hence sum of lengths Σ from i=1 to k √ (x'(p_i)Δt)² +(y'(q_i)Δt)² =Δt Σ √(x'(p_i))² +(y'(q_i))²
as k tends to infinity and Δt tends to 0 and by definition of integral
∫ from t_0 to t
√[(x(τ)² + y(τ)²] . dτ
= s(t)
DEF 2.7.1 UNIT SPEED PARAMETRIZATION
A unit-speed parametrization of a curve is one such that
all the tangent vectors have length 1.
(choose origin of curve, calc ARC LENGTH FUNCTION and use its inverse as the repara funct Ф = s-1 = REPARA BY ARC LENGTH)
THEOREM 2.7.2
When does a unit speed repara exist
Let γ: ] α , β [ to R^2 be a smooth para. Then γ has
a unit speed re-parametrization
if and only if it is regular.
example: arc lengths for
*γ_1: R to R^2 , t to (t, t+1)
from (0,1) = γ_1(0) to (1,2)= γ_1(1)
*γ_2: R to R^2 , t to (cost, sint)
from γ_2(0) to γ_2(pi)
arc lengths for
*γ_1: R to R^2 , t to (t, t+1)
from (0,1) = γ_1(0) to (1,2)= γ_1(1)
∫ from 0 to 1 of ||(1,1)|| . du = √2
*γ_2: R to R^2 , t to (cost, sint)
from γ_2(0) to γ_2(pi)
∫ from 0 to pi of ||(-sinu,cosu)|| . du = pi
PROOF
THEOREM 2.7.2
When does a unit speed repara exist
PROOF
IMPLICATION:Assume γ has a unit speed repara, γ~= γ○Ф:
1 = ||γ~dot(t~)||
=||γdot(Ф(t~))Ф’(t~)||
=||γdot(Ф(t~))|| |Ф’(t~)| as Ф’(t~) is scalar
hence ||γdot(Ф(t~))|| ≠ 0 for all t~ ∈] α ~, β ~[
||γdot(t)|| ≠ 0 for all t ∈] α , β [
CONVERSE:
Assume γ is Regular: γdot(t) ≠ (0,0) for all t ∈] α , β [.
For t_0 ∈] α , β [ take arc length:
s(t) = ∫ from t_0 to t of || γ DOT(τ)|| .dτ.
giving s: ] α , β [ to R.
so by the fundamental theorem of calculus:
s’(t) = ||γ DOT(t)|| ≠ 0 as γ is Regular
(using the inverse function theorem)
the arc length function has inverse
s-1 : ]α~ , β~[ to ]α, β[ which is smooth.
now using this s-1 as a reparametrization function: Ф(t~) = s-1
γ~(t~) = γ(s-1(t~)) ie γ~(s(t)) = γ(t)
||γdot(t)|| = ||γ~dot(s(t)) s'(t)|| = ||γ~dot(s(t))|| | s'(t)| = ||γ~dot(s(t))|| ||γ DOT(t)|| s'(t) = ||γ DOT(t)|| ≠ 0 dividing ............................... So ||γ~dot(s(t))|| = 1 and || ||γ~dot(t~)|| = 1 for all t~ as t~ =s(s-1(t~)). So γ~ is a unit speed parametrization.
DEF 2.8.1 ROTATION
Let J : R² to R² be the 90o, anti-clockwise rotation centred
at 0; this has matrix
[0 -1]
[1 0]
(x,y) to (-y,x)
.
PREFERRED UNIT NORMAL VECTOR
For a parametrized curve γ define n(t), the preferred unit normal vector
at γ(t), to be the 90o, anti-clockwise rotation of the tangent vector γ˙(t):
n(t) := J(γ˙^(t)).
where γ˙^(t) is the UNIT TANGENT
ndot dot n = 0
LEMMA 2.8.2 small h, deviation from the tangent line for a unit speed parametrised curve
Suppose that γ is a unit-speed parametrized curve. For small h, the
point γ(t + h) deviates from the tangent line at γ(t) by
(1/2)n(t) · γ¨(t) h^2 + O(h^3).
PROOF
by Taylor’s theorem
γ(t+h) = γ(t) +γdot(t)h + 0.5 γ¨(t) h^2 + Oh^3
vector γ(t+h) - γ(t) = 0.5 γdot(t) h^2 + Oh^3
Direction we are looking for is how much γ has moved in the normal Direction:
(γ(t+h) - γ(t)) dot product _n(t)
= n(t)•γdot(t)h + 0.5 n(t)• γ¨(t) h^2 + n(t)• Oh^3
orthogonal term: def of J rotation
=0.5 n(t)• γ¨(t) h^2 + Oh^3
DEF 2.8.3
CURVATURE
SIGNED?
Let γ: ]α , β[ to R²
be a unit-speed parametrized curve.
The signed curvature κ(t) of γ at γ(t) is defined to be
κ(t) := n(t) · γ¨(t) for all t in ]α , β[ .
The unsigned curvature is then |κ(t)| := ||γ¨(t)|| for all t in ]α , β[
(double dot derivative)
curvature is how much it deviates from the tangent
the more curved the quicker it moves away from the tangent line
the inverse function theorem
suppose s: ]α , β[ to R^2 is differentiable and derivative
s’(t) is bigger than 0 for all t ∈ ]α , β [.
Then s is injective onto image ]α~ , β~[ for α~ , β~ in R and has differentiable inverse s-1 : ]α~ , β~[ to ]α, β[
PROP 2.9.1
Curvature of a parametrised curve without having to find unit speed para
. Let γ = (x, y): ]α , β[ to R^2 be a parametrized curve. Then
the curvature of γ at t is given by
k(t) = γ¨(t) · J(γ˙(t)) / ||γ˙(t)||^3 ,
or more explicitly by
= [x’(t)y’‘(t) - x’‘(t)y’(t) ] / [x’(t)^2 + y’(t)^2] ^3/2
for all t in ]]α , β[.
- curvature doesn’t depend on the choice of unit speed parametrization only the direction
- a parametrized curve has countlessly many unit speed reparametrization
- units speed reparametrization functions are t to t +c ( plus a constant)
DEF 2.9.3 TURNING ANGLE
Let γ : ]α , β[ to R2 be a unit-speed parametrized curve,
and choose t₀ in ]a, b[ and ϴ₀ such that γ(t₀)=(cos(ϴ₀), sin(ϴ₀)). The
turning angle of γ determined by ϴ₀ = ϴ(t₀) is the unique smooth function
ϴ : ]α , β[ to R such that ϴ(t₀) = ϴ₀ and
γ˙(t)=(cos(ϴ(t)), sin(ϴ(t))) for all t in ]α , β[
ϴ unique up to adding 2npi n in integers
PROP 2.9.4 TURNING ANGLE
Let γ be a unit-speed parametrized curve and let ϴ be a turning angle for γ.
Then the curvature of γ is given by
k = ϴ’
thus the signed curvature is equal to the rate at which the tangent vector of the curve rotates
proof:…
DEF 2.10.1 TRANSLATION
ROTATION
The translation by (a, b) in R^2 is the map
T(a,b) : R2 to R2
T(a,b)(x, y)=(x + a, y + b).
Rotation about origin by angle ϴ∈R anticlockwise is the map R_θ : R² to R²
R_θ = (xcosθ - ysinθ, xsinθ + ycosθ)
PROP 2.10.2
Curvature after rotation and translation
Let γ be a parametrized curve and Φ a rotation or a translation.
Then the curvature of γ˜ = Φ ○ γ at Φ (γ(t)) equals the curvature of γ at γ(t)
proof:…
THEOREM 2.11.1
Recipe from smooth function
Given any vector v in R², any constant c, any smooth function
κ : ] α , β [ to R and any t₀ in ] α , β [
there exists exactly one unit-speed parametrized
curve γ: ] α , β [ to R² with
γ(t₀ ) = v, γ˙(t₀ )=(cos c, sin c), and curvature function κ. ****** This is done by setting *θ(t) = c + (t₀,t) ∫ of κ(u)du (1)
- δ(t) = (cos θ(t), sin θ(t)) (2)
- γ(t) = v + ∫ over ( t₀,t) of δ(u)du. (3)
(v is bold)
CIRCLE WITH CENTRE (x_1,x_2) and radius r
γ: R to R^2 γ=(Rcost +x_1, Rsint + x_2)
and interval length β -α larger than 2 pi
EXAMPLE: is
γ~ : R to R2 t~ to (cos2πt~, sin2πt~)
a repara of γ : R to R2 t to (cost, sint)
γ~ : R to R2 t~ to (cos2πt~, sin2πt~)
is a repara of γ : R to R2 t to (cost, sint)
1) by using repara funct
Ф(t~) = 2πt~
checking Ф-1(t) = t/2π is continuous and bijective
2) γ(t) = γ~(t/2π) = γ~○Ф-1
3) γ~(t~) = γ(2πt~)= γ○Ф
4) γ~(Ф-1 (t)) = γ(Ф(Ф-1 (t))) = γ(t) for all t in ] α , β [
hence is a repara
example : re-parametrize by arc length the circle with Radius 3 centre (1 ,- 1)
γ : R to R2 t to (1+3cost, -1+3sint)
pick t₀= 0
ARC LENGTH FUNCT = t₀ to t ∫ || (-3sinτ, 3cosτ)||.dτ = 3t
s(t) = 3t so s-1 (t~) = t~/3
take Ф(t~) = t~/3
we have a unit speed para γ~(t~) = γ( t~/3)
γ~ : R to R2 t to (1+3cos(t~/3), -1+3sin(t~/3) )
|| γ~dot (t~)||=1 checked
example: consider the logarithmic spiral
show that a unit speed parametrization is
Consider the logarithmic spiral γ: R → R^2 given by
γ(t) = e^{kt}(cos t, sin t) for k ∈ R
Show that a unit-speed parametrization is
γ˜(t) = (( [kt/√[k^2 + 1]] + 1 )* cos({1/k}ln([kt/√[k^2 + 1]]+ 1)),
([kt/√[k^2 + 1]] + 1)* sin({1/k}ln([kt/√k^2 + 1]+ 1)))
Even though this example shows that unit-speed parametrizations are not
that useful for practical calculations, their existence will often facilitate the
more abstract analyses of curves
*calculate γ˜(t) dot and show magnitude = 1
*or choose t_0=0 and find arc length function to repara by, its in terms of exp so inverse has the lns we require
this is long and complicated
we then use fundamental theorem of calculus to show the speed for our new gamma dot
UNIT TANGENT γ˙^(t)
γ˙^(t) = γ˙(t) / ( || γ˙(t)|| )
example: circle radius r centre (0,0) is a unit speed parametrization show that the signed curvature is
κ
γ:R to R^2 γ(t)=(Rcos(t/R), Rsin(t/R)) is a unit speed repara. Show that κ(t)=1/R for all t in R
preferred unit vector n(t)= J(-sin(t/R),cos(t/R))
=(-cos(t/R),-sin(t/R))
thus κ(t)=(-cos(t/R), -sin(t/R)) • (-(1/R)cos(t/R),-(1/R)sin(t/R))
= 1/R
This agrees with our intuition that small circles should have large curvature
and large circles small curva
LET γ: ]α, β[ → R^2 be a unit speed parametrized curve. and γ˜ = γ ◦ ϕ with ϕ: ]α−c, β− c[ → ]α, β[, ϕ(t) = t + c, a unit speed reparametrization of γ.
Show that κ_γ˜(t) = κ_γ(t + c) for all t ∈ ]α − c, β − c[.
CURVATURE DOES NOT DEPEND ON PARAMETERIZATION ONLY DIRECTION
κ_γ˜(t)=J(γ˜•(t)) • γ˜••(t)
=[J((γ(ϕ(t)))’]•[γ(ϕ(t)))’’]
=[J(γ•(ϕ(t))ϕ’(t))]•[γ••(ϕ(t))ϕ’^2(t)+γ•(ϕ(t))ϕ’‘(t)]
= J(γ•(ϕ(t))ϕ'(t)) • γ••(ϕ(t))ϕ'^2(t) = κ_γ(ϕ(t)) = κ_γ(t + c)
**ϕ(t)= t+c, so ϕ’(t)=1, ϕ’‘(t)=0
LET γ: ]α, β[ → R^2 be a unit-speed parametrized curve and γ˜ = γ ◦ ϕ
with
ϕ: ]−β − c, −α − c[ → ]α, β[,
ϕ(t) = −t + c, a unit-speed reparametrization of γ.
Show that κ_γ˜(t) = −κ_γ(−t + c) for all t ∈ ]−β − c, −α − c[.
CURVATURE DOES NOT DEPEND ON PARAMETERIZATION ONLY DIRECTION
κ_γ˜(t)=J(γ˜•(t)) • γ˜••(t)
=[J((γ(ϕ(t)))’]•[γ(ϕ(t)))’’]
=[J(γ•(ϕ(t))ϕ’(t))]•[γ••(ϕ(t))ϕ’^2(t)+γ•(ϕ(t))ϕ’‘(t)]
= J(γ•(ϕ(t))ϕ'(t)) • γ••(ϕ(t))ϕ'^2(t) =J(γ•(ϕ(t))(-1)) • γ••(ϕ(t))(-1)^2(t) +0 = -κ_γ(ϕ(t)) = -κ_γ(t + c)
**ϕ(t)= -t+c, so ϕ’(t)=-1, ϕ’‘(t)=0
**CHAIN RULE
γ˜••(t˜)= γ••(ϕ(t˜)) ϕ’^2(t˜) + γ•(ϕ(t˜))ϕ’‘(t˜)
γ˜(t˜)= γ(ϕ(t˜)) = γ(t)
proof
PROP 2.9.1
Curvature of a parametrised curve without unit tangent
proof
Let γ˜= γ○ϕ be a unit speed repara of γ, S= ϕ^-1
κ_γ˜ (t˜) =
J(γ•(ϕ(t˜)) ϕ’(t˜)) • (γ˜••(t˜))
***
chain rules used, facts of scalars, using orthogonal,
setting ϕ(t~)= 1/s’(ϕ(t~)) = 1/s’(t)= 1/ ||γ•(t)||
ftoc and thus κ value
Example: a circle on R^2 with the radius r centre (a,b) can be parametrized by…
we know that curvature is…
γ:R to R^2 γ(t)= (a +Rcost, b+Rsint)
κ_γ(t) = 1/R for any t and by prop 2.9.1 circle can be parametrised by γ(t)= (a+Rcost, b-Rsint)
giving κ_γ(t)= - 1/R..
in general, curvature κ is +ve/-ve if curve bends left/right as t increases
diagrams of two circles moving a) anti and b)clockwise
a) first para, curves towards preferred normal, which is towards middle and κ= 1/R
b) 2nd para, curves away from preferred normal, which is outwards and κ= -1/R
what does
THEOREM 2.11.1
Recipe from smooth function
give us?
- *gives a way to construct a unit speed Curve γ with prescribed curvature function κ and initial position v̲ initial velocity at time t₀
- *can be applied to any smooth function, this means that any smooth function is the curvature of some Curve
proof of
THEOREM 2.11.1
Recipe from smooth function
** EXISTENCE
given the values that we use we show that are gamma has the required properties e.g turning angle curvature and initial value
**UNIQUENESS
we suppose that we have a curve which satisfies the three properties and that there is a unique rotation such that a differential equation is given from what we know of the turning angle than there is a unique solution to a first order differential equation with the given initial condition of initial vector V and the solution we gave prev.
example applied to the 0 function
deduce that the plane Curve 0 curvature are Straight Lines
κ=0 R to R v=(a,b) t_0=0, c’ constant
used on recipe and checked
γ(t)=(a-t_0cosc’, b-t_0sinc’)
+
t(cost,sint)
eq of straight line
(a-t_0cosc’, b-t_0sinc’) is a constant
example applied to the non zero constant functions
deduce that the plane curves with constant non-zero curvature are precisely circles
used on recipe and checked
standard integral of a non zero funct which is constant k
γ(t)=(a-{1/k}sinc’, b- {1/k}cosc’)
+
(1/k)(sin(c+kt), -cos (c+kt))
eq of straight line
(a-t_0cosc’, bsinc’) is a constant
COROLLARY 2.11.4 ( CLASSIFICATION OF PLANE CURVES)
If two unit speed parametrized curves γ₁, γ₂ : R to R²
have the same curvature function, then either one can be obtained from the other by the combination of a translation and a rotation
proof:…
using recipe of smooth function with the same curvature, initial direction (γ•(t_0)) and initial position to recover curve
Let R:R^2 to R^2 be the rotation st R(γ•_1(t_0))= γ•_2(t_0)
This is possible as
||γ•_1(t_0)||=1 = ||γ•_2(t_0)||
Take T to be the translation st
T(R(γ_1(t_0))= γ_2(t_0)
now define γ_3= T○R○ γ_1 : R to R^2
now γ_3 and γ_1 have the same curvature by 2.10.2
γ_3(t_0)= (T○R○ γ_1 )(t_0)= γ_2(t_0)
and γ•_3(t_0)= Rγ•_1(t_0)= γ•_2(t_0)
so by uniqueness of 2.1.11 γ_3 =γ_2 ie γ_2 is obtained from γ_1 by a rotation and a translation
chain rules
γ˜••(t˜)= γ••(ϕ(t˜)) ϕ’^2(t˜) + γ•(ϕ(t˜))ϕ’‘(t˜)
γ˜•(t˜)= γ•(ϕ(t˜)) ϕ’(t˜) +
