Chapter 2:curves in R2 and curvature

Question 1

Vector valued function

Answer 1

Function with values in R²

Eg vector(x) = (x(t),y(t))

Components are functions of t

Question 2

Derivative of vector(x)

Answer 2

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

Question 3

Dot product of two vector function

Answer 3

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)

Question 4

Theorem 2.1.4: Derivative of dot

Answer 4

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

Question 5

Derivatives and orthogonality

by considering

derivative of x dot x (itself)

when is x orthogonal to xdot

Answer 5

||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

Question 6

DEF 2.2.1 LEVEL CURVE

trajectory traced by a point

Answer 6

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

Question 7

DEF 2.2.2 A Parametrized curve

Answer 7

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)

Question 8

Parametrization for line and line segment

Answer 8

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)?

Question 9

parametrization for unit circle

Answer 9

• 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

Question 10

Parametrization for Parabola

and parabolic arc

Answer 10

• 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

Question 11

general graph parametrisation

proving

Answer 11

*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

Question 12

stereographic projection

Answer 12

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

Question 13

TRIG IDENTITIES

Answer 13

tan^2 + 1 = sec^2
1+cot^2 =cosec^2
cosx = sin(pi/2 - x) etc

Question 14

DEF 2.3.1 Re-parametrization

Answer 14

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˜) “

Question 15

REPARA properties

Answer 15

*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

Question 16

DEF 2.4.1 TANGENT VECTOR

and SPEED

Answer 16

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)

Question 17

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)

Answer 17

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

Question 18

DEF 2.5.1 SMOOTH

Answer 18

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?

Question 19

DEF 2.5.1 REGULAR

Answer 19

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

Question 20

EXAMPLE check if smooth and regular

γ :R to R^2 γ(t) =(t^2, t^3)

Answer 20

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)
|/
--
|\

Question 21

ASSUMPTION

Answer 21

our parametrized curves will always be smooth and regular, unless
stated otherwise.

Question 22

THEOREM 2.5.3

SMOOTH AND REGULAR

(relationship between parameterizations being so)

Answer 22

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

Question 23

Relationship between tangent vectors to the curve and parametrisations

Answer 23

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

Question 24

DEF 2.6.2 ARC LENGTH

Answer 24

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

Question 25

ARC LENGTH derivation

Answer 25

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

Question 26

LEMMA 2.6.1

Taking an appropriate limit and asum tendsto

Answer 26

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

Question 27

ARC LENGTH PROPERTIES

Answer 27

• 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

Question 28

PROP 2.6.3 arc length and. Reparametrization

Answer 28

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.

Question 29

ARC LENGTH PROOF

Answer 29

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)

Question 30

DEF 2.7.1 UNIT SPEED PARAMETRIZATION

Answer 30

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)

Question 31

THEOREM 2.7.2

When does a unit speed repara exist

Answer 31

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.

Question 32

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)

Answer 32

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

Question 33

PROOF

THEOREM 2.7.2

When does a unit speed repara exist

Answer 33

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.

Question 34

DEF 2.8.1 ROTATION

Answer 34

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)

.

Question 35

PREFERRED UNIT NORMAL VECTOR

Answer 35

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

Question 36

LEMMA 2.8.2 small h, deviation from the tangent line for a unit speed parametrised curve

Answer 36

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

Question 37

DEF 2.8.3
CURVATURE

SIGNED?

Answer 37

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

Question 38

the inverse function theorem

Answer 38

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[

Question 39

PROP 2.9.1

Curvature of a parametrised curve without having to find unit speed para

Answer 39

. 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)

Question 40

DEF 2.9.3 TURNING ANGLE

Answer 40

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

Question 41

PROP 2.9.4 TURNING ANGLE

Answer 41

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:…

Question 42

DEF 2.10.1 TRANSLATION

ROTATION

Answer 42

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)

Question 43

PROP 2.10.2

Curvature after rotation and translation

Answer 43

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:…

Question 44

THEOREM 2.11.1

Recipe from smooth function

Answer 44

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)

Question 45

CIRCLE WITH CENTRE (x_1,x_2) and radius r

Answer 45

γ: R to R^2 γ=(Rcost +x_1, Rsint + x_2)

and interval length β -α larger than 2 pi

Question 46

EXAMPLE: is
γ~ : R to R2 t~ to (cos2πt~, sin2πt~)
a repara of γ : R to R2 t to (cost, sint)

Answer 46

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

Question 47

example : re-parametrize by arc length the circle with Radius 3 centre (1 ,- 1)

Answer 47

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

Question 48

example: consider the logarithmic spiral

show that a unit speed parametrization is

Answer 48

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

Question 49

UNIT TANGENT γ˙^(t)

Answer 49

γ˙^(t) = γ˙(t) / ( || γ˙(t)|| )

Question 50

example: circle radius r centre (0,0) is a unit speed parametrization show that the signed curvature is

κ

Answer 50

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

Question 51

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[.

Answer 51

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

Question 52

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[.

Answer 52

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)

Question 53

proof

PROP 2.9.1

Curvature of a parametrised curve without unit tangent

Answer 53

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

Question 54

Example: a circle on R^2 with the radius r centre (a,b) can be parametrized by…

we know that curvature is…

Answer 54

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

Question 55

what does
THEOREM 2.11.1

Recipe from smooth function
give us?

Answer 55

• *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

Question 56

proof of

THEOREM 2.11.1

Recipe from smooth function

Answer 56

** 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.

Question 57

example applied to the 0 function

deduce that the plane Curve 0 curvature are Straight Lines

Answer 57

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

Question 58

example applied to the non zero constant functions

deduce that the plane curves with constant non-zero curvature are precisely circles

Answer 58

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

Question 59

COROLLARY 2.11.4 ( CLASSIFICATION OF PLANE CURVES)

Answer 59

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

Question 60

chain rules

Answer 60

γ˜••(t˜)= γ••(ϕ(t˜)) ϕ’^2(t˜) + γ•(ϕ(t˜))ϕ’‘(t˜)

γ˜•(t˜)= γ•(ϕ(t˜)) ϕ’(t˜) +