Mathematics, Course MST30070B Mathematics Introduction ...2.2 Archimedean spiral as helix involute . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.3 involute of cycloid

University College Dublin

Mathematics, Course

MST30070B Mathematics

Introduction to Differential Geometry(evening first semester 2008 )

http://maths.ucd.ie/courses/mst30070B

December 6, 2008

Dr. J. Brendan Quigley

prepared using LATEXrunning under Redhat Linuxdrawings prepared usinggnuplot

comments to:-

Dr J.Brendan.QuigleyDepartment of MathematicsUniversity College Dublin, Belfieldph. 716–2584, 716–2580; fax 716–1196email [email protected].

ii

c©jbquig-UCD December 6, 2008

Contents

0 Organization of mst30070B 1

1 curves,surfaces and solids 31.1 image, graph and contour . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 31.2 curves inR2 andR3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8

1.2.1 the circle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 81.2.2 the Archimedean spiral . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 91.2.3 the cycloid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 101.2.4 the helix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . 11

1.3 problem set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12

2 the geometry of curves 132.1 velocity, speed and acceleration . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 13

2.1.1 velocity, speed and acceleration inR2 . . . . . . . . . . . . . . . . . . . . . . . . . 132.1.2 velocity speed and acceleration inR3 . . . . . . . . . . . . . . . . . . . . . . . . . 15

2.2 arclength . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 162.2.1 arclength inR2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 162.2.2 arclength inR3 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

2.3 curvature inR2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 192.3.1 curvature of the circle . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 192.3.2 curvature of a general curveΓ ⊂ R2 . . . . . . . . . . . . . . . . . . . . . . . . . . 20

2.4 tangent and normal (Serret-Frenet frame) . . . . . . . . . . . .. . . . . . . . . . . . . . . 222.4.1 unit tangent and normal vectors inR2 . . . . . . . . . . . . . . . . . . . . . . . . . 222.4.2 tangent vector in 3-space . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 24

2.5 involute . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 252.6 evolute . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 292.7 problem set . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 32

3 functions of two variables 353.1 nondegenerate stationary points . . . . . . . . . . . . . . . . . . .. . . . . . . . . . . . . 35

3.1.1 single variable, a review . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 353.2 several variables . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . 35

3.2.1 examples . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 383.3 the Hessian . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 42

3.3.1 example, closest approach of 2 lines inR3 . . . . . . . . . . . . . . . . . . . . . . . 433.3.2 example, elastic band and two rings . . . . . . . . . . . . . . . .. . . . . . . . . . 453.3.3 example, critical approach of line and circle . . . . . . .. . . . . . . . . . . . . . . 46

I Answers to problem sets of part I 51

4 Answers to questions in Chapter1 534.1 question and answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 53

iii

iv CONTENTS

4.1.1 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 534.1.2 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 564.1.3 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 564.1.4 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 56

4.2 question and answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 564.2.1 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 564.2.2 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 564.2.3 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 574.2.4 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 574.2.5 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 57

5 Answers to questions in Chapter2 595.1 question and answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 59

5.1.1 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 595.1.2 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 595.1.3 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 605.1.4 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 605.1.5 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 605.1.6 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 605.1.7 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 605.1.8 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 605.1.9 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 605.1.10 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 615.1.11 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 61

5.2 question and answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 615.2.1 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 615.2.2 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 615.2.3 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 615.2.4 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 615.2.5 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 615.2.6 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 615.2.7 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 615.2.8 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 625.2.9 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 625.2.10 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 625.2.11 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 625.2.12 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 62

5.3 question and answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . 635.3.1 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 635.3.2 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 635.3.3 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 635.3.4 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 635.3.5 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 635.3.6 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 635.3.7 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 635.3.8 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 635.3.9 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. 645.3.10 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 645.3.11 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 645.3.12 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 655.3.13 answer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . 65


List of Figures

1.1 parabolay = x2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 graph and contour diagram ofy = x2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.3 taut unwinding, Archimededean spiral . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 91.4 wheel and cycloid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 101.5 the helix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . 11

2.1 involute of involute of circle i.e. involute of Archimedean Spiral . . . . . . . . . . . . . . . 262.2 Archimedean spiral as helix involute . . . . . . . . . . . . . . . .. . . . . . . . . . . . . . 282.3 involute of cycloid is the cycloid translated . . . . . . . . .. . . . . . . . . . . . . . . . . . 29

2.4

(

(i) envolute of cycloid is the cycloid translated(ii) both inv/env-olute of the cycloidΓ are translates ofΓ

)

. . . . . . . . . . . . . . . . 32

3.1 (i) local minimum point (ii) local maximum point . . . . . . .. . . . . . . . . . . . . 373.2 (i) saddle point . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . 373.3 z= f (x,y) = x3 +y3−3x−3y (i) contour diagram (ii) graph . . . . . . . . . . . . . . 38

4.1 contours (i) wherex2−y2 = 0 (ii) wherex2−y2 = 1 . . . . . . . . . . . . . . . . . . . . . 534.2 contour (i) contour wherex2−y2 = −1 (i) contour diagram off (x,y) = x2−y2 . . . . . . . 544.3 saddle, graph off (x,y) = x2−y2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.4 cylinder and screw surfaces . . . . . . . . . . . . . . . . . . . . . . . .. . . . . . . . . . 554.5 helix curve formed as cylinder cuts screw . . . . . . . . . . . . .. . . . . . . . . . . . . . 554.6 (i) label with diagonal , (ii) soup can with helix . . . . . . .. . . . . . . . . . . . . . . 56

5.1 (i) circle as in Ch02:q01 , (ii) cardioid . . . . . . . . . . . . . .. . . . . . . . . . . . 595.2 cardioid with involute (big) and evolute (small) . . . . . .. . . . . . . . . . . . . . . . . . 60

v

vi LIST OF FIGURES


Chapter 0

Organization of mst30070B

Times

day time venue lecturer topic

Tue 07 : 30−08 : 20 TR−math−SB JBQ IntroDi f fGeomTue 08 : 30−09 : 20 TR−math−SB JBQ IntroDi f fGeom?whatday ?whattime ?whatroom ?who Tutorial

JBQ=J.Brendan QuigleySB=Science Building Belfield, EB=Engineering Building Belfield

LecturerJ.Brendan Quigley

room:- 35, Science Buildingph:- 716-2584, 716-2560(sec)email:- [email protected]

Books

see bibliography at end

urls(universal resource locators)

http://maths.ucd.ie/courses/mst30070B

1

2 CHAPTER 0. ORGANIZATION OF MST30070B

...............


Chapter 1

curves,surfaces and solids

lectureTue 01/247:30pm Tue09/09/2008

1.1 image, graph and contour

We are interested in geometric objects called curves and surfaces and solids. These objects arise as image,graph and contour sets associated with various functions. We begin by reviewing the theory of sets andfunctions.

The definitions of set and function are assumed known. as are the concepts of injective, surjective andbijective function. Ths definition of product set is assumedknown.

We define 5 useful sets related to a function. The definitions may seem difficult but see the examples below.

Letf : A → B

a 7→ b = f (a)

be a function.The setA is called thedomainof the f .The setB is called thecodomainof f .The imageset of f (a subset of the codomain) is defined as follows.

Im( f )

= f (A)

={

b∈ B∣

∣

∣∃ a∈ A such thatf (a) = b}

={

f (a)∣

∣

∣a∈ A}

⊂ B

Thegraphof f is a subset of the product of the domain and codomain sets and is defined as follows.

Gr( f )

={

(a,b)∣

∣

∣a∈ A, b∈ B andb = f (a)}

={

(a, f (a))∣

∣

∣a∈ A}

⊂ A×B

Let b∈ B, the codomain. Thecontour, preimageor level seton which f takes constant valueb is a subset ofthe domainA and is defined as follows.

f−1(b)

3

4 CHAPTER 1. CURVES,SURFACES AND SOLIDS

={

a∈ A∣

∣

∣ f (a) = b}

⊂ A

Note thatf−1(b) 6= /0 iff b∈ Im( f ).

In this course the domainA and codomainB will generally beR,R2 or R3 (the line, plane or Euclidean3-space). Thusf will be a scalar or vector valued function of one or several real variables.

example. with finite setsLet A = {1,2,3,4,5,6,7}, B = {a,b,c,d} and f be the function

f : A → B1 7→ a2 7→ b3 7→ a4 7→ d5 7→ a6 7→ b

Im f = {a,b,d} ⊂ B

Gr( f )

= {(1,a)(2,b)(3,a)(4,d)(5,a)(6,b)}⊂ A×B

=

(1,d) (2,d) (3,d) (4,d) (5,d) (6,d)(1,c) (2,c) (3,c) (4,c) (5,c) (6,c)(1,b) (2,b) (3,b) (4,b) (5,b) (6,b)(1,a) (2,a) (3,a) (4,a) (5,a) (6,a)

There are four level sets or contours

f−1(a) = {1,3,5} ⊂ A

f−1(b) = {2,6} ⊂ A

f−1(c) = /0 ⊂ A

f−1(d) = {4} ⊂ A

Most sets in this course will be geometric in nature i.e. curves, surfaces or solid regions inR,R2 or R3.Nonetheless the curent purely set theoretic example well illustrates underlying concepts from set and functiontheory.

For example, the domain is the disjoint union of all non emptycontours; this is referred to as thecontourdiagram. A = f−1(a)

S

f−1(b)S

f−1(d)f−1(y) 6= /0 iff y∈ Im( f ).The graph appears above, boldfaced inA×B. Project the graph sidewise to see the image. Slice the graphhorizontally to see the contours.

example. image, graph and contour sets off (x) = x2

Considerf : R ∋ x 7→ f(x) = y = x2 ∈ R

The image set off is

Im( f ) ={

f (x)∣

∣

∣x∈ A}

={

x2∣

∣

∣x∈ R}

= [0,∞)


1.1. IMAGE, GRAPH AND CONTOUR 5

ThusIm( f ) = [0,∞) ⊂ B = R. Another way to put this is, “f takes only positive values”.

Next we turn to the graphGr( f ) ⊂ A×B= R×R = R2 which lies in the plane.

Gr( f )

={

(x,y)∣

∣

∣x∈ A,y∈ B andy = f (x)}

={

(x,y)∣

∣

∣y = x2, x∈ R}

⊂ A×B

= R2

This graph is parabolic curve, see figure(1.1), with explicit equationy = x2. Next we compute the level set

Figure 1.1: parabolay = x2

f−1(4) ⊂ A = R, on which f takes constant value 4.

f−1(4)

={

x∈ R∣

∣

∣ f (x) = 4}

={

x∈ R∣

∣

∣x2 = 4}

= {2,−2}⊂ A

= R

I leave the reader to see thatf−1(9) = {3,−3}, f−1(100) = {10,−10} and in general ifa> 0 then f−1(a) ={±√

a}.

The casef−1(0) is different.

f−1(0)

={

x∈ R∣

∣

∣x2 = 0}

= {0}

December 6, 2008 c©jbquig-UCD


The casef−1(−4) is different again.

f−1(−4)

={

x∈ R∣

∣

∣x2 = −4}

= /0

I leave the reader to see that in general ifa < 0 then f−1(a) = /0.

The image and contours off are not much use but the graph is the important parabolic curve whose explicitequation isy = x2. The iamage can be visualised by projecting the graph horizontally onto thez-axis. Thelevel sets f−1(4), f−1(9), f−1(100), f−1(0), f−1(−4), appear when th graph is sliced horizontally atheights 4,9,100,0,−9 respectively.

example. image, graph and contours off (x,y) = x2 +y2

Let A = R2 andB = R and let f be the function

f : R2 ∋(

xy

)

7→ f (x,y) = z= x2 +y2 ∈ R

The image set off is

Im( f ) =

{

f (x,y)

∣

∣

∣

∣

(

xy

)

∈ A = R2}

={

x2 +y2∣

∣

∣x,y∈ R

}

= [0,∞)

ThusIm( f ) = [0,∞) ⊂ B = R; i.e. f takes only positive values.

The graph,Gr( f ) ⊂ A×B= R2×R = R3, we will see, is a surface lying in Euclidean 3-spaceR3.

Gr( f )

=

xyz

∣

∣

∣

∣

∣

∣

(

xy

)

∈ A = R2,z∈ B = R andz= f (x,y)

=

xyz

∣

∣

∣

∣

∣

∣

z= x2 +y2, x,y∈ R

⊂ A×B

= R3

This graph is well known as a surface in the shape of a parabolic bowl, see figure (1.2)lying inR3. The levelset on whichf takes constant value 4 is

f−1(4)

=

{ (

xy

)

∈ R2

∣

∣

∣

∣

f (x,y) = 4

}

=

{ (

xy

)

∈ R2

∣

∣

∣

∣

x2 +y2 = 4

}

= the curveC(2,0) ⊂ R2, being the circle, center the origin, radius 2.

⊂ A

= R2

The reader can check thatf−1(9) =C(0,3)⊂R2, f−1(100) =C(0,10) and in general ifa> 0 thenf−1(a) =C(0,

√a).


1.1. IMAGE, GRAPH AND CONTOUR 7

z

-4 -2 0 2 4 -4-2

02

4147

101316

Figure 1.2: graph and contour diagram ofy = x2

The contourf−1(0) is different.

f−1(0)

=

{ (

xy

)

∈ R2

∣

∣

∣

∣

x2 +y2 = 0

}

= {(

00

)

}

This is not a curve but a set containing a single point, the origin.The contourf−1(−4) is different again.

f−1(−4)

=

{ (

xy

)

∈ R2

∣

∣

∣

∣

x2 +y2 = −4

}

= /0

The reader can check that in general ifa < 0 then f−1(a) = /0.

The graph is the useful parabolic surface with explicit eqnz= x2 +y2. The contours are interesting circularcurves. The image set is not useful. This example illustrates thecontour diagramconcept:- the domain set isa disjoint union of contours, see figure(1.2).

R2 =D

[

a∈R

f−1(a)

The plane is composed of concentric circles and a point at their common center.

The graph can be created by elevating each contourf−1(a) to heightz = a. The contourf−1(a) can beobtained by horizontal slicing of the graph at heightz= a.



1.2 curves in R2 and R3

1.2.1 the circle

Let a > 0 andC = C(a,0)⊂ R2 denote the circle with centre the origin0 and radiusa. C is our first exampleof a curve. We first presentC as a level set or contour of a functionf : R2 → R; this approach will yield theimplicit equation ofC. Second we presentC as the graph of a functiong : R → R;this approach will yieldtheexplicit equation ofC. Third and last we presentC as the image of a functionh : [−π,π] ⊂ R → R2; thisapproach will yield theparametricequation ofC. In this course the parametric equation of a curve will provemore useful than either the implicit or the explicit equation. The circle is our first curve, it is rather simplebut nonetheless is archytypal.

contour presentation and implicit equation of circleConsider the function

f : R2 ∋(

xy

)

7→ z= f (x,y) = x2 +y2 ∈ R (1.1)

Then

C = f−1(a2) =

{(

xy

) ∣

∣

∣

∣

f (x,y) = a2}

=

{(

xy

) ∣

∣

∣

∣

x2 +y2 = a2}

⊂ R2

a contour or level set off . Theimplicit equation of the curveC is

x2 +y2 = a2 (1.2)

We may isolatey on the left of the latter equationy = ±√

a2−x2, −a < x < a. This leads to our secondpresentation of the circle curve.

graphical presentation and explicit equation of circleConsider the function

g : [−a,a]⊂ R ∋ x 7→ y = g(x) =√

a2−x2 ∈ R (1.3)

Then

C+ = Gr(g) =

{(

xy

) ∣

∣

∣

∣

y =√

a2−x2, −a < x < a

}

⊂ R2

The graph ofG is the curveC, more precisely only the upper halfC2 of C. The implicit equation of the curveC+ is

y =√

a2−x2, −a < x < a (1.4)

image presentation and parametric equation of circleConsider the function

γ : [−π,π]⊂ R → R2

t 7→ γ(t) =

(

x(t)y(t)

)

=

(

acostasint

)

(1.5)

The image of the functionγ is

Im(γ)

={

γ(t)∣

∣

∣t ∈ [−π,π]}

=

{(

x(t)y(t)

) ∣

∣

∣

∣

t ∈ [−π,π]

}

=

{(

acos(t)asin(t)

) ∣

∣

∣

∣

t ∈ [−π,π]

}

= C(a,0)

⊂ R2


1.2. CURVES INR2 AND R3 9

Theparametricequation(s) of the circle curve is

x(t) = acos(t)y(t) = asin(t)

}

, −π ≤ t ≤ π (1.6)

One might check that the coordinates in the parametric equation satisfy both the explicit and implicit equa-tions. Parametrization of the circle has been compared to winding a thread of length 2πa onto a spool ofradiusa.

1.2.2 the Archimedean spiral

TheArchimedean spiralis a curveΓ lying in the planeR2. It is the curve traced out by the end of a threadtautly unwound from a circular spool, see figure (1.3).

Let C =C(1,0)⊂ R2 denote the circle with explicit equationx2+y2 = 1. Consider the standard parametriza-tion of C

c : [0,∞) ∋ t 7→ c(t) =

(

cos(t)sin(t)

)

∈ C ⊂ R2

One can think of this parametrization ofC as winding a thread[0,∞) onto the spoolC starting at the point

c(0) = i =

(

10

)

∈ C. Alternatively one can envisage a point traveling round andround the circle being at

point c(t) at timet. Note that the arc distance (measured alongC) from the start pointc(0) to the generalpoint c(t) is exactlyt. Unwind a lengtht of thread, holding it taut, the thread departs from the pointc(t) onthe curveC in a line of lengtht and directiond being reverse tangential, see figure(1.3). Letγ(t) denote theposition of the end of the thread. Note thatd is perpendicular to the radial vectorc(t). Thus

γ(t)= c(t)+ td

=

(

cos(t)sin(t)

)

+ t

(

sin(t)−cos(t)

)

We have found the parametrization of the Archimedian spiralΓ.

Figure 1.3: taut unwinding, Archimededean spiral

γ : [0,∞) ∋ t 7→ γ(t) =

(

x(t)y(t)

)

=

(

cos(t)+ t sin(t)sin(t)− t cos(t

)

∈ Γ ⊂ R2 (1.7)



1.2.3 the cycloid

Consider a circular wheel rolling along a straight line. ThecurveΓ traced out by a point on the circumferenceis known as thecycloid. As a bicycle moves along the wheel valves trace out cycloids. We will parameterizethe cycloid.

Remind yourself of the (very useful) matrix of rotation of the plane, anticlockwise by anglet.

S(t) =

(

cos(t) −sin(t)sin(t) cos(t)

)

: R2 → R2

(

xy

)

7→ S(t)

(

xy

)

=

(

xcos(t)−ysin(t)xsin(t)+ycos(t)

) (1.8)

We are ready to parametrize the cycloid. Consider a circle ofradius 1 sitting on thex-axis with point ofcontact the origin and center atj , see figure (1.4). Now roll the circle along thex-axis a distancet. The pointof contact moves from0 to t i and the center moves fromj to j + t i, i.e. the center is always directly above thepoint of contact.

From the point of view of a bug sitting on the center the initial point of contact is at position−j and the wheelturns through angle−t (clockwise by anglet). From the bug’s point of view the initial point of contact hasmoved toS(−t)(− j).

Adding the position of the center (the bug’s position) to thelatter we obtain the actual position of the initialpoint of contact after timet.

j + t i +S(−t)(− j)

=

(

t1

)

+

(

cos(−t) −sin(−t)sin(−t) cos(−t)

)(

0−1

)

=

(

t1

)

+

(

cos(t) sin(t)−sin(t) cos(t)

)(

0−1

)

=

(

t1

)

+

(

−sin(t)−cos(t

)

=

(

t −sin(t)1−cos(t

)

Parameterization of the cycloid, see figure(1.4), is

γ : R ∋ t 7→ γ(t) =

(

x(t)y(t)

)

=

(

t −sint1−cost

)

∈ Γ ⊂ R2 (1.9)

Figure 1.4: wheel and cycloid


1.2. CURVES INR2 AND R3 11

1.2.4 the helix

The helix curve lies inR3. Phone handset cables are helices. Screw threads are helices. Banister rails ofcircular staircases are helices. An electron moving freelyin a constant magnetic field travels on a helix. Thehelix appears from circular motion around thez axis combined with forward motion in thez-direction, seefigure (1.5). We can parameterize one turn of the helix by adding az(t) parameter to the parameterization ofthe circle. Leta,b > 0.

x

y

z

Figure 1.5: the helix

γ : [0,2π] ∋ t 7→ γ(t) = x(t) =

x(t)y(t)z(t)

=

acos(t)asin(t)

bt

∈ R3 (1.10)

In 2π seconds the pointγ(t) = x(t) passes once round the circlex2 +y2 = a2 (of radiusa) while simultane-ously ascending to height 2πb. Check that

γ(0) =

a00

and γ(2π) =

a0

2πb



1.3 problem set

curves, surfaces, solids1. Let f be the function

f : R2 ∋(

xy

)

7→ f (x,y) = x2−y2 ∈ R

(i) Sketch the contoursf−1(−1), f−1(0), f−1(1). Each is a curve in the plane.

(ii) Now you have the general idea. Roughly sketch the contours f−1(d) ford = −16,−9,−4,−1,0,1,4,9,16.

(iii) Draw the graphGr( f ) ⊂ R3.

(iv) Put a common sense name on the graph.

[This question is very instructive but rated quite difficult. Start by looking at the easier case off (x,y) = x2 +y2 as treated in the notes.]

2. Let Γ ⊂ R3 be the helix curve with parameterization

c : [0,2π] ∋ t 7→ x(t) =

x(t)y(t)z(t)

=

acos(t)asin(t)

b(t)

∈ Γ ⊂ R3

(i) Sketch the cylindrical surfaceC⊂ R3 with implicit equationx2 +y2 = a2.

(ii) Sketch the screw surfaceS⊂ R3 with explicit equationz= bθ = barctan(y/x).

(iii) In a sketch showΓ = C∩S.

(iv) ProveΓ =C∩Sby showing that the parametric equation ofΓ satisfies both the equation ofC andthe equation ofS.

(v) Take a can of Heinz baked beans (a cylindrical surface). Remove the label in one piece (a rectan-gle). Emulating the genius of Andy Warhol make a single well thought out stroke on this label.Put the label back on. If you have the correct insight your stroke should become the helix. Bymeasuring the length of the stroke on the label you willl havethe total arc length of the helix.


Chapter 2

the geometry of curves

We will study curves inR2. Associated with such we will study studyvelocityandaccelerationvectors,speedandarc-length scalars, tangentandnormalvectors and thecurvaturescalar. Armed with these toolwill study involuteandevolutecurves of a given curve.

2.1 velocity, speed and acceleration

We will study velocity, speed and acceleration for curves inthe plane and in 3-space.

2.1.1 velocity, speed and acceleration in R2

Let

γ : [a,b] ∋ t 7→ γ(t) = x(t) =

(

x(t)y(t)

)

∈ Γ ⊂ R2

be a parameterization of the curveΓ. Under this parameterization,thevelocityvector at timet is

ddt

γ(t) = γ(t) = x(t) =

(

x(t)y(t)

)

(2.1)

thespeedscalar at time (i.e. the magnitude of velocity) is

||γ(t)|| = ||x(t)|| =√

(x(t))2 +(y(t))2

theaccelerationvector at timet is

d2

dt2γ(t) = γ(t) = x(t) =

(

x(t)y(t)

)

(2.2)

example. velocity, acceleration, speed on the circle

c : [0,2π/ω] → C(ai +bj , r)⊂ R2

t 7→(

x(t)y(t)

)

=

(

ab

)

+ r

(

cos(ωt)sin(ωt)

)

parameterizes (once round) the circleC(ai +bj , r) with center

(

ab

)

and radiusr > 0. The parametrization

is carried out at angular speedω.

x(t) = ωr

(

−sin(ωt)cos(ωt)

)

, ||x(t)|| = ωr , x(t) = −ω2r

(

cos(ωt)sin(ωt)

)

13

14 CHAPTER 2. THE GEOMETRY OF CURVES

example. velocity, acceleration, speed on the Archimedean spiralΓRecall the position vectorx(t), 0≤ t ≤ 6π (go 3 times round) on the Archimedean spiralΓ, see figure(1.3).

γ : [0,∞) ∋ t 7→ x(t) =

(

x(t)y(t)

)

=

(


)

∈ Γ ⊂ R2

Velocity underγ is

x(t) =

(

−sin(t)+sin(t)+ t cos(t)cos(t)−cos(t)+ t sin(t)

)

=

(

t cos(t)t sin(t)

)

= t

(

cos(t)sin(t)

)

Acceleration underγ is

x(t) =

(

cos(t)− t sin(t)sin(t)+ t cos(t)

)

=

(

cos(t)sin(t)

)

+ t

(

−sin(t)cos(t)

)

Speed is

||x(t)||

=√

x(t)2 + y()2

=

√

t2cos2(t)+ t2sin2(t)

= t√

cos2(t)+sin2(t)

= t

example. velocity, acceleration, speed on the cycloid curveRecall the position vectorx(t), 0≤ t ≤ 2π (once round) on the cycloidΓ, see figure(1.4).

γ : [0,2π) ∋ t 7→ x(t) =

(

x(t)y(t)

)

=

(

t −sin(t)1−cos(t)

)

∈ Γ ⊂ R2

Velocity and acceleration underγ are

x(t) =

(

1−cos(t)sin(t)

)

, x(t) =

(

sin(t)cos(t)

)

Speed is

||x(t)||

=√

x(t)2 + y(t)2

=

√

(1−cos(t))2 +sin2(t)

=

√

1−2cos(t)+cos2(t)+sin2(t)

=√

1−2cos(t)+1

=√

2−2cos(t)

=

√

4sin2( t

2

)

, use D.A.F cos2x = 1−2sin2x ⇔ 1−cos2x = 2sin2 x

= 2sin( t

2

)


2.1. VELOCITY, SPEED AND ACCELERATION 15

For example||(||x(0)) = 0 = ||(||x(2π)) and ||(||x(2π)) = 2; for bycle travelling at constant speed 1, thewheel valve is moving at speed 0 at its lowest point (whent = 0,t = 2π) and at speed 2 at its highest point(whent = π).

2.1.2 velocity speed and acceleration in R3

Formulae(2.1) and (2.2) are readily generalized ifγ parameterizes a curve in 3-space.

x(t) =

x(t)y(t)z(t)

, x(t) =

x(t)y(t)z(t)

, x(t) =

x(t)y(t)z(t)

, ||x(t)|| =√

x2(t)+ y2(t)+ z2(t) (2.3)

example. velocity, acceleration, speed on the helixRecall the position vectorx(t), 0≤ t ≤ 2π (once round) on the helixΓ ⊂ R3, see figure(1.5).

γ : [0,2π) ∋ t 7→ x(t) =

x(t)y(t)z(t)

=

acos(t)asin(t)

bt

∈ Γ ⊂ R3

Velocity and acceleration underγ are

x(t) =

−asin(t)acos(t)

b

, x(t) =

−acos(t)−asin(t)

0

Speed is constant

||x(t)||

=√

x(t)2 + y(t)2 + z(t)2

=

√

a2sin2(t)+a2cos2(t)+b2

=

√

a2(sin2(t)+cos2(t))+b2

=√

a2 +b2



2.2 arclength

2.2.1 arclength in R2

Consider the parameterization

γ : [a,b] ∋ t 7→ x(t) =

(

x(t)y(t)

)

∈ Γ ⊂ R2

of the curveΓ in the plane. The distance,measured along the curve, from the start pointγ(a) = x(a) to thegeneral pointγ(t) = x(t) is calledarc-lengthand denoteds(t). The toal distance,measured along the curvefrom startx(a) to endx(b) is called thetotal arc lengthof Γ and is denotedL = L(Γ) = s(b). We will findformulae for general and total arc length.

Consider two close together times,t andt +dt . The corresponding points onΓ are

γ(t) = x(t) =

(

xy

)

and γ(t +dt) = x(t +dt) =

(

x+dxy+dy

)

These points being close together, the distance betweends, measured along the arc ofΓ, is approximatelyequal to the straight line distance between. Using Pythagoras theorem

ds ≈√

dx2 +dy2

Total Arc lengthis the integral

L

= s(b)

=

Z b

ads

=Z b

a

dsdt

dt

=

Z b

a

√

dx2 +dy2

dtdt

=

Z b

a

√

(

dxdt

)2

+

(

dydt

)2

dt, i.e. speed× time = distance

Thustotal arc length is

L = s(b) =

Z b

a

√

x(t)2 + y(t)2dt =

Z b

a||x||dt (2.4)

and partialarclengthis

s(t) =

Z t

a

√

x(τ)2 + y(τ)2 dτ =

Z t

a||x||dτ (2.5)

example. arc length on the circleConsider the parameterization of the circleC = C(a,0) ⊂ R2.

γ : [0,2π] ∋ t 7→(

x(t)y(t)

)

=

(

acos(t)asin(t)

)

∈C⊂ R2

We saw in example(2.1.1) that||x(t)|| = a.We compute the total arc length.

L =Z 2π

0||x(t)||dt =

Z 2π

0adt = at

∣

∣

∣

2π

t=0= 2πa−0 = 2πa


2.2. ARCLENGTH 17

Next we compute the arc distances(t) from γ(0) =

(

10

)

to γ(t) = a

(

cos(t)sin(t)

)

.

s(t) =Z t

0||x(τ)||dτ =

Z t

0adτ = aτ

∣

∣

∣

τ=t

τ=0= a · t−a ·0 = at

Of course you probably knew already that the circumference of a circle of radiusa is 2πa and the arc lengthsubtended by radial anglet is at. But what about the distance travelled on the archimedean spiral by the endof a thread tautly unwound from a spool? And what about the distance travelled on the cycloid by the valveof a bicycle?

example. arc length on the Archimedean spiralConsider the parameterization (thrice round) the Archimedean spiralΓ ⊂ R2, figure(1.3).

γ : [0,6π] ∋ t 7→(

x(t)y(t)

)

=

(

cos(t)+ t sin(t)sin(t)− t cos(t)

)

∈ Γ ⊂ R2

See example(2.1.1),

||x(t)|| =√

x(t)2 + y(t)2 = t

We compute the total arc length (3 times round).

L =

Z 6π

0

√

x(t)2 + y(t)2dt =

Z 6π

0t dt =

t2

2

∣

∣

∣

∣

6π

t=0=

12

(

(6π)2−02) = 18π2

While unwinding 6π of thread your hand spirals through a distance 18π2 > 175.Next we compute the arc distances(t) from γ(0) to γ(t) measured along the spiralΓ.

s(t) =

Z t

0

√

x(τ)2 + y(τ)2 dτ =

Z t

0τdτ =

τ2

2

∣

∣

∣

∣

τ=t

τ=0=

t2

2

example. arc length on the cycloidConsider the parameterization of one period of the cycloidΓ ⊂ R2, see figure(1.4).

γ : [0,2π] ∋ t 7→(

x(t)y(t)

)

=

(

t +sin(t)1−cos(t)

)

∈ Γ ⊂ R2

See example(2.1.1),

||x(t)|| =√

x(t)2 + y(t)2 = 2sint2

We compute the total arc length of one loop of the cycloid

L =

Z 2π

0||x(t)||dt =

Z 2π

02sin

t2

dt = −4cost2

∣

∣

∣

2π

t=0= −4

[

cos(π)−cos(0)]

= −4(−1−1) = 8

While a wheel rotate once, a bicycle move forward 2π, but a valve travels arc distance of 8 along a cycloidcurve.Next we compute the arc distances(t) from γ(0) to γ(t) measured along the cycloidΓ.

s(t) =

Z t

0||x(τ)||dτ =

Z t

02sin

τ2

dτ = −4cosτ2

∣

∣

∣

τ=t

τ=0= −4

[

cost2−cos0

]

= 4−4cost2

= 8sin2( t

4

)

.

For examples(π) = 8sin2(π/4) = 8(1/√

2)2 = 4 .



2.2.2 arclength in R3

The formulae forarc lengthandtotal arc lengthin R3 are similar to those forR2

s(t) =

Z t

a

√

x2(τ)+ y2(τ)+ z2(τ)dτ =

Z t

a||x||dτ (2.6)

L = s(b) =

Z b

a

√

x2(t)+ y2(t)+ z2(t)dt =

Z b

a||x||dt (2.7)

example. arc length on the helixRecall from example(2.1.2) that parameterization of the helix which turns on the cylinderx2 + y2 = a2 andrises a distance 2πb in each turn is

γ : [0,2π) ∋ t 7→ x(t) =

x(t)y(t)z(t)

=

acos(t)asin(t)

bt

∈ Γ ⊂ R3

see figure(1.5). In example(2.1.2) we saw that speed was constant at||x|| =√

a2 +b2. Thus

L =

Z 2π

0||x||dt =

Z 2π

0

√

a2 +b2dt = 2π√

a2 +b2

and

s(t) =Z t

0||x(τ)||dτ =

Z t

0

√

a2+b2dτ = t√

a2 +b2


2.3. CURVATURE INR2 19

2.3 curvature in R2

What do we mean by thecurvatureof the simplest curve, the circle inR2. A circle with large radius is lesscurved than a circle of small radius. Wetentativelydefine curvature

κ =1r

wherer > 0 is the radius

But one traverses some circles ACW and some CW, we say curvature is either positive or negative. Wedeclare that, for the circle, an appropriate definition of curvature is

κ = ±1r

(2.8)

A large circle of radius 100 traversed ACW has curvatureκ = 1/100; a small circle of radius 1/29 traversedCW has curvatureκ = −29. The straight line is considered to be a degenerate circleof radiusr = ∞ andcurvatureκ = 1/∞ = 0.

Curvature is constant on a circle but for a general curveΓ curvatureκ(x) varies from point to pointx ∈ Γ.Curvatureκ(x) at the pointx is defined to be that of the circle which best fitsΓ near the pointx. The bestfitting circle is known as theosculatingcircle.

2.3.1 curvature of the circle

The most general circleC⊂ R2 has center at some point

(

ab

)

and radiusr > 0 and implicit equation

(x−a)2+(y−b)2 = r2 (2.9)

The following is a quite general parametrization of this circle; the reader can verify that indeedx(t) andy(t)satisfy the implicit equation(2.9) .

c : R ∋ t 7→(

x(t)y(t)

)

=

(

a+ r cos(ωt)b+ r sin(ωt)

)

∈C⊂ R2 (2.10)

We will soon have use forx, ||x|| andx so we compute them right now

c(t) =

(

x(t)y(t)

)

=

(

−ωr sin(ωt)ωr cos(ωt)

)

(2.11)

c(t) =

(

x(t)y(t)

)

=

(

−ω2r cos(ωt)−ω2r sin(ωt)

)

(2.12)

||c(t)|| =√

x2(t)+ y2(t) = r√

ω2 = r |ω | . (2.13)

Hereω controls the speeed and sense (ACW/CW) with which the parameterized pointc(t) goes around thecircle.C is traversed ACW orCW according asω > 0 or ω < 0, i.e. according as

sgn(ω) =ω|ω | =

ω√ω2

= ±1

In terms of the parameterizationc we have a formula for the curvature of the circleC

κ =x(t)y(t) − y(t)x(t)

||x||3 (2.14)

We will prove (2.14), see (2.15). The reader might well ask, “Why use such an elaborate formula when wehave the simple formula (2.8) above?”. The answer is that (2.14) works, not only for the circle but for themost general curve, such as the spiral or cycloid.



proof

x(t)y(t) − y(t)x(t)||x||3 (2.15)

=(−ωr sin(ωt))(−ω2r sin(ωt)) − (ωr cos(ωt))(−ω2r cos(ωt))

(r |ω |)3 (2.16)

=r2ω3(sin2(ωt) + cos2(ωt))

r3 |ω |3(2.17)

=1r

(sgn(ω))3 , since sin2+cos2 = 1 and sgn(ω) = ω|ω | (2.18)

=1r

(sgn(ω)) , note sgn3 = sgn since sgn= ±1 (2.19)

= κ (2.20)

2.3.2 curvature of a general curveΓ ⊂ R2

Let

γ : R ∋ t 7→ x(t) =

(

x(t)y(t)

)

∈ Γ ⊂ R2 (2.21)

be a parameterization of the curveΓ ⊂ R2. We definecurvature at the pointx(t) ∈ Γ to be the curvature oftheosculating circle, i.e. of the circleC whichbest fitsthe curveΓ near the pointx(t). The precise meaningof best fitsis thatc andγ agree att up to the second order of differentiation, i.e

c(t) = γ(t) , c(t) = γ(t) , c(t) = γ(t) (2.22)

This can be stated that, at timet, position, velocity and acceleration on the osculation circle and on the curveΓ are equal.

It is now straightforward to give a formula for the curvatureκ(t) at the pointx(t) ∈ Γ. All the hard work hasalready been done during study of the circle above.

κ(t) =det(x, x)

||x||3 =

det

(

x xy y

)

||(

xy

)

||3=

x(t)y(t) − y(t)x(t)(

x2(t)+ y2(t))3/2

(2.23)

Formula (2.23) follows immediately from formula (2.14) using properties (2.22).

example. curvature of the Archimedean spiralΓSee example(2.1.1), we have forΓ, (see figure(1.3))(

x(t)y(t)

)

=

(


)

,

(

x(t)y(t)

)

= t

(

cos(t)sin(t)

)

,

(

x(t)y(t)

)

=

(


)

, ||x(t)||= t

Thus

κ(t)

=

det

((

xy

)

,

(

xy

))

||x||3


2.3. CURVATURE INR2 21

=

det

((

t costt sint

)

,

(

cost − t sintsint + t cost

))

t3

=

det

((

t costt sint

)

,

(

−t sint+t cost

))

t3 , took (1/t) times column 1 from column 2

=

t · t ·det

((

costsint

)

,

(

−sint+cost

))

t3 , divided col 1 and col 2 byt

=t2 · (1)

t3

=1t

The farther out one goes on the spiral the less the curvature,κ(t) = 1/t, and the greater the radius, 1/κ = t,of the osculating circle.

example. curvature of the cycloidΓSee example(2.1.1), we have forΓ, (see figure(1.4))

(

x(t)y(t)

)

=

(


)

,

(

xy

)

=

(

1−cos(t)sin(t)

)

,

(

xy

)

=

(

sin(t)cos(t)

)

, ||x(t)|| = 2sin( t

2

)

Thus

κ(t)

=

det

(

x xy y

)

||x||3

=

det

(

1−cost sintsint cost

)

(2sin(t/2))3

=cost −cos2 t −sin2 t

(2sin(t/2))3

=cost −1

8sin3 (t/2)

=−2sin2 (t/2)

8sin3 (t/2)

=−1

4sin(t/2)



2.4 tangent and normal (Serret-Frenet frame)

2.4.1 unit tangent and normal vectors in R2

The velocity vectorx(t) is tangential to the curveΓ ⊂ R2 at the pointx(t) ∈ Γ. A curve has many parame-terizations andx depends on which particular parameterization is used. In particular the magnitude (speed)||x|| of (velocity) x is parametrization dependent. However the unit vector

t(t) =x

||x|| (t) =xi + yj

√

x2 + y2(t) (2.24)

called theunit tangentvector depends only on the curveΓ and positionx(t) thereon and is independent ofparameterization.

The unit vectorn(t) perpendicular tot(t) is called theunit normal. A moment’s thought shows that thisdefinition is ambiguous; there are two possibilities forn. We give a concise definition. InR2 theunit normalvectorn(t) is obtained from the unit tangent vectort(t) by ACW rotation through the angleπ/2. Recall thatThe 2×2 rotation matrix is

S(π/2) =

(

cos(π/2) −sin(π/2)sin(π/2) cos(π/2)

)

=

(

0 −11 0

)

Thus

n(t) = S(π/2)t(t) =

(

0 −11 0

)

1√

x2 + y2

(

xy

)

=1

√

x2 + y2

(

−yx

)

To sum up, the pair{t(t),n(t)} is called theSerret-Frenetframe at the pointx(t) ∈ Γ ⊂ R2 and

t =x

||x|| , n = S(π/2)t , {t,n} =

{

1√

x2 + y2

(

xy

)

,1

√

x2 + y2

(

−yx

)

}

(2.25)

example. tangent and normal to the circleFor the circle, we saw in example(2.1.1)

x(t) =

(

r cos(ωt)r sin(ωt)

)

, x(t) = ωr

(

−sin(ωt)cos(ωt)

)

, ||x(t)|| = ωr

Thus

t(t)

=1

||x(t)||

(

x(t)y(t)

)

=1

|ω | r

(

−ωr sin(ωt)ωr cos(ωt)

)

= sgn(ω)

(

−sin(ωt)cos(ωt)

)

And

n(t)

=1

||x(t)||

(

−y(t)x(t)

)

=1

|ω | r

(

−ωr cos(ωt)−ωr sin(ωt)

)

= sgn(ω)

(

−cos(ωt)sin(ωt)

)


2.4. TANGENT AND NORMAL (SERRET-FRENET FRAME) 23

The Serret-Frenet frame is{

t , n}

=

{

sgn(ω)

(

−sin(ωt)cos(ωt)

)

, sgn(ω)

(

−cos(ωt)−sin(ωt)

)}

example. tangent and normal to the Archimedean spiralFor the spiral (see figure(1.3)), we saw in example(2.1.1)

x(t) =

(


)

, x(t) =

(

t cos(t)t sin(t)

)

, ||x(t)|| = t

Thus

t(t)

=1

||x(t)||

(

x(t)y(t)

)

=1

ωt

(

t cos(t)t sin(t)

)

=

(

cos(t)sin(t)

)

And

n(t)

=1

||x(t)||

(

−y(t)x(t)

)

=1t

(

−t sin(t)t cos(t)

)

=

(

−sin(t)cos(t)

)

The Serret-Frenet frame is{

t , n}

=

{(

cos(t)sin(t)

)

,

(

−sin(t)cos(t)

)}

example. tangent and normal to the cycloidFor the cycloid (see figure(1.4)), we saw in example(2.1.1)

x(t) =

(


)

, x(t) =

(

1−cos(t)sin(t)

)

, ||x(t)|| = 2sin( t

2

)

Thus

t(t)

=1

||x(t)||

(

x(t)y(t)

)

=1

2sin(t/2)

(

1−cos(t)sin(t)

)

=1

2sin(t/2)

(

2sin2(t/2)2sin(t/2)cos(t/2)

)

=

(

sin(t/2)cos(t/2)

)



And

n(t)

=

(

0 −11 0

)

t(t)

=

(

0 −11 0

)(

sin(t/2)cos(t/2)

)

=

(

−cos(t/2)sin(t/2)

)

The Serret-Frenet frame is

{

t , n}

=

{(

sin(t/2)cos(t/2)

)

,

(

−cos(t/2)sin(t/2)

)}

2.4.2 tangent vector in 3-space

We will compute only the unit tangent vectort(t) to a curve inR3. Time does not permit us to investigate thequestion of the unit normal. The definition oft is similar to that inR2.

t =x

||x|| =1

√

x2 + y2 + z2

xyz

(2.26)

example. unit tangent on the helixFor the helix curve (see figure(1.5))Γ ⊂ R3 we saw in example (2.1.2)

x(t) =

acos(t)asin(t)

bt

, x(t) =

−asin(t)acos(t)

b

, ||x(t)|| =√

a2 +b2

Thus

t(t)

=1

||x(t)||

x(t)y(t)z(t)

=1√

a2 +b2

−asin(t)acos(t)

b


2.5. INVOLUTE 25

2.5 involute

In this section we will put ideas defined in§(2.2 and2.4) to work. Given a regular curveΓ, using the originalparameterization, arclength and the unit tangent vector, we will build a new curve called denoted Inv(Γ) andcalled theinvoluteof Γ. The involute is the curve traced out by the end a taut thread peeled offΓ.

We have already seen the Archimedean Spiral described as theinvolute of the circleC, see§§(1.2.2). Givena parametrization

γ : [a,b] ∋ t 7→ x(t) ∈ Γ ⊂ R2

of a regular curveΓ we will find a formula for parametrization

i : [a,b] ∋ t 7→ i(t) ∈ Inv(Γ)

of the involute curve Inv(Γ).

Starting atx(a) a thread is laid along the curveΓ. It is peeled off being held taut at all times. If the threaddepartsΓ from the pointx(t) its direction is the reverse tangent−t(t) and its length is the arclengths(t) fromx(a) to x(t). Denote the position of the end of the thread byi(t).

i(t) (2.27)

= x(t) − s(t)t(t) (2.28)

(2.29)

= (2.30)

(

x(t)y(t)

)

−

Z t

a

√

x2(t)+ y2(t)dτ√

x2(t)+ y2(t)

(

x(t)y(t)

)

, for R2 (2.31)

(2.32)

= (2.33)

x(t)y(t)z(t)

−

Z t

a

√

x2(t)+ y2(t)+ z2(t)dτ√

x2(t)+ y2(t)+ z(t)

x(t)y(t)z(t)

, for R3 (2.34)

example. involute of the circleConsider the parametrization of the unit circleC⊂ R2

γ : [0,2π] ∋ t 7→ x(t) =

(

x(t)y(t)

)

=

(

costsint

)

∈C

with velocity and speed

x(t) =

(

x(t)y(t)

)

=

(

−sintcost

)

, ||x(t)|| =√

x2(2)+ y2(t) = 1

and arclength

s(t) =

Z t

0||x(τ)||dτ =

Z t

01dτ = t

Applying formula2.31 we obtain the parmeterization of Inv(C) ⊂ R2, the involute of the circle.

i(t)

=

(

x(t)y(t)

)

−

Z t

a

√

x2(t)+ y2(t)dτ√

x2(t)+ y2(t)

(

x(t)y(t)

)



=

(

cos(t)sin(t)

)

− t1

(

−sin(t)cos(t)

)

=

(


)

Of course we have seen this all before in§§(1.2.2). The involute of the circle is the Archimedean spiral;figure(1.3) illustrates this fact.

example. involute of the Archimedean spiralΓ ⊂ R2

In example(2.1.1) we saw that for this spiral

(

x(t)y(t)

)

=

(


)

,

(

x(t)y(t)

)

= t

(

cos(t)sin(t)

)

, ||x(t)|| = t

In example(2.2.1) we also founds(t) = t2/2

Applying formula2.31 we obtain the parmeterization of Inv(Γ)⊂ R2, the involute of the Archimedean spiral,see figure(2.1).

Figure 2.1: involute of involute of circle i.e. involute of Archimedean Spiral

i(t)

=

(

x(t)y(t)

)

−

Z t

a

√

x2(t)+ y2(t)dτ√

x2(t)+ y2(t)

(

x(t)y(t)

)

=

(


)

− t2/2t

(

t cos(t)t sin(t)

)

= (1− t2

2)

(

cos(t)sin(t)

)

+ t

(

sin(t)−cos(t)

)

example. involute of the helixΓ ⊂ R3

Consider the parametrization of the helixΓ ⊂ R3

γ : [0,2π] ∋ t 7→ x(t) =

x(t)y(t)z(t)

=

acostasint

bt

∈ Γ


2.5. INVOLUTE 27

with velocity and speed

x(t) =

x(t)y(t)z(t)

=

−asintacost

b

, ||x(t)|| =√

x2(2)+ y2(t)+ z2(t) =√

a2 +b2

and arclength

s(t) =Z t

0||x(τ)||dτ =

Z t

0

√

a2 +b2dτ = t√

a2 +b2

Applying formula2.31 we obtain the parmeterization of Inv(Γ) ⊂ R2, the involute of the helix curve.

i(t)

=

xyz

−

Z t

a

√

x2 + y2+ z2

√

x2 + y2 + z2

xyz

=

acos(t)asin(t)

bt

− t√

a2+b2√

a2 +b2

−asin(t)acos(t)

b

=

acos(t)asin(t)

bt

− t

−asin(t)acos(t)

b

=

acos(t)− tasin(t)asin(t)+ tacos(t)

0

= a


0

The zero third coordinate means that this involute curve lies in the plane. We have discovered that the involutecurve of the 3-dimensional helix (seefigure(2.2)) is a 2 dimensional Archimedean spiral (usually created byunwinding thread from a spool of radiusa).

Consider a child swinging round a lamppost. The lampost is a cylinder, the rope under tension must assumea helical shape (wind a taut cord round a can of beans and you’ll get a perfect helix). The child’s feet traceout an everwidening Archimedean spiral, the child is safe ashe/she remains on the ground.

example. involute of the cycloidFrom example(2.1.1) we have

x(t) =

(

x(t)y(t)

)

=

(


)

,

(

x(t)y(t)

)

=

(

1−cos(t)sin(t)

)

, ||x(t)|| = 2sin(t/2)

From example(2.2.1) we haves(t) = 4−4cos(t/2)

HOWEVER it has been found that in the case of the cycloid it is better to ’peel’ starting from half way alongthe cycloid loop. So we subtract 4 (half the total arc length). This way we get a ’nice’ involute, if we proceedwith s(t) as usual we get a ’nasty’ involute. So take

s(t) = −4cos(t/2)

Applying formula(2.31) we obtain the parmeterization of Inv(Γ) ⊂ R2, the involute of the cycloid.

i(t)



Figure 2.2: Archimedean spiral as helix involute

=

(

x(t)y(t)

)

− s(t)√

x2(t)+ y2(t)

(

x(t)y(t)

)

=

(


)

− −4cos(t/2)

2sin(t/2)

(

1−cos(t)sin(t)

)

=

(


)

+2cos(t/2)

sin(t/2)

(

1−cos(t)sin(t)

)

=

(


)

+2cos(t/2)

sin(t/2)

(


)

=

(


)

+

(

4sin(t/2)cos(t/2)4cos2(t/2)

)

=

(


)

+

(

2sin(t)2+2cos(t)

)

=

(

t +sin(t)3+cos(t)

)

next we massage this into a nicer form

=

(

(t −π)+sin(t)1+cos(t)

)

+

(

π2

)

=

(

(t −π)−sin(t −π)1−cos(t −π)

)

+

(

π2

)

= x(t −π)+ πi +2j

We find that, see figure(2.3), theinvolute of the cycloid is itselfbut translated in space byπi +2j and with atime delayt = t −π.


2.6. EVOLUTE 29

Figure 2.3: involute of cycloid is the cycloid translated

2.6 evolute

In this section we will put ideas defined in§(2.3 and2.4) to work. Given a regular curveΓ ⊂ R2, using theoriginal parameterization, curvature and the unit normal vector, we will build a new curve denoted Env(Γ)and called theevoluteof Γ. The evolute is the curve traced out by the center of the osculating circle

Given a parametrizationγ : [a,b] ∋ t 7→ x(t) ∈ Γ ⊂ R2

of a regular curveΓ we will find a formula for parametrization

e : [a,b] ∋ t 7→ e(t) ∈ Env(Γ)

of the evolute curve Env(Γ).

e(t)

= x(t) +1

κ(t)n(t)

=

x(t) +

( ||x(t)||3det(x(t), x(t)

)

·(

1||x(t)||

(

−y(t)x(t)

) )

=

x(t) +||x(t)||2

det(x(t), x(t)

(

−y(t)x(t)

)

=(

x(t)y(t)

)

+x2(t)+ y2(t)

x(t)y(t)− y(t)x(t)

(

−y(t)x(t)

)

We summarize

e = x +nκ

=

(

xy

)

+x2 + y2

xy− yx

(

−yx

)



example. evolute of the circleFor the circleC of radiusr > 0 parameterized by

c : [0,2π] ∋ t 7→(

x(t)y(t)

)

=

(


)

∈C

The curvature is constant atκ = sgn(ω)/r and from example(2.4.1)

n = sgn(ω)

(


)

Thus

e(t)

= x(t) + n(t)/κ(t)

=

(


)

+r

sgn(ω)sgn(ω)

(


)

=

(


)

+ r

(


)

=

(

00

)

The evolute of the circle is a degenerate one point curve,Γ = {(

00

)

}, the single point is of course the

center ofC.

example. evolute of the Archimedean spiralΓFor the spiral parametrized by

γ : [0,6π] ∋ t 7→(

x(t)y(t)

)

=

(


)

∈ γ

from example(2.3.2)

κ(t) =1t

and from example(2.4.1)

n(t) =

(

−sin(t)cos(t)

)

Thus

e(t)

= x(t) + n(t)/κ(t)

=

(


)

+1

1/t

(

−sin(t)cos(t)

)

=

(


)

+

(

−t sin(t)t cos(t)

)

=

(

cos(t)sin(t)

)

But this is the parameterization of the unit circle. We summarize


2.6. EVOLUTE 31

The involute of the unit circle is the Archimedean Spiral.The evolute of the Archimedean spiral is the unit circle.Looking again, we see that figure(1.3) displays both phenomena.

example. evolute of the cycloidFrom example(2.1.1) we have

x(t) =

(

x(t)y(t)

)

=

(


)

,

(

x(t)y(t)

)

=

(

1−cos(t)sin(t)

)

, ||x(t)|| = 2sin(t/2)

From example(2.3.2) we have

κ(t) =−1

4sin(t/2)

and from example(2.4.1)

n(t) =

(

−cos(t/2)sin(t/2)

)

Thus

e(t)

= x(t) + n(t)/κ(t)

=

(


)

− 4sin(t/2)

(

−cos(t/2)sin(t/2)

)

=

(


)

+ 2

(

2sin(t/2)cos(t/2)

−2sin2(t/2)

)

=

(


)

+ 2

(

sin(t)cos(t)−1

)

=

(

t +sin(t)−1+cos(t)

)

we massage this into a nicer form

=

(

(t −π)+sin(t)1+cos(t)

)

+

(

π−2

)

=

(

(t −π)−sin(t −π)1−cos(t −π)

)

+

(

π−2

)

= x(t −π) +

(

π−2

)

We conclude thati(t) = x(t) except for trivial details. The evolute of the cycloid, see figure(2.4), is again thesame cycloid except displaced byπi −2j and with a time shiftt = t −π.

Both the evolute and the involute of the cycloid are copies (translates) of the original cycloid, see figure(2.4).



,

Figure 2.4:

(

(i) envolute of cycloid is the cycloid translated(ii) both inv/env-olute of the cycloidΓ are translates ofΓ

)

2.7 problem set

geometry of curves1. Let C⊂ R2 be the circle with center 2i + j , radius 5 and parameterization (with angular speed 3)

c : [0,2π] ∋ t 7→ c(t) =

(

21

)

+

(

5cos(3t)5sin(3t)

)

∈ R2

(i) Sketch the curve.

(ii) Write down the position vector.

(iii) Compute the velocity vector.

(iv) Compute the acceleration vector.

(v) Compute the speed scalar.

(vi) Compute the unit tangent vector.

(vii) Compute the unit normal vector.

(viii) Compute the arc length functions(t),0≤ t ≤ 2π.

(ix) Compute the curvature functionκ(t).

(x) Parameterize the involute curve (i.e. computei(t)).

(xi) Parameterize the evolute curve (i.e. computee(t)).

(xii) Sketch the involute curve.

(xiii) Sketch the evolute curve.

2. Let C⊂ R2 be the cycloid curve with parameterization.

c : [0,6π] ∋ t 7→ c(t) =

(

t −sint1−cos(t)

)

∈ R2


2.7. PROBLEM SET 33














3. [** hard question]LetC⊂ R2 be the cardoid curve obtained by rolling a circle of radius 1 around a circle of radius 1 andfollowing the initial point of contact. The parameterization is

c : [0,2π] ∋ t 7→ c(t) =

(

2cos(t)−cos(2t)2sin(t)−sin(2t)

)

∈ R2


(ii) Write down the position vector. Explain the parameterization.















Chapter 3

functions of two variables

In a first calculus course the nondegenerate stationary points of a functionf (x) of one variable are found andthen classified as either maximum or minimum, see§(3.1.1). In this chapter we carry out a similar programfor functions f (x,y) of two variables. We find that non degenerate stationary points are of three varieties,maximum, minimum or saddle. Saddle point do not occur for functions of one variable.

3.1 nondegenerate stationary points

3.1.1 single variable, a review

We review the theory of nondegenerate stationary (critical) points of a functionf (x) of one variable. Neara(a fixed point)f can be approximated to the second order by Taylor’s polynomial

f (x) ≈ f (a) + f ′(a) · (x−a) +12

f ′′(a) · (x−a)2 (3.1)

The pointa is stationary ifff ′(a)= 0. In the approximation (3.1) the termf (a) can be ignored being constant,the linear term is zero, leaving only the quadratic term. Thus neara f(x) behaves like

f ′′(a) · (x−a)2, nearx = a

i.e. likef ′(a) ·x2, nearx = 0

Next we consider only the casef ′′(a) 6= 0, i.e. only non-degenerate critical points. Then eitherf ′′(a) > 0 orf ′′(a) < 0. Neara f(x) behaves like±x2 nearx = 0.If f ′′(a) > 0 we have a minimum point.If f ′′(a) < 0 we have a maximum point.

3.2 several variables

Consider a function of 2 variables,

f : U ⊂ R3 → R

x =

(

xy

)

7→ f (x,y)(3.2)

HereU is a suitable subset ofR2 called thedomainof f . We say thatf is a scalar real valued function of

two real variables. Corresponding to approximation(3.1),near a fixed pointp =

(

ab

)

∈ R2, we have the

35

36 CHAPTER 3. FUNCTIONS OF TWO VARIABLES

second order approximation

f (x,y) (3.3)

≈ f (a,b) + D f (a,b)

(

x−ay−b

)

+12

(

x−a,y−b)

D2 f (a,b)

(

x−ay−b

)

(3.4)

= f (a,b) + ( fx , fy )

(

x−ay−b

)

+12

(x−a, y−b)

(

fxx fxy

fyx fyy

)(

x−ay−b

)

(3.5)

= f (a,b) (3.6)

+ (3.7)

fx · (x−a) + fy · (y−b) (3.8)

+ (3.9)12

[

fxx · (x−a)2 + 2 fxy · (x−a) · (y−b) + fyx · (y−b2)]

(3.10)

[all partial derivatives taken atp =

(

ab

)

].

At first glance (3.3) seems complicated. Let us tease out the details. The first orderdifferential matrixD f (a,b) of f at p is the 1×2 matrix or row vector made up of two first order partial derivatives of f

D f (a,b) = ( fx , fy )(a,b)

The second order differential matrix orHessian D2 f (a,b) of f atp is the 2×2 matrix made up of four secondorder partial derivatives off

D2 f (a,b) =

(

fxx fxy

fyx fyy

)

(a,b)

The only other terms appearing in (3.3) are the function value f (a,b) and the vector

(

x−ay−b

)

and its

transpose. The various matrices and vectors are multipliedout in the standard manner.

The pointp is said to be astationaryor critical point for f iff

D f (a,b) =(

0,0)

⇔ fx(a,b) = 0 = fy(a,b) (3.11)

The stationary pointp is said to benondegenerateiff

det(D f 2(a,b)) 6= 0 (3.12)

Assume then thatp is a nondegenerate stationary point forf . In the approximation(3.3) we can ignore themere constantf (a,b), the linear term involvingD f is zero. The behaviour off nearp is determined by theonly remaining term, the quadratic expression

12

(

x−a,y−b) 1

2D2 f

(

x−ay−b

)

, nearp =

(

ab

)

(3.13)

i.e. by the behaviour of

(

x,y) 1

2D2 f (a,b)

(

xy

)

= fxxx2 +2 fxyxy+ fyyy

2, near

(

xy

)

=

(

00

)

(3.14)

We will show that (after a change of variables) the matrixD2 f can be assumed diagonal, i.e. of the form

D2 f (a,b) =

(

A 00 B

)

and the local behaviour off is like that ofz= Ax2 +By2 nearx = 0 = y. Thus all nondegenerate stationarystationary points fall into one of only three categories.


3.2. SEVERAL VARIABLES 37

,

Figure 3.1: (i) local minimum point (ii) local maximum point

Figure 3.2: (i) saddle point



(i) Both a > 0 andb > 0.Rescalingx andy, locally f behaves likez = x2 + y2 nearx = 0 = y with bowl shaped graph, seefigure(3.1).

p =

(

ab

)

is aminimum point.

(ii) Both a < 0 andb < 0.Rescalingx andy, locally f behaves likez= −x2−y2 nearx= 0= y with mountain shaped graph, seefigure(3.1).

p =

(

ab

)

is amaximum point.

(iii) a > 0 butb < 0.Rescalingx andy, locally f behaves likez= x2 − y2 nearx = 0 = y with saddle shaped graph, see

figure(3.2).p =

(

ab

)

is asaddlepoint.

We are ready to do some examples.

3.2.1 examples

,

Figure 3.3:z= f (x,y) = x3 +y3−3x−3y (i) contour diagram (ii) graph

example. f (x,y) = x3 +y3−3x−3yLet f be the function

f : R2 ∋(

xy

)

7→ z= f (x,y) = x3 +y3−3x−3y∈ R

(i) Find all stationary (critical) points.

(ii) Find the corresponding stationary values.

(iii) Decide which stationary points are non-degenerate.

(iv) Classify each non-degenerate stationary point as local maximum, local minimum, or saddle.



We begin by computing two first order partial derivatives.

fx = 3x2−3 , fy = 3y2−3

Next we compute four second order partial derivatives

fxx = 6x fxy = 0fyx = 0 fyy = 6y

• finding stationary pointsWe must simultaneously solve two equations

fx = 0 , fy = 0

⇔ 3x2−3 = 0 , 3y2−3 = 0

We obtain four solutions(

xy

)

=

(

11

)

,

(

−1−1

)

,

(

1−1

)

,

(

−1−1

)

• computing stationary values

f (1,1) = −4 f (−1,−1) = 4f (1,−1) = 0 f (−1,1) = 0

• non-degeneracyRecall

D2( f )(a,b) =

(

fxx fxy

fyx fyy

)

=

(

6x 00 6y

)

Thus

det(D2 f (1,1)) = det

(

6 00 6

)

= 36 6= 0, non−degenerate

det(D2 f (−1,−1)) = det

(

−6 00 −6

)

= 36 6= 0, non−degenerate

det(D2 f (1,−1)) = det

(

6 00 −6

)

= −36 6= 0, non−degenerate

det(D2 f (−1,1)) = det

(

−6 00 6

)

= −36 6= 0, non−degenerate

All four critical points are non-degenerate

• classificationClassification is easy since all fourD2 f Hessian matrices are diagonal. Also, see figure(3.3). See thenext example for a more difficult classification with non diagonal Hessians.

Near a non degenerate critical point

(

xy

)

=

(

ab

)

the functionf behaves like(

x,y)

(

fxx fxy

fyx fyy

)

(a,b)

(

xy

)

near

(

xy

)

=

(

00

)

.

(i) Near

(

11

)

f behaves like 6(x2 +y2).

(

11

)

is a local minimum point

(ii) Near

(

−1−1

)

f behaves like 6(−x2−y2).

(

−11

)

is a local maximum point



(iii) Near

(

1−1

)

f behaves like 6(x2−y2).

(

−11

)

is a saddle point

(iv) Near

(

−11

)

f behaves like 6(−x2 +y2).

(

1−1

)

is a saddle point

example. g(x,y) = 2x3 +6xy2−6xLet g be the function

g : R2 ∋(

xy

)

7→ z= g(x,y) = 2x3 +6xy2−6x∈ R

(i) Find all stationary (critical) points.

(ii) Find the corresponding stationary values.

(iii) Decide which stationary points are non-degenerate.

(iv) Classify each non-degenerate stationary point as local maximum, local minimum, or saddle.

We begin by computing two first order partial derivatives.

gx = 6x2 +6y2−6 , gy = 12xy

Next we compute four second order partial derivatives

gxx = 12x gxy = 12ygyx = 12y gyy = 12x

• finding stationary pointsWe must simultaneously solve two equations

gx = 0 , gy = 0

⇔ 6(x2 +y2−1) = 0 , 12xy= 0

⇔ x2 +y2 = 1 , xy= 0

From the second equation, eitherx = 0 ory = 0. If x = 0 from the first equationy = ±1; if y = 0 fromthe first equationx = ±1. Thus we have four solutions, giving the following four stationary points.

(

xy

)

=

(

01

)

,

(

0−1

)

,

(

10

)

,

(

−10

)

• computing stationary valuesg(0,1) = 0 g(0,−1) = 0

g(1,0) = −4 g(−1,0) = 4

• non-degeneracyRecall

D2(g)(a,b) =

(

gxx gxy

gyx gyy

)

=

(

12x 12y12y 12x

)

Thusdet(D2g(x,y)) = (12x)(12x)− (12y)(12y) = 144(x2−y2)

anddet(D2g(0,1)) = 144(0−1) = −144 6= 0, non-degenerate

det(D2g(0,−1)) = 144(0−1) = −144 6= 0, non-degenerate

det(D2g(1,0)) = 144(1−0) = 144 6= 0, non-degenerate

det(D2g(−1,0)) = 144(1−0) = 144 6= 0, non-degenerate

All four stationary (critical) points are non-degenerate



• classificationFor two only of the stationary points classification is easy since the two Hessian matrices are diagonal.

(i) Near

(

10

)

g behaves like

(

x,y)

(

fxx fxy

fyx fyy

)

(a,b)

(

xy

)

=(

x,y)

(

12x 12y12y 12x

)

(1,0)

(

xy

)

=(

x,y)

(

12 00 12

)(

xy

)

= 12(x2 +y2)(

10

)

is a localminimum point

(ii) Near

(

−10

)

g behaves like

(

x,y)

(

12x 12y12y 12x

)

(1,0)

(

xy

)

=(

x,y)

(

−12 00 −12

)(

xy

)

= −12(x2+y2)

(

−10

)

is a localmaximum point

• Near

(

01

)

g behaves like

(

x,y)

(

12x 12y12y 12x

)

(1,0)

(

xy

)

=(

x,y)

(

0 1212 0

)(

xy

)

This is the first time we have met a non diagonal Hessian. The theory of §(3.2) is extended below,see§(3.3), to deal with a non diagonal Hessian. The reader might like to look ahead to§(3.3) to helpunderstand the next computation.

The characteristic polynomial ofD2g(0,1) is

det(λI −D2 f (0,1))

= det(λ(

1 00 1

)

− 0 1212 0

)

= det(λ 12

12 λ )

= λ2−144

The roots of the characteristic polynomial (i.e. the eigenvalues ofD2g(01)) are

λ = α = 12 , λ = β = −12

According to§(3.3),g behaves near

(

01

)

like

(

x,y)

(

α 00 β

)(

xy

)

=(

x,y)

(

12 00 −12

)(

xy

)

= 12(x2−y2)

(

01

)

is asaddlepoint



• Near

(

0−1

)

g behaves like

(

x,y)

(

12x 12y12y 12x

)

(1,0)

(

xy

)

=(

x,y)

(

0 −12−12 0

)(

xy

)

This is a non diagonal Hessian.

The characteristic polynomial ofD2g(0,−1) is

det(λI −D2 f (0,−1))

= det(λ(

1 00 1

)

−(

0 −12−12 0

)

)

= det

(

λ 1212 λ

)

= λ2−144

The roots of the characteristic polynomial (i.e. the eigenvalues ofD2g(0,−1)) are

λ = α = 12 , λ = β = −12

g behaves near

(

0−1

)

like 12(x2−y2).

(

0−1

)

is again asaddlepoint

3.3 the Hessian

The classificatios in§§(3.2.1) involved very simple diagonal Hessians. We now discuss classification involv-ing the general hessian. Happily the general Hessian can be made diagonal.

The Hessian 2× 2 matrix D2 f (a,b) if not diagonal can be made diagonal by a change of variable. Thediagonal form ofD2 f (a,b) is

(

α 00 β

)

whereα andβ are the eigenvalues of the 2×2 matrixD2 f (a,b). See a course in Linear Algebra.

The behaviour off near

(

xy

)

=

(

ab

)

is like that of

(

x, y)

(

α 00 β

)(

xy

)

= αx2 + βy2

near

(

xy

)

=

(

00

)

. There are 3 cases

• α > 0 andβ > 0 local minimum

• α < 0 andβ < 0 local maximum

• one ofα,β is positive, the other negative saddle

Another way to look at this is

• αβ > 0 andα+ β > 0 local minimum


3.3. THE HESSIAN 43

• αβ > 0 andα+ β < 0 local maximum

• αβ < 0 saddle

But

αβ = detD2 f (a,b) = fxx fyy− f 2xy and α+ β = trD2 f (a,b) = fxx+ fyy (3.15)

(3.15) is proven as follows

λ2− (α+ β)λ + αβ= (λ−α)(λ−β)

= det(λI −D2 f (a,b))

= det

(

λ− fxx fxy

fyx λ− fxx

)

= (λ− fxx)(λ− fyy)− ( fxy)2

= λ2− ( fxx+ fyy)λ +( fxx fyy− f 2xy)

= λ2− trD2 f (a,b)+detD2 f (a,b)

Equation(3.15) follows by comparing coefficients in the first and last lines of the latter.

At last we have a computationally undemanding fast method toclassify nondegenerate stationary points. Weemphasise that that this method applies to all Hessians especially those which are not diagonal. We use (3.15)to translate (3.3)

• fxx fyy > f 2xy and fxx+ fyy > 0 local minimum

• fxx fyy > f 2xy and fxx+ fyy < 0 local maximum

• fxx fyy < f 2xy saddle

3.3.1 example, closest approach of 2 lines in R3

PlaneA is travelling at constant velocityα from positionx = a at timet = 0;PlaneB is travelling at constant velocityβ from positionx = b at times= 0;Thus both flight paths are straight lines.

Given that

a =

125

, α =

111

, b =

032

, β =

101

(i) Find the shortest distance between these flight paths.

(ii) Find the points of closest approach on the two flight paths.

We can parameterize the path of planeA

c : R ∋ t 7→ c(t) = a+ tα ∈ R3

and the path of planeB

c : R ∋ s 7→ d(s) = b+sβ ∈ R3



We wish to minimize the distance squared||d(s)− c(t)||2. (the distance squared is computationally easierthan the distance). Our problem is to minimize the function

f (s, t)

= ||d(s)−c(t)||2

= ||(b−a)+ βs−αt||2

= ||

0−13−22−5

+ s

101

− t

111

||2

= ||

−1+s− t1− t

−3+s− t

||2

= (−1+s− t)2+(1− t)2+(−3+s− t)2

= 11−8s+6t +2s2−4st+3t2

We compute first order partials

fs = −8+4s−4t , ft = 6−4s+6t

To find stationary points we must solve the simultaneous equations fs = 0 and ft = 0 , i.e.

s− t = 2 and 2s−3t = 3

These are straightforward to solve, the (unique) solution is

s= 3 , t = 1

Thus we have one and only one stationary point

(

st

)

=

(

31

)

of the distance squared functionf (s,t).

Next we compute the second order partials of f

D2 f (s, t) =

(

fss fst

fts ftt

)

=

(

4 −4−4 6

)

The Hessian is constant with

detD2(3,1) = det

(

4 −4−4 6

)

= 8 > 0 , trD2 f = 4+6= 10> 0

The unique stationary point

(

st

)

=

(

31

)

is nondegenerate and is a minimum point off the distance

squared function. The shortest distance between the two flight paths is

√

f (3,1) =√

11−8(3)+6(1)+2(32)−4(3 ·1)+3(1)2 =√

11−24+6+18−12+3 =√

2

The point of closest approach on the flight path of aeroplaneA is

a+1 ·α =

125

+1

111

=

236

The point of closest approach on the flight path of aeroplaneB is

b+3 ·β =

032

+3

101

=

335


3.3. THE HESSIAN 45

3.3.2 example, elastic band and two rings

The two ends of an elatic band slide on two smooth circular rings with equations

(x+2)2+y2 = 1 , (x−2)2+y2 = 1

(i) Sketch the situation.

(ii) Find the equilibrium positions of the elastic band.

(iii) For each such position determine whether it is stable,unstable or metastable.

First we explain the jargon (used in physics). The energy locked up in the band is proportional to its lengthsquared. Equilibrium means a stationary point of the energy. Stability means an energy minimum point.Unstability means an energy maximum point. Metastable means an energy saddle point.

We parameterize the first ring.

c : R ∋ t 7→ c(t) =

(

−2+cos(t)sin(t)

)

∈ R2

We parameterize the second ring.

d : R ∋ s 7→ d(s) =

(

2+cos(s)sin(s)

)

∈ R2

The energy or length-squared function is

f (s, t)

= ||d(s)−c(t||2

= ||(

2+cosssins

)

−(

−2+costsint

)

||

= ||(

4+coss−costsins−sint

)

||

= (4+coss−cost)2 +(sins−sint)2

= (16+cos2s+cos2 t +8coss−8cost −2cosscost)+ (sin2s+sin2 t −2sinssint)

= 16+(cos2s+sin2s)+ (cos2 t +sin2 t)−2(cosscost −sinssint)+8(coss−cost)

= 16+1+1−2cos(s− t)+8(coss−cos(t))

= 18−2cos(s− t)+8(coss−cos(t))

We compute the first order partial derivatives

fs = 2sin(s− t)−8sin(s) , ft = −2sin(s− t)+8sin(t)

We must simultaneously solve the equations

fs = 0 , ft = 0

⇔ sin(s− t) = 4sin(s) , sin(s− t) = 4sin(t)

Thus sin(s) = sin(t) and eithers= t or s= π− t.In the cases= t we have

sin(s) = sin(t) = sin(s− t) = sin(0) = 0

In the cases= π− t we have

4sin(t) = sin(π− t− t) = sin(π−2t) = sin(2t) = 2sin(t)cos(t)



and thussin(t)[2−cos(t)] = 0

Since cos(t) = 2 is impossible this leaves us with 0= sin(t) = sin(s).Thus in all cases sin(s) = sin(t) = 0. There are four stationary points.

(

st

)

=

(

00

)

,

(

ππ

)

,

(

0π

)

,

(

π0

)

Next we compute the second partial derivatives

D2 f =

(

fss fst

fts ftt

)

=

(

2cos(s− t)−8cos(s) −2cos(s− t)−2cos(s− t) 2cos(s− t)+8cos(t)

)

For the four stationary points the HessiansD2 f are in order(

−6 −2−2 10

)

,

(

10 −2−2 −6

)

,

(

−10 22 −10

)

,

(

6 22 6

)

The corresponding determinants are

−64 , −64 , 96 , 32 ,

showing that all four stationary points are nondegenerate.The corresponding traces are

4 , 4 , −20 , 12 ,

We conclude that these stationary points are in ordersaddle, saddle, local maximum, local minimum

In terms of physics these equilibrium positions are respectively metastable, metastable, unstable, stableequilibriae of the stretched elastic band. A simple sketch shows that all this corresponds to common sense,sketch done in class.

3.3.3 example, critical approach of line and circle

Let f (s,t) be the distance squared function between the general point

c(s) =

(

x(s)y(s)

)

=

(

5cos(s)5sin(s)

)

on the circleC with center

(

xy

)

=

(

00

)

and radius 5. and the general point

d(t) =

(

x(t)y(t)

)

=

(

5−3t10+4t

)

on the lineL through the point

(

510

)

in the direction

(

−34

)

.

(i) Computef (s, t).

(ii) Compute the first order partial derivativesfs and ft

(iii) Find all stationary points.

(iv) Compute all four second order partial derivatives.

(v) ComputeD2 f at each stationary point.

(vi) Which stationary points are degenerate, which nondegenerate?


3.3. THE HESSIAN 47

(vii) Classify nondegenerate saddle points as loc.max., loc.min. or saddle.

(viii) For each stationary point

(

st

)

show the line segment (of critical length)c(s)d(t) in a sketch of

C andL in R2.

computation of f (s, t)

f (s, t)

= < d(t)−c(s),d(t)−c(s) >

= <

(

5−3t−5cos(s)10+4t−5sin(s)

)

,

(

5−3t−5cos(s)10+4t−5sin(s)

)

>

= (5−3t−5cos(s))2 + (10+4t−5sin(s))2

first order partial derivatives

fs = 10(

5−3t−5cos(s))

sin(s) − 10(

10+4t−5sin(s))

cos(s)

ft = −6(

5−3t−5cos(s))

+ 8(

10+4t−5sin(s))

stationary pointsWe must solve two simultaneous equations

fs = 0 , ft = 0

⇔(

5−3t−5cos(s))

sin(s) =(

10+4t−5sin(s))

cos(s)

and

6(

5−3t−5cos(s))

= 8(

10+4t−5sin(s))

’Dividing’ the second equation by the first we get

sin(s)6

=cos(s)

8⇔ tan(s) =

34

Plugging sin(s) = 34 cos(s) into the ft = 0 equation we get

6(

5−3t−5cos(s))

= 8(

10+4t−5(3/4)cos(s))

⇔ 24(

5−3t−5cos(s))

= 8(

40+16t−15cos(s))

⇔ 3(

5−3t−5cos(s))

=(

40+16t−15cos(s))

⇔ 15−9t−15cos(s) = 40+16t−15cos(s)

⇔ 25+25t = 0

⇔ t = −1

We sum up. If

(

st

)

is a stationary point then tan(s) = 3/4 andt =−1. This leads to two stationary points.

(

s1

t1

)

=

(

arctan(3/4)−1

)

,

(

s2

t2

)

=

(

π +arctan(3/4)−1

)

This is more conveniently expressed

cos(s1) =45

, sin(s1) =35

, t1 = −1

cos(s2) = −45

, sin(s2) = −35

, t2 = −1



second order partial derivativesThree out of four second order partial derivatives can computed without much work. Since

ft = −6(

5−3t−5cos(s))

+ 8(

10+4t−5sin(s))

we find thatftt = (−6)(−3)+ (8)(4) = 50

and thatfst = fts = −30sin(s))−40cos(s)

We obtain the last p.d. by a slog differentiation w.r.t.s from

fs = 10(

5−3t−5cos(s))

sin(s) − 10(

10+4t−5sin(s))

cos(s)

fss

= 10(

5−3t−5cos(s))

cos(s)+10(5sin(s))sin(s)

+

10(

10+4t−5sin(s))

sin(s) + 10(5cos(s))cos(s)

= (50cos(s)−30t cos(s)−50cos2(s)) + (100sin(s)+40t sin(s)−50sin2(3)) + 50(cos2(s)+sin2(s))

= (50cos(s)−30t cos(s)) + (100sin(s)+40t sin(s))

We summarize

D2 f (s, t)

=

(

fss fst

fts ftt

)

=

(

(50cos(s)−30t cos(s)) + (100sin(s)+40t sin(s)) −30sin(s))−40cos(s)−30sin(s))−40cos(s) 50

)

D2 f at stationary points

D2 f (s1, t1)

= D2 f (arctan(3/4),−1)

=

(

(50(4/5)−30(−1)(4/5)) + (100(3/5)+40(−1)(3/5)) −30(3/5)−40(4/5)−30(3/5)−40(4/5) 50

)

=

(

100 −50−50 50

)

D2 f (s2, t2)

= D2 f (arctan(3/4)+ π,−1)

=

(

(50(−4/5)−30(−1)(−4/5)) + (100(−3/5)+40(−1)(−3/5)) −30(−3/5)−40(−4/5)−30(−3/5)−40(−4/5) 50

)

=

(

−100 5050 50

)

nondegeneracy

det(D2 f (s1, t1)) = det

(

100 −50−50 50

)

= 2500 6= 0


3.3. THE HESSIAN 49

and

det(D2 f (s2,t2)) = det

(

−100 5050 50

)

= −7500 6= 0

Both stationary points are non-degenerate.

classification of stationary pointsdet(D2 f (s1, t1)) = 2500> 0 and tr(D2 f (s1,t1)) = −100< 0, local minimum pointdet(D2 f (s2, t2)) = −7500< 0 saddle point

sketch

c(s1) = 5

(

cos(s1)sin(s1)

)

= 5

(

4/53/5

)

=

(

43

)

c(s2) = 5

(

cos(s2)sin(s2)

)

= 5

(

−4/5−3/5

)

= −(

43

)

d(t1) = d(t2) =

(

5−3(−1)10+4(−1)

)

=

(

86

)

The sketch showing the circleC and the lineL in R2 and showing line segments

(

43

)(

86

)

and

(

−4−3

)(

86

)

was drawn in class.




Part I

Answers to problem sets of part I

51

Chapter 4

Answers to questions in Chapter1

Figure 4.1: contours (i) wherex2−y2 = 0 (ii) wherex2−y2 = 1

4.1 question and answer

Let f be the function

f : R2 ∋(

xy

)

7→ f (x,y) = x2−y2 ∈ R

(i) Sketch the contoursf−1(−1), f−1(0), f−1(1).Each is a curve in the plane.

(ii) Now you have the general idea. Roughly sketch thecontoursf−1(d) ford = −16,−9,−4,−1,0,1,4,9,16.

(iii) Draw the graphGr( f ) ⊂ R3.

(iv) Put a common sense name on the graph.

[This question is very instructive but rated quite difficult.Start by looking at the easier case off (x,y) = x2 +y2 astreated in the notes.]

4.1.1 answerThe contour

f−1(0)

= {x | f (x,y) = 0 }=

{

x∣

∣

∣x2−y2 = 0}

= {x |(x+y)(x−y) = 0 }= {x |x+y = 0 or x−y = 0 }= {x |y = −x }

[

{x |y = x }

This contour is the union of 2 straight lines, seefigure(4.1).

The contour

f−1(1)

= {x | f (x,y) = 1 }

53

54 CHAPTER 4. ANSWERS TO QUESTIONS IN CHAPTER1

Figure 4.2: contour (i) contour wherex2−y2 = −1 (i) contour diagram off (x,y) = x2−y2

Figure 4.3: saddle, graph off (x,y) = x2−y2


4.1. QUESTION AND ANSWER 55

Figure 4.4: cylinder and screw surfaces

Figure 4.5: helix curve formed as cylinder cuts screw



1.0

1.0

0.75

0.5

0.5

0.25

0.0

0 1 0.02 3 -0.54 5 6-1.0-1.0

-0.50.0

0.51.07.07.58.08.59.0

0.0

0.25

0.5

0.75

1.0

Figure 4.6: (i) label with diagonal , (ii) soup can with helix

={

x∣

∣

∣x2−y2 = 1}

This contour is an hyperbola which consists of twodisjoint parts with explicit equations

x = ±√

1+y2

see figure(4.1).

The contour

f−1(−1)

= {x | f (x,y) = −1 }=

{

x∣

∣

∣x2−y2 = −1}

This contour is a different hyperbola which again consistsof two disjoint parts with explicit equations

y = ±√

1+x2

see figure(4.2).

4.1.2 answer

See figure(4.2)

4.1.3 answer

See figure(4.3);

4.1.4 answer

The graph is called thesaddle. Indeed it is the archetypalsaddle. Saddles play an important role in chapter(3) of thiscourse.


Let Γ ⊂ R3 be the helix curve with parameterization

c : [0,2π]∋ t 7→ x(t) =

x(t)y(t)z(t)

=

acos(t)asin(t)

b(t)

∈Γ⊂R3

(i) Sketch the cylindrical surfaceC ⊂ R3 with implicitequationx2 +y2 = a2.

(ii) Sketch the screw surfaceS⊂ R3 with explicitequationz= bθ = barctan(y/x).

(iii) In a sketch showΓ = C∩S.

(iv) ProveΓ = C∩Sby showing that the parametricequation ofΓ satisfies both the equation ofC andthe equation ofS.

(v) Take a can of Heinz baked beans (a cylindricalsurface). Remove the label in one piece (arectangle). Emulating the genius of Andy Warholmake a single well thought out stroke on this label.Put the label back on. If you have the correctinsight your stroke should become the helix. Bymeasuring the length of the stroke on the label youwilll have the total arc length of the helix.

4.2.1 answer

See figure(4.4);

4.2.2 answer

See figure(4.4);



4.2.3 answerSee figure(4.5);

4.2.4 answerFirst

x2(t)+y2(t)

= a2 cos2(t)+a2 sin2(t)

= a2(cos2(t)+sin2(t))

= a2 · (1)

= a2

The parametric equation of the helix curve satisfies theexplicit equationx2 +y2 = a2 of the cylinder surface.

Second

b arctany(t)x(t)

= b arctanasin(t)acos(t)

= b arctansin(t)cos(t)

= b arctan(tan(t))

= bt

= z

The parametric equation of the helix curve satisfies theexplicit equationz= barctan(y/x) of the screw surface. ofthe cylinder surface.

4.2.5 answer

Take the label off the can. The label is a rectangle withhorizontal side of length 2πa and vertical side of length2πb. Draw the diagonal which, by Pythagoras’ theoremhas length 2π

√a2 +b2. Stick the label back on the can.

The diagonal becomes the helix and the length of the helixis 2π

√a2 +b2. See figure(4.6)(i) and figure(4.6(ii).

We have found a slick way to draw a helix on a cylinder.But there is an even slicker method. Wind a cord roundthe soup can. It forms a helix when pulled taut. If you usea carpenter’s chalk line, pluck it and leave a helix drawnon the can (usually in red chalk).




Chapter 5

Answers to questions in Chapter2

K3 K2 K1 0 1

K2

K1

1

2

Figure 5.1: (i) circle as in Ch02:q01 , (ii) cardioid


Let C⊂ R2 be the circle with center 2i + j , radius 5 andparameterization (with angular speed 3)

c : [0,2π] ∋ t 7→ c(t) =

(

21

)

+

(

5cos(3t)5sin(3t)

)

∈ R2














5.1.1 answerSee figure(5.1).

5.1.2 answerposition

c(t) =

(

x(t)y(t)

)

=

(

2+5cos(3t)1+5sin(3t)

)

59


K4 K2 0 2 4 6 8

K6

K4

K2

2

4

6

Figure 5.2: cardioid with involute (big) and evolute (small)

5.1.3 answervelocity

c(t) =

(

x(t)y(t)

)

=

(

−15sin(3t)15cos(3t)

)

5.1.4 answeracceleration

c(t) =

(

x(t)y(t)

)

=

(

−45cos(3t)−45sin(3t)

)

5.1.5 answerspeed (scalar)

||c(t)||

=√

x2(t) + y2(t)

=

√

225sin2(t)+225cos2(t)

=√

225

= 15

5.1.6 answerunit tangent

t(t) =c(t)

||c(t)|| =115

(

−15sin(3t)15cos(3t)

)

=

(

−sin(3t)cos(3t)

)

5.1.7 answerunit normal

n(t)

= Sπ/2t(t)

=

(

0 −11 0

)(

−sin(3t)cos(3t)

)

=

(

−cos(3t)−sin(3t)

)

5.1.8 answer

arc-length

s(t)

=Z t

0||c(τ)||dτ

=Z t

015dτ

= 15t

5.1.9 answer

This circle has radiusr = 5 and is positively traversed, i.e.

A.C.W.. The curvature of this circle is thusκ = +1r

=15

.

Let us also findκ using the general formula for curvatureof a curve at a point thereon.

κ(t)

=xy− yx

(x2 + y2)3/2

=(−15sin(3t))(−45sin(3t))− (15cos(3t))(−45cos(3t))

(225)3/2

=(15)(45)(cos2(3t)+sin2(3t))

153

=3(15)2(1)

153



=3(15)2(1)

153

=15

5.1.10 answerinvolute

i

c(t) − s(t)t(t)

=

(


)

− (15t)

(

−sin(3t)cos(3t)

)

=

(

2+5cos(3t)+15sin(3t)1+5sin(3t)−15cos(3t)

)

5.1.11 answerevolute

e(t)

= c(t) +n(t)κ(t)

=

(


)

+1

1/5

(

−cos(3t)−sin(3t)

)

=

(


)

+

(

−5cos(3t)−5sin(3t)

)

=

(

21

)

As expected, the evolute curve is a degenerate one pointcurve, located at the center of the original circle.

5.2 question and answerLet C⊂ R2 be the cycloid curve with parameterization.

c : [0,6π] ∋ t 7→ c(t) =

(

t −sint1−cos(t)

)

∈ R2














5.2.1 answer

See figure(1.4)

5.2.2 answer

position

c(t) =

(

t1

)

+S(−t)

(

0−1

)

=

(


)

5.2.3 answer

velocity

c(t) =

(

x(t)y(t)

)

=

(

1−cos(t)sin(t)

)

5.2.4 answer

acceleration

c(t) =

(

x(t)y(t)

)

=

(

sin(t)cos(t)

)

5.2.5 answer

speed

||c(t)||

=√

x2(t)+ y2(t)

=

√

(1+cos(t))2 +sin2(t))

=√

2+2cos(t)

=

√

4sin2(t/2)

= 2sin(t/2)

5.2.6 answer

unit tangent

t

=1

||c(t)|| c(t)

=1

2sin(t/2)

(

1−cos(t)sin(t)

)

=1

2sin(t/2)

(


)

=

(

sin(t/2)cos(t/2)

)

5.2.7 answer

unit normal

n(t)

= Sπ/2t(t)



=

(

0 −11 0

)(

sin(t/2)cos(t/2)

)

=

(

−cos(t/2)sin(t/2)

)

5.2.8 answer

arc-length

s(t)

=

Z t

0||c′(p)||dp

=Z t

02sin(p/2)dp

= −4cos(p2

)∣

∣

∣

t

0

= 4−4cos(t/2)

5.2.9 answer

We wish to computeκ(t) the curvature. First we need

det(c, c)

= det

(

1−cost sintsint cost

)

= cos(t)−1

= −2sin2(t/2)

Now we can compute the curvature

κ(t)

=detc, c||c(t)||3

=−2sin2(t/2)

8sin3 t/2

=−1

4sin(t/2)

5.2.10 answer

We wish to parametrize the involute curve. The cycloidhas irregular points atc(0) andc(2π). To be safe wemeasure arc length from the regular (half way) pointc(π).We suitably adjusts(t) = 4−4cos(t/2) to bes(t) = −4cos(t/2).

i(t)

= c(t)−s(t)tt(t)

=

(


)

+4cos(t/2)

2sin(t/2)

(

1−cos(t)sin(t)

)

=

(


)

+4cos(t/2)

2sin(t/2)

(


)

=

(


)

+2cos(t/2)

(

2sin(t/2)2cos(t/2)

)

=

(


)

+2

(

2sin(t/2)cos(t/2)2cos2(t/2)

)

=

(


)

+2

(

sin(t)1+cos(t)

)

=

(

t +sin(t)3+cos(t)

)


e(t)

= c(t)+n(t)κ(t)

=

(


)

+ −4sin(t/2)1

2sin(t/2)

(

−sin(t)1−cos(t)

)

=

(


)

+ 2

(

sin(t)cos(t)−1

)

=

(

t +sin(t)cos(t)−1

)

5.2.12 answerWe will present the cycloid, its involute and its evolutetogether in one drawing, see figure2.4

We begin by displaying clearly the three parametrizations

c(t) =

(


)

e(t) =

(

t +sin(t)cos(t)−1

)

i(t) =

(

t +sin(t)3+cos(t)

)

It seems likely that bothe andi are variants onc i.e. thatbothEnv(Γ) andInv(Γ) are cycloids: but the details aretricky.

Note that

c(t +π)

=

(

t +π−sin(t +π)1−cos(t +π)

)

=

(

π2

)

+

(

t +sin(t)cos(t)−1

)

= e(t) +

(

π2

)

i.e.

e(t) = c(t +π) −(

π2

)

Conclusion, the evoluteEnv(Γ) is the original cycloidtranslated by the vectorπi +2j ; the origin of time (theparametert) here was shifted byπ.

Argue similarly for the involute curveInv(Γ), check that

i(t) = c(t +π) +

(

−π2

)



Animated drawings of all three curves can be found on ourhome page athttp://mathsci.ucd.ie/courses/mst30070B

5.3 question and answer[** hard question]Let C⊂ R2 be the cardoid curve obtained by rolling acircle of radius 1 around a circle of radius 1 and followingthe initial point of contact. The parameterization is

c : [0,2π] ∋ t 7→ c(t) =

(


)

∈ R2


(ii) Write down the position vector. Explain theparameterization.












5.3.1 answer

See figure(5.1).

5.3.2 answer

position

c(t) =

(

x(t)y(t)

)

=

(


)

5.3.3 answer

velocity

c(t)

=

(

x(t)y(t)

)

=

(

−2sin(t)+2sin(2t)2cos(t)−2cos(2t)

)

= 4sin(t/2)

(

cos(3t/2)sin(3t/2)

)

5.3.4 answeracceleration

c(t) =

(

x(t)y(t)

)

=

(

−2cos(t)+4cos(2t)−2sin(t)+4sin(2t)

)

5.3.5 answerspeed scalar

||c(t)||

=

√

x2 + y2

=

√

(4sin2(t/2))(cos2(3t/2)+sin2(3t/2))

=

√

(4sin2(t/2))(1)

= 4sin(t/2)

5.3.6 answerunit tangent

t(t)

=1

||c(t)|| c(t)

=1

4sin(t/2)4sin(t/2)

(

cos(3t/2)sin(3t/2)

)

=

(

cos(3t/2)sin(3t/2)

)

5.3.7 answerunit normal

n(t)

=

(

0 −11 0

)

t(t)c(t)

=

(

0 −11 0

)

t(t)(

cos(3t/2)sin(3t/2)

)

=

(

−sin(3t/2)cos(3t/2)

)

5.3.8 answerarc-length.When we compute arc length we do not start, as is usual,

at c(0) =

(

10

)

but half way around the cardoid i.e we

start atc(π) =

(

−30

)

. This will later yield a nice

cycloid curve. The usual measure of arc length yields amessy meaningless involute.

s(t)

=Z t

π||c′(p)||dp,

note the first limit of inegration isπ not 0



=

Z t

π4sin(p/2)dp

= −8cos(p2

)∣

∣

∣

t

π= −8cos(t/2)

5.3.9 answer

curvature

κ(t)

=xy − yx

(x2, y2)3/2

∣

∣

∣

∣

t

=

(−2sin(t)+2sin(2t))(−2sin(t)+4sin(2t))−

(2cos(t)−2cos(2t))(−2cos(t)+4cos(2t))

(4sin(t/2))3

=

4

(−sin(t)+sin(2t))(−sin(t)+2sin(2t))−

(cos(t)−cos(2t))(−cos(t)+2cos(2t))

64sin3(t/2)

=

4

sin2 t −3sint sin(2t)+2sin2(2t)−

−cos2 t +3cost cos(2t)−2cos2(2t)

64sin3(t/2)

=

4

sin2 t +cos2 t−

3(cos(2t)cost +sin(2t)sint)+

2sin2(2t)+2cos2(2t)

64sin3(t/2)

=

4

1−

3cos(2t − t)+2

64sin3(t/2)

=4·3(1−cos(t))

64sin3(t/2)

=

12(2sin2(t/2))

64sin3(t/2)

=

24sin2(t/2)

64sin3(t/2)

=3

8sin(t/2)

5.3.10 answerinvolute

i(t)

= c(t)−s(t)t(t)

=

(


)

− (−8cos(t/2)

(

cos(3t/2)sin(3t/2)

)

=

(


)

+ 4

(

2cos(3t/2)cos(t/2)2sin(3t/2)cos(t/2)

)

=

(


)

+ 4

(

cos(2t)+cos(t)sin(2t)+sin(t)

)

= 3

(

2cos(t)+cos(2t)2sin(t)+sin(2t)

)

Comparingi(t) to the originalc(t) we see that the involuteis some kind(?) of cardioid scaled up by a factor 3. In factwe can manipulate the details and get

i(t)

= −3

(

2cos(t −π)−cos(2(t −π))2sin(t −π)−sin(2(t −π)

)

= −3c(t −π)

Thus the involute is the original cardioid but

• scaled by 3

• and reflected in the origin (to explain the ’−’)

• and time delayed byπ


e(t)

= c(t)+n(t)κ(t)

=

(


)

+1

3/8sin(t/2)

(


)

=

(


)

+8sin(t/2)

3

(


)

=

(


)

+43

(

−2sin(3t/2)sin(t/2)2cos(3t/2)sin(t/2)

)

=

(


)

+43

(

cos(2t)−cos(t)sin(2t)−sin(t)

)

=13

(

6cos(t)−3cos(2t)6sin(t)−3sin(2t)

)

+13

(

4cos(2t)−4cos(t)4sin(2t)−4sin(t)

)



=13

(

2cos(t)+cos(2t)2sin(t)+sin(2t)

)

We see thate(t) = i(t)/9 = −1/3c(t −π). The descriptionof the evolute is similar to that of the involute above butscaled,not by 3, but by 1/3.

5.3.12 answer

sketch the involuteSee figure(5.2)

5.3.13 answer

sketch the evoluteSee figure(5.2)


Documents

Mathematics, Course MST30070B Mathematics Introduction ...2.2 Archimedean spiral as helix involute . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 28 2.3 involute of cycloid