Homework sheet Exercises done during the lecture of March 12, 2014

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1 EXERCISE SESSION 2A FOR THE COURSE GÉOMÉTRIE EUCLIDIENNE, NON EUCLIDIENNE ET PROJECTIVE MATTEO TOMMASINI Homework shee Exercises done during he lecure of March 2, 204 Exercise 2 Is i rue ha he parameerized curve γ() := ( 2, 4 ) in R 2 describes he parabola y = x 2? Moivae your answer. I his parameerized curve regular? Le us fix any inverval I of he real line and any curve γ : I R 2. Moreover, le us fix any se S in R 2. To say ha γ describes he se S means ha where Im(γ) = S, Im(γ) := {γ() s.. I}. So in our case we have o prove or disprove ha we have an ideniy of he following 2 ses in R 2 : (0.) {( 2, 4 ) s.. R}? = {(x, y) R 2 s.. y = x 2 }. (Noe: we wrie? = all he imes ha we would like o prove somehing, bu we have no he proof ye (or we don know if he saemen is rue or no). This serves you as a reminder: you CANNOT use any equaliy wih a sign? = in order o prove somehing in he nex lines. Acually, wha we wan o prove is exacly ha? = is = or o prove ha? = is and we canno use i as an hypohesis!) Firs of all, le us ake any poin ( 2, 4 ) R 2 (wih coordinaes (x, y)) for some R. Then i saisfies y = x 2 since 4 = ( 2 ) 2. So we have proved ha (0.2) {( 2, 4 ) s.. R} {(x, y) R 2 s.. y = x 2 }. Conversely, le us fix any poin (x, y) R 2, such ha y = x 2. If x is non-negaive, we se := x. Then we have:

2 2 MATTEO TOMMASINI ( 2, 4 ) = (( x) 2, ( x) 4 ) = (x, x 2 ) = (x, y). So if x is non-negaive, hen he poin (x, y) can be obained as a poin of he form γ(). HOWEVER, WHAT DOES IT HAPPEN IF x is negaive? In his case, we would like o find a poin R, such ha ( 2, 4 )? = (x, y). Since is a real number, hen 2 is always non-negaive. Therefore, here is no such. For example, he poin (, ) belongs o he se on he righ of (0.2) bu no o he se on he lef of (0.2). So he inclusion opposie o (0.2) does NOT hold. Therefore he 2 ses in (0.) are differen. To be more precise: he parabola y = x 2 consiss of 2 pars: he righ par (he one for x 0) and he lef par ) (he one for x < 0); he firs par coincides wih Im(γ) (and no any poin of he second par belongs o Im(γ)). So he answer o he firs quesion is NO. The second quesion of he Exercise asks if he curve γ is regular. In order o answer o his quesion, we recall he definiion of angen vecor of γ for a given poin R. This is he vecor ( dγ (s) γ() := ds, dγ ) 2(s) ds R 2. In our case, γ (s) = s 2 and γ 2 (s) = s 4, so for any R we have So for = 0 we have γ() = (2, 4 3 ) R 2. γ(0) = (0, 0) R 2. Therefore, he curve γ is NOT regular (we recall ha regular means ha for every poin (in he domain of definiion of γ) we have γ() 0). Roughly speaking, he idea is ha for = 0 he curve has an inversion poin, i.e. for < 0 he poin γ() is on he righ par of he parabola and approaches he poin (0, 0) when goes o 0. Then insead of coninuing on he lef side of he parabola, for > 0 he poin γ() reurns on he righ par of he parabola. Exercise 3 Calculae he angen vecor o he curve γ() := (cos, sin ) a = π. Since γ() = (cos, sin ), hen for each poin R we have

3 EXERCISE SESSION 2A 3 ( d(cos s) γ() := ds, In paricular, for = π we have: d(sin s) ds ) = ( sin, cos ). γ(π) = ( sin π, cos π) = (0, ). Exercise 4 Show ha he following curves in R 3 are uni-speed: ( (i) γ() := ( + 3 )3/2, ( 3 )3/2, 2 ); (ii) γ() := ( 4 cos, sin, 3 cos ). 5 5 Le us sar wih (i): firs of all, he naural domain of definiion of γ is he inerval I := [, ] = { s.. } (if we ask γ o be smooh, hen he domain mus be equal o ], [). Indeed, for < he quaniy ( + ) 3/2 is no defined; for >, he quaniy ( ) 3/2 is no defined. So γ : I R 3. For every I we have o compue he angen vecor of γ a. This is he vecor in R 3 given as follows: = ( dγ (s) γ() := ds, dγ 2(s) ds, dγ ) 3(s) ds = ( ( + )/2, ( )/2, Therefore, for every I we have 2 ) = ( 2 ( + )/2, 2 ( )/2, γ() = 4 ( + ) + 4 ( ) + 2 = =. So he curve γ in (i) is a uni-speed curve. ). 2 Le us consider now he curve γ in (ii). In his case, he curve is defined for every R. For every such, we have: ( dγ (s) γ() := ds, dγ 2(s) ds, dγ ) 3(s) ds = ( 45 sin, cos, 35 ) sin. Therefore, for every R we have:

4 4 MATTEO TOMMASINI 6 γ() = 25 (sin )2 + (cos ) (sin )2 = (sin ) 2 + (cos ) 2 = =. So also he curve γ in (ii) is a uni-speed curve. Find he lengh of he following curve in R 3 : Exercise 5 γ : [, ] ( R 3 ( + 3 )3/2, ( 3 )3/2, In Exercise 4 we have already proved ha γ is a uni-speed curve, i.e. we have γ() = for all I = [, ]. Therefore, he lengh of he curve γ is given as follows: S(γ) = γ() d = Find he lengh of he following curve in R 2 : Exercise 6 d = 2. γ : [0, ] ( R 3 ), 2 2, /2. For each [0, ] we have: ( dγ (s) γ() := ds, dγ 2(s) ds, dγ ) 3(s) ds = Therefore, for each [0, ] we have: 2 ) (,, 2 /2 ). γ() = = ( + ) 2 = +. Therefore, he lengh of he curve γ is given as follows: S(γ) = 0 γ() d = 0 + d = Since a primiive for + is given by + 2 /2, we have 0. ( + )d. S(γ) = ( + 2 /2) = ( + 2 /2) =0 = + 2 = 3 2.

5 EXERCISE SESSION 2A 5 Exercise 8 Find a parameric curve γ : R R 2 whose image coincides wih he subse y 3 x 2 = 0 of R 2. y 3 = x 2 We have o find γ() such ha Firs of all, we define Im(γ) = {(x, y) R 2 s.. y 3 = x 2 }. γ() := ( 3, 2 ) for all R. Clearly (γ ()) 2 = 6 = (γ 2 ()) 3, so we have Im(γ) {(x, y) R 2 s.. y 3 = x 2 }. In order o conclude, we need also o prove ha he opposie inclusion holds. So le us fix any pair (x, y) such ha y 3 = x 2. Since x 2 is non-negaive, hen y 3 is non-negaive, hence y is non-negaive. So i makes sense o consider ( y) 3 = y 3 R. Since y 3 = x 2, we have: and So we have: x = y 3 = ( y) 3 if x 0 x = y 3 = y 3 = ( y) 3 if x < 0.

6 6 MATTEO TOMMASINI (x, y) = (( y) 3, ( y) 2 ) = γ( y) if x 0 and (x, y) = (( y) 3, ( y) 2 ) = γ( y) if x < 0. So his proves ha every poin (x, y) such ha y 3 = x 2 is of he form γ() for some poin R. So we have proved ha γ is such ha Im(γ) = {(x, y) R 2 s.. y 3 = x 2 }. Exercise 9 Find a parameric curve γ : R R 2 whose image lies in he se (0.3) {(x, y) R 2 s.. y m x n = 0}, where m and n are arbirary naural numbers. Taking ino accoun he previous Exercise above, we define γ() := ( m, n ) for all R. Since ( n ) m = ( m ) n, we ge ha γ() has values in he se (0.3). Moreover, γ() is smooh, so we are done. NOTE: in general he image of γ does no cover he whole se (0.3). To be more precise, le us consider he following cases: Case m odd and n even. The se is symmeric wih respec o he y axis (i.e. by replacing x wih x):

7 EXERCISE SESSION 2A 7 y m = x n for m odd and n even In his case he image of γ covers all he se (0.3). Indeed, le us fix any poin (x, y) in (0.3). Since n is even, hen x n is non-negaive, so y m is non-negaive. Since m is odd, his implies ha y is non-negaive. So i makes sense o consider n y. Since y m = x n, we have: x = n y m = ( n y) m if x 0 (noe ha n y m = ( n y) m because y is non-negaive) and (because m is odd). So we have: x = n y m = ( n y) m = ( n y) m if x < 0 (x, y) = (( n y) m, ( n y) n ) = γ( n y) if x 0 Moreover, since n is even, we have ( y) n = ( y) n = y, so (x, y) = (( n y) m, ( n y) n ) = γ( n y) if x < 0. So his proves ha he se (0.3) is compleely covered by he image of γ. Case m even and n odd. replacing y wih y): The se is symmeric wih respec o he x axis (i.e. by

8 8 MATTEO TOMMASINI y m = x n for m even and n odd Also in his case he image of γ covers all he se (0.3). Indeed, le us fix any poin (x, y) in (0.3). Since m is even, hen y m is non-negaive, so x n is non-negaive. Since n is odd, his implies ha x is non-negaive. So i makes sense o consider m x. Since y m = x n, we have: y = m x n = ( m x) n if y 0 (noe ha m x n = ( m x) n because x is non-negaive) and (because n is odd). So we have: y = m x n = ( m x) n = ( m x) n if y < 0 (x, y) = ( ( m x) m, ( m x) n) = γ( m x) if y 0. Moreover, since m is even, ( m x) m = ( m x) m = x, so (x, y) = ( ( m x) m, ( m x) n) = γ( m x) if y < 0. So his proves ha he se (0.3) is compleely covered by he image of γ. Case m odd and n odd. The se is symmeric wih respec o he origin (i.e. by replacing he pair (x, y) wih ( x, y)):

9 EXERCISE SESSION 2A 9 y m = x n for m odd and n odd Also in his case he image of γ covers all he se (0.3). Indeed, le us fix any poin (x, y) in (0.3). Since boh m and n are odd, we can ake boh m-h roos and n-h roos wihou having o check if x or y are non-negaive. So for example i makes sense o consider m x and we have we have: y = m x n = ( m x) n So we have: (x, y) = ( ( m x) n, ( m x) m) = γ( m x). This proves ha he se (0.3) is compleely covered by he image of γ. Case m even and n even. The se is symmeric wih respec o he x axis and o he y axiss (i.e. boh by replacing x wih x and by replacing y wih y), hence he se is symmeric also wih respec o he origin of R 2 :

10 0 MATTEO TOMMASINI y m = x n for m odd and n odd In his case he curve γ covers only he par of (0.3) ha lies in he firs quadran because boh m and n are non-negaive (since boh m and n are even naural numbers). In his case, we are able o cover separaely any of he pars of (0.3) in anoher quadran by chosing a differen curve. For example, if we wan o cover he par of (0.3) in he second quadran, we can consider he curve η() := ( m, n ) for all R. Anyway, here is no curve ha can cover he whole se (0.3) if boh m and n are even. address: maeo.ommasini2@gmail.com, maeo.ommasini@uni.lu Mahemaics Research Uni Universiy of Luxembourg 6, rue Richard Coudenhove-Kalergi L-359 Luxembourg