Natural parameterization of the following curve
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The given curve is: $ 27 x^2 = 4 y^3 $.
I have to find the natural parameterization.
First, I parameterized the curve:
$$ 27 x^2 = 4 y^3 $$
$$ x = pm sqrtfrac427 y^3 $$
What should I do when I have two possibilities?
I took one possibility, for example: $$ x = sqrtfrac427 y^3 $$
Parameterization: $$ alpha (t) = (sqrtfrac427 t^3,t) $$
The other steps:
$$ alpha ' (t) = (fracsqrttsqrt3,1) $$
$$ || alpha ' (t) || = frac1sqrt3 sqrtt+3 $$
$$s(t) = frac1sqrt3 int_0^t sqrtu+3 du = frac1sqrt3(frac23(t+3)^frac32- 2sqrt3)$$
Did I go wrong at some point? It became too complicated, so I need help. Thanks!
geometry differential
asked Aug 16 at 10:12
stakindmidl
574
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up vote
2
down vote
favorite
The given curve is: $ 27 x^2 = 4 y^3 $.
I have to find the natural parameterization.
First, I parameterized the curve:
$$ 27 x^2 = 4 y^3 $$
$$ x = pm sqrtfrac427 y^3 $$
What should I do when I have two possibilities?
I took one possibility, for example: $$ x = sqrtfrac427 y^3 $$
Parameterization: $$ alpha (t) = (sqrtfrac427 t^3,t) $$
The other steps:
$$ alpha ' (t) = (fracsqrttsqrt3,1) $$
$$ || alpha ' (t) || = frac1sqrt3 sqrtt+3 $$
$$s(t) = frac1sqrt3 int_0^t sqrtu+3 du = frac1sqrt3(frac23(t+3)^frac32- 2sqrt3)$$
Did I go wrong at some point? It became too complicated, so I need help. Thanks!
geometry differential
asked Aug 16 at 10:12
stakindmidl
574
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
The given curve is: $ 27 x^2 = 4 y^3 $.
I have to find the natural parameterization.
First, I parameterized the curve:
$$ 27 x^2 = 4 y^3 $$
$$ x = pm sqrtfrac427 y^3 $$
What should I do when I have two possibilities?
I took one possibility, for example: $$ x = sqrtfrac427 y^3 $$
Parameterization: $$ alpha (t) = (sqrtfrac427 t^3,t) $$
The other steps:
$$ alpha ' (t) = (fracsqrttsqrt3,1) $$
$$ || alpha ' (t) || = frac1sqrt3 sqrtt+3 $$
$$s(t) = frac1sqrt3 int_0^t sqrtu+3 du = frac1sqrt3(frac23(t+3)^frac32- 2sqrt3)$$
Did I go wrong at some point? It became too complicated, so I need help. Thanks!
geometry differential
asked Aug 16 at 10:12
stakindmidl
574
The given curve is: $ 27 x^2 = 4 y^3 $.
I have to find the natural parameterization.
First, I parameterized the curve:
$$ 27 x^2 = 4 y^3 $$
$$ x = pm sqrtfrac427 y^3 $$
What should I do when I have two possibilities?
I took one possibility, for example: $$ x = sqrtfrac427 y^3 $$
Parameterization: $$ alpha (t) = (sqrtfrac427 t^3,t) $$
The other steps:
$$ alpha ' (t) = (fracsqrttsqrt3,1) $$
$$ || alpha ' (t) || = frac1sqrt3 sqrtt+3 $$
$$s(t) = frac1sqrt3 int_0^t sqrtu+3 du = frac1sqrt3(frac23(t+3)^frac32- 2sqrt3)$$
Did I go wrong at some point? It became too complicated, so I need help. Thanks!
geometry differential
asked Aug 16 at 10:12
stakindmidl
574
asked Aug 16 at 10:12
stakindmidl
574
asked Aug 16 at 10:12
stakindmidl
574
asked Aug 16 at 10:12
stakindmidl
574
574
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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up vote
1
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What you did fine. Now, you invert $s$:$$s^-1(t)=frac3sqrt[3](t+2)^2sqrt[3]4-3.$$So, a natural parametrisation of your curve is$$tmapstoalphabigl(s^-1(t)bigr)=left(fracleft(sqrt[3]2(t+2)^2-2right)^frac32sqrt2,left(frac3sqrt[3](t+2)^2sqrt[3]4-3right)^2right)$$
answered Aug 16 at 10:48
José Carlos Santos
117k1699179
one question, what should I do with the $ pm $ ? Should I do both cases?
– stakindmidl
Aug 16 at 11:44
1
Note the your curve has acorner at $(0,0)$. THerefore, there is na way you can obtain the whole curve by this process. But you can get a natural parametrization of the other half adding a $-$ sign to the second component of my answer.
– José Carlos Santos
Aug 16 at 11:46
@ChristianBlatter Right: cusp. I couldn't remember the right word and that's why I used “corner†instead. The $x$ was typo, which I have edited. Thank you.
– José Carlos Santos
Aug 16 at 13:51
The curve is $y=|x|^2/3$ $>(-infty<x<infty)$. It has a characteristic cusp, not a "corner", at the origin. Concerning your comment: The curve is lying completely in the upper half plane.
– Christian Blatter
Aug 16 at 14:13
@ChristianBlatter It's a pity that I can't edit that comment. The $-$ sign should be put on the first component, of course.
– José Carlos Santos
Aug 16 at 14:55
add a comment |Â
up vote
3
down vote
If we write $x=2t^3$ then we have $$dfracy^327=dfracx^24=t^6implies y=3t^2.$$
Thus we get $$alpha(t)=(2t^3,3t^2),$$ which is a nice parametrization to get the length of the curve.
answered Aug 16 at 10:45
mfl
24.7k12040
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
What you did fine. Now, you invert $s$:$$s^-1(t)=frac3sqrt[3](t+2)^2sqrt[3]4-3.$$So, a natural parametrisation of your curve is$$tmapstoalphabigl(s^-1(t)bigr)=left(fracleft(sqrt[3]2(t+2)^2-2right)^frac32sqrt2,left(frac3sqrt[3](t+2)^2sqrt[3]4-3right)^2right)$$
answered Aug 16 at 10:48
José Carlos Santos
117k1699179
one question, what should I do with the $ pm $ ? Should I do both cases?
– stakindmidl
Aug 16 at 11:44
1
Note the your curve has acorner at $(0,0)$. THerefore, there is na way you can obtain the whole curve by this process. But you can get a natural parametrization of the other half adding a $-$ sign to the second component of my answer.
– José Carlos Santos
Aug 16 at 11:46
@ChristianBlatter Right: cusp. I couldn't remember the right word and that's why I used “corner†instead. The $x$ was typo, which I have edited. Thank you.
– José Carlos Santos
Aug 16 at 13:51
The curve is $y=|x|^2/3$ $>(-infty<x<infty)$. It has a characteristic cusp, not a "corner", at the origin. Concerning your comment: The curve is lying completely in the upper half plane.
– Christian Blatter
Aug 16 at 14:13
@ChristianBlatter It's a pity that I can't edit that comment. The $-$ sign should be put on the first component, of course.
– José Carlos Santos
Aug 16 at 14:55
add a comment |Â
up vote
1
down vote
accepted
What you did fine. Now, you invert $s$:$$s^-1(t)=frac3sqrt[3](t+2)^2sqrt[3]4-3.$$So, a natural parametrisation of your curve is$$tmapstoalphabigl(s^-1(t)bigr)=left(fracleft(sqrt[3]2(t+2)^2-2right)^frac32sqrt2,left(frac3sqrt[3](t+2)^2sqrt[3]4-3right)^2right)$$
answered Aug 16 at 10:48
José Carlos Santos
117k1699179
one question, what should I do with the $ pm $ ? Should I do both cases?
– stakindmidl
Aug 16 at 11:44
1
Note the your curve has acorner at $(0,0)$. THerefore, there is na way you can obtain the whole curve by this process. But you can get a natural parametrization of the other half adding a $-$ sign to the second component of my answer.
– José Carlos Santos
Aug 16 at 11:46
@ChristianBlatter Right: cusp. I couldn't remember the right word and that's why I used “corner†instead. The $x$ was typo, which I have edited. Thank you.
– José Carlos Santos
Aug 16 at 13:51
The curve is $y=|x|^2/3$ $>(-infty<x<infty)$. It has a characteristic cusp, not a "corner", at the origin. Concerning your comment: The curve is lying completely in the upper half plane.
– Christian Blatter
Aug 16 at 14:13
@ChristianBlatter It's a pity that I can't edit that comment. The $-$ sign should be put on the first component, of course.
– José Carlos Santos
Aug 16 at 14:55
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
What you did fine. Now, you invert $s$:$$s^-1(t)=frac3sqrt[3](t+2)^2sqrt[3]4-3.$$So, a natural parametrisation of your curve is$$tmapstoalphabigl(s^-1(t)bigr)=left(fracleft(sqrt[3]2(t+2)^2-2right)^frac32sqrt2,left(frac3sqrt[3](t+2)^2sqrt[3]4-3right)^2right)$$
answered Aug 16 at 10:48
José Carlos Santos
117k1699179
What you did fine. Now, you invert $s$:$$s^-1(t)=frac3sqrt[3](t+2)^2sqrt[3]4-3.$$So, a natural parametrisation of your curve is$$tmapstoalphabigl(s^-1(t)bigr)=left(fracleft(sqrt[3]2(t+2)^2-2right)^frac32sqrt2,left(frac3sqrt[3](t+2)^2sqrt[3]4-3right)^2right)$$
answered Aug 16 at 10:48
José Carlos Santos
117k1699179
edited Aug 16 at 13:49
answered Aug 16 at 10:48
José Carlos Santos
117k1699179
answered Aug 16 at 10:48
José Carlos Santos
117k1699179
answered Aug 16 at 10:48
José Carlos Santos
117k1699179
117k1699179
one question, what should I do with the $ pm $ ? Should I do both cases?
– stakindmidl
Aug 16 at 11:44
1
Note the your curve has acorner at $(0,0)$. THerefore, there is na way you can obtain the whole curve by this process. But you can get a natural parametrization of the other half adding a $-$ sign to the second component of my answer.
– José Carlos Santos
Aug 16 at 11:46
@ChristianBlatter Right: cusp. I couldn't remember the right word and that's why I used “corner†instead. The $x$ was typo, which I have edited. Thank you.
– José Carlos Santos
Aug 16 at 13:51
The curve is $y=|x|^2/3$ $>(-infty<x<infty)$. It has a characteristic cusp, not a "corner", at the origin. Concerning your comment: The curve is lying completely in the upper half plane.
– Christian Blatter
Aug 16 at 14:13
@ChristianBlatter It's a pity that I can't edit that comment. The $-$ sign should be put on the first component, of course.
– José Carlos Santos
Aug 16 at 14:55
add a comment |Â
one question, what should I do with the $ pm $ ? Should I do both cases?
– stakindmidl
Aug 16 at 11:44
1
Note the your curve has acorner at $(0,0)$. THerefore, there is na way you can obtain the whole curve by this process. But you can get a natural parametrization of the other half adding a $-$ sign to the second component of my answer.
– José Carlos Santos
Aug 16 at 11:46
@ChristianBlatter Right: cusp. I couldn't remember the right word and that's why I used “corner†instead. The $x$ was typo, which I have edited. Thank you.
– José Carlos Santos
Aug 16 at 13:51
The curve is $y=|x|^2/3$ $>(-infty<x<infty)$. It has a characteristic cusp, not a "corner", at the origin. Concerning your comment: The curve is lying completely in the upper half plane.
– Christian Blatter
Aug 16 at 14:13
@ChristianBlatter It's a pity that I can't edit that comment. The $-$ sign should be put on the first component, of course.
– José Carlos Santos
Aug 16 at 14:55
one question, what should I do with the $ pm $ ? Should I do both cases?
– stakindmidl
Aug 16 at 11:44
one question, what should I do with the $ pm $ ? Should I do both cases?
– stakindmidl
Aug 16 at 11:44
1
1
Note the your curve has acorner at $(0,0)$. THerefore, there is na way you can obtain the whole curve by this process. But you can get a natural parametrization of the other half adding a $-$ sign to the second component of my answer.
– José Carlos Santos
Aug 16 at 11:46
Note the your curve has acorner at $(0,0)$. THerefore, there is na way you can obtain the whole curve by this process. But you can get a natural parametrization of the other half adding a $-$ sign to the second component of my answer.
– José Carlos Santos
Aug 16 at 11:46
@ChristianBlatter Right: cusp. I couldn't remember the right word and that's why I used “corner†instead. The $x$ was typo, which I have edited. Thank you.
– José Carlos Santos
Aug 16 at 13:51
@ChristianBlatter Right: cusp. I couldn't remember the right word and that's why I used “corner†instead. The $x$ was typo, which I have edited. Thank you.
– José Carlos Santos
Aug 16 at 13:51
The curve is $y=|x|^2/3$ $>(-infty<x<infty)$. It has a characteristic cusp, not a "corner", at the origin. Concerning your comment: The curve is lying completely in the upper half plane.
– Christian Blatter
Aug 16 at 14:13
The curve is $y=|x|^2/3$ $>(-infty<x<infty)$. It has a characteristic cusp, not a "corner", at the origin. Concerning your comment: The curve is lying completely in the upper half plane.
– Christian Blatter
Aug 16 at 14:13
@ChristianBlatter It's a pity that I can't edit that comment. The $-$ sign should be put on the first component, of course.
– José Carlos Santos
Aug 16 at 14:55
@ChristianBlatter It's a pity that I can't edit that comment. The $-$ sign should be put on the first component, of course.
– José Carlos Santos
Aug 16 at 14:55
add a comment |Â
up vote
3
down vote
If we write $x=2t^3$ then we have $$dfracy^327=dfracx^24=t^6implies y=3t^2.$$
Thus we get $$alpha(t)=(2t^3,3t^2),$$ which is a nice parametrization to get the length of the curve.
answered Aug 16 at 10:45
mfl
24.7k12040
add a comment |Â
up vote
3
down vote
If we write $x=2t^3$ then we have $$dfracy^327=dfracx^24=t^6implies y=3t^2.$$
Thus we get $$alpha(t)=(2t^3,3t^2),$$ which is a nice parametrization to get the length of the curve.
answered Aug 16 at 10:45
mfl
24.7k12040
add a comment |Â
up vote
3
down vote
up vote
3
down vote
If we write $x=2t^3$ then we have $$dfracy^327=dfracx^24=t^6implies y=3t^2.$$
Thus we get $$alpha(t)=(2t^3,3t^2),$$ which is a nice parametrization to get the length of the curve.
answered Aug 16 at 10:45
mfl
24.7k12040
If we write $x=2t^3$ then we have $$dfracy^327=dfracx^24=t^6implies y=3t^2.$$
Thus we get $$alpha(t)=(2t^3,3t^2),$$ which is a nice parametrization to get the length of the curve.
answered Aug 16 at 10:45
mfl
24.7k12040
answered Aug 16 at 10:45
mfl
24.7k12040
answered Aug 16 at 10:45
mfl
24.7k12040
answered Aug 16 at 10:45
mfl
24.7k12040
24.7k12040
add a comment |Â
add a comment |Â
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