Mixing
Check if a parametrisation is suitable
Clash Royale CLAN TAG#URR8PPP
up vote
0
down vote
favorite
I parametrised the curve C: $y^2=x^3+x^2$ by $r(t) = (x(t), y(t)) = (t^2−1, t^3−t)$
Is there a way to check that $r(t)$ lies on the curve C for all t? How can I be sure that the parametrisation reaches all points of C?
Thanks
calculus parametric parametrization
asked Jul 17 at 0:43
Relax295
849
add a comment |Â
up vote
0
down vote
favorite
I parametrised the curve C: $y^2=x^3+x^2$ by $r(t) = (x(t), y(t)) = (t^2−1, t^3−t)$
Is there a way to check that $r(t)$ lies on the curve C for all t? How can I be sure that the parametrisation reaches all points of C?
Thanks
calculus parametric parametrization
asked Jul 17 at 0:43
Relax295
849
Simplest is to just plot them both.
– David G. Stork
Jul 17 at 0:55
2
Could you not just plug $r(t)$ into your curve $C$ and demonstrate that the equation holds for all $t$ (that is, that the equality reduces to a tautology)? Seems like the most straightforward approach to me.
– Ispil
Jul 17 at 1:08
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I parametrised the curve C: $y^2=x^3+x^2$ by $r(t) = (x(t), y(t)) = (t^2−1, t^3−t)$
Is there a way to check that $r(t)$ lies on the curve C for all t? How can I be sure that the parametrisation reaches all points of C?
Thanks
calculus parametric parametrization
asked Jul 17 at 0:43
Relax295
849
I parametrised the curve C: $y^2=x^3+x^2$ by $r(t) = (x(t), y(t)) = (t^2−1, t^3−t)$
Is there a way to check that $r(t)$ lies on the curve C for all t? How can I be sure that the parametrisation reaches all points of C?
Thanks
calculus parametric parametrization
asked Jul 17 at 0:43
Relax295
849
asked Jul 17 at 0:43
Relax295
849
asked Jul 17 at 0:43
Relax295
849
asked Jul 17 at 0:43
Relax295
849
849
Simplest is to just plot them both.
– David G. Stork
Jul 17 at 0:55
2
Could you not just plug $r(t)$ into your curve $C$ and demonstrate that the equation holds for all $t$ (that is, that the equality reduces to a tautology)? Seems like the most straightforward approach to me.
– Ispil
Jul 17 at 1:08
add a comment |Â
Simplest is to just plot them both.
– David G. Stork
Jul 17 at 0:55
2
Could you not just plug $r(t)$ into your curve $C$ and demonstrate that the equation holds for all $t$ (that is, that the equality reduces to a tautology)? Seems like the most straightforward approach to me.
– Ispil
Jul 17 at 1:08
Simplest is to just plot them both.
– David G. Stork
Jul 17 at 0:55
Simplest is to just plot them both.
– David G. Stork
Jul 17 at 0:55
2
2
Could you not just plug $r(t)$ into your curve $C$ and demonstrate that the equation holds for all $t$ (that is, that the equality reduces to a tautology)? Seems like the most straightforward approach to me.
– Ispil
Jul 17 at 1:08
Could you not just plug $r(t)$ into your curve $C$ and demonstrate that the equation holds for all $t$ (that is, that the equality reduces to a tautology)? Seems like the most straightforward approach to me.
– Ispil
Jul 17 at 1:08
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
0
down vote
Yes indeed, your parameterization just works fine!
Technically you probably derived the parameterization too and hence there shouldn't be any doubt, but if you were given a curve and a parameterization, then, as @Ispil already pointed out, you can just simply plug in and check whether the equality holds or not, so for $tinmathbb R$:
$$
y^2=x^3+x^2iff (t^3-t)^2=(t^2-1)^3+(t^2-1)^2
$$
which holds.
Anyway, the idea behind the parameteriztion of yours is if you start by
$$
y^2=x^3+x^2=x^2(x+1)
$$
which would yield to a root like $y=xsqrtx+1$ so we want $(x+1)$ actually non-negative - to be well defined - so we take $x=t^2-1$ which gives $y=(t^2-1)sqrt(t^2-1)+1=t^3-t$ and we are done.
answered Jul 17 at 20:32
user190080
3,14621327
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Yes indeed, your parameterization just works fine!
Technically you probably derived the parameterization too and hence there shouldn't be any doubt, but if you were given a curve and a parameterization, then, as @Ispil already pointed out, you can just simply plug in and check whether the equality holds or not, so for $tinmathbb R$:
$$
y^2=x^3+x^2iff (t^3-t)^2=(t^2-1)^3+(t^2-1)^2
$$
which holds.
Anyway, the idea behind the parameteriztion of yours is if you start by
$$
y^2=x^3+x^2=x^2(x+1)
$$
which would yield to a root like $y=xsqrtx+1$ so we want $(x+1)$ actually non-negative - to be well defined - so we take $x=t^2-1$ which gives $y=(t^2-1)sqrt(t^2-1)+1=t^3-t$ and we are done.
answered Jul 17 at 20:32
user190080
3,14621327
add a comment |Â
up vote
0
down vote
Yes indeed, your parameterization just works fine!
Technically you probably derived the parameterization too and hence there shouldn't be any doubt, but if you were given a curve and a parameterization, then, as @Ispil already pointed out, you can just simply plug in and check whether the equality holds or not, so for $tinmathbb R$:
$$
y^2=x^3+x^2iff (t^3-t)^2=(t^2-1)^3+(t^2-1)^2
$$
which holds.
Anyway, the idea behind the parameteriztion of yours is if you start by
$$
y^2=x^3+x^2=x^2(x+1)
$$
which would yield to a root like $y=xsqrtx+1$ so we want $(x+1)$ actually non-negative - to be well defined - so we take $x=t^2-1$ which gives $y=(t^2-1)sqrt(t^2-1)+1=t^3-t$ and we are done.
answered Jul 17 at 20:32
user190080
3,14621327
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Yes indeed, your parameterization just works fine!
Technically you probably derived the parameterization too and hence there shouldn't be any doubt, but if you were given a curve and a parameterization, then, as @Ispil already pointed out, you can just simply plug in and check whether the equality holds or not, so for $tinmathbb R$:
$$
y^2=x^3+x^2iff (t^3-t)^2=(t^2-1)^3+(t^2-1)^2
$$
which holds.
Anyway, the idea behind the parameteriztion of yours is if you start by
$$
y^2=x^3+x^2=x^2(x+1)
$$
which would yield to a root like $y=xsqrtx+1$ so we want $(x+1)$ actually non-negative - to be well defined - so we take $x=t^2-1$ which gives $y=(t^2-1)sqrt(t^2-1)+1=t^3-t$ and we are done.
answered Jul 17 at 20:32
user190080
3,14621327
Yes indeed, your parameterization just works fine!
Technically you probably derived the parameterization too and hence there shouldn't be any doubt, but if you were given a curve and a parameterization, then, as @Ispil already pointed out, you can just simply plug in and check whether the equality holds or not, so for $tinmathbb R$:
$$
y^2=x^3+x^2iff (t^3-t)^2=(t^2-1)^3+(t^2-1)^2
$$
which holds.
Anyway, the idea behind the parameteriztion of yours is if you start by
$$
y^2=x^3+x^2=x^2(x+1)
$$
which would yield to a root like $y=xsqrtx+1$ so we want $(x+1)$ actually non-negative - to be well defined - so we take $x=t^2-1$ which gives $y=(t^2-1)sqrt(t^2-1)+1=t^3-t$ and we are done.
answered Jul 17 at 20:32
user190080
3,14621327
answered Jul 17 at 20:32
user190080
3,14621327
answered Jul 17 at 20:32
user190080
3,14621327
answered Jul 17 at 20:32
user190080
3,14621327
3,14621327
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2854013%2fcheck-if-a-parametrisation-is-suitable%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Simplest is to just plot them both.
– David G. Stork
Jul 17 at 0:55
2
Could you not just plug $r(t)$ into your curve $C$ and demonstrate that the equation holds for all $t$ (that is, that the equality reduces to a tautology)? Seems like the most straightforward approach to me.
– Ispil
Jul 17 at 1:08