Given a cartesian equation get points in the plane












1












$begingroup$


I hope this make sense. I'm trying to understand (I'm very newbie with curves) how could I get points from a cartesian equation.



For example, given $(x^2+y^2)^2-2a^2cdot(x^2-y^2)-a^4+c^4=0$ that is the cartesian equation for the Cassian Ovals:



...



I would like to know how to know the points (like in this image). (I'm doing a program that represent points on a texture)



I know that resolving the parametric equation of this (I don't know if this can be done) I can give as much points as iterations done.



But I'm not sure if I can achieve that only with the cartesian equation.



Maybe, if do x=<something> and I try to resolve for this?






parametric implicit-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    there are infinitely many points. What do you mean how to "get" the points?
    $endgroup$
    – dezdichado
    Dec 13 '18 at 18:55










  • $begingroup$
    Well, yes I only want to know the x and y (point) for a serie of number (maybe from 0 to 100). For a circle is easy: $x=asin(t) and $y=acos(t) where a is the squared radius and and t the current iteration. I would like to do something similar but with cartesian equations.
    $endgroup$
    – z3nth10n
    Dec 13 '18 at 19:19
















1












$begingroup$


I hope this make sense. I'm trying to understand (I'm very newbie with curves) how could I get points from a cartesian equation.



For example, given $(x^2+y^2)^2-2a^2cdot(x^2-y^2)-a^4+c^4=0$ that is the cartesian equation for the Cassian Ovals:



...



I would like to know how to know the points (like in this image). (I'm doing a program that represent points on a texture)



I know that resolving the parametric equation of this (I don't know if this can be done) I can give as much points as iterations done.



But I'm not sure if I can achieve that only with the cartesian equation.



Maybe, if do x=<something> and I try to resolve for this?






parametric implicit-function







share|cite|improve this question











$endgroup$












  • $begingroup$
    there are infinitely many points. What do you mean how to "get" the points?
    $endgroup$
    – dezdichado
    Dec 13 '18 at 18:55










  • $begingroup$
    Well, yes I only want to know the x and y (point) for a serie of number (maybe from 0 to 100). For a circle is easy: $x=asin(t) and $y=acos(t) where a is the squared radius and and t the current iteration. I would like to do something similar but with cartesian equations.
    $endgroup$
    – z3nth10n
    Dec 13 '18 at 19:19














1












1








1





$begingroup$


I hope this make sense. I'm trying to understand (I'm very newbie with curves) how could I get points from a cartesian equation.



For example, given $(x^2+y^2)^2-2a^2cdot(x^2-y^2)-a^4+c^4=0$ that is the cartesian equation for the Cassian Ovals:



...



I would like to know how to know the points (like in this image). (I'm doing a program that represent points on a texture)



I know that resolving the parametric equation of this (I don't know if this can be done) I can give as much points as iterations done.



But I'm not sure if I can achieve that only with the cartesian equation.



Maybe, if do x=<something> and I try to resolve for this?






parametric implicit-function







share|cite|improve this question











$endgroup$




I hope this make sense. I'm trying to understand (I'm very newbie with curves) how could I get points from a cartesian equation.



For example, given $(x^2+y^2)^2-2a^2cdot(x^2-y^2)-a^4+c^4=0$ that is the cartesian equation for the Cassian Ovals:



...



I would like to know how to know the points (like in this image). (I'm doing a program that represent points on a texture)



I know that resolving the parametric equation of this (I don't know if this can be done) I can give as much points as iterations done.



But I'm not sure if I can achieve that only with the cartesian equation.



Maybe, if do x=<something> and I try to resolve for this?






parametric implicit-function




parametric implicit-function






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 13 '18 at 22:30









Tianlalu

3,08421038




3,08421038










asked Dec 13 '18 at 18:44









z3nth10nz3nth10n

1085




1085












  • $begingroup$
    there are infinitely many points. What do you mean how to "get" the points?
    $endgroup$
    – dezdichado
    Dec 13 '18 at 18:55










  • $begingroup$
    Well, yes I only want to know the x and y (point) for a serie of number (maybe from 0 to 100). For a circle is easy: $x=asin(t) and $y=acos(t) where a is the squared radius and and t the current iteration. I would like to do something similar but with cartesian equations.
    $endgroup$
    – z3nth10n
    Dec 13 '18 at 19:19


















  • $begingroup$
    there are infinitely many points. What do you mean how to "get" the points?
    $endgroup$
    – dezdichado
    Dec 13 '18 at 18:55










  • $begingroup$
    Well, yes I only want to know the x and y (point) for a serie of number (maybe from 0 to 100). For a circle is easy: $x=asin(t) and $y=acos(t) where a is the squared radius and and t the current iteration. I would like to do something similar but with cartesian equations.
    $endgroup$
    – z3nth10n
    Dec 13 '18 at 19:19
















$begingroup$
there are infinitely many points. What do you mean how to "get" the points?
$endgroup$
– dezdichado
Dec 13 '18 at 18:55




$begingroup$
there are infinitely many points. What do you mean how to "get" the points?
$endgroup$
– dezdichado
Dec 13 '18 at 18:55












$begingroup$
Well, yes I only want to know the x and y (point) for a serie of number (maybe from 0 to 100). For a circle is easy: $x=asin(t) and $y=acos(t) where a is the squared radius and and t the current iteration. I would like to do something similar but with cartesian equations.
$endgroup$
– z3nth10n
Dec 13 '18 at 19:19




$begingroup$
Well, yes I only want to know the x and y (point) for a serie of number (maybe from 0 to 100). For a circle is easy: $x=asin(t) and $y=acos(t) where a is the squared radius and and t the current iteration. I would like to do something similar but with cartesian equations.
$endgroup$
– z3nth10n
Dec 13 '18 at 19:19










1 Answer
1






active

oldest

votes


















1












$begingroup$

With the present curve, after the change of variables



$$
x = rcostheta\
y = rsintheta
$$



we arrive to



$$
f(r,theta) = c^4+r^4-a^4 - 2 a^2 r^2 cos (2 theta )= 0
$$



and solving for $r$



$$
r(theta) = pmsqrt{a^2 cos (2 theta )pmfrac{sqrt{a^4 cos (4 theta )+3 a^4-2 c^4}}{sqrt{2}}}
$$



NOTE



Also another way to do that is to find the explicit dependence $y = g(x)$



Calling $x_2 = x^2, y_2 = y^2$ we have



$$
(x_2 + y_2)^2 - 2 a^2 (x_2 - y_2) - a^4 + c^4 = 0
$$



now solving for $y_2$



$$
y_2 = sqrt{2 a^4+4 a^2 x_2-c^4}-x_2-a^2
$$



then finally



$$
y = pmsqrt{sqrt{2 a^4+4 a^2 x^2-c^4}-x^2-a^2}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Well, this is the polar equation... But with this I would need to make a circular sweep to get the radius on each degree... I would like to know if there is any method to get the representation (a integer point) on the plane without getting the parametric equation. Thanks.
    $endgroup$
    – z3nth10n
    Dec 13 '18 at 23:02










  • $begingroup$
    @z3nth10n Please. See attached note.
    $endgroup$
    – Cesareo
    Dec 14 '18 at 0:13










  • $begingroup$
    Yep, this is what I did: wolframalpha.com/input/… as you can see: i.gyazo.com/0553fee37d290d178d2a2ab0f6957261.png thanks!
    $endgroup$
    – z3nth10n
    Dec 14 '18 at 0:15













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1 Answer
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active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

With the present curve, after the change of variables



$$
x = rcostheta\
y = rsintheta
$$



we arrive to



$$
f(r,theta) = c^4+r^4-a^4 - 2 a^2 r^2 cos (2 theta )= 0
$$



and solving for $r$



$$
r(theta) = pmsqrt{a^2 cos (2 theta )pmfrac{sqrt{a^4 cos (4 theta )+3 a^4-2 c^4}}{sqrt{2}}}
$$



NOTE



Also another way to do that is to find the explicit dependence $y = g(x)$



Calling $x_2 = x^2, y_2 = y^2$ we have



$$
(x_2 + y_2)^2 - 2 a^2 (x_2 - y_2) - a^4 + c^4 = 0
$$



now solving for $y_2$



$$
y_2 = sqrt{2 a^4+4 a^2 x_2-c^4}-x_2-a^2
$$



then finally



$$
y = pmsqrt{sqrt{2 a^4+4 a^2 x^2-c^4}-x^2-a^2}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Well, this is the polar equation... But with this I would need to make a circular sweep to get the radius on each degree... I would like to know if there is any method to get the representation (a integer point) on the plane without getting the parametric equation. Thanks.
    $endgroup$
    – z3nth10n
    Dec 13 '18 at 23:02










  • $begingroup$
    @z3nth10n Please. See attached note.
    $endgroup$
    – Cesareo
    Dec 14 '18 at 0:13










  • $begingroup$
    Yep, this is what I did: wolframalpha.com/input/… as you can see: i.gyazo.com/0553fee37d290d178d2a2ab0f6957261.png thanks!
    $endgroup$
    – z3nth10n
    Dec 14 '18 at 0:15


















1












$begingroup$

With the present curve, after the change of variables



$$
x = rcostheta\
y = rsintheta
$$



we arrive to



$$
f(r,theta) = c^4+r^4-a^4 - 2 a^2 r^2 cos (2 theta )= 0
$$



and solving for $r$



$$
r(theta) = pmsqrt{a^2 cos (2 theta )pmfrac{sqrt{a^4 cos (4 theta )+3 a^4-2 c^4}}{sqrt{2}}}
$$



NOTE



Also another way to do that is to find the explicit dependence $y = g(x)$



Calling $x_2 = x^2, y_2 = y^2$ we have



$$
(x_2 + y_2)^2 - 2 a^2 (x_2 - y_2) - a^4 + c^4 = 0
$$



now solving for $y_2$



$$
y_2 = sqrt{2 a^4+4 a^2 x_2-c^4}-x_2-a^2
$$



then finally



$$
y = pmsqrt{sqrt{2 a^4+4 a^2 x^2-c^4}-x^2-a^2}
$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Well, this is the polar equation... But with this I would need to make a circular sweep to get the radius on each degree... I would like to know if there is any method to get the representation (a integer point) on the plane without getting the parametric equation. Thanks.
    $endgroup$
    – z3nth10n
    Dec 13 '18 at 23:02










  • $begingroup$
    @z3nth10n Please. See attached note.
    $endgroup$
    – Cesareo
    Dec 14 '18 at 0:13










  • $begingroup$
    Yep, this is what I did: wolframalpha.com/input/… as you can see: i.gyazo.com/0553fee37d290d178d2a2ab0f6957261.png thanks!
    $endgroup$
    – z3nth10n
    Dec 14 '18 at 0:15
















1












1








1





$begingroup$

With the present curve, after the change of variables



$$
x = rcostheta\
y = rsintheta
$$



we arrive to



$$
f(r,theta) = c^4+r^4-a^4 - 2 a^2 r^2 cos (2 theta )= 0
$$



and solving for $r$



$$
r(theta) = pmsqrt{a^2 cos (2 theta )pmfrac{sqrt{a^4 cos (4 theta )+3 a^4-2 c^4}}{sqrt{2}}}
$$



NOTE



Also another way to do that is to find the explicit dependence $y = g(x)$



Calling $x_2 = x^2, y_2 = y^2$ we have



$$
(x_2 + y_2)^2 - 2 a^2 (x_2 - y_2) - a^4 + c^4 = 0
$$



now solving for $y_2$



$$
y_2 = sqrt{2 a^4+4 a^2 x_2-c^4}-x_2-a^2
$$



then finally



$$
y = pmsqrt{sqrt{2 a^4+4 a^2 x^2-c^4}-x^2-a^2}
$$






share|cite|improve this answer











$endgroup$



With the present curve, after the change of variables



$$
x = rcostheta\
y = rsintheta
$$



we arrive to



$$
f(r,theta) = c^4+r^4-a^4 - 2 a^2 r^2 cos (2 theta )= 0
$$



and solving for $r$



$$
r(theta) = pmsqrt{a^2 cos (2 theta )pmfrac{sqrt{a^4 cos (4 theta )+3 a^4-2 c^4}}{sqrt{2}}}
$$



NOTE



Also another way to do that is to find the explicit dependence $y = g(x)$



Calling $x_2 = x^2, y_2 = y^2$ we have



$$
(x_2 + y_2)^2 - 2 a^2 (x_2 - y_2) - a^4 + c^4 = 0
$$



now solving for $y_2$



$$
y_2 = sqrt{2 a^4+4 a^2 x_2-c^4}-x_2-a^2
$$



then finally



$$
y = pmsqrt{sqrt{2 a^4+4 a^2 x^2-c^4}-x^2-a^2}
$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 14 '18 at 0:12

























answered Dec 13 '18 at 19:50









CesareoCesareo

9,0863516




9,0863516












  • $begingroup$
    Well, this is the polar equation... But with this I would need to make a circular sweep to get the radius on each degree... I would like to know if there is any method to get the representation (a integer point) on the plane without getting the parametric equation. Thanks.
    $endgroup$
    – z3nth10n
    Dec 13 '18 at 23:02










  • $begingroup$
    @z3nth10n Please. See attached note.
    $endgroup$
    – Cesareo
    Dec 14 '18 at 0:13










  • $begingroup$
    Yep, this is what I did: wolframalpha.com/input/… as you can see: i.gyazo.com/0553fee37d290d178d2a2ab0f6957261.png thanks!
    $endgroup$
    – z3nth10n
    Dec 14 '18 at 0:15




















  • $begingroup$
    Well, this is the polar equation... But with this I would need to make a circular sweep to get the radius on each degree... I would like to know if there is any method to get the representation (a integer point) on the plane without getting the parametric equation. Thanks.
    $endgroup$
    – z3nth10n
    Dec 13 '18 at 23:02










  • $begingroup$
    @z3nth10n Please. See attached note.
    $endgroup$
    – Cesareo
    Dec 14 '18 at 0:13










  • $begingroup$
    Yep, this is what I did: wolframalpha.com/input/… as you can see: i.gyazo.com/0553fee37d290d178d2a2ab0f6957261.png thanks!
    $endgroup$
    – z3nth10n
    Dec 14 '18 at 0:15


















$begingroup$
Well, this is the polar equation... But with this I would need to make a circular sweep to get the radius on each degree... I would like to know if there is any method to get the representation (a integer point) on the plane without getting the parametric equation. Thanks.
$endgroup$
– z3nth10n
Dec 13 '18 at 23:02




$begingroup$
Well, this is the polar equation... But with this I would need to make a circular sweep to get the radius on each degree... I would like to know if there is any method to get the representation (a integer point) on the plane without getting the parametric equation. Thanks.
$endgroup$
– z3nth10n
Dec 13 '18 at 23:02












$begingroup$
@z3nth10n Please. See attached note.
$endgroup$
– Cesareo
Dec 14 '18 at 0:13




$begingroup$
@z3nth10n Please. See attached note.
$endgroup$
– Cesareo
Dec 14 '18 at 0:13












$begingroup$
Yep, this is what I did: wolframalpha.com/input/… as you can see: i.gyazo.com/0553fee37d290d178d2a2ab0f6957261.png thanks!
$endgroup$
– z3nth10n
Dec 14 '18 at 0:15






$begingroup$
Yep, this is what I did: wolframalpha.com/input/… as you can see: i.gyazo.com/0553fee37d290d178d2a2ab0f6957261.png thanks!
$endgroup$
– z3nth10n
Dec 14 '18 at 0:15




















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