how to make parametric equation of cube












-1












$begingroup$


this is my hyperbolic equation



$y = x^2$



then convert to paramteric equation, like this:



$x = u$



$y = u^2$



so i insert the equation into x and y Axis Generator



x: u
y: u
z: 0


the output is: 2D hyperbolic object



when i insert v parameter into z Axis



the output is: 3D hyperbolic object



my question is how to make 3D cube using parametric equation and how to insert it into the Generator.



any help thank youu :-) I LOVE YOU



another exemple:



x: cos(u)
y: sin(u)
z: v


the output is: pipe



where:



u minimal: 0
u maximal: 6.28

v minimal: 0
v maximal: 1









share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
    $endgroup$
    – Matti P.
    Dec 14 '18 at 7:52










  • $begingroup$
    i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
    $endgroup$
    – vaiana
    Dec 14 '18 at 8:00












  • $begingroup$
    thank you for your respon. what should i do with those equations? because hard to read the equation without u and v parameter. @MohammadZuhairKhan
    $endgroup$
    – vaiana
    Dec 14 '18 at 8:16


















-1












$begingroup$


this is my hyperbolic equation



$y = x^2$



then convert to paramteric equation, like this:



$x = u$



$y = u^2$



so i insert the equation into x and y Axis Generator



x: u
y: u
z: 0


the output is: 2D hyperbolic object



when i insert v parameter into z Axis



the output is: 3D hyperbolic object



my question is how to make 3D cube using parametric equation and how to insert it into the Generator.



any help thank youu :-) I LOVE YOU



another exemple:



x: cos(u)
y: sin(u)
z: v


the output is: pipe



where:



u minimal: 0
u maximal: 6.28

v minimal: 0
v maximal: 1









share|cite|improve this question











$endgroup$












  • $begingroup$
    Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
    $endgroup$
    – Matti P.
    Dec 14 '18 at 7:52










  • $begingroup$
    i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
    $endgroup$
    – vaiana
    Dec 14 '18 at 8:00












  • $begingroup$
    thank you for your respon. what should i do with those equations? because hard to read the equation without u and v parameter. @MohammadZuhairKhan
    $endgroup$
    – vaiana
    Dec 14 '18 at 8:16
















-1












-1








-1





$begingroup$


this is my hyperbolic equation



$y = x^2$



then convert to paramteric equation, like this:



$x = u$



$y = u^2$



so i insert the equation into x and y Axis Generator



x: u
y: u
z: 0


the output is: 2D hyperbolic object



when i insert v parameter into z Axis



the output is: 3D hyperbolic object



my question is how to make 3D cube using parametric equation and how to insert it into the Generator.



any help thank youu :-) I LOVE YOU



another exemple:



x: cos(u)
y: sin(u)
z: v


the output is: pipe



where:



u minimal: 0
u maximal: 6.28

v minimal: 0
v maximal: 1









share|cite|improve this question











$endgroup$




this is my hyperbolic equation



$y = x^2$



then convert to paramteric equation, like this:



$x = u$



$y = u^2$



so i insert the equation into x and y Axis Generator



x: u
y: u
z: 0


the output is: 2D hyperbolic object



when i insert v parameter into z Axis



the output is: 3D hyperbolic object



my question is how to make 3D cube using parametric equation and how to insert it into the Generator.



any help thank youu :-) I LOVE YOU



another exemple:



x: cos(u)
y: sin(u)
z: v


the output is: pipe



where:



u minimal: 0
u maximal: 6.28

v minimal: 0
v maximal: 1






parametric






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 14 '18 at 7:47







vaiana

















asked Dec 14 '18 at 7:35









vaianavaiana

32




32












  • $begingroup$
    Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
    $endgroup$
    – Matti P.
    Dec 14 '18 at 7:52










  • $begingroup$
    i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
    $endgroup$
    – vaiana
    Dec 14 '18 at 8:00












  • $begingroup$
    thank you for your respon. what should i do with those equations? because hard to read the equation without u and v parameter. @MohammadZuhairKhan
    $endgroup$
    – vaiana
    Dec 14 '18 at 8:16




















  • $begingroup$
    Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
    $endgroup$
    – Matti P.
    Dec 14 '18 at 7:52










  • $begingroup$
    i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
    $endgroup$
    – vaiana
    Dec 14 '18 at 8:00












  • $begingroup$
    thank you for your respon. what should i do with those equations? because hard to read the equation without u and v parameter. @MohammadZuhairKhan
    $endgroup$
    – vaiana
    Dec 14 '18 at 8:16


















$begingroup$
Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 7:52




$begingroup$
Well, you could start by considering the walls of the cube. How would you make the floor? Something like $z=0$ ?
$endgroup$
– Matti P.
Dec 14 '18 at 7:52












$begingroup$
i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
$endgroup$
– vaiana
Dec 14 '18 at 8:00






$begingroup$
i make floor using simple parametric u and v. but hard to find how make the walls. is it possible make the walls as same as when i make circle. i don't know, is my problem about parametric or not.
$endgroup$
– vaiana
Dec 14 '18 at 8:00














$begingroup$
thank you for your respon. what should i do with those equations? because hard to read the equation without u and v parameter. @MohammadZuhairKhan
$endgroup$
– vaiana
Dec 14 '18 at 8:16






$begingroup$
thank you for your respon. what should i do with those equations? because hard to read the equation without u and v parameter. @MohammadZuhairKhan
$endgroup$
– vaiana
Dec 14 '18 at 8:16












1 Answer
1






active

oldest

votes


















1












$begingroup$

The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:



$|x|le a/2\|y|le a/2\|z|le a/2$



Parameterize it as:



$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 20 '18 at 3:56










  • $begingroup$
    Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
    $endgroup$
    – Shubham Johri
    Dec 20 '18 at 4:53












  • $begingroup$
    still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 24 '18 at 6:13











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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:



$|x|le a/2\|y|le a/2\|z|le a/2$



Parameterize it as:



$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 20 '18 at 3:56










  • $begingroup$
    Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
    $endgroup$
    – Shubham Johri
    Dec 20 '18 at 4:53












  • $begingroup$
    still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 24 '18 at 6:13
















1












$begingroup$

The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:



$|x|le a/2\|y|le a/2\|z|le a/2$



Parameterize it as:



$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 20 '18 at 3:56










  • $begingroup$
    Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
    $endgroup$
    – Shubham Johri
    Dec 20 '18 at 4:53












  • $begingroup$
    still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 24 '18 at 6:13














1












1








1





$begingroup$

The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:



$|x|le a/2\|y|le a/2\|z|le a/2$



Parameterize it as:



$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$






share|cite|improve this answer











$endgroup$



The equation of a solid cube of side-length $ainBbb R^+$ centred at the origin would be:



$|x|le a/2\|y|le a/2\|z|le a/2$



Parameterize it as:



$x=u, |u|le a/2\y=v, |v|le a/2\z=w, |w|le a/2$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 10:33

























answered Dec 14 '18 at 8:07









Shubham JohriShubham Johri

5,204718




5,204718












  • $begingroup$
    i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 20 '18 at 3:56










  • $begingroup$
    Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
    $endgroup$
    – Shubham Johri
    Dec 20 '18 at 4:53












  • $begingroup$
    still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 24 '18 at 6:13


















  • $begingroup$
    i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 20 '18 at 3:56










  • $begingroup$
    Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
    $endgroup$
    – Shubham Johri
    Dec 20 '18 at 4:53












  • $begingroup$
    still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
    $endgroup$
    – vaiana
    Dec 24 '18 at 6:13
















$begingroup$
i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
$endgroup$
– vaiana
Dec 20 '18 at 3:56




$begingroup$
i cannot read your equation. i input x = v ≤ a/2 and y = v ≤a/2 and then z = a/2 and still doesn't work @ShubhamJohri
$endgroup$
– vaiana
Dec 20 '18 at 3:56












$begingroup$
Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
$endgroup$
– Shubham Johri
Dec 20 '18 at 4:53






$begingroup$
Carefully not the modulus sign around $v. y=v, |v|le a/2implies -a/2le vle a/2$, in other words $y=v,v_{min}=-a/2,v_{max}=a/2$
$endgroup$
– Shubham Johri
Dec 20 '18 at 4:53














$begingroup$
still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
$endgroup$
– vaiana
Dec 24 '18 at 6:13




$begingroup$
still not work well sir, .. i use free Blender software to make the object.. still confuse for weeks @ShubhamJohri
$endgroup$
– vaiana
Dec 24 '18 at 6:13


















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