Easier way to parametrize an ellipse.












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$begingroup$


I want to calculate the parametrization of the curve $$E={(x,y,z):z=x^2+y^2,x+y+z=1}$$
To do so I did a change of variables taking as basis the vectors $frac{1}{sqrt{2}}(-1,1,0),frac{1}{sqrt{6}}(-1,-1,2),frac{1}{sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=sqrt{3}/3$, and $$begin{cases}x=-x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3}\y=x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3} \z=2y'/sqrt{6}+z'/sqrt{3}end{cases}$$
Now using this change of variables into $E$ I have that $$frac{(x')^2}{frac{1}{(3^{(1/4)})^2}}+frac{(y'-sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}costheta,;y'=sqrt{2/3}+3^{1/4}sintheta,; z'=sqrt{3}/3,; 0leq theta leq2pi$$



As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?






multivariable-calculus parametrization







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  • $begingroup$
    There are other parametrization options. Do you want this particular one?
    $endgroup$
    – Andrei
    Jan 16 at 14:01
















1












$begingroup$


I want to calculate the parametrization of the curve $$E={(x,y,z):z=x^2+y^2,x+y+z=1}$$
To do so I did a change of variables taking as basis the vectors $frac{1}{sqrt{2}}(-1,1,0),frac{1}{sqrt{6}}(-1,-1,2),frac{1}{sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=sqrt{3}/3$, and $$begin{cases}x=-x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3}\y=x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3} \z=2y'/sqrt{6}+z'/sqrt{3}end{cases}$$
Now using this change of variables into $E$ I have that $$frac{(x')^2}{frac{1}{(3^{(1/4)})^2}}+frac{(y'-sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}costheta,;y'=sqrt{2/3}+3^{1/4}sintheta,; z'=sqrt{3}/3,; 0leq theta leq2pi$$



As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?






multivariable-calculus parametrization







share|cite|improve this question











$endgroup$












  • $begingroup$
    There are other parametrization options. Do you want this particular one?
    $endgroup$
    – Andrei
    Jan 16 at 14:01














1












1








1


1



$begingroup$


I want to calculate the parametrization of the curve $$E={(x,y,z):z=x^2+y^2,x+y+z=1}$$
To do so I did a change of variables taking as basis the vectors $frac{1}{sqrt{2}}(-1,1,0),frac{1}{sqrt{6}}(-1,-1,2),frac{1}{sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=sqrt{3}/3$, and $$begin{cases}x=-x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3}\y=x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3} \z=2y'/sqrt{6}+z'/sqrt{3}end{cases}$$
Now using this change of variables into $E$ I have that $$frac{(x')^2}{frac{1}{(3^{(1/4)})^2}}+frac{(y'-sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}costheta,;y'=sqrt{2/3}+3^{1/4}sintheta,; z'=sqrt{3}/3,; 0leq theta leq2pi$$



As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?






multivariable-calculus parametrization







share|cite|improve this question











$endgroup$




I want to calculate the parametrization of the curve $$E={(x,y,z):z=x^2+y^2,x+y+z=1}$$
To do so I did a change of variables taking as basis the vectors $frac{1}{sqrt{2}}(-1,1,0),frac{1}{sqrt{6}}(-1,-1,2),frac{1}{sqrt{3}}(1,1,1)$ which are the normal vector of the plane, and two other perpendicular vecors of the plane. With this basis the plane $x+y+z=1$ becomes $z'=sqrt{3}/3$, and $$begin{cases}x=-x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3}\y=x'/sqrt{2}-y'/sqrt{6}+z'/sqrt{3} \z=2y'/sqrt{6}+z'/sqrt{3}end{cases}$$
Now using this change of variables into $E$ I have that $$frac{(x')^2}{frac{1}{(3^{(1/4)})^2}}+frac{(y'-sqrt{2/3})^2}{(3^{(1/4)})^2}=1$$ and therefore the parametrization of the ellipse is $$x'=3^{1/4}costheta,;y'=sqrt{2/3}+3^{1/4}sintheta,; z'=sqrt{3}/3,; 0leq theta leq2pi$$



As I checked with wolfram alpha, this is correct. However it requires a lot of calculations to just parametrize this curve, is there any faster way to do it?






multivariable-calculus parametrization




multivariable-calculus parametrization






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 16 at 13:46









user376343

3,8133828




3,8133828










asked Jan 16 at 13:37









John KeeperJohn Keeper

532315




532315












  • $begingroup$
    There are other parametrization options. Do you want this particular one?
    $endgroup$
    – Andrei
    Jan 16 at 14:01


















  • $begingroup$
    There are other parametrization options. Do you want this particular one?
    $endgroup$
    – Andrei
    Jan 16 at 14:01
















$begingroup$
There are other parametrization options. Do you want this particular one?
$endgroup$
– Andrei
Jan 16 at 14:01




$begingroup$
There are other parametrization options. Do you want this particular one?
$endgroup$
– Andrei
Jan 16 at 14:01










2 Answers
2






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1












$begingroup$

You can parametrize in the original coordinates. Solving the plane for $z$ you have:
$$x+y+z=1 iff z = 1-x-y tag{$star$}$$
so substituting into the paraboloid and rewriting gives:
$$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
$$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
to get:
$$gamma : [0,2pi]tomathbb{R}^3: t mapsto
left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$






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    0












    $begingroup$

    $$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$



    Solve for $rho$ in terms of $theta$ and you are done.






    share|cite|improve this answer









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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      You can parametrize in the original coordinates. Solving the plane for $z$ you have:
      $$x+y+z=1 iff z = 1-x-y tag{$star$}$$
      so substituting into the paraboloid and rewriting gives:
      $$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
      This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
      $$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
      to get:
      $$gamma : [0,2pi]tomathbb{R}^3: t mapsto
      left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
      sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
      2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$






      share|cite|improve this answer











      $endgroup$


















        1












        $begingroup$

        You can parametrize in the original coordinates. Solving the plane for $z$ you have:
        $$x+y+z=1 iff z = 1-x-y tag{$star$}$$
        so substituting into the paraboloid and rewriting gives:
        $$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
        This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
        $$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
        to get:
        $$gamma : [0,2pi]tomathbb{R}^3: t mapsto
        left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
        sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
        2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$






        share|cite|improve this answer











        $endgroup$
















          1












          1








          1





          $begingroup$

          You can parametrize in the original coordinates. Solving the plane for $z$ you have:
          $$x+y+z=1 iff z = 1-x-y tag{$star$}$$
          so substituting into the paraboloid and rewriting gives:
          $$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
          This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
          $$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
          to get:
          $$gamma : [0,2pi]tomathbb{R}^3: t mapsto
          left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
          sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
          2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$






          share|cite|improve this answer











          $endgroup$



          You can parametrize in the original coordinates. Solving the plane for $z$ you have:
          $$x+y+z=1 iff z = 1-x-y tag{$star$}$$
          so substituting into the paraboloid and rewriting gives:
          $$z=x^2+y^2 implies 1-x-y=x^2+y^2 iff left(x+tfrac{1}{2}right)^2+left(y+tfrac{1}{2}right)^2=tfrac{3}{2}$$
          This is easily parametrized as $x+tfrac{1}{2}=sqrt{tfrac{3}{2}}cos t$ and $y+tfrac{1}{2}=sqrt{tfrac{3}{2}}sin t$ and $z$ follows from $(star)$:
          $$z = 1-x-y=1-left(sqrt{tfrac{3}{2}}cos t-tfrac{1}{2}right)-left(sqrt{tfrac{3}{2}}sin t-tfrac{1}{2}right)=2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t$$
          to get:
          $$gamma : [0,2pi]tomathbb{R}^3: t mapsto
          left( sqrt{tfrac{3}{2}}cos t-tfrac{1}{2} ,
          sqrt{tfrac{3}{2}}sin t-tfrac{1}{2} ,
          2-sqrt{tfrac{3}{2}}cos t-sqrt{tfrac{3}{2}}sin t right)$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Jan 16 at 14:29

























          answered Jan 16 at 14:17









          StackTDStackTD

          22.9k2152




          22.9k2152























              0












              $begingroup$

              $$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$



              Solve for $rho$ in terms of $theta$ and you are done.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$



                Solve for $rho$ in terms of $theta$ and you are done.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$



                  Solve for $rho$ in terms of $theta$ and you are done.






                  share|cite|improve this answer









                  $endgroup$



                  $$z=x^2+y^2$$ hints the use of polar coordinates and the equation of the plane becomes $$rho^2+rho(costheta+sintheta)-1=0.$$



                  Solve for $rho$ in terms of $theta$ and you are done.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 16 at 14:50









                  Yves DaoustYves Daoust

                  128k675227




                  128k675227






























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