Find the tangent space of Ellipsoid $M = {(x,y,z)|frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1}$












0












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Find the tangent space of
$$M = {(x,y,z)|frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1}$$




So I know the formula of tangent space for a manifold represnted by $F$ such that $F=0$: it is $ker (DF)$.



So I'll define - $F = frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} - 1$ and of course $F=0$.



By definition, $DF = (frac{2x}{a^2},frac{2y}{b^2},frac{2z}{c^2})$ and we just need to find $ker (DF)$.



Besides $x=y=z=0$, the solutions are $(x,y,(-frac{x^2}{a^2} -frac{y^2}{b^2} )c^2)$, $(x,(-frac{x^2}{a^2} -frac{z^2}{c^2})b^2,z)$ and $(-frac{y^2}{b^2} -frac{z^2}{c^2} )a^2,y,z)$.



But what is the final tangent space that is spanned by these solutions?






calculus multivariable-calculus differential-geometry manifolds tangent-spaces







share|cite|improve this question











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    0












    $begingroup$



    Find the tangent space of
    $$M = {(x,y,z)|frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1}$$




    So I know the formula of tangent space for a manifold represnted by $F$ such that $F=0$: it is $ker (DF)$.



    So I'll define - $F = frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} - 1$ and of course $F=0$.



    By definition, $DF = (frac{2x}{a^2},frac{2y}{b^2},frac{2z}{c^2})$ and we just need to find $ker (DF)$.



    Besides $x=y=z=0$, the solutions are $(x,y,(-frac{x^2}{a^2} -frac{y^2}{b^2} )c^2)$, $(x,(-frac{x^2}{a^2} -frac{z^2}{c^2})b^2,z)$ and $(-frac{y^2}{b^2} -frac{z^2}{c^2} )a^2,y,z)$.



    But what is the final tangent space that is spanned by these solutions?






    calculus multivariable-calculus differential-geometry manifolds tangent-spaces







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$



      Find the tangent space of
      $$M = {(x,y,z)|frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1}$$




      So I know the formula of tangent space for a manifold represnted by $F$ such that $F=0$: it is $ker (DF)$.



      So I'll define - $F = frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} - 1$ and of course $F=0$.



      By definition, $DF = (frac{2x}{a^2},frac{2y}{b^2},frac{2z}{c^2})$ and we just need to find $ker (DF)$.



      Besides $x=y=z=0$, the solutions are $(x,y,(-frac{x^2}{a^2} -frac{y^2}{b^2} )c^2)$, $(x,(-frac{x^2}{a^2} -frac{z^2}{c^2})b^2,z)$ and $(-frac{y^2}{b^2} -frac{z^2}{c^2} )a^2,y,z)$.



      But what is the final tangent space that is spanned by these solutions?






      calculus multivariable-calculus differential-geometry manifolds tangent-spaces







      share|cite|improve this question











      $endgroup$





      Find the tangent space of
      $$M = {(x,y,z)|frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2}=1}$$




      So I know the formula of tangent space for a manifold represnted by $F$ such that $F=0$: it is $ker (DF)$.



      So I'll define - $F = frac{x^2}{a^2}+frac{y^2}{b^2}+frac{z^2}{c^2} - 1$ and of course $F=0$.



      By definition, $DF = (frac{2x}{a^2},frac{2y}{b^2},frac{2z}{c^2})$ and we just need to find $ker (DF)$.



      Besides $x=y=z=0$, the solutions are $(x,y,(-frac{x^2}{a^2} -frac{y^2}{b^2} )c^2)$, $(x,(-frac{x^2}{a^2} -frac{z^2}{c^2})b^2,z)$ and $(-frac{y^2}{b^2} -frac{z^2}{c^2} )a^2,y,z)$.



      But what is the final tangent space that is spanned by these solutions?






      calculus multivariable-calculus differential-geometry manifolds tangent-spaces




      calculus multivariable-calculus differential-geometry manifolds tangent-spaces






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Dec 8 '18 at 17:22









      Nosrati

      26.5k62354




      26.5k62354










      asked Dec 5 '18 at 22:23









      ChikChakChikChak

      757418




      757418






















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          $begingroup$

          At a point $p in M$, the tangent space to $M$ is given by $ker(DF(p))$, as you said. So, in your case, it is the set of points $(x,y,z) in mathbb{R^3}$ such that (I divided the $2$ which comes from differentiating)



          begin{align*}
          begin{pmatrix}
          dfrac{p_1}{a^2} & dfrac{p_2}{b^2} & dfrac{p_3}{c^2}
          end{pmatrix}
          cdot
          begin{pmatrix}
          x \
          y \
          z
          end{pmatrix}
          &= 0 \
          dfrac{p_1}{a^2}x + dfrac{p_2}{b^2}y + dfrac{p_3}{c^2}z &= 0
          end{align*}



          In other words the tangent space of $M$ at $p$ is the plane through the origin given by the equation above.



          (If you want the actual tangent plane to $M$ at the point $p$, you simply have to translate the plane to pass through $p$, by replacing $(x,y,z)$ with $(x-p_1, y-p_2, z-p_3)$)






          share|cite|improve this answer









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            $begingroup$

            At a point $p in M$, the tangent space to $M$ is given by $ker(DF(p))$, as you said. So, in your case, it is the set of points $(x,y,z) in mathbb{R^3}$ such that (I divided the $2$ which comes from differentiating)



            begin{align*}
            begin{pmatrix}
            dfrac{p_1}{a^2} & dfrac{p_2}{b^2} & dfrac{p_3}{c^2}
            end{pmatrix}
            cdot
            begin{pmatrix}
            x \
            y \
            z
            end{pmatrix}
            &= 0 \
            dfrac{p_1}{a^2}x + dfrac{p_2}{b^2}y + dfrac{p_3}{c^2}z &= 0
            end{align*}



            In other words the tangent space of $M$ at $p$ is the plane through the origin given by the equation above.



            (If you want the actual tangent plane to $M$ at the point $p$, you simply have to translate the plane to pass through $p$, by replacing $(x,y,z)$ with $(x-p_1, y-p_2, z-p_3)$)






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              At a point $p in M$, the tangent space to $M$ is given by $ker(DF(p))$, as you said. So, in your case, it is the set of points $(x,y,z) in mathbb{R^3}$ such that (I divided the $2$ which comes from differentiating)



              begin{align*}
              begin{pmatrix}
              dfrac{p_1}{a^2} & dfrac{p_2}{b^2} & dfrac{p_3}{c^2}
              end{pmatrix}
              cdot
              begin{pmatrix}
              x \
              y \
              z
              end{pmatrix}
              &= 0 \
              dfrac{p_1}{a^2}x + dfrac{p_2}{b^2}y + dfrac{p_3}{c^2}z &= 0
              end{align*}



              In other words the tangent space of $M$ at $p$ is the plane through the origin given by the equation above.



              (If you want the actual tangent plane to $M$ at the point $p$, you simply have to translate the plane to pass through $p$, by replacing $(x,y,z)$ with $(x-p_1, y-p_2, z-p_3)$)






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                At a point $p in M$, the tangent space to $M$ is given by $ker(DF(p))$, as you said. So, in your case, it is the set of points $(x,y,z) in mathbb{R^3}$ such that (I divided the $2$ which comes from differentiating)



                begin{align*}
                begin{pmatrix}
                dfrac{p_1}{a^2} & dfrac{p_2}{b^2} & dfrac{p_3}{c^2}
                end{pmatrix}
                cdot
                begin{pmatrix}
                x \
                y \
                z
                end{pmatrix}
                &= 0 \
                dfrac{p_1}{a^2}x + dfrac{p_2}{b^2}y + dfrac{p_3}{c^2}z &= 0
                end{align*}



                In other words the tangent space of $M$ at $p$ is the plane through the origin given by the equation above.



                (If you want the actual tangent plane to $M$ at the point $p$, you simply have to translate the plane to pass through $p$, by replacing $(x,y,z)$ with $(x-p_1, y-p_2, z-p_3)$)






                share|cite|improve this answer









                $endgroup$



                At a point $p in M$, the tangent space to $M$ is given by $ker(DF(p))$, as you said. So, in your case, it is the set of points $(x,y,z) in mathbb{R^3}$ such that (I divided the $2$ which comes from differentiating)



                begin{align*}
                begin{pmatrix}
                dfrac{p_1}{a^2} & dfrac{p_2}{b^2} & dfrac{p_3}{c^2}
                end{pmatrix}
                cdot
                begin{pmatrix}
                x \
                y \
                z
                end{pmatrix}
                &= 0 \
                dfrac{p_1}{a^2}x + dfrac{p_2}{b^2}y + dfrac{p_3}{c^2}z &= 0
                end{align*}



                In other words the tangent space of $M$ at $p$ is the plane through the origin given by the equation above.



                (If you want the actual tangent plane to $M$ at the point $p$, you simply have to translate the plane to pass through $p$, by replacing $(x,y,z)$ with $(x-p_1, y-p_2, z-p_3)$)







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 28 '18 at 19:21









                peek-a-boopeek-a-boo

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