Intuitive idea tangent plane to a surface












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Given a function $f:mathbb{R}^2tomathbb{R}$ I'm trying to reach, intuitively, to the definition of tangent plane to the surface $G(f)$ defined by $z=f(x,y)$ at the point $p=(a,b,f(a,b))$. In the book I'm using it is said that, to get a satisfactory definition, we require that the set ${(v,D_vf(a,b)):vin mathbb{R}^2}$ is a vector subspace and ${(v,D_vf(a,b)):vin mathbb{R}^2} = T_p(G(f))$. My first question is: what does this set ${(v,D_vf(a,b)):vin mathbb{R}^2}$ represents intuitively?






calculus derivatives







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  • $begingroup$
    Did you mean $f(a,b)$ instead of $f(a)$?
    $endgroup$
    – J_Psi
    Nov 30 '18 at 8:39










  • $begingroup$
    yes, sorry, I have edited
    $endgroup$
    – Seven
    Nov 30 '18 at 9:07










  • $begingroup$
    What is your definition of $D_v$?
    $endgroup$
    – Christian Blatter
    Nov 30 '18 at 15:18
















0












$begingroup$


Given a function $f:mathbb{R}^2tomathbb{R}$ I'm trying to reach, intuitively, to the definition of tangent plane to the surface $G(f)$ defined by $z=f(x,y)$ at the point $p=(a,b,f(a,b))$. In the book I'm using it is said that, to get a satisfactory definition, we require that the set ${(v,D_vf(a,b)):vin mathbb{R}^2}$ is a vector subspace and ${(v,D_vf(a,b)):vin mathbb{R}^2} = T_p(G(f))$. My first question is: what does this set ${(v,D_vf(a,b)):vin mathbb{R}^2}$ represents intuitively?






calculus derivatives







share|cite|improve this question











$endgroup$












  • $begingroup$
    Did you mean $f(a,b)$ instead of $f(a)$?
    $endgroup$
    – J_Psi
    Nov 30 '18 at 8:39










  • $begingroup$
    yes, sorry, I have edited
    $endgroup$
    – Seven
    Nov 30 '18 at 9:07










  • $begingroup$
    What is your definition of $D_v$?
    $endgroup$
    – Christian Blatter
    Nov 30 '18 at 15:18














0












0








0





$begingroup$


Given a function $f:mathbb{R}^2tomathbb{R}$ I'm trying to reach, intuitively, to the definition of tangent plane to the surface $G(f)$ defined by $z=f(x,y)$ at the point $p=(a,b,f(a,b))$. In the book I'm using it is said that, to get a satisfactory definition, we require that the set ${(v,D_vf(a,b)):vin mathbb{R}^2}$ is a vector subspace and ${(v,D_vf(a,b)):vin mathbb{R}^2} = T_p(G(f))$. My first question is: what does this set ${(v,D_vf(a,b)):vin mathbb{R}^2}$ represents intuitively?






calculus derivatives







share|cite|improve this question











$endgroup$




Given a function $f:mathbb{R}^2tomathbb{R}$ I'm trying to reach, intuitively, to the definition of tangent plane to the surface $G(f)$ defined by $z=f(x,y)$ at the point $p=(a,b,f(a,b))$. In the book I'm using it is said that, to get a satisfactory definition, we require that the set ${(v,D_vf(a,b)):vin mathbb{R}^2}$ is a vector subspace and ${(v,D_vf(a,b)):vin mathbb{R}^2} = T_p(G(f))$. My first question is: what does this set ${(v,D_vf(a,b)):vin mathbb{R}^2}$ represents intuitively?






calculus derivatives




calculus derivatives






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share|cite|improve this question













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share|cite|improve this question








edited Nov 30 '18 at 8:49







Seven

















asked Nov 30 '18 at 8:33









SevenSeven

989




989












  • $begingroup$
    Did you mean $f(a,b)$ instead of $f(a)$?
    $endgroup$
    – J_Psi
    Nov 30 '18 at 8:39










  • $begingroup$
    yes, sorry, I have edited
    $endgroup$
    – Seven
    Nov 30 '18 at 9:07










  • $begingroup$
    What is your definition of $D_v$?
    $endgroup$
    – Christian Blatter
    Nov 30 '18 at 15:18


















  • $begingroup$
    Did you mean $f(a,b)$ instead of $f(a)$?
    $endgroup$
    – J_Psi
    Nov 30 '18 at 8:39










  • $begingroup$
    yes, sorry, I have edited
    $endgroup$
    – Seven
    Nov 30 '18 at 9:07










  • $begingroup$
    What is your definition of $D_v$?
    $endgroup$
    – Christian Blatter
    Nov 30 '18 at 15:18
















$begingroup$
Did you mean $f(a,b)$ instead of $f(a)$?
$endgroup$
– J_Psi
Nov 30 '18 at 8:39




$begingroup$
Did you mean $f(a,b)$ instead of $f(a)$?
$endgroup$
– J_Psi
Nov 30 '18 at 8:39












$begingroup$
yes, sorry, I have edited
$endgroup$
– Seven
Nov 30 '18 at 9:07




$begingroup$
yes, sorry, I have edited
$endgroup$
– Seven
Nov 30 '18 at 9:07












$begingroup$
What is your definition of $D_v$?
$endgroup$
– Christian Blatter
Nov 30 '18 at 15:18




$begingroup$
What is your definition of $D_v$?
$endgroup$
– Christian Blatter
Nov 30 '18 at 15:18










3 Answers
3






active

oldest

votes


















1












$begingroup$

What do you seek beyond the familiar graphical representation?



enter image description here






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    The set $S={(v,D_vf(a,b)):vin mathbb{R}^2}$ can be mapped to the tangent plane $T_p$ at point $p = (a,b,f(a,b))$ with the mapping:



    $S rightarrow T_P: (v_x, v_y, D_vf) rightarrow (a+v_x, b+v_y, f(a,b)+D_vf)$



    $S$ will be a vector subspace of $mathbb{R}^3$ provided:



    1) $0 in S$ : this is fine as long as we define $D_0 f(a,b)=0 space forall (a,b) in mathbb{R}^2$



    2) $s in S Rightarrow lambda s in S space forall lambda in mathbb{R}$ : this requires $D_{lambda v}f=lambda D_vf$



    3) $s_1,s_2 in S Rightarrow s_1+s_2 in S$ : this requires $D_{v_1+v_2}f=D_{v_1}f+D_{v_2}f$



    Properties (2) and (3) are simply saying that $D_vf$ must be linear in its argumentv $v$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Properties 1) and 2) always verify, I think
      $endgroup$
      – Seven
      Dec 2 '18 at 5:41










    • $begingroup$
      I can't see why the function is well defined
      $endgroup$
      – Seven
      Dec 2 '18 at 5:57



















    0












    $begingroup$

    The vectors $v_x(1,0,f'_x(a,b) )$ and $v_y(0,1,f'_y(a,b) )$ are tangent to G . Then their cross product is normal to G at point $(a,b) $






    share|cite|improve this answer









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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      What do you seek beyond the familiar graphical representation?



      enter image description here






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        What do you seek beyond the familiar graphical representation?



        enter image description here






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          What do you seek beyond the familiar graphical representation?



          enter image description here






          share|cite|improve this answer









          $endgroup$



          What do you seek beyond the familiar graphical representation?



          enter image description here







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 30 '18 at 8:55









          David G. StorkDavid G. Stork

          10.2k21332




          10.2k21332























              1












              $begingroup$

              The set $S={(v,D_vf(a,b)):vin mathbb{R}^2}$ can be mapped to the tangent plane $T_p$ at point $p = (a,b,f(a,b))$ with the mapping:



              $S rightarrow T_P: (v_x, v_y, D_vf) rightarrow (a+v_x, b+v_y, f(a,b)+D_vf)$



              $S$ will be a vector subspace of $mathbb{R}^3$ provided:



              1) $0 in S$ : this is fine as long as we define $D_0 f(a,b)=0 space forall (a,b) in mathbb{R}^2$



              2) $s in S Rightarrow lambda s in S space forall lambda in mathbb{R}$ : this requires $D_{lambda v}f=lambda D_vf$



              3) $s_1,s_2 in S Rightarrow s_1+s_2 in S$ : this requires $D_{v_1+v_2}f=D_{v_1}f+D_{v_2}f$



              Properties (2) and (3) are simply saying that $D_vf$ must be linear in its argumentv $v$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Properties 1) and 2) always verify, I think
                $endgroup$
                – Seven
                Dec 2 '18 at 5:41










              • $begingroup$
                I can't see why the function is well defined
                $endgroup$
                – Seven
                Dec 2 '18 at 5:57
















              1












              $begingroup$

              The set $S={(v,D_vf(a,b)):vin mathbb{R}^2}$ can be mapped to the tangent plane $T_p$ at point $p = (a,b,f(a,b))$ with the mapping:



              $S rightarrow T_P: (v_x, v_y, D_vf) rightarrow (a+v_x, b+v_y, f(a,b)+D_vf)$



              $S$ will be a vector subspace of $mathbb{R}^3$ provided:



              1) $0 in S$ : this is fine as long as we define $D_0 f(a,b)=0 space forall (a,b) in mathbb{R}^2$



              2) $s in S Rightarrow lambda s in S space forall lambda in mathbb{R}$ : this requires $D_{lambda v}f=lambda D_vf$



              3) $s_1,s_2 in S Rightarrow s_1+s_2 in S$ : this requires $D_{v_1+v_2}f=D_{v_1}f+D_{v_2}f$



              Properties (2) and (3) are simply saying that $D_vf$ must be linear in its argumentv $v$.






              share|cite|improve this answer









              $endgroup$













              • $begingroup$
                Properties 1) and 2) always verify, I think
                $endgroup$
                – Seven
                Dec 2 '18 at 5:41










              • $begingroup$
                I can't see why the function is well defined
                $endgroup$
                – Seven
                Dec 2 '18 at 5:57














              1












              1








              1





              $begingroup$

              The set $S={(v,D_vf(a,b)):vin mathbb{R}^2}$ can be mapped to the tangent plane $T_p$ at point $p = (a,b,f(a,b))$ with the mapping:



              $S rightarrow T_P: (v_x, v_y, D_vf) rightarrow (a+v_x, b+v_y, f(a,b)+D_vf)$



              $S$ will be a vector subspace of $mathbb{R}^3$ provided:



              1) $0 in S$ : this is fine as long as we define $D_0 f(a,b)=0 space forall (a,b) in mathbb{R}^2$



              2) $s in S Rightarrow lambda s in S space forall lambda in mathbb{R}$ : this requires $D_{lambda v}f=lambda D_vf$



              3) $s_1,s_2 in S Rightarrow s_1+s_2 in S$ : this requires $D_{v_1+v_2}f=D_{v_1}f+D_{v_2}f$



              Properties (2) and (3) are simply saying that $D_vf$ must be linear in its argumentv $v$.






              share|cite|improve this answer









              $endgroup$



              The set $S={(v,D_vf(a,b)):vin mathbb{R}^2}$ can be mapped to the tangent plane $T_p$ at point $p = (a,b,f(a,b))$ with the mapping:



              $S rightarrow T_P: (v_x, v_y, D_vf) rightarrow (a+v_x, b+v_y, f(a,b)+D_vf)$



              $S$ will be a vector subspace of $mathbb{R}^3$ provided:



              1) $0 in S$ : this is fine as long as we define $D_0 f(a,b)=0 space forall (a,b) in mathbb{R}^2$



              2) $s in S Rightarrow lambda s in S space forall lambda in mathbb{R}$ : this requires $D_{lambda v}f=lambda D_vf$



              3) $s_1,s_2 in S Rightarrow s_1+s_2 in S$ : this requires $D_{v_1+v_2}f=D_{v_1}f+D_{v_2}f$



              Properties (2) and (3) are simply saying that $D_vf$ must be linear in its argumentv $v$.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 30 '18 at 11:17









              gandalf61gandalf61

              7,951625




              7,951625












              • $begingroup$
                Properties 1) and 2) always verify, I think
                $endgroup$
                – Seven
                Dec 2 '18 at 5:41










              • $begingroup$
                I can't see why the function is well defined
                $endgroup$
                – Seven
                Dec 2 '18 at 5:57


















              • $begingroup$
                Properties 1) and 2) always verify, I think
                $endgroup$
                – Seven
                Dec 2 '18 at 5:41










              • $begingroup$
                I can't see why the function is well defined
                $endgroup$
                – Seven
                Dec 2 '18 at 5:57
















              $begingroup$
              Properties 1) and 2) always verify, I think
              $endgroup$
              – Seven
              Dec 2 '18 at 5:41




              $begingroup$
              Properties 1) and 2) always verify, I think
              $endgroup$
              – Seven
              Dec 2 '18 at 5:41












              $begingroup$
              I can't see why the function is well defined
              $endgroup$
              – Seven
              Dec 2 '18 at 5:57




              $begingroup$
              I can't see why the function is well defined
              $endgroup$
              – Seven
              Dec 2 '18 at 5:57











              0












              $begingroup$

              The vectors $v_x(1,0,f'_x(a,b) )$ and $v_y(0,1,f'_y(a,b) )$ are tangent to G . Then their cross product is normal to G at point $(a,b) $






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The vectors $v_x(1,0,f'_x(a,b) )$ and $v_y(0,1,f'_y(a,b) )$ are tangent to G . Then their cross product is normal to G at point $(a,b) $






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The vectors $v_x(1,0,f'_x(a,b) )$ and $v_y(0,1,f'_y(a,b) )$ are tangent to G . Then their cross product is normal to G at point $(a,b) $






                  share|cite|improve this answer









                  $endgroup$



                  The vectors $v_x(1,0,f'_x(a,b) )$ and $v_y(0,1,f'_y(a,b) )$ are tangent to G . Then their cross product is normal to G at point $(a,b) $







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 30 '18 at 8:43









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