Finding parametric equations of the tangent line to a curve of intersection












3












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The question asks to find the parametric equations of the tangent line to the curve of intersection of the surface $z=2sqrt{9-frac{x^2}{2}-y^2}$ and the plane $x=2$ at the point $(2,sqrt{3},4)$



I've seen solutions to questions similar to this one but they use the gradient notation (the triangle) which I am unfamiliar with. How do you do this question? I was thinking of starting with the partial derivatives w.r.t $x$ and $y$ but I'm not sure where to go from there.



Then I thought of ignoring the partial derivative of $x$ since the intersection involves $x=2$ so $x$ is unchanging, so there's that.



Thanks!






calculus multivariable-calculus parametric curves







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  • 2




    $begingroup$
    Your curve is $alpha=(2,y,2sqrt{7-y^2})$ which is a half circle on the $yz$ plane. Do yo know how to construct the tangent line from here?
    $endgroup$
    – Dog_69
    Dec 30 '18 at 22:02












  • $begingroup$
    @Dog_69 isn't it a half ellipse?
    $endgroup$
    – user376343
    Dec 30 '18 at 22:26










  • $begingroup$
    @user376343 Probably haha. Well done.
    $endgroup$
    – Dog_69
    Dec 31 '18 at 9:50
















3












$begingroup$


The question asks to find the parametric equations of the tangent line to the curve of intersection of the surface $z=2sqrt{9-frac{x^2}{2}-y^2}$ and the plane $x=2$ at the point $(2,sqrt{3},4)$



I've seen solutions to questions similar to this one but they use the gradient notation (the triangle) which I am unfamiliar with. How do you do this question? I was thinking of starting with the partial derivatives w.r.t $x$ and $y$ but I'm not sure where to go from there.



Then I thought of ignoring the partial derivative of $x$ since the intersection involves $x=2$ so $x$ is unchanging, so there's that.



Thanks!






calculus multivariable-calculus parametric curves







share|cite|improve this question









$endgroup$








  • 2




    $begingroup$
    Your curve is $alpha=(2,y,2sqrt{7-y^2})$ which is a half circle on the $yz$ plane. Do yo know how to construct the tangent line from here?
    $endgroup$
    – Dog_69
    Dec 30 '18 at 22:02












  • $begingroup$
    @Dog_69 isn't it a half ellipse?
    $endgroup$
    – user376343
    Dec 30 '18 at 22:26










  • $begingroup$
    @user376343 Probably haha. Well done.
    $endgroup$
    – Dog_69
    Dec 31 '18 at 9:50














3












3








3





$begingroup$


The question asks to find the parametric equations of the tangent line to the curve of intersection of the surface $z=2sqrt{9-frac{x^2}{2}-y^2}$ and the plane $x=2$ at the point $(2,sqrt{3},4)$



I've seen solutions to questions similar to this one but they use the gradient notation (the triangle) which I am unfamiliar with. How do you do this question? I was thinking of starting with the partial derivatives w.r.t $x$ and $y$ but I'm not sure where to go from there.



Then I thought of ignoring the partial derivative of $x$ since the intersection involves $x=2$ so $x$ is unchanging, so there's that.



Thanks!






calculus multivariable-calculus parametric curves







share|cite|improve this question









$endgroup$




The question asks to find the parametric equations of the tangent line to the curve of intersection of the surface $z=2sqrt{9-frac{x^2}{2}-y^2}$ and the plane $x=2$ at the point $(2,sqrt{3},4)$



I've seen solutions to questions similar to this one but they use the gradient notation (the triangle) which I am unfamiliar with. How do you do this question? I was thinking of starting with the partial derivatives w.r.t $x$ and $y$ but I'm not sure where to go from there.



Then I thought of ignoring the partial derivative of $x$ since the intersection involves $x=2$ so $x$ is unchanging, so there's that.



Thanks!






calculus multivariable-calculus parametric curves




calculus multivariable-calculus parametric curves






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 30 '18 at 21:56









ChristopherChristopher

373




373








  • 2




    $begingroup$
    Your curve is $alpha=(2,y,2sqrt{7-y^2})$ which is a half circle on the $yz$ plane. Do yo know how to construct the tangent line from here?
    $endgroup$
    – Dog_69
    Dec 30 '18 at 22:02












  • $begingroup$
    @Dog_69 isn't it a half ellipse?
    $endgroup$
    – user376343
    Dec 30 '18 at 22:26










  • $begingroup$
    @user376343 Probably haha. Well done.
    $endgroup$
    – Dog_69
    Dec 31 '18 at 9:50














  • 2




    $begingroup$
    Your curve is $alpha=(2,y,2sqrt{7-y^2})$ which is a half circle on the $yz$ plane. Do yo know how to construct the tangent line from here?
    $endgroup$
    – Dog_69
    Dec 30 '18 at 22:02












  • $begingroup$
    @Dog_69 isn't it a half ellipse?
    $endgroup$
    – user376343
    Dec 30 '18 at 22:26










  • $begingroup$
    @user376343 Probably haha. Well done.
    $endgroup$
    – Dog_69
    Dec 31 '18 at 9:50








2




2




$begingroup$
Your curve is $alpha=(2,y,2sqrt{7-y^2})$ which is a half circle on the $yz$ plane. Do yo know how to construct the tangent line from here?
$endgroup$
– Dog_69
Dec 30 '18 at 22:02






$begingroup$
Your curve is $alpha=(2,y,2sqrt{7-y^2})$ which is a half circle on the $yz$ plane. Do yo know how to construct the tangent line from here?
$endgroup$
– Dog_69
Dec 30 '18 at 22:02














$begingroup$
@Dog_69 isn't it a half ellipse?
$endgroup$
– user376343
Dec 30 '18 at 22:26




$begingroup$
@Dog_69 isn't it a half ellipse?
$endgroup$
– user376343
Dec 30 '18 at 22:26












$begingroup$
@user376343 Probably haha. Well done.
$endgroup$
– Dog_69
Dec 31 '18 at 9:50




$begingroup$
@user376343 Probably haha. Well done.
$endgroup$
– Dog_69
Dec 31 '18 at 9:50










2 Answers
2






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$begingroup$

The parametric equation of the curve is
$$
vec v=(x,y,z)=(2,t,2sqrt{7-t^2})
$$
so, deriving with respect to $t$, we find



$$
frac {dvec v}{dt}=left(0,1,frac{-2t}{sqrt{7-t^2}}right)
$$
and the tangent vector at the point $(2,sqrt{3},4)$ is $vec u=(0,1,-sqrt{3})$.



From this you can find the equation of the line with this direction and passing thorough the given point.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    The gradient will give a vector normal to the surface (when you treat it as a level surface.) But you can get normal vectors in other ways. A vector normal to the plane $x=2$ is $mathbf{i}$. A vector normal to the surface $z=f(x,y)$ is given by $-f_xmathbf{i}-f_ymathbf{j}+mathbf{k}$ evaluated at your point.



    A vector parallel to the tangent line is the cross product of these vectors.






    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









      2












      $begingroup$

      The parametric equation of the curve is
      $$
      vec v=(x,y,z)=(2,t,2sqrt{7-t^2})
      $$
      so, deriving with respect to $t$, we find



      $$
      frac {dvec v}{dt}=left(0,1,frac{-2t}{sqrt{7-t^2}}right)
      $$
      and the tangent vector at the point $(2,sqrt{3},4)$ is $vec u=(0,1,-sqrt{3})$.



      From this you can find the equation of the line with this direction and passing thorough the given point.






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        The parametric equation of the curve is
        $$
        vec v=(x,y,z)=(2,t,2sqrt{7-t^2})
        $$
        so, deriving with respect to $t$, we find



        $$
        frac {dvec v}{dt}=left(0,1,frac{-2t}{sqrt{7-t^2}}right)
        $$
        and the tangent vector at the point $(2,sqrt{3},4)$ is $vec u=(0,1,-sqrt{3})$.



        From this you can find the equation of the line with this direction and passing thorough the given point.






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          The parametric equation of the curve is
          $$
          vec v=(x,y,z)=(2,t,2sqrt{7-t^2})
          $$
          so, deriving with respect to $t$, we find



          $$
          frac {dvec v}{dt}=left(0,1,frac{-2t}{sqrt{7-t^2}}right)
          $$
          and the tangent vector at the point $(2,sqrt{3},4)$ is $vec u=(0,1,-sqrt{3})$.



          From this you can find the equation of the line with this direction and passing thorough the given point.






          share|cite|improve this answer









          $endgroup$



          The parametric equation of the curve is
          $$
          vec v=(x,y,z)=(2,t,2sqrt{7-t^2})
          $$
          so, deriving with respect to $t$, we find



          $$
          frac {dvec v}{dt}=left(0,1,frac{-2t}{sqrt{7-t^2}}right)
          $$
          and the tangent vector at the point $(2,sqrt{3},4)$ is $vec u=(0,1,-sqrt{3})$.



          From this you can find the equation of the line with this direction and passing thorough the given point.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 22:20









          Emilio NovatiEmilio Novati

          52.3k43574




          52.3k43574























              0












              $begingroup$

              The gradient will give a vector normal to the surface (when you treat it as a level surface.) But you can get normal vectors in other ways. A vector normal to the plane $x=2$ is $mathbf{i}$. A vector normal to the surface $z=f(x,y)$ is given by $-f_xmathbf{i}-f_ymathbf{j}+mathbf{k}$ evaluated at your point.



              A vector parallel to the tangent line is the cross product of these vectors.






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                The gradient will give a vector normal to the surface (when you treat it as a level surface.) But you can get normal vectors in other ways. A vector normal to the plane $x=2$ is $mathbf{i}$. A vector normal to the surface $z=f(x,y)$ is given by $-f_xmathbf{i}-f_ymathbf{j}+mathbf{k}$ evaluated at your point.



                A vector parallel to the tangent line is the cross product of these vectors.






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  The gradient will give a vector normal to the surface (when you treat it as a level surface.) But you can get normal vectors in other ways. A vector normal to the plane $x=2$ is $mathbf{i}$. A vector normal to the surface $z=f(x,y)$ is given by $-f_xmathbf{i}-f_ymathbf{j}+mathbf{k}$ evaluated at your point.



                  A vector parallel to the tangent line is the cross product of these vectors.






                  share|cite|improve this answer









                  $endgroup$



                  The gradient will give a vector normal to the surface (when you treat it as a level surface.) But you can get normal vectors in other ways. A vector normal to the plane $x=2$ is $mathbf{i}$. A vector normal to the surface $z=f(x,y)$ is given by $-f_xmathbf{i}-f_ymathbf{j}+mathbf{k}$ evaluated at your point.



                  A vector parallel to the tangent line is the cross product of these vectors.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Dec 30 '18 at 22:43









                  B. GoddardB. Goddard

                  20.2k21543




                  20.2k21543






























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