Does parametrizing a function of three independent variables reduce the function to one independent variable?












0












$begingroup$


For example, $f(x,y,z)= x^2 + 2y^2 + 3z^2$



Now, if we parametrize the independent variables $(x,y,z)$ in terms of $u$ and it happens to be $x=u, y=u-1, z=u+1$ (just an assumption), then the independent variables can now be related as;



$x=u,$



$y= u-1 = x-1 ⇒ 1 = x-y$



$z=u+1 ⇒ z = (x)+(x-y)= 2x-y = 2x-(x-1) = x+1 $



Is the function reduced to a one independent variable?
What am I missing that leads me to this confusion?
I am a beginner for multi variable calculus.



https://youtu.be/nIJQPX5kxp4?list=PLDesaqWTN6ESk16YRmzuJ8f6-rnuy0Ry7&t=1053 source of origin of confusion (17:32 to 20:28).






calculus multivariable-calculus parametrization change-of-variable







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    0












    $begingroup$


    For example, $f(x,y,z)= x^2 + 2y^2 + 3z^2$



    Now, if we parametrize the independent variables $(x,y,z)$ in terms of $u$ and it happens to be $x=u, y=u-1, z=u+1$ (just an assumption), then the independent variables can now be related as;



    $x=u,$



    $y= u-1 = x-1 ⇒ 1 = x-y$



    $z=u+1 ⇒ z = (x)+(x-y)= 2x-y = 2x-(x-1) = x+1 $



    Is the function reduced to a one independent variable?
    What am I missing that leads me to this confusion?
    I am a beginner for multi variable calculus.



    https://youtu.be/nIJQPX5kxp4?list=PLDesaqWTN6ESk16YRmzuJ8f6-rnuy0Ry7&t=1053 source of origin of confusion (17:32 to 20:28).






    calculus multivariable-calculus parametrization change-of-variable







    share|cite|improve this question











    $endgroup$















      0












      0








      0





      $begingroup$


      For example, $f(x,y,z)= x^2 + 2y^2 + 3z^2$



      Now, if we parametrize the independent variables $(x,y,z)$ in terms of $u$ and it happens to be $x=u, y=u-1, z=u+1$ (just an assumption), then the independent variables can now be related as;



      $x=u,$



      $y= u-1 = x-1 ⇒ 1 = x-y$



      $z=u+1 ⇒ z = (x)+(x-y)= 2x-y = 2x-(x-1) = x+1 $



      Is the function reduced to a one independent variable?
      What am I missing that leads me to this confusion?
      I am a beginner for multi variable calculus.



      https://youtu.be/nIJQPX5kxp4?list=PLDesaqWTN6ESk16YRmzuJ8f6-rnuy0Ry7&t=1053 source of origin of confusion (17:32 to 20:28).






      calculus multivariable-calculus parametrization change-of-variable







      share|cite|improve this question











      $endgroup$




      For example, $f(x,y,z)= x^2 + 2y^2 + 3z^2$



      Now, if we parametrize the independent variables $(x,y,z)$ in terms of $u$ and it happens to be $x=u, y=u-1, z=u+1$ (just an assumption), then the independent variables can now be related as;



      $x=u,$



      $y= u-1 = x-1 ⇒ 1 = x-y$



      $z=u+1 ⇒ z = (x)+(x-y)= 2x-y = 2x-(x-1) = x+1 $



      Is the function reduced to a one independent variable?
      What am I missing that leads me to this confusion?
      I am a beginner for multi variable calculus.



      https://youtu.be/nIJQPX5kxp4?list=PLDesaqWTN6ESk16YRmzuJ8f6-rnuy0Ry7&t=1053 source of origin of confusion (17:32 to 20:28).






      calculus multivariable-calculus parametrization change-of-variable




      calculus multivariable-calculus parametrization change-of-variable






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      share|cite|improve this question













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      edited Dec 2 '18 at 23:40







      Abbas Miya

















      asked Dec 2 '18 at 23:37









      Abbas MiyaAbbas Miya

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          $begingroup$

          It's not so much that the function is 'reduced' per say, it is still a function of three variables, it's just that you have isolated a subset of this function, and restricted your attention there. One way to understand what's going on here is to think of your function as a graph of a function in $mathbb{R}^4$. You've got three independent variables and one dependent variable. What you have done here is slice this graph by hypersurfaces in $y$ and $z$, since you set $u = x$.



          So the graph of the function is a $3D$ object living in $4D$ space, and you cut it with two other $3D$ objects. To understand what's going on here, think about what happens in the case of one dimension lower. If you have a surface graph in $mathbb{R}^3$, and you intersect it with another surface, in general, you will get a curve as their intersection. Here, intersecting two $3D$ objects in general will give a surface. Doing another such intersection gives you a curve.



          So the two further equations you are imposing, $y = x - 1$ and $z = x + 1$ can be thought of as 'cutting out' a curve in this $3D$ graph.






          share|cite|improve this answer









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            $begingroup$

            It's not so much that the function is 'reduced' per say, it is still a function of three variables, it's just that you have isolated a subset of this function, and restricted your attention there. One way to understand what's going on here is to think of your function as a graph of a function in $mathbb{R}^4$. You've got three independent variables and one dependent variable. What you have done here is slice this graph by hypersurfaces in $y$ and $z$, since you set $u = x$.



            So the graph of the function is a $3D$ object living in $4D$ space, and you cut it with two other $3D$ objects. To understand what's going on here, think about what happens in the case of one dimension lower. If you have a surface graph in $mathbb{R}^3$, and you intersect it with another surface, in general, you will get a curve as their intersection. Here, intersecting two $3D$ objects in general will give a surface. Doing another such intersection gives you a curve.



            So the two further equations you are imposing, $y = x - 1$ and $z = x + 1$ can be thought of as 'cutting out' a curve in this $3D$ graph.






            share|cite|improve this answer









            $endgroup$


















              1












              $begingroup$

              It's not so much that the function is 'reduced' per say, it is still a function of three variables, it's just that you have isolated a subset of this function, and restricted your attention there. One way to understand what's going on here is to think of your function as a graph of a function in $mathbb{R}^4$. You've got three independent variables and one dependent variable. What you have done here is slice this graph by hypersurfaces in $y$ and $z$, since you set $u = x$.



              So the graph of the function is a $3D$ object living in $4D$ space, and you cut it with two other $3D$ objects. To understand what's going on here, think about what happens in the case of one dimension lower. If you have a surface graph in $mathbb{R}^3$, and you intersect it with another surface, in general, you will get a curve as their intersection. Here, intersecting two $3D$ objects in general will give a surface. Doing another such intersection gives you a curve.



              So the two further equations you are imposing, $y = x - 1$ and $z = x + 1$ can be thought of as 'cutting out' a curve in this $3D$ graph.






              share|cite|improve this answer









              $endgroup$
















                1












                1








                1





                $begingroup$

                It's not so much that the function is 'reduced' per say, it is still a function of three variables, it's just that you have isolated a subset of this function, and restricted your attention there. One way to understand what's going on here is to think of your function as a graph of a function in $mathbb{R}^4$. You've got three independent variables and one dependent variable. What you have done here is slice this graph by hypersurfaces in $y$ and $z$, since you set $u = x$.



                So the graph of the function is a $3D$ object living in $4D$ space, and you cut it with two other $3D$ objects. To understand what's going on here, think about what happens in the case of one dimension lower. If you have a surface graph in $mathbb{R}^3$, and you intersect it with another surface, in general, you will get a curve as their intersection. Here, intersecting two $3D$ objects in general will give a surface. Doing another such intersection gives you a curve.



                So the two further equations you are imposing, $y = x - 1$ and $z = x + 1$ can be thought of as 'cutting out' a curve in this $3D$ graph.






                share|cite|improve this answer









                $endgroup$



                It's not so much that the function is 'reduced' per say, it is still a function of three variables, it's just that you have isolated a subset of this function, and restricted your attention there. One way to understand what's going on here is to think of your function as a graph of a function in $mathbb{R}^4$. You've got three independent variables and one dependent variable. What you have done here is slice this graph by hypersurfaces in $y$ and $z$, since you set $u = x$.



                So the graph of the function is a $3D$ object living in $4D$ space, and you cut it with two other $3D$ objects. To understand what's going on here, think about what happens in the case of one dimension lower. If you have a surface graph in $mathbb{R}^3$, and you intersect it with another surface, in general, you will get a curve as their intersection. Here, intersecting two $3D$ objects in general will give a surface. Doing another such intersection gives you a curve.



                So the two further equations you are imposing, $y = x - 1$ and $z = x + 1$ can be thought of as 'cutting out' a curve in this $3D$ graph.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 2 '18 at 23:49









                Alfred YergerAlfred Yerger

                10.3k2148




                10.3k2148






























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