Does parametrizing a function of three independent variables reduce the function to one independent variable?
$begingroup$
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
asked Dec 2 '18 at 23:37
Abbas MiyaAbbas Miya
1338
$endgroup$
add a comment |
$begingroup$
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
asked Dec 2 '18 at 23:37
Abbas MiyaAbbas Miya
1338
$endgroup$
add a comment |
$begingroup$
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
asked Dec 2 '18 at 23:37
Abbas MiyaAbbas Miya
1338
$endgroup$
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
asked Dec 2 '18 at 23:37
Abbas MiyaAbbas Miya
1338
asked Dec 2 '18 at 23:37
Abbas MiyaAbbas Miya
1338
edited Dec 2 '18 at 23:40
Abbas Miya
asked Dec 2 '18 at 23:37
Abbas MiyaAbbas Miya
1338
asked Dec 2 '18 at 23:37
Abbas MiyaAbbas Miya
1338
asked Dec 2 '18 at 23:37
Abbas MiyaAbbas Miya
1338
1338
add a comment |
add a comment |
1 Answer
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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.
answered Dec 2 '18 at 23:49
Alfred YergerAlfred Yerger
10.3k2148
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1 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.
answered Dec 2 '18 at 23:49
Alfred YergerAlfred Yerger
10.3k2148
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 2 '18 at 23:49
Alfred YergerAlfred Yerger
10.3k2148
$endgroup$
add a comment |
$begingroup$
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.
answered Dec 2 '18 at 23:49
Alfred YergerAlfred Yerger
10.3k2148
$endgroup$
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.
answered Dec 2 '18 at 23:49
Alfred YergerAlfred Yerger
10.3k2148
answered Dec 2 '18 at 23:49
Alfred YergerAlfred Yerger
10.3k2148
answered Dec 2 '18 at 23:49
Alfred YergerAlfred Yerger
10.3k2148
answered Dec 2 '18 at 23:49
Alfred YergerAlfred Yerger
10.3k2148
10.3k2148
add a comment |
add a comment |
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