Mixing
Notation doubt: Is $nabla f(x(u,v), y(u,v))$ the same as $nabla f(u, v)$?
1
$begingroup$
Is $nabla f(x(u,v), y(u,v))$ the same as $nabla f(u, v)$?
Background: Was solving this assignment, problem E-7.
Kinda confused here, would appreciate help.
multivariable-calculus partial-derivative
edited Feb 1 at 3:42
Blue
49.5k870158
asked Feb 1 at 3:35
StayAheadOfLifeStayAheadOfLife
264
$endgroup$
add a comment |
1
$begingroup$
Is $nabla f(x(u,v), y(u,v))$ the same as $nabla f(u, v)$?
Background: Was solving this assignment, problem E-7.
Kinda confused here, would appreciate help.
multivariable-calculus partial-derivative
edited Feb 1 at 3:42
Blue
49.5k870158
asked Feb 1 at 3:35
StayAheadOfLifeStayAheadOfLife
264
$endgroup$
$begingroup$
Yes, the idea is to verify that changing coordinates from $(x,y)$ to $(u,v)$ can be accomplished through a usual tool.
$endgroup$
– David Kraemer
Feb 1 at 3:40
1
$begingroup$
In principle, those are not the same. $f(x(u,v),y(u,v))$ is a different function of $(u,v)$ and you should call it something else, like $g(u,v)$. $g$ is the composition of $f$ with your change of variables $(u,v)to(x,y)$. The exercise is on the chain rule for functions of two variables, and you end up with things like $g_u=f_xcdot x_u+f_ycdot y_u$ and similar for the derivative with respect to $v$.
$endgroup$
– GReyes
Feb 1 at 3:49
1
$begingroup$
The distinction is very clear for mathematicians. However, in Physics or engineering books sometimes they keep the same name for the function. This is because physicists and engineers look at these as quantities, not as functions in the mathematical sense of the term. So they say $f(u,v)$, meaning that we look at the same quantity as before, but as a function of $(u,v)$ instead of $(x,y)$.
$endgroup$
– GReyes
Feb 1 at 3:52
add a comment |
1
1
1
$begingroup$
Is $nabla f(x(u,v), y(u,v))$ the same as $nabla f(u, v)$?
Background: Was solving this assignment, problem E-7.
Kinda confused here, would appreciate help.
multivariable-calculus partial-derivative
edited Feb 1 at 3:42
Blue
49.5k870158
asked Feb 1 at 3:35
StayAheadOfLifeStayAheadOfLife
264
$endgroup$
Is $nabla f(x(u,v), y(u,v))$ the same as $nabla f(u, v)$?
Background: Was solving this assignment, problem E-7.
Kinda confused here, would appreciate help.
multivariable-calculus partial-derivative
multivariable-calculus partial-derivative
edited Feb 1 at 3:42
Blue
49.5k870158
asked Feb 1 at 3:35
StayAheadOfLifeStayAheadOfLife
264
edited Feb 1 at 3:42
Blue
49.5k870158
asked Feb 1 at 3:35
StayAheadOfLifeStayAheadOfLife
264
edited Feb 1 at 3:42
Blue
49.5k870158
edited Feb 1 at 3:42
Blue
49.5k870158
edited Feb 1 at 3:42
Blue
49.5k870158
49.5k870158
asked Feb 1 at 3:35
StayAheadOfLifeStayAheadOfLife
264
asked Feb 1 at 3:35
StayAheadOfLifeStayAheadOfLife
264
asked Feb 1 at 3:35
StayAheadOfLifeStayAheadOfLife
264
264
$begingroup$
Yes, the idea is to verify that changing coordinates from $(x,y)$ to $(u,v)$ can be accomplished through a usual tool.
$endgroup$
– David Kraemer
Feb 1 at 3:40
1
$begingroup$
In principle, those are not the same. $f(x(u,v),y(u,v))$ is a different function of $(u,v)$ and you should call it something else, like $g(u,v)$. $g$ is the composition of $f$ with your change of variables $(u,v)to(x,y)$. The exercise is on the chain rule for functions of two variables, and you end up with things like $g_u=f_xcdot x_u+f_ycdot y_u$ and similar for the derivative with respect to $v$.
$endgroup$
– GReyes
Feb 1 at 3:49
1
$begingroup$
The distinction is very clear for mathematicians. However, in Physics or engineering books sometimes they keep the same name for the function. This is because physicists and engineers look at these as quantities, not as functions in the mathematical sense of the term. So they say $f(u,v)$, meaning that we look at the same quantity as before, but as a function of $(u,v)$ instead of $(x,y)$.
$endgroup$
– GReyes
Feb 1 at 3:52
add a comment |
$begingroup$
Yes, the idea is to verify that changing coordinates from $(x,y)$ to $(u,v)$ can be accomplished through a usual tool.
$endgroup$
– David Kraemer
Feb 1 at 3:40
1
$begingroup$
In principle, those are not the same. $f(x(u,v),y(u,v))$ is a different function of $(u,v)$ and you should call it something else, like $g(u,v)$. $g$ is the composition of $f$ with your change of variables $(u,v)to(x,y)$. The exercise is on the chain rule for functions of two variables, and you end up with things like $g_u=f_xcdot x_u+f_ycdot y_u$ and similar for the derivative with respect to $v$.
$endgroup$
– GReyes
Feb 1 at 3:49
1
$begingroup$
The distinction is very clear for mathematicians. However, in Physics or engineering books sometimes they keep the same name for the function. This is because physicists and engineers look at these as quantities, not as functions in the mathematical sense of the term. So they say $f(u,v)$, meaning that we look at the same quantity as before, but as a function of $(u,v)$ instead of $(x,y)$.
$endgroup$
– GReyes
Feb 1 at 3:52
$begingroup$
Yes, the idea is to verify that changing coordinates from $(x,y)$ to $(u,v)$ can be accomplished through a usual tool.
$endgroup$
– David Kraemer
Feb 1 at 3:40
$begingroup$
Yes, the idea is to verify that changing coordinates from $(x,y)$ to $(u,v)$ can be accomplished through a usual tool.
$endgroup$
– David Kraemer
Feb 1 at 3:40
1
1
$begingroup$
In principle, those are not the same. $f(x(u,v),y(u,v))$ is a different function of $(u,v)$ and you should call it something else, like $g(u,v)$. $g$ is the composition of $f$ with your change of variables $(u,v)to(x,y)$. The exercise is on the chain rule for functions of two variables, and you end up with things like $g_u=f_xcdot x_u+f_ycdot y_u$ and similar for the derivative with respect to $v$.
$endgroup$
– GReyes
Feb 1 at 3:49
$begingroup$
In principle, those are not the same. $f(x(u,v),y(u,v))$ is a different function of $(u,v)$ and you should call it something else, like $g(u,v)$. $g$ is the composition of $f$ with your change of variables $(u,v)to(x,y)$. The exercise is on the chain rule for functions of two variables, and you end up with things like $g_u=f_xcdot x_u+f_ycdot y_u$ and similar for the derivative with respect to $v$.
$endgroup$
– GReyes
Feb 1 at 3:49
1
1
$begingroup$
The distinction is very clear for mathematicians. However, in Physics or engineering books sometimes they keep the same name for the function. This is because physicists and engineers look at these as quantities, not as functions in the mathematical sense of the term. So they say $f(u,v)$, meaning that we look at the same quantity as before, but as a function of $(u,v)$ instead of $(x,y)$.
$endgroup$
– GReyes
Feb 1 at 3:52
$begingroup$
The distinction is very clear for mathematicians. However, in Physics or engineering books sometimes they keep the same name for the function. This is because physicists and engineers look at these as quantities, not as functions in the mathematical sense of the term. So they say $f(u,v)$, meaning that we look at the same quantity as before, but as a function of $(u,v)$ instead of $(x,y)$.
$endgroup$
– GReyes
Feb 1 at 3:52
add a comment |
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$begingroup$
Yes, the idea is to verify that changing coordinates from $(x,y)$ to $(u,v)$ can be accomplished through a usual tool.
$endgroup$
– David Kraemer
Feb 1 at 3:40
1
$begingroup$
In principle, those are not the same. $f(x(u,v),y(u,v))$ is a different function of $(u,v)$ and you should call it something else, like $g(u,v)$. $g$ is the composition of $f$ with your change of variables $(u,v)to(x,y)$. The exercise is on the chain rule for functions of two variables, and you end up with things like $g_u=f_xcdot x_u+f_ycdot y_u$ and similar for the derivative with respect to $v$.
$endgroup$
– GReyes
Feb 1 at 3:49
1
$begingroup$
The distinction is very clear for mathematicians. However, in Physics or engineering books sometimes they keep the same name for the function. This is because physicists and engineers look at these as quantities, not as functions in the mathematical sense of the term. So they say $f(u,v)$, meaning that we look at the same quantity as before, but as a function of $(u,v)$ instead of $(x,y)$.
$endgroup$
– GReyes
Feb 1 at 3:52