Mixing
Partial derivatives are 0 iff the function is constant
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Let $f:mathbb R^2rightarrow mathbb R$ has first partial derivatives and $frac{partial f}{partial x}=frac{partial f}{partial y}=0, text{for all } (x,y)inmathbb R^2$. Show that $f$ is constant. (Hint: Show that the restriction of $f$ to a line parallel to one of the coordinate axes is constant).
My attempt:
Following the hint, consider the restriction of $f$ to a line $y=c$, then $frac{partial f}{partial x} (x,c) = f_x(x,c)=0$. Pick 2 points a and b, then by mean value theorem there is $x_0$, such that $f_x(x,c) = frac{f(b,c)-f(a,c)}{b-a}=0 implies f(b,c)-f(a,c)=0 implies f(b,c)=f(a,c)implies f(x,c)=const.$
We can show the same way that $f(c,y)=const$.
Am I thinking in the right direction? If so, can I just combine $f(x,c)=const$ and $f(c,y)=const$ to get $f(x,y)=const$? Did I miss something?
real-analysis multivariable-calculus partial-derivative
asked Feb 1 at 19:35
dxdydzdxdydz
49110
$endgroup$
add a comment |
$begingroup$
Let $f:mathbb R^2rightarrow mathbb R$ has first partial derivatives and $frac{partial f}{partial x}=frac{partial f}{partial y}=0, text{for all } (x,y)inmathbb R^2$. Show that $f$ is constant. (Hint: Show that the restriction of $f$ to a line parallel to one of the coordinate axes is constant).
My attempt:
Following the hint, consider the restriction of $f$ to a line $y=c$, then $frac{partial f}{partial x} (x,c) = f_x(x,c)=0$. Pick 2 points a and b, then by mean value theorem there is $x_0$, such that $f_x(x,c) = frac{f(b,c)-f(a,c)}{b-a}=0 implies f(b,c)-f(a,c)=0 implies f(b,c)=f(a,c)implies f(x,c)=const.$
We can show the same way that $f(c,y)=const$.
Am I thinking in the right direction? If so, can I just combine $f(x,c)=const$ and $f(c,y)=const$ to get $f(x,y)=const$? Did I miss something?
real-analysis multivariable-calculus partial-derivative
asked Feb 1 at 19:35
dxdydzdxdydz
49110
$endgroup$
1
$begingroup$
That's pretty good. But how do you know all these constants are the same?
$endgroup$
– Matthew Leingang
Feb 1 at 19:39
$begingroup$
@MatthewLeingang Oh, I didn't think about that. How can I show that then?
$endgroup$
– dxdydz
Feb 1 at 19:41
$begingroup$
Maybe try to connect every point $(x,y)$ to $(0,0)$ along particular lines?
$endgroup$
– Matthew Leingang
Feb 1 at 19:46
add a comment |
$begingroup$
Let $f:mathbb R^2rightarrow mathbb R$ has first partial derivatives and $frac{partial f}{partial x}=frac{partial f}{partial y}=0, text{for all } (x,y)inmathbb R^2$. Show that $f$ is constant. (Hint: Show that the restriction of $f$ to a line parallel to one of the coordinate axes is constant).
My attempt:
Following the hint, consider the restriction of $f$ to a line $y=c$, then $frac{partial f}{partial x} (x,c) = f_x(x,c)=0$. Pick 2 points a and b, then by mean value theorem there is $x_0$, such that $f_x(x,c) = frac{f(b,c)-f(a,c)}{b-a}=0 implies f(b,c)-f(a,c)=0 implies f(b,c)=f(a,c)implies f(x,c)=const.$
We can show the same way that $f(c,y)=const$.
Am I thinking in the right direction? If so, can I just combine $f(x,c)=const$ and $f(c,y)=const$ to get $f(x,y)=const$? Did I miss something?
real-analysis multivariable-calculus partial-derivative
asked Feb 1 at 19:35
dxdydzdxdydz
49110
$endgroup$
Let $f:mathbb R^2rightarrow mathbb R$ has first partial derivatives and $frac{partial f}{partial x}=frac{partial f}{partial y}=0, text{for all } (x,y)inmathbb R^2$. Show that $f$ is constant. (Hint: Show that the restriction of $f$ to a line parallel to one of the coordinate axes is constant).
My attempt:
Following the hint, consider the restriction of $f$ to a line $y=c$, then $frac{partial f}{partial x} (x,c) = f_x(x,c)=0$. Pick 2 points a and b, then by mean value theorem there is $x_0$, such that $f_x(x,c) = frac{f(b,c)-f(a,c)}{b-a}=0 implies f(b,c)-f(a,c)=0 implies f(b,c)=f(a,c)implies f(x,c)=const.$
We can show the same way that $f(c,y)=const$.
Am I thinking in the right direction? If so, can I just combine $f(x,c)=const$ and $f(c,y)=const$ to get $f(x,y)=const$? Did I miss something?
real-analysis multivariable-calculus partial-derivative
real-analysis multivariable-calculus partial-derivative
asked Feb 1 at 19:35
dxdydzdxdydz
49110
asked Feb 1 at 19:35
dxdydzdxdydz
49110
asked Feb 1 at 19:35
dxdydzdxdydz
49110
asked Feb 1 at 19:35
dxdydzdxdydz
49110
asked Feb 1 at 19:35
dxdydzdxdydz
49110
49110
1
$begingroup$
That's pretty good. But how do you know all these constants are the same?
$endgroup$
– Matthew Leingang
Feb 1 at 19:39
$begingroup$
@MatthewLeingang Oh, I didn't think about that. How can I show that then?
$endgroup$
– dxdydz
Feb 1 at 19:41
$begingroup$
Maybe try to connect every point $(x,y)$ to $(0,0)$ along particular lines?
$endgroup$
– Matthew Leingang
Feb 1 at 19:46
add a comment |
1
$begingroup$
That's pretty good. But how do you know all these constants are the same?
$endgroup$
– Matthew Leingang
Feb 1 at 19:39
$begingroup$
@MatthewLeingang Oh, I didn't think about that. How can I show that then?
$endgroup$
– dxdydz
Feb 1 at 19:41
$begingroup$
Maybe try to connect every point $(x,y)$ to $(0,0)$ along particular lines?
$endgroup$
– Matthew Leingang
Feb 1 at 19:46
1
1
$begingroup$
That's pretty good. But how do you know all these constants are the same?
$endgroup$
– Matthew Leingang
Feb 1 at 19:39
$begingroup$
That's pretty good. But how do you know all these constants are the same?
$endgroup$
– Matthew Leingang
Feb 1 at 19:39
$begingroup$
@MatthewLeingang Oh, I didn't think about that. How can I show that then?
$endgroup$
– dxdydz
Feb 1 at 19:41
$begingroup$
@MatthewLeingang Oh, I didn't think about that. How can I show that then?
$endgroup$
– dxdydz
Feb 1 at 19:41
$begingroup$
Maybe try to connect every point $(x,y)$ to $(0,0)$ along particular lines?
$endgroup$
– Matthew Leingang
Feb 1 at 19:46
$begingroup$
Maybe try to connect every point $(x,y)$ to $(0,0)$ along particular lines?
$endgroup$
– Matthew Leingang
Feb 1 at 19:46
add a comment |
1 Answer
1
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votes
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Select two points $(x_1,y_1)$ and $(x_2,y_2)$. Use your argument about the partial derivatives to show that
$$f(x_1,y_1) = f(x_2,y_1) quad text{and} quad f(x_2,y_2) = f(x_2,y_1)$$ and in particular, $f(x_1,y_1) = f(x_2,y_2)$.
answered Feb 1 at 19:48
Umberto P.Umberto P.
40.4k13370
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add a comment |
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$begingroup$
Select two points $(x_1,y_1)$ and $(x_2,y_2)$. Use your argument about the partial derivatives to show that
$$f(x_1,y_1) = f(x_2,y_1) quad text{and} quad f(x_2,y_2) = f(x_2,y_1)$$ and in particular, $f(x_1,y_1) = f(x_2,y_2)$.
answered Feb 1 at 19:48
Umberto P.Umberto P.
40.4k13370
$endgroup$
add a comment |
$begingroup$
Select two points $(x_1,y_1)$ and $(x_2,y_2)$. Use your argument about the partial derivatives to show that
$$f(x_1,y_1) = f(x_2,y_1) quad text{and} quad f(x_2,y_2) = f(x_2,y_1)$$ and in particular, $f(x_1,y_1) = f(x_2,y_2)$.
answered Feb 1 at 19:48
Umberto P.Umberto P.
40.4k13370
$endgroup$
add a comment |
$begingroup$
Select two points $(x_1,y_1)$ and $(x_2,y_2)$. Use your argument about the partial derivatives to show that
$$f(x_1,y_1) = f(x_2,y_1) quad text{and} quad f(x_2,y_2) = f(x_2,y_1)$$ and in particular, $f(x_1,y_1) = f(x_2,y_2)$.
answered Feb 1 at 19:48
Umberto P.Umberto P.
40.4k13370
$endgroup$
Select two points $(x_1,y_1)$ and $(x_2,y_2)$. Use your argument about the partial derivatives to show that
$$f(x_1,y_1) = f(x_2,y_1) quad text{and} quad f(x_2,y_2) = f(x_2,y_1)$$ and in particular, $f(x_1,y_1) = f(x_2,y_2)$.
answered Feb 1 at 19:48
Umberto P.Umberto P.
40.4k13370
answered Feb 1 at 19:48
Umberto P.Umberto P.
40.4k13370
answered Feb 1 at 19:48
Umberto P.Umberto P.
40.4k13370
answered Feb 1 at 19:48
Umberto P.Umberto P.
40.4k13370
40.4k13370
add a comment |
add a comment |
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1
$begingroup$
That's pretty good. But how do you know all these constants are the same?
$endgroup$
– Matthew Leingang
Feb 1 at 19:39
$begingroup$
@MatthewLeingang Oh, I didn't think about that. How can I show that then?
$endgroup$
– dxdydz
Feb 1 at 19:41
$begingroup$
Maybe try to connect every point $(x,y)$ to $(0,0)$ along particular lines?
$endgroup$
– Matthew Leingang
Feb 1 at 19:46