Is the differential forms perspective on $dx$ incompatible with the technique of implicit differentiation?
$begingroup$
Suppose $$x^2 + y^2 = 5^2.$$ We're trying to find $dy/dx$ at $(3,4).$
Applying $d$ to both sides: $$2x dx + 2y dy = 0$$
Or in other words:
$$2x dx + 2y dy = 0dx + 0dy$$
Since the covectors $dx_p$ and $dy_p$ form a basis for the cotangent space at any $p in mathbb{R}^2$, hence $2x = 0$ and $2y = 0.$ Hence $x = 0$ and $y = 0$. Ergo $0^2 + 0^2 = 5^2$, a contradiction.
Question. Does this mean that the differential forms perspective on $dx$ is incompatible with the technique of implicit differentiation?
If not, why not?
If so, what definition of $dx$ can be used to avoid this issue?
calculus differential-geometry differential-topology differential-forms implicit-differentiation
edited Jan 2 at 0:46
David G. Stork
10.7k31332
asked Jan 2 at 0:33
goblingoblin
36.8k1159193
$endgroup$
add a comment |
$begingroup$
Suppose $$x^2 + y^2 = 5^2.$$ We're trying to find $dy/dx$ at $(3,4).$
Applying $d$ to both sides: $$2x dx + 2y dy = 0$$
Or in other words:
$$2x dx + 2y dy = 0dx + 0dy$$
Since the covectors $dx_p$ and $dy_p$ form a basis for the cotangent space at any $p in mathbb{R}^2$, hence $2x = 0$ and $2y = 0.$ Hence $x = 0$ and $y = 0$. Ergo $0^2 + 0^2 = 5^2$, a contradiction.
Question. Does this mean that the differential forms perspective on $dx$ is incompatible with the technique of implicit differentiation?
If not, why not?
If so, what definition of $dx$ can be used to avoid this issue?
calculus differential-geometry differential-topology differential-forms implicit-differentiation
edited Jan 2 at 0:46
David G. Stork
10.7k31332
asked Jan 2 at 0:33
goblingoblin
36.8k1159193
$endgroup$
$begingroup$
The equation $x^2+y^2=5^2$ describes a curve. What do yo mean by "taking the gradient" of a curve?
$endgroup$
– Jackozee Hakkiuz
Jan 2 at 0:40
$begingroup$
@JackozeeHakkiuz, edited.
$endgroup$
– goblin
Jan 2 at 0:41
$begingroup$
$2xdx+2ydy=(2x,2y)(dx,dy)^t$ is a dot product and will be $0$ whenever $(dx,dy)$ and $(2x,2y)$ are orthogonal. This will be the case for all directions tangent to the circle because the gradient of a level curve is normal to that curve.
$endgroup$
– John Douma
Jan 2 at 0:51
add a comment |
$begingroup$
Suppose $$x^2 + y^2 = 5^2.$$ We're trying to find $dy/dx$ at $(3,4).$
Applying $d$ to both sides: $$2x dx + 2y dy = 0$$
Or in other words:
$$2x dx + 2y dy = 0dx + 0dy$$
Since the covectors $dx_p$ and $dy_p$ form a basis for the cotangent space at any $p in mathbb{R}^2$, hence $2x = 0$ and $2y = 0.$ Hence $x = 0$ and $y = 0$. Ergo $0^2 + 0^2 = 5^2$, a contradiction.
Question. Does this mean that the differential forms perspective on $dx$ is incompatible with the technique of implicit differentiation?
If not, why not?
If so, what definition of $dx$ can be used to avoid this issue?
calculus differential-geometry differential-topology differential-forms implicit-differentiation
edited Jan 2 at 0:46
David G. Stork
10.7k31332
asked Jan 2 at 0:33
goblingoblin
36.8k1159193
$endgroup$
Suppose $$x^2 + y^2 = 5^2.$$ We're trying to find $dy/dx$ at $(3,4).$
Applying $d$ to both sides: $$2x dx + 2y dy = 0$$
Or in other words:
$$2x dx + 2y dy = 0dx + 0dy$$
Since the covectors $dx_p$ and $dy_p$ form a basis for the cotangent space at any $p in mathbb{R}^2$, hence $2x = 0$ and $2y = 0.$ Hence $x = 0$ and $y = 0$. Ergo $0^2 + 0^2 = 5^2$, a contradiction.
Question. Does this mean that the differential forms perspective on $dx$ is incompatible with the technique of implicit differentiation?
If not, why not?
If so, what definition of $dx$ can be used to avoid this issue?
calculus differential-geometry differential-topology differential-forms implicit-differentiation
calculus differential-geometry differential-topology differential-forms implicit-differentiation
edited Jan 2 at 0:46
David G. Stork
10.7k31332
asked Jan 2 at 0:33
goblingoblin
36.8k1159193
edited Jan 2 at 0:46
David G. Stork
10.7k31332
asked Jan 2 at 0:33
goblingoblin
36.8k1159193
edited Jan 2 at 0:46
David G. Stork
10.7k31332
edited Jan 2 at 0:46
David G. Stork
10.7k31332
edited Jan 2 at 0:46
David G. Stork
10.7k31332
10.7k31332
asked Jan 2 at 0:33
goblingoblin
36.8k1159193
asked Jan 2 at 0:33
goblingoblin
36.8k1159193
asked Jan 2 at 0:33
goblingoblin
36.8k1159193
36.8k1159193
$begingroup$
The equation $x^2+y^2=5^2$ describes a curve. What do yo mean by "taking the gradient" of a curve?
$endgroup$
– Jackozee Hakkiuz
Jan 2 at 0:40
$begingroup$
@JackozeeHakkiuz, edited.
$endgroup$
– goblin
Jan 2 at 0:41
$begingroup$
$2xdx+2ydy=(2x,2y)(dx,dy)^t$ is a dot product and will be $0$ whenever $(dx,dy)$ and $(2x,2y)$ are orthogonal. This will be the case for all directions tangent to the circle because the gradient of a level curve is normal to that curve.
$endgroup$
– John Douma
Jan 2 at 0:51
add a comment |
$begingroup$
The equation $x^2+y^2=5^2$ describes a curve. What do yo mean by "taking the gradient" of a curve?
$endgroup$
– Jackozee Hakkiuz
Jan 2 at 0:40
$begingroup$
@JackozeeHakkiuz, edited.
$endgroup$
– goblin
Jan 2 at 0:41
$begingroup$
$2xdx+2ydy=(2x,2y)(dx,dy)^t$ is a dot product and will be $0$ whenever $(dx,dy)$ and $(2x,2y)$ are orthogonal. This will be the case for all directions tangent to the circle because the gradient of a level curve is normal to that curve.
$endgroup$
– John Douma
Jan 2 at 0:51
$begingroup$
The equation $x^2+y^2=5^2$ describes a curve. What do yo mean by "taking the gradient" of a curve?
$endgroup$
– Jackozee Hakkiuz
Jan 2 at 0:40
$begingroup$
The equation $x^2+y^2=5^2$ describes a curve. What do yo mean by "taking the gradient" of a curve?
$endgroup$
– Jackozee Hakkiuz
Jan 2 at 0:40
$begingroup$
@JackozeeHakkiuz, edited.
$endgroup$
– goblin
Jan 2 at 0:41
$begingroup$
@JackozeeHakkiuz, edited.
$endgroup$
– goblin
Jan 2 at 0:41
$begingroup$
$2xdx+2ydy=(2x,2y)(dx,dy)^t$ is a dot product and will be $0$ whenever $(dx,dy)$ and $(2x,2y)$ are orthogonal. This will be the case for all directions tangent to the circle because the gradient of a level curve is normal to that curve.
$endgroup$
– John Douma
Jan 2 at 0:51
$begingroup$
$2xdx+2ydy=(2x,2y)(dx,dy)^t$ is a dot product and will be $0$ whenever $(dx,dy)$ and $(2x,2y)$ are orthogonal. This will be the case for all directions tangent to the circle because the gradient of a level curve is normal to that curve.
$endgroup$
– John Douma
Jan 2 at 0:51
add a comment |
1 Answer
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$begingroup$
What you're missing is that this computation does not take place in $mathbb{R}^2$: it takes place in the submanifold $M$ of $mathbb{R}^2$ defined by the equation $x^2+y^2=5^2$! We can consider $x$ and $y$ as smooth functions on $M$, and these smooth functions satisfy $x^2+y^2=5^2$ at every point of $M$. So, we can differentiate and deduce that $$2xdx+2ydy=0$$ at every point of $M$. This is not a contradiction, because $dx$ and $dy$ here are sections of the cotangent bundle of $M$, not the cotangent bundle of $mathbb{R}^2$ (they are the pullback of the usual $dx$ and $dy$ on $mathbb{R}^2$ to $M$).
Note that on $mathbb{R}^2$, you cannot differentiate the equation $x^2+y^2=5^2$ since this is not actually an equation of functions, it is just an equation that is true at some points of $mathbb{R}^2$. Even at a point where the equation is true, you cannot expect the differential of $x^2+y^2$ to be equal to the differential of $5^2$, since there are nearby points where the two functions are not equal. But we don't have this problem when we restrict to $M$, since $x^2+y^2$ and $5^2$ really are the same function on $M$.
(The elementary calculus justification for this implicit differentiation is to say we are not differentiating on $mathbb{R}^2$ but that instead we are considering $x$ and $y$ as functions of some other variable $t$ via a parametrization of our curve, and then differentiating with respect to $t$. (Or, if we're finding $dy/dx$ as you suggest, we're treating $y$ as a function of $x$ locally on the curve so the parameter is $x$ itself.) This is in fact exactly the same as the manifold approach when you unwrap all the definitions, since the differential on $M$ is defined in terms of differentiating in a local parametrization.)
answered Jan 2 at 0:39
Eric WofseyEric Wofsey
183k13209337
$endgroup$
$begingroup$
That makes sense. Thanks!
$endgroup$
– goblin
Jan 2 at 0:41
add a comment |
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1 Answer
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$begingroup$
What you're missing is that this computation does not take place in $mathbb{R}^2$: it takes place in the submanifold $M$ of $mathbb{R}^2$ defined by the equation $x^2+y^2=5^2$! We can consider $x$ and $y$ as smooth functions on $M$, and these smooth functions satisfy $x^2+y^2=5^2$ at every point of $M$. So, we can differentiate and deduce that $$2xdx+2ydy=0$$ at every point of $M$. This is not a contradiction, because $dx$ and $dy$ here are sections of the cotangent bundle of $M$, not the cotangent bundle of $mathbb{R}^2$ (they are the pullback of the usual $dx$ and $dy$ on $mathbb{R}^2$ to $M$).
Note that on $mathbb{R}^2$, you cannot differentiate the equation $x^2+y^2=5^2$ since this is not actually an equation of functions, it is just an equation that is true at some points of $mathbb{R}^2$. Even at a point where the equation is true, you cannot expect the differential of $x^2+y^2$ to be equal to the differential of $5^2$, since there are nearby points where the two functions are not equal. But we don't have this problem when we restrict to $M$, since $x^2+y^2$ and $5^2$ really are the same function on $M$.
(The elementary calculus justification for this implicit differentiation is to say we are not differentiating on $mathbb{R}^2$ but that instead we are considering $x$ and $y$ as functions of some other variable $t$ via a parametrization of our curve, and then differentiating with respect to $t$. (Or, if we're finding $dy/dx$ as you suggest, we're treating $y$ as a function of $x$ locally on the curve so the parameter is $x$ itself.) This is in fact exactly the same as the manifold approach when you unwrap all the definitions, since the differential on $M$ is defined in terms of differentiating in a local parametrization.)
answered Jan 2 at 0:39
Eric WofseyEric Wofsey
183k13209337
$endgroup$
$begingroup$
That makes sense. Thanks!
$endgroup$
– goblin
Jan 2 at 0:41
add a comment |
$begingroup$
What you're missing is that this computation does not take place in $mathbb{R}^2$: it takes place in the submanifold $M$ of $mathbb{R}^2$ defined by the equation $x^2+y^2=5^2$! We can consider $x$ and $y$ as smooth functions on $M$, and these smooth functions satisfy $x^2+y^2=5^2$ at every point of $M$. So, we can differentiate and deduce that $$2xdx+2ydy=0$$ at every point of $M$. This is not a contradiction, because $dx$ and $dy$ here are sections of the cotangent bundle of $M$, not the cotangent bundle of $mathbb{R}^2$ (they are the pullback of the usual $dx$ and $dy$ on $mathbb{R}^2$ to $M$).
Note that on $mathbb{R}^2$, you cannot differentiate the equation $x^2+y^2=5^2$ since this is not actually an equation of functions, it is just an equation that is true at some points of $mathbb{R}^2$. Even at a point where the equation is true, you cannot expect the differential of $x^2+y^2$ to be equal to the differential of $5^2$, since there are nearby points where the two functions are not equal. But we don't have this problem when we restrict to $M$, since $x^2+y^2$ and $5^2$ really are the same function on $M$.
(The elementary calculus justification for this implicit differentiation is to say we are not differentiating on $mathbb{R}^2$ but that instead we are considering $x$ and $y$ as functions of some other variable $t$ via a parametrization of our curve, and then differentiating with respect to $t$. (Or, if we're finding $dy/dx$ as you suggest, we're treating $y$ as a function of $x$ locally on the curve so the parameter is $x$ itself.) This is in fact exactly the same as the manifold approach when you unwrap all the definitions, since the differential on $M$ is defined in terms of differentiating in a local parametrization.)
answered Jan 2 at 0:39
Eric WofseyEric Wofsey
183k13209337
$endgroup$
$begingroup$
That makes sense. Thanks!
$endgroup$
– goblin
Jan 2 at 0:41
add a comment |
$begingroup$
What you're missing is that this computation does not take place in $mathbb{R}^2$: it takes place in the submanifold $M$ of $mathbb{R}^2$ defined by the equation $x^2+y^2=5^2$! We can consider $x$ and $y$ as smooth functions on $M$, and these smooth functions satisfy $x^2+y^2=5^2$ at every point of $M$. So, we can differentiate and deduce that $$2xdx+2ydy=0$$ at every point of $M$. This is not a contradiction, because $dx$ and $dy$ here are sections of the cotangent bundle of $M$, not the cotangent bundle of $mathbb{R}^2$ (they are the pullback of the usual $dx$ and $dy$ on $mathbb{R}^2$ to $M$).
Note that on $mathbb{R}^2$, you cannot differentiate the equation $x^2+y^2=5^2$ since this is not actually an equation of functions, it is just an equation that is true at some points of $mathbb{R}^2$. Even at a point where the equation is true, you cannot expect the differential of $x^2+y^2$ to be equal to the differential of $5^2$, since there are nearby points where the two functions are not equal. But we don't have this problem when we restrict to $M$, since $x^2+y^2$ and $5^2$ really are the same function on $M$.
(The elementary calculus justification for this implicit differentiation is to say we are not differentiating on $mathbb{R}^2$ but that instead we are considering $x$ and $y$ as functions of some other variable $t$ via a parametrization of our curve, and then differentiating with respect to $t$. (Or, if we're finding $dy/dx$ as you suggest, we're treating $y$ as a function of $x$ locally on the curve so the parameter is $x$ itself.) This is in fact exactly the same as the manifold approach when you unwrap all the definitions, since the differential on $M$ is defined in terms of differentiating in a local parametrization.)
answered Jan 2 at 0:39
Eric WofseyEric Wofsey
183k13209337
$endgroup$
What you're missing is that this computation does not take place in $mathbb{R}^2$: it takes place in the submanifold $M$ of $mathbb{R}^2$ defined by the equation $x^2+y^2=5^2$! We can consider $x$ and $y$ as smooth functions on $M$, and these smooth functions satisfy $x^2+y^2=5^2$ at every point of $M$. So, we can differentiate and deduce that $$2xdx+2ydy=0$$ at every point of $M$. This is not a contradiction, because $dx$ and $dy$ here are sections of the cotangent bundle of $M$, not the cotangent bundle of $mathbb{R}^2$ (they are the pullback of the usual $dx$ and $dy$ on $mathbb{R}^2$ to $M$).
Note that on $mathbb{R}^2$, you cannot differentiate the equation $x^2+y^2=5^2$ since this is not actually an equation of functions, it is just an equation that is true at some points of $mathbb{R}^2$. Even at a point where the equation is true, you cannot expect the differential of $x^2+y^2$ to be equal to the differential of $5^2$, since there are nearby points where the two functions are not equal. But we don't have this problem when we restrict to $M$, since $x^2+y^2$ and $5^2$ really are the same function on $M$.
(The elementary calculus justification for this implicit differentiation is to say we are not differentiating on $mathbb{R}^2$ but that instead we are considering $x$ and $y$ as functions of some other variable $t$ via a parametrization of our curve, and then differentiating with respect to $t$. (Or, if we're finding $dy/dx$ as you suggest, we're treating $y$ as a function of $x$ locally on the curve so the parameter is $x$ itself.) This is in fact exactly the same as the manifold approach when you unwrap all the definitions, since the differential on $M$ is defined in terms of differentiating in a local parametrization.)
answered Jan 2 at 0:39
Eric WofseyEric Wofsey
183k13209337
edited Jan 2 at 0:44
answered Jan 2 at 0:39
Eric WofseyEric Wofsey
183k13209337
answered Jan 2 at 0:39
Eric WofseyEric Wofsey
183k13209337
answered Jan 2 at 0:39
Eric WofseyEric Wofsey
183k13209337
183k13209337
$begingroup$
That makes sense. Thanks!
$endgroup$
– goblin
Jan 2 at 0:41
add a comment |
$begingroup$
That makes sense. Thanks!
$endgroup$
– goblin
Jan 2 at 0:41
$begingroup$
That makes sense. Thanks!
$endgroup$
– goblin
Jan 2 at 0:41
$begingroup$
That makes sense. Thanks!
$endgroup$
– goblin
Jan 2 at 0:41
add a comment |
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$begingroup$
The equation $x^2+y^2=5^2$ describes a curve. What do yo mean by "taking the gradient" of a curve?
$endgroup$
– Jackozee Hakkiuz
Jan 2 at 0:40
$begingroup$
@JackozeeHakkiuz, edited.
$endgroup$
– goblin
Jan 2 at 0:41
$begingroup$
$2xdx+2ydy=(2x,2y)(dx,dy)^t$ is a dot product and will be $0$ whenever $(dx,dy)$ and $(2x,2y)$ are orthogonal. This will be the case for all directions tangent to the circle because the gradient of a level curve is normal to that curve.
$endgroup$
– John Douma
Jan 2 at 0:51