Mixing
How to check if a 2 dimensional vector field is irrotational (curl=0)?
$begingroup$
I need to check if a this vector field is irrotational and conservative:
$F=langle e^{x cos y} cos y,-x e^{x cos y}sin{y} rangle$
curl of F should be => $dfrac{dQ}{dx} - dfrac{dP}{y}=0$ (where 'd' = partial derivative).
With some calculation, I found out that the partials are equal :
$-e^{x cos y}sin y[1+x cos y]$
But here is my confusion, does that mean that the vector field is irratational? or conservative? My doubt arises from the fact that the definition of curl I know is only for 3 dimensional vector field.
calculus linear-algebra multivariable-calculus vector-fields
edited Jan 14 at 14:45
amWhy
1
asked Jan 14 at 13:50
NPLSNPLS
7112
$endgroup$
add a comment |
$begingroup$
I need to check if a this vector field is irrotational and conservative:
$F=langle e^{x cos y} cos y,-x e^{x cos y}sin{y} rangle$
curl of F should be => $dfrac{dQ}{dx} - dfrac{dP}{y}=0$ (where 'd' = partial derivative).
With some calculation, I found out that the partials are equal :
$-e^{x cos y}sin y[1+x cos y]$
But here is my confusion, does that mean that the vector field is irratational? or conservative? My doubt arises from the fact that the definition of curl I know is only for 3 dimensional vector field.
calculus linear-algebra multivariable-calculus vector-fields
edited Jan 14 at 14:45
amWhy
1
asked Jan 14 at 13:50
NPLSNPLS
7112
$endgroup$
$begingroup$
Your field is the gradient of $e^{x cos y}$, confirming it is conservative
$endgroup$
– kmm
Jan 14 at 14:26
$begingroup$
yes, but how do I check if it is irrotational ?
$endgroup$
– NPLS
Jan 14 at 14:27
add a comment |
$begingroup$
I need to check if a this vector field is irrotational and conservative:
$F=langle e^{x cos y} cos y,-x e^{x cos y}sin{y} rangle$
curl of F should be => $dfrac{dQ}{dx} - dfrac{dP}{y}=0$ (where 'd' = partial derivative).
With some calculation, I found out that the partials are equal :
$-e^{x cos y}sin y[1+x cos y]$
But here is my confusion, does that mean that the vector field is irratational? or conservative? My doubt arises from the fact that the definition of curl I know is only for 3 dimensional vector field.
calculus linear-algebra multivariable-calculus vector-fields
edited Jan 14 at 14:45
amWhy
1
asked Jan 14 at 13:50
NPLSNPLS
7112
$endgroup$
I need to check if a this vector field is irrotational and conservative:
$F=langle e^{x cos y} cos y,-x e^{x cos y}sin{y} rangle$
curl of F should be => $dfrac{dQ}{dx} - dfrac{dP}{y}=0$ (where 'd' = partial derivative).
With some calculation, I found out that the partials are equal :
$-e^{x cos y}sin y[1+x cos y]$
But here is my confusion, does that mean that the vector field is irratational? or conservative? My doubt arises from the fact that the definition of curl I know is only for 3 dimensional vector field.
calculus linear-algebra multivariable-calculus vector-fields
calculus linear-algebra multivariable-calculus vector-fields
edited Jan 14 at 14:45
amWhy
1
asked Jan 14 at 13:50
NPLSNPLS
7112
edited Jan 14 at 14:45
amWhy
1
asked Jan 14 at 13:50
NPLSNPLS
7112
edited Jan 14 at 14:45
amWhy
1
edited Jan 14 at 14:45
amWhy
1
edited Jan 14 at 14:45
amWhy
1
1
asked Jan 14 at 13:50
NPLSNPLS
7112
asked Jan 14 at 13:50
NPLSNPLS
7112
asked Jan 14 at 13:50
NPLSNPLS
7112
7112
$begingroup$
Your field is the gradient of $e^{x cos y}$, confirming it is conservative
$endgroup$
– kmm
Jan 14 at 14:26
$begingroup$
yes, but how do I check if it is irrotational ?
$endgroup$
– NPLS
Jan 14 at 14:27
add a comment |
$begingroup$
Your field is the gradient of $e^{x cos y}$, confirming it is conservative
$endgroup$
– kmm
Jan 14 at 14:26
$begingroup$
yes, but how do I check if it is irrotational ?
$endgroup$
– NPLS
Jan 14 at 14:27
$begingroup$
Your field is the gradient of $e^{x cos y}$, confirming it is conservative
$endgroup$
– kmm
Jan 14 at 14:26
$begingroup$
Your field is the gradient of $e^{x cos y}$, confirming it is conservative
$endgroup$
– kmm
Jan 14 at 14:26
$begingroup$
yes, but how do I check if it is irrotational ?
$endgroup$
– NPLS
Jan 14 at 14:27
$begingroup$
yes, but how do I check if it is irrotational ?
$endgroup$
– NPLS
Jan 14 at 14:27
add a comment |
1 Answer
1
active
oldest
votes
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You are right that the usual notion of curl is only defined for 3 dimensions. To prove a two-dimensional vector field is conservative, you can use Green's theorem which says that the following identity holds for a region $D$ bounded by a curve $C$
$$ oint (P,dx+Q,dy)=iint _{D}left({frac {partial Q}{partial x}}-{frac {partial P}{partial y}}right),dx,dy$$
A conservative vector field is a vector field which can be expressed as the gradient of a scalar function. Remember that a conservative vector field has the property that the result of a line integral is path independent, implying the line integral over a simple closed path is always zero (there are some subtleties involving non-simply connected spaces). Hence the left integral will be zero if and only if the vector field $vec{F} = P hat{e}_x + Q hat{e}_y $ is conservative. So to check if a field is conservative, we can simply check if the integrand of integral on the right hand side is zero. This expression is the two-dimensional analog of curl.
$$ curl vec{F} = frac {partial Q}{partial x}-frac {partial P}{partial y} $$
This is the formula you wrote down. In two-dimensions, the curl is a scalar, not a vector field itself. To get a completely general description of the concept of curl, you need differential forms, but this a relatively complex matter.
Another way to see it is that your expression is the z-component of the 3-dimensional curl for a vector field which does not depend on the z-direction.
answered Jan 14 at 14:17
kmmkmm
1637
$endgroup$
$begingroup$
So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition)
$endgroup$
– NPLS
Jan 15 at 8:18
$begingroup$
It's irrotational if this two-dimensional analog of the curl vanishes. A conservative field will always be irrotational. To see this, consider a gradient $vec{F} = (frac{partial f}{partial x}, frac{partial f}{partial y})$ . The 2D curl will be $ frac{partial }{partial x}frac{partial f}{partial y} - frac{partial }{partial y}frac{partial f}{partial x} $ which will always be zero as partial derivatives commute. So conservativeness implies irrotationality
$endgroup$
– kmm
Jan 15 at 23:10
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are right that the usual notion of curl is only defined for 3 dimensions. To prove a two-dimensional vector field is conservative, you can use Green's theorem which says that the following identity holds for a region $D$ bounded by a curve $C$
$$ oint (P,dx+Q,dy)=iint _{D}left({frac {partial Q}{partial x}}-{frac {partial P}{partial y}}right),dx,dy$$
A conservative vector field is a vector field which can be expressed as the gradient of a scalar function. Remember that a conservative vector field has the property that the result of a line integral is path independent, implying the line integral over a simple closed path is always zero (there are some subtleties involving non-simply connected spaces). Hence the left integral will be zero if and only if the vector field $vec{F} = P hat{e}_x + Q hat{e}_y $ is conservative. So to check if a field is conservative, we can simply check if the integrand of integral on the right hand side is zero. This expression is the two-dimensional analog of curl.
$$ curl vec{F} = frac {partial Q}{partial x}-frac {partial P}{partial y} $$
This is the formula you wrote down. In two-dimensions, the curl is a scalar, not a vector field itself. To get a completely general description of the concept of curl, you need differential forms, but this a relatively complex matter.
Another way to see it is that your expression is the z-component of the 3-dimensional curl for a vector field which does not depend on the z-direction.
answered Jan 14 at 14:17
kmmkmm
1637
$endgroup$
$begingroup$
So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition)
$endgroup$
– NPLS
Jan 15 at 8:18
$begingroup$
It's irrotational if this two-dimensional analog of the curl vanishes. A conservative field will always be irrotational. To see this, consider a gradient $vec{F} = (frac{partial f}{partial x}, frac{partial f}{partial y})$ . The 2D curl will be $ frac{partial }{partial x}frac{partial f}{partial y} - frac{partial }{partial y}frac{partial f}{partial x} $ which will always be zero as partial derivatives commute. So conservativeness implies irrotationality
$endgroup$
– kmm
Jan 15 at 23:10
add a comment |
$begingroup$
You are right that the usual notion of curl is only defined for 3 dimensions. To prove a two-dimensional vector field is conservative, you can use Green's theorem which says that the following identity holds for a region $D$ bounded by a curve $C$
$$ oint (P,dx+Q,dy)=iint _{D}left({frac {partial Q}{partial x}}-{frac {partial P}{partial y}}right),dx,dy$$
A conservative vector field is a vector field which can be expressed as the gradient of a scalar function. Remember that a conservative vector field has the property that the result of a line integral is path independent, implying the line integral over a simple closed path is always zero (there are some subtleties involving non-simply connected spaces). Hence the left integral will be zero if and only if the vector field $vec{F} = P hat{e}_x + Q hat{e}_y $ is conservative. So to check if a field is conservative, we can simply check if the integrand of integral on the right hand side is zero. This expression is the two-dimensional analog of curl.
$$ curl vec{F} = frac {partial Q}{partial x}-frac {partial P}{partial y} $$
This is the formula you wrote down. In two-dimensions, the curl is a scalar, not a vector field itself. To get a completely general description of the concept of curl, you need differential forms, but this a relatively complex matter.
Another way to see it is that your expression is the z-component of the 3-dimensional curl for a vector field which does not depend on the z-direction.
answered Jan 14 at 14:17
kmmkmm
1637
$endgroup$
$begingroup$
So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition)
$endgroup$
– NPLS
Jan 15 at 8:18
$begingroup$
It's irrotational if this two-dimensional analog of the curl vanishes. A conservative field will always be irrotational. To see this, consider a gradient $vec{F} = (frac{partial f}{partial x}, frac{partial f}{partial y})$ . The 2D curl will be $ frac{partial }{partial x}frac{partial f}{partial y} - frac{partial }{partial y}frac{partial f}{partial x} $ which will always be zero as partial derivatives commute. So conservativeness implies irrotationality
$endgroup$
– kmm
Jan 15 at 23:10
add a comment |
$begingroup$
You are right that the usual notion of curl is only defined for 3 dimensions. To prove a two-dimensional vector field is conservative, you can use Green's theorem which says that the following identity holds for a region $D$ bounded by a curve $C$
$$ oint (P,dx+Q,dy)=iint _{D}left({frac {partial Q}{partial x}}-{frac {partial P}{partial y}}right),dx,dy$$
A conservative vector field is a vector field which can be expressed as the gradient of a scalar function. Remember that a conservative vector field has the property that the result of a line integral is path independent, implying the line integral over a simple closed path is always zero (there are some subtleties involving non-simply connected spaces). Hence the left integral will be zero if and only if the vector field $vec{F} = P hat{e}_x + Q hat{e}_y $ is conservative. So to check if a field is conservative, we can simply check if the integrand of integral on the right hand side is zero. This expression is the two-dimensional analog of curl.
$$ curl vec{F} = frac {partial Q}{partial x}-frac {partial P}{partial y} $$
This is the formula you wrote down. In two-dimensions, the curl is a scalar, not a vector field itself. To get a completely general description of the concept of curl, you need differential forms, but this a relatively complex matter.
Another way to see it is that your expression is the z-component of the 3-dimensional curl for a vector field which does not depend on the z-direction.
answered Jan 14 at 14:17
kmmkmm
1637
$endgroup$
You are right that the usual notion of curl is only defined for 3 dimensions. To prove a two-dimensional vector field is conservative, you can use Green's theorem which says that the following identity holds for a region $D$ bounded by a curve $C$
$$ oint (P,dx+Q,dy)=iint _{D}left({frac {partial Q}{partial x}}-{frac {partial P}{partial y}}right),dx,dy$$
A conservative vector field is a vector field which can be expressed as the gradient of a scalar function. Remember that a conservative vector field has the property that the result of a line integral is path independent, implying the line integral over a simple closed path is always zero (there are some subtleties involving non-simply connected spaces). Hence the left integral will be zero if and only if the vector field $vec{F} = P hat{e}_x + Q hat{e}_y $ is conservative. So to check if a field is conservative, we can simply check if the integrand of integral on the right hand side is zero. This expression is the two-dimensional analog of curl.
$$ curl vec{F} = frac {partial Q}{partial x}-frac {partial P}{partial y} $$
This is the formula you wrote down. In two-dimensions, the curl is a scalar, not a vector field itself. To get a completely general description of the concept of curl, you need differential forms, but this a relatively complex matter.
Another way to see it is that your expression is the z-component of the 3-dimensional curl for a vector field which does not depend on the z-direction.
answered Jan 14 at 14:17
kmmkmm
1637
edited Jan 14 at 14:28
answered Jan 14 at 14:17
kmmkmm
1637
answered Jan 14 at 14:17
kmmkmm
1637
answered Jan 14 at 14:17
kmmkmm
1637
1637
$begingroup$
So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition)
$endgroup$
– NPLS
Jan 15 at 8:18
$begingroup$
It's irrotational if this two-dimensional analog of the curl vanishes. A conservative field will always be irrotational. To see this, consider a gradient $vec{F} = (frac{partial f}{partial x}, frac{partial f}{partial y})$ . The 2D curl will be $ frac{partial }{partial x}frac{partial f}{partial y} - frac{partial }{partial y}frac{partial f}{partial x} $ which will always be zero as partial derivatives commute. So conservativeness implies irrotationality
$endgroup$
– kmm
Jan 15 at 23:10
add a comment |
$begingroup$
So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition)
$endgroup$
– NPLS
Jan 15 at 8:18
$begingroup$
It's irrotational if this two-dimensional analog of the curl vanishes. A conservative field will always be irrotational. To see this, consider a gradient $vec{F} = (frac{partial f}{partial x}, frac{partial f}{partial y})$ . The 2D curl will be $ frac{partial }{partial x}frac{partial f}{partial y} - frac{partial }{partial y}frac{partial f}{partial x} $ which will always be zero as partial derivatives commute. So conservativeness implies irrotationality
$endgroup$
– kmm
Jan 15 at 23:10
$begingroup$
So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition)
$endgroup$
– NPLS
Jan 15 at 8:18
$begingroup$
So what I can say about the irrotationality of my vector field? Does that mean that conservativeness implies irrotationality? (in the domain of definition)
$endgroup$
– NPLS
Jan 15 at 8:18
$begingroup$
It's irrotational if this two-dimensional analog of the curl vanishes. A conservative field will always be irrotational. To see this, consider a gradient $vec{F} = (frac{partial f}{partial x}, frac{partial f}{partial y})$ . The 2D curl will be $ frac{partial }{partial x}frac{partial f}{partial y} - frac{partial }{partial y}frac{partial f}{partial x} $ which will always be zero as partial derivatives commute. So conservativeness implies irrotationality
$endgroup$
– kmm
Jan 15 at 23:10
$begingroup$
It's irrotational if this two-dimensional analog of the curl vanishes. A conservative field will always be irrotational. To see this, consider a gradient $vec{F} = (frac{partial f}{partial x}, frac{partial f}{partial y})$ . The 2D curl will be $ frac{partial }{partial x}frac{partial f}{partial y} - frac{partial }{partial y}frac{partial f}{partial x} $ which will always be zero as partial derivatives commute. So conservativeness implies irrotationality
$endgroup$
– kmm
Jan 15 at 23:10
add a comment |
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$begingroup$
Your field is the gradient of $e^{x cos y}$, confirming it is conservative
$endgroup$
– kmm
Jan 14 at 14:26
$begingroup$
yes, but how do I check if it is irrotational ?
$endgroup$
– NPLS
Jan 14 at 14:27