Mixing
What does do Carmo mean here and also what is a curve parametrized by arc length?
$begingroup$
From do Carmo Differential Geometry:
In the third paragraph, Do Carmo says:
it can happen that the parameter $t$ is already the arc length
measured from some point.
What does he mean by this?
Also, what does he mean by curves parametrized by arc length?
calculus differential-geometry
asked Jan 19 at 2:00
WolfgangWolfgang
4,29943377
$endgroup$
add a comment |
$begingroup$
From do Carmo Differential Geometry:
In the third paragraph, Do Carmo says:
it can happen that the parameter $t$ is already the arc length
measured from some point.
What does he mean by this?
Also, what does he mean by curves parametrized by arc length?
calculus differential-geometry
asked Jan 19 at 2:00
WolfgangWolfgang
4,29943377
$endgroup$
add a comment |
$begingroup$
From do Carmo Differential Geometry:
In the third paragraph, Do Carmo says:
it can happen that the parameter $t$ is already the arc length
measured from some point.
What does he mean by this?
Also, what does he mean by curves parametrized by arc length?
calculus differential-geometry
asked Jan 19 at 2:00
WolfgangWolfgang
4,29943377
$endgroup$
From do Carmo Differential Geometry:
In the third paragraph, Do Carmo says:
it can happen that the parameter $t$ is already the arc length
measured from some point.
What does he mean by this?
Also, what does he mean by curves parametrized by arc length?
calculus differential-geometry
calculus differential-geometry
asked Jan 19 at 2:00
WolfgangWolfgang
4,29943377
asked Jan 19 at 2:00
WolfgangWolfgang
4,29943377
asked Jan 19 at 2:00
WolfgangWolfgang
4,29943377
asked Jan 19 at 2:00
WolfgangWolfgang
4,29943377
asked Jan 19 at 2:00
WolfgangWolfgang
4,29943377
4,29943377
add a comment |
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
A curve $c: [0,1] to mathbb{R}^3$ can be parametrized in many different ways. All parametrizations will have the same image $c([0,1])$ in $mathbb{R}^3$ (or more generally $mathbb{R}^n$) as subsets of the ambient space but they can have different velocity vectors at different points. For example, you can switch $t$ with $2t-1$ over $[0,1]$ and you'll get a parametrization that is twice faster because of the chain rule.
In general, assuming that our curve is at least twice continuously differentiable, when we want to compute curvature or torsion, it's easier to deal with curves having the constant speed $1$. Think about acceleration which is related to curvature, for example. Physically, we know that a change in velocity exerts a force on us by Newton's second law of motion $vec{F}=mvec{a}$. But a change in velocity can be caused by a change in its magnitude or a change in its direction. In geometry, we're interested mostly in the later kind of change which depends on the shape of our path (curve) rather than the speed we're traveling at.
So, we want to have $forall t:| alpha'(t) | = 1$. It can be shown that for any regular curve $(|alpha'(t)|neq 0)$ that is continuously differentiable, such a parametrization exists: a parametrization that gives the same curve as the image of the function $alpha: [0,1] to mathbb{R}^3$ while it has the property that $forall t: | alpha'(t) | = 1$. This parametrization is sometimes called the standard parametrization and it's defined by reparametrization of the curve by its arc-length.
Note that the arc-length of a curve from its starting point up to $t$ is defined as:
$$s(t) = int_0^t | alpha'(u)| du$$
Because $| alpha'(u)|>0$ due to the regularity of our curve, $s(t)$ is a strictly increasing function. So, $s(t)$ has an inverse function. Let's call the inverse $t(s)$. Now define $beta(s)=alpha(t(s))$. Now observe that we have $|beta'(s)|=1$ by the chain rule. This is our desired parametrization.
answered Jan 19 at 2:19
stressed outstressed out
6,0491838
$endgroup$
add a comment |
$begingroup$
The phrases "conversely" and "i.e." are important. The converse of $Pto Q$ is $Qto P$. And "$P_1$ i.e. $P_2$" means that $P_1$ and $P_2$ mean the same thing. Now, the paragraph takes the following form:
If $t$ is already the arc length measured from some point, then $Q$. Conversely, if $Q$ then $P$; i.e., $t$ is already the arc length measured from some point.
This means that we are to take "$t$ is already the arc length measured from some point" to be the same statement as $P$. And $P$ is the statement that $s=t-t_0$.
In the following paragraph, "curves parametrized by arc length" means curves that satisfy either of the equivalent conditions in the preceding paragraph. Explicitly, a curve is parametrized by arc length if $|alpha'(t)|equiv1$, or equivalently if $s=t-t_0$.
answered Jan 19 at 2:12
Chris CulterChris Culter
21.4k43887
$endgroup$
add a comment |
$begingroup$
I will attempt to give an intuitive explanation instead of an analytical one: Parameterizing a curve essentially means identifying the position along the curve for a given value of a parameter - that is, relating the curve (in physical space) with a copy of the real line (the parameter space). The parameter can have various physical interpretations and units.
For example, when one drives along a road, possible options to identify the position can be:
- where one is after driving for 20 minutes;
- where one is after driving for 20 kilometers.
In the first case, the parameter is the time since starting, measured through standard methods - but different drivers can be in different places after driving for the exact same amount of time.
In the second case the parameter is the distance driven, or the arclength of the road from the starting point until that position. That is usually identified by markers along the road, and no matter who drives, 20 km from the start point is going to be the same place.
The first approach emphasizes the parameter, outside the curve (hence extrinsic), the second emphasizes the curve (hence intrinsic).
The arclength parameterization is essentially a way of identifying the position after moving a certain distance along the curve from a fixed starting point.
answered Jan 19 at 2:36
Catalin ZaraCatalin Zara
3,737514
$endgroup$
add a comment |
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3 Answers
3
active
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3 Answers
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active
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$begingroup$
A curve $c: [0,1] to mathbb{R}^3$ can be parametrized in many different ways. All parametrizations will have the same image $c([0,1])$ in $mathbb{R}^3$ (or more generally $mathbb{R}^n$) as subsets of the ambient space but they can have different velocity vectors at different points. For example, you can switch $t$ with $2t-1$ over $[0,1]$ and you'll get a parametrization that is twice faster because of the chain rule.
In general, assuming that our curve is at least twice continuously differentiable, when we want to compute curvature or torsion, it's easier to deal with curves having the constant speed $1$. Think about acceleration which is related to curvature, for example. Physically, we know that a change in velocity exerts a force on us by Newton's second law of motion $vec{F}=mvec{a}$. But a change in velocity can be caused by a change in its magnitude or a change in its direction. In geometry, we're interested mostly in the later kind of change which depends on the shape of our path (curve) rather than the speed we're traveling at.
So, we want to have $forall t:| alpha'(t) | = 1$. It can be shown that for any regular curve $(|alpha'(t)|neq 0)$ that is continuously differentiable, such a parametrization exists: a parametrization that gives the same curve as the image of the function $alpha: [0,1] to mathbb{R}^3$ while it has the property that $forall t: | alpha'(t) | = 1$. This parametrization is sometimes called the standard parametrization and it's defined by reparametrization of the curve by its arc-length.
Note that the arc-length of a curve from its starting point up to $t$ is defined as:
$$s(t) = int_0^t | alpha'(u)| du$$
Because $| alpha'(u)|>0$ due to the regularity of our curve, $s(t)$ is a strictly increasing function. So, $s(t)$ has an inverse function. Let's call the inverse $t(s)$. Now define $beta(s)=alpha(t(s))$. Now observe that we have $|beta'(s)|=1$ by the chain rule. This is our desired parametrization.
answered Jan 19 at 2:19
stressed outstressed out
6,0491838
$endgroup$
add a comment |
$begingroup$
A curve $c: [0,1] to mathbb{R}^3$ can be parametrized in many different ways. All parametrizations will have the same image $c([0,1])$ in $mathbb{R}^3$ (or more generally $mathbb{R}^n$) as subsets of the ambient space but they can have different velocity vectors at different points. For example, you can switch $t$ with $2t-1$ over $[0,1]$ and you'll get a parametrization that is twice faster because of the chain rule.
In general, assuming that our curve is at least twice continuously differentiable, when we want to compute curvature or torsion, it's easier to deal with curves having the constant speed $1$. Think about acceleration which is related to curvature, for example. Physically, we know that a change in velocity exerts a force on us by Newton's second law of motion $vec{F}=mvec{a}$. But a change in velocity can be caused by a change in its magnitude or a change in its direction. In geometry, we're interested mostly in the later kind of change which depends on the shape of our path (curve) rather than the speed we're traveling at.
So, we want to have $forall t:| alpha'(t) | = 1$. It can be shown that for any regular curve $(|alpha'(t)|neq 0)$ that is continuously differentiable, such a parametrization exists: a parametrization that gives the same curve as the image of the function $alpha: [0,1] to mathbb{R}^3$ while it has the property that $forall t: | alpha'(t) | = 1$. This parametrization is sometimes called the standard parametrization and it's defined by reparametrization of the curve by its arc-length.
Note that the arc-length of a curve from its starting point up to $t$ is defined as:
$$s(t) = int_0^t | alpha'(u)| du$$
Because $| alpha'(u)|>0$ due to the regularity of our curve, $s(t)$ is a strictly increasing function. So, $s(t)$ has an inverse function. Let's call the inverse $t(s)$. Now define $beta(s)=alpha(t(s))$. Now observe that we have $|beta'(s)|=1$ by the chain rule. This is our desired parametrization.
answered Jan 19 at 2:19
stressed outstressed out
6,0491838
$endgroup$
add a comment |
$begingroup$
A curve $c: [0,1] to mathbb{R}^3$ can be parametrized in many different ways. All parametrizations will have the same image $c([0,1])$ in $mathbb{R}^3$ (or more generally $mathbb{R}^n$) as subsets of the ambient space but they can have different velocity vectors at different points. For example, you can switch $t$ with $2t-1$ over $[0,1]$ and you'll get a parametrization that is twice faster because of the chain rule.
In general, assuming that our curve is at least twice continuously differentiable, when we want to compute curvature or torsion, it's easier to deal with curves having the constant speed $1$. Think about acceleration which is related to curvature, for example. Physically, we know that a change in velocity exerts a force on us by Newton's second law of motion $vec{F}=mvec{a}$. But a change in velocity can be caused by a change in its magnitude or a change in its direction. In geometry, we're interested mostly in the later kind of change which depends on the shape of our path (curve) rather than the speed we're traveling at.
So, we want to have $forall t:| alpha'(t) | = 1$. It can be shown that for any regular curve $(|alpha'(t)|neq 0)$ that is continuously differentiable, such a parametrization exists: a parametrization that gives the same curve as the image of the function $alpha: [0,1] to mathbb{R}^3$ while it has the property that $forall t: | alpha'(t) | = 1$. This parametrization is sometimes called the standard parametrization and it's defined by reparametrization of the curve by its arc-length.
Note that the arc-length of a curve from its starting point up to $t$ is defined as:
$$s(t) = int_0^t | alpha'(u)| du$$
Because $| alpha'(u)|>0$ due to the regularity of our curve, $s(t)$ is a strictly increasing function. So, $s(t)$ has an inverse function. Let's call the inverse $t(s)$. Now define $beta(s)=alpha(t(s))$. Now observe that we have $|beta'(s)|=1$ by the chain rule. This is our desired parametrization.
answered Jan 19 at 2:19
stressed outstressed out
6,0491838
$endgroup$
A curve $c: [0,1] to mathbb{R}^3$ can be parametrized in many different ways. All parametrizations will have the same image $c([0,1])$ in $mathbb{R}^3$ (or more generally $mathbb{R}^n$) as subsets of the ambient space but they can have different velocity vectors at different points. For example, you can switch $t$ with $2t-1$ over $[0,1]$ and you'll get a parametrization that is twice faster because of the chain rule.
In general, assuming that our curve is at least twice continuously differentiable, when we want to compute curvature or torsion, it's easier to deal with curves having the constant speed $1$. Think about acceleration which is related to curvature, for example. Physically, we know that a change in velocity exerts a force on us by Newton's second law of motion $vec{F}=mvec{a}$. But a change in velocity can be caused by a change in its magnitude or a change in its direction. In geometry, we're interested mostly in the later kind of change which depends on the shape of our path (curve) rather than the speed we're traveling at.
So, we want to have $forall t:| alpha'(t) | = 1$. It can be shown that for any regular curve $(|alpha'(t)|neq 0)$ that is continuously differentiable, such a parametrization exists: a parametrization that gives the same curve as the image of the function $alpha: [0,1] to mathbb{R}^3$ while it has the property that $forall t: | alpha'(t) | = 1$. This parametrization is sometimes called the standard parametrization and it's defined by reparametrization of the curve by its arc-length.
Note that the arc-length of a curve from its starting point up to $t$ is defined as:
$$s(t) = int_0^t | alpha'(u)| du$$
Because $| alpha'(u)|>0$ due to the regularity of our curve, $s(t)$ is a strictly increasing function. So, $s(t)$ has an inverse function. Let's call the inverse $t(s)$. Now define $beta(s)=alpha(t(s))$. Now observe that we have $|beta'(s)|=1$ by the chain rule. This is our desired parametrization.
answered Jan 19 at 2:19
stressed outstressed out
6,0491838
edited Jan 19 at 2:40
answered Jan 19 at 2:19
stressed outstressed out
6,0491838
answered Jan 19 at 2:19
stressed outstressed out
6,0491838
answered Jan 19 at 2:19
stressed outstressed out
6,0491838
6,0491838
add a comment |
add a comment |
$begingroup$
The phrases "conversely" and "i.e." are important. The converse of $Pto Q$ is $Qto P$. And "$P_1$ i.e. $P_2$" means that $P_1$ and $P_2$ mean the same thing. Now, the paragraph takes the following form:
If $t$ is already the arc length measured from some point, then $Q$. Conversely, if $Q$ then $P$; i.e., $t$ is already the arc length measured from some point.
This means that we are to take "$t$ is already the arc length measured from some point" to be the same statement as $P$. And $P$ is the statement that $s=t-t_0$.
In the following paragraph, "curves parametrized by arc length" means curves that satisfy either of the equivalent conditions in the preceding paragraph. Explicitly, a curve is parametrized by arc length if $|alpha'(t)|equiv1$, or equivalently if $s=t-t_0$.
answered Jan 19 at 2:12
Chris CulterChris Culter
21.4k43887
$endgroup$
add a comment |
$begingroup$
The phrases "conversely" and "i.e." are important. The converse of $Pto Q$ is $Qto P$. And "$P_1$ i.e. $P_2$" means that $P_1$ and $P_2$ mean the same thing. Now, the paragraph takes the following form:
If $t$ is already the arc length measured from some point, then $Q$. Conversely, if $Q$ then $P$; i.e., $t$ is already the arc length measured from some point.
This means that we are to take "$t$ is already the arc length measured from some point" to be the same statement as $P$. And $P$ is the statement that $s=t-t_0$.
In the following paragraph, "curves parametrized by arc length" means curves that satisfy either of the equivalent conditions in the preceding paragraph. Explicitly, a curve is parametrized by arc length if $|alpha'(t)|equiv1$, or equivalently if $s=t-t_0$.
answered Jan 19 at 2:12
Chris CulterChris Culter
21.4k43887
$endgroup$
add a comment |
$begingroup$
The phrases "conversely" and "i.e." are important. The converse of $Pto Q$ is $Qto P$. And "$P_1$ i.e. $P_2$" means that $P_1$ and $P_2$ mean the same thing. Now, the paragraph takes the following form:
If $t$ is already the arc length measured from some point, then $Q$. Conversely, if $Q$ then $P$; i.e., $t$ is already the arc length measured from some point.
This means that we are to take "$t$ is already the arc length measured from some point" to be the same statement as $P$. And $P$ is the statement that $s=t-t_0$.
In the following paragraph, "curves parametrized by arc length" means curves that satisfy either of the equivalent conditions in the preceding paragraph. Explicitly, a curve is parametrized by arc length if $|alpha'(t)|equiv1$, or equivalently if $s=t-t_0$.
answered Jan 19 at 2:12
Chris CulterChris Culter
21.4k43887
$endgroup$
The phrases "conversely" and "i.e." are important. The converse of $Pto Q$ is $Qto P$. And "$P_1$ i.e. $P_2$" means that $P_1$ and $P_2$ mean the same thing. Now, the paragraph takes the following form:
If $t$ is already the arc length measured from some point, then $Q$. Conversely, if $Q$ then $P$; i.e., $t$ is already the arc length measured from some point.
This means that we are to take "$t$ is already the arc length measured from some point" to be the same statement as $P$. And $P$ is the statement that $s=t-t_0$.
In the following paragraph, "curves parametrized by arc length" means curves that satisfy either of the equivalent conditions in the preceding paragraph. Explicitly, a curve is parametrized by arc length if $|alpha'(t)|equiv1$, or equivalently if $s=t-t_0$.
answered Jan 19 at 2:12
Chris CulterChris Culter
21.4k43887
answered Jan 19 at 2:12
Chris CulterChris Culter
21.4k43887
answered Jan 19 at 2:12
Chris CulterChris Culter
21.4k43887
answered Jan 19 at 2:12
Chris CulterChris Culter
21.4k43887
21.4k43887
add a comment |
add a comment |
$begingroup$
I will attempt to give an intuitive explanation instead of an analytical one: Parameterizing a curve essentially means identifying the position along the curve for a given value of a parameter - that is, relating the curve (in physical space) with a copy of the real line (the parameter space). The parameter can have various physical interpretations and units.
For example, when one drives along a road, possible options to identify the position can be:
- where one is after driving for 20 minutes;
- where one is after driving for 20 kilometers.
In the first case, the parameter is the time since starting, measured through standard methods - but different drivers can be in different places after driving for the exact same amount of time.
In the second case the parameter is the distance driven, or the arclength of the road from the starting point until that position. That is usually identified by markers along the road, and no matter who drives, 20 km from the start point is going to be the same place.
The first approach emphasizes the parameter, outside the curve (hence extrinsic), the second emphasizes the curve (hence intrinsic).
The arclength parameterization is essentially a way of identifying the position after moving a certain distance along the curve from a fixed starting point.
answered Jan 19 at 2:36
Catalin ZaraCatalin Zara
3,737514
$endgroup$
add a comment |
$begingroup$
I will attempt to give an intuitive explanation instead of an analytical one: Parameterizing a curve essentially means identifying the position along the curve for a given value of a parameter - that is, relating the curve (in physical space) with a copy of the real line (the parameter space). The parameter can have various physical interpretations and units.
For example, when one drives along a road, possible options to identify the position can be:
- where one is after driving for 20 minutes;
- where one is after driving for 20 kilometers.
In the first case, the parameter is the time since starting, measured through standard methods - but different drivers can be in different places after driving for the exact same amount of time.
In the second case the parameter is the distance driven, or the arclength of the road from the starting point until that position. That is usually identified by markers along the road, and no matter who drives, 20 km from the start point is going to be the same place.
The first approach emphasizes the parameter, outside the curve (hence extrinsic), the second emphasizes the curve (hence intrinsic).
The arclength parameterization is essentially a way of identifying the position after moving a certain distance along the curve from a fixed starting point.
answered Jan 19 at 2:36
Catalin ZaraCatalin Zara
3,737514
$endgroup$
add a comment |
$begingroup$
I will attempt to give an intuitive explanation instead of an analytical one: Parameterizing a curve essentially means identifying the position along the curve for a given value of a parameter - that is, relating the curve (in physical space) with a copy of the real line (the parameter space). The parameter can have various physical interpretations and units.
For example, when one drives along a road, possible options to identify the position can be:
- where one is after driving for 20 minutes;
- where one is after driving for 20 kilometers.
In the first case, the parameter is the time since starting, measured through standard methods - but different drivers can be in different places after driving for the exact same amount of time.
In the second case the parameter is the distance driven, or the arclength of the road from the starting point until that position. That is usually identified by markers along the road, and no matter who drives, 20 km from the start point is going to be the same place.
The first approach emphasizes the parameter, outside the curve (hence extrinsic), the second emphasizes the curve (hence intrinsic).
The arclength parameterization is essentially a way of identifying the position after moving a certain distance along the curve from a fixed starting point.
answered Jan 19 at 2:36
Catalin ZaraCatalin Zara
3,737514
$endgroup$
I will attempt to give an intuitive explanation instead of an analytical one: Parameterizing a curve essentially means identifying the position along the curve for a given value of a parameter - that is, relating the curve (in physical space) with a copy of the real line (the parameter space). The parameter can have various physical interpretations and units.
For example, when one drives along a road, possible options to identify the position can be:
- where one is after driving for 20 minutes;
- where one is after driving for 20 kilometers.
In the first case, the parameter is the time since starting, measured through standard methods - but different drivers can be in different places after driving for the exact same amount of time.
In the second case the parameter is the distance driven, or the arclength of the road from the starting point until that position. That is usually identified by markers along the road, and no matter who drives, 20 km from the start point is going to be the same place.
The first approach emphasizes the parameter, outside the curve (hence extrinsic), the second emphasizes the curve (hence intrinsic).
The arclength parameterization is essentially a way of identifying the position after moving a certain distance along the curve from a fixed starting point.
answered Jan 19 at 2:36
Catalin ZaraCatalin Zara
3,737514
answered Jan 19 at 2:36
Catalin ZaraCatalin Zara
3,737514
answered Jan 19 at 2:36
Catalin ZaraCatalin Zara
3,737514
answered Jan 19 at 2:36
Catalin ZaraCatalin Zara
3,737514
3,737514
add a comment |
add a comment |
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