How to tell whether a curve has a regular parametrization?
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A parametrization of a 1-dimensional curve is called regular if its velocity is always positive. For example, the following parametrization:
$$x(t)=t^3, y(t)=t^6$$
is not regular because its velocity is 0 in $t=0$.
But, this same curve can be re-parametrized as:
$$x(t)=t, y(t)=t^2$$
and this second parametrization is regular because its velocity anywhere is at least 1.
So, my question is: given a non-regular parametrization of a curve, is there an algorithm to tell whether the curve has a regular parametrization, and find it if it exists?
differential-geometry plane-curves
asked Jul 20 '15 at 5:59
Erel Segal-Halevi
4,14711757
add a comment |Â
up vote
8
down vote
favorite
A parametrization of a 1-dimensional curve is called regular if its velocity is always positive. For example, the following parametrization:
$$x(t)=t^3, y(t)=t^6$$
is not regular because its velocity is 0 in $t=0$.
But, this same curve can be re-parametrized as:
$$x(t)=t, y(t)=t^2$$
and this second parametrization is regular because its velocity anywhere is at least 1.
So, my question is: given a non-regular parametrization of a curve, is there an algorithm to tell whether the curve has a regular parametrization, and find it if it exists?
differential-geometry plane-curves
asked Jul 20 '15 at 5:59
Erel Segal-Halevi
4,14711757
1
What about natural parametrization?
– Michael Galuza
Jul 20 '15 at 6:03
1
If we perform a reparametrization of any curve with parameter $t$ to parameter $tau = tau(t)$ then $v_tau = v_t left|fracdtdtauright|$. In your case $v_t = 3t^2sqrt1 + 4t^6$ so we can for example cancel the first factor by picking $3t^2 = fracdtaudtto tau = t^3$. If $tau = int v_t dt$ is a valid parametriation then $v_tau = 1$ everywhere.
– Winther
Jul 20 '15 at 6:46
@Winther what do you mean by "if $tau$ is a valid parametrization"? How can I tell whether it is a valid parametrization?
– Erel Segal-Halevi
Jul 20 '15 at 8:26
1
I just added that since I don't know too much about this. I was thinking the integral might have singularities, it might have branches so that it can only describe parts of the curve etc. I don't know if this is a show-stopper. But $tau(t)$ as described above seems to be closely related to what Michael mentioned above namely the so-called natural parametrisation (as mentioned here).
– Winther
Jul 20 '15 at 8:42
add a comment |Â
up vote
8
down vote
favorite
up vote
8
down vote
favorite
A parametrization of a 1-dimensional curve is called regular if its velocity is always positive. For example, the following parametrization:
$$x(t)=t^3, y(t)=t^6$$
is not regular because its velocity is 0 in $t=0$.
But, this same curve can be re-parametrized as:
$$x(t)=t, y(t)=t^2$$
and this second parametrization is regular because its velocity anywhere is at least 1.
So, my question is: given a non-regular parametrization of a curve, is there an algorithm to tell whether the curve has a regular parametrization, and find it if it exists?
differential-geometry plane-curves
asked Jul 20 '15 at 5:59
Erel Segal-Halevi
4,14711757
A parametrization of a 1-dimensional curve is called regular if its velocity is always positive. For example, the following parametrization:
$$x(t)=t^3, y(t)=t^6$$
is not regular because its velocity is 0 in $t=0$.
But, this same curve can be re-parametrized as:
$$x(t)=t, y(t)=t^2$$
and this second parametrization is regular because its velocity anywhere is at least 1.
So, my question is: given a non-regular parametrization of a curve, is there an algorithm to tell whether the curve has a regular parametrization, and find it if it exists?
differential-geometry plane-curves
asked Jul 20 '15 at 5:59
Erel Segal-Halevi
4,14711757
asked Jul 20 '15 at 5:59
Erel Segal-Halevi
4,14711757
asked Jul 20 '15 at 5:59
Erel Segal-Halevi
4,14711757
asked Jul 20 '15 at 5:59
Erel Segal-Halevi
4,14711757
4,14711757
1
What about natural parametrization?
– Michael Galuza
Jul 20 '15 at 6:03
1
If we perform a reparametrization of any curve with parameter $t$ to parameter $tau = tau(t)$ then $v_tau = v_t left|fracdtdtauright|$. In your case $v_t = 3t^2sqrt1 + 4t^6$ so we can for example cancel the first factor by picking $3t^2 = fracdtaudtto tau = t^3$. If $tau = int v_t dt$ is a valid parametriation then $v_tau = 1$ everywhere.
– Winther
Jul 20 '15 at 6:46
@Winther what do you mean by "if $tau$ is a valid parametrization"? How can I tell whether it is a valid parametrization?
– Erel Segal-Halevi
Jul 20 '15 at 8:26
1
I just added that since I don't know too much about this. I was thinking the integral might have singularities, it might have branches so that it can only describe parts of the curve etc. I don't know if this is a show-stopper. But $tau(t)$ as described above seems to be closely related to what Michael mentioned above namely the so-called natural parametrisation (as mentioned here).
– Winther
Jul 20 '15 at 8:42
add a comment |Â
1
What about natural parametrization?
– Michael Galuza
Jul 20 '15 at 6:03
1
If we perform a reparametrization of any curve with parameter $t$ to parameter $tau = tau(t)$ then $v_tau = v_t left|fracdtdtauright|$. In your case $v_t = 3t^2sqrt1 + 4t^6$ so we can for example cancel the first factor by picking $3t^2 = fracdtaudtto tau = t^3$. If $tau = int v_t dt$ is a valid parametriation then $v_tau = 1$ everywhere.
– Winther
Jul 20 '15 at 6:46
@Winther what do you mean by "if $tau$ is a valid parametrization"? How can I tell whether it is a valid parametrization?
– Erel Segal-Halevi
Jul 20 '15 at 8:26
1
I just added that since I don't know too much about this. I was thinking the integral might have singularities, it might have branches so that it can only describe parts of the curve etc. I don't know if this is a show-stopper. But $tau(t)$ as described above seems to be closely related to what Michael mentioned above namely the so-called natural parametrisation (as mentioned here).
– Winther
Jul 20 '15 at 8:42
1
1
What about natural parametrization?
– Michael Galuza
Jul 20 '15 at 6:03
What about natural parametrization?
– Michael Galuza
Jul 20 '15 at 6:03
1
1
If we perform a reparametrization of any curve with parameter $t$ to parameter $tau = tau(t)$ then $v_tau = v_t left|fracdtdtauright|$. In your case $v_t = 3t^2sqrt1 + 4t^6$ so we can for example cancel the first factor by picking $3t^2 = fracdtaudtto tau = t^3$. If $tau = int v_t dt$ is a valid parametriation then $v_tau = 1$ everywhere.
– Winther
Jul 20 '15 at 6:46
If we perform a reparametrization of any curve with parameter $t$ to parameter $tau = tau(t)$ then $v_tau = v_t left|fracdtdtauright|$. In your case $v_t = 3t^2sqrt1 + 4t^6$ so we can for example cancel the first factor by picking $3t^2 = fracdtaudtto tau = t^3$. If $tau = int v_t dt$ is a valid parametriation then $v_tau = 1$ everywhere.
– Winther
Jul 20 '15 at 6:46
@Winther what do you mean by "if $tau$ is a valid parametrization"? How can I tell whether it is a valid parametrization?
– Erel Segal-Halevi
Jul 20 '15 at 8:26
@Winther what do you mean by "if $tau$ is a valid parametrization"? How can I tell whether it is a valid parametrization?
– Erel Segal-Halevi
Jul 20 '15 at 8:26
1
1
I just added that since I don't know too much about this. I was thinking the integral might have singularities, it might have branches so that it can only describe parts of the curve etc. I don't know if this is a show-stopper. But $tau(t)$ as described above seems to be closely related to what Michael mentioned above namely the so-called natural parametrisation (as mentioned here).
– Winther
Jul 20 '15 at 8:42
I just added that since I don't know too much about this. I was thinking the integral might have singularities, it might have branches so that it can only describe parts of the curve etc. I don't know if this is a show-stopper. But $tau(t)$ as described above seems to be closely related to what Michael mentioned above namely the so-called natural parametrisation (as mentioned here).
– Winther
Jul 20 '15 at 8:42
add a comment |Â
1 Answer
1
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$newcommandRealsmathbfR$Let $I$ be a non-empty open interval of real numbers, and $gamma:I to Reals^n$ a continuously-differentiable path. (This hypothesis is arguably substantial, but appears not out of line with the OP's intent.)
At each point where $gamma$ is regular, define the unit tangent field
$$
T(t) = fracgamma'(t)gamma'(t).
$$
By continuity, the zero set of $gamma'$ is a closed subset of $I$; that is, the domain of $T$ is an open subset of $I$. Without loss of generality, we may assume the domain of $T$ is the complement of a discrete set. (Loosely, if there are intervals on which $gamma' = 0$, "excise them and push their endpoints together").
Claim: The following are equivalent:
(i) $gamma$ has a regular $C^1$ parametrization.
(ii) $T$ has a continuous extension to $I$.
(i) implies (ii): The unit tangent field is independent of monotone reparametrization in the sense that if $Gamma(t) = gamma(tau(t))$ for some differentiable function $tau$ with positive derivative, then the unit tangent field of $Gamma$ at $t$ is the unit tangent field of $gamma$ at $tau(t)$. If $gamma$ has a regular $C^1$ reparametrization $Gamma$, then the unit tangent field of $Gamma$ continuously extends the unit tangent field of $T$.
(ii) implies (i). If $T$ has a continuous extension to $I$ (which we continue to denote $T$), then
$$
Gamma(s) = int_0^s T(t), dt
$$
is a regular parametrization of $gamma$.
For the curve in question, we have
$$
gamma(t) = (t^3, t^6),qquad
gamma'(t) = (3t^2, 6t^5),qquad
T(t) = frac(3t^2, 6t^5)sqrt9t^4 + 36t^10
= frac(1, 2t^3)sqrt1 + 4t^6,
$$
which has a continuous extension at $t = 0$ (given by the same formula).
For a cusp, say, we have
$$
gamma(t) = (t^2, t^3),qquad
gamma'(t) = (2t, 3t^2),qquad
T(t) = frac(2t, 3t^2)sqrt4t^2 + 9t^6
= fract(2, 3t)t,
$$
which does not extend continuously to $0$.
answered Oct 9 '16 at 22:20
Andrew D. Hwang
51.5k445110
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
7
down vote
$newcommandRealsmathbfR$Let $I$ be a non-empty open interval of real numbers, and $gamma:I to Reals^n$ a continuously-differentiable path. (This hypothesis is arguably substantial, but appears not out of line with the OP's intent.)
At each point where $gamma$ is regular, define the unit tangent field
$$
T(t) = fracgamma'(t)gamma'(t).
$$
By continuity, the zero set of $gamma'$ is a closed subset of $I$; that is, the domain of $T$ is an open subset of $I$. Without loss of generality, we may assume the domain of $T$ is the complement of a discrete set. (Loosely, if there are intervals on which $gamma' = 0$, "excise them and push their endpoints together").
Claim: The following are equivalent:
(i) $gamma$ has a regular $C^1$ parametrization.
(ii) $T$ has a continuous extension to $I$.
(i) implies (ii): The unit tangent field is independent of monotone reparametrization in the sense that if $Gamma(t) = gamma(tau(t))$ for some differentiable function $tau$ with positive derivative, then the unit tangent field of $Gamma$ at $t$ is the unit tangent field of $gamma$ at $tau(t)$. If $gamma$ has a regular $C^1$ reparametrization $Gamma$, then the unit tangent field of $Gamma$ continuously extends the unit tangent field of $T$.
(ii) implies (i). If $T$ has a continuous extension to $I$ (which we continue to denote $T$), then
$$
Gamma(s) = int_0^s T(t), dt
$$
is a regular parametrization of $gamma$.
For the curve in question, we have
$$
gamma(t) = (t^3, t^6),qquad
gamma'(t) = (3t^2, 6t^5),qquad
T(t) = frac(3t^2, 6t^5)sqrt9t^4 + 36t^10
= frac(1, 2t^3)sqrt1 + 4t^6,
$$
which has a continuous extension at $t = 0$ (given by the same formula).
For a cusp, say, we have
$$
gamma(t) = (t^2, t^3),qquad
gamma'(t) = (2t, 3t^2),qquad
T(t) = frac(2t, 3t^2)sqrt4t^2 + 9t^6
= fract(2, 3t)t,
$$
which does not extend continuously to $0$.
answered Oct 9 '16 at 22:20
Andrew D. Hwang
51.5k445110
add a comment |Â
up vote
7
down vote
$newcommandRealsmathbfR$Let $I$ be a non-empty open interval of real numbers, and $gamma:I to Reals^n$ a continuously-differentiable path. (This hypothesis is arguably substantial, but appears not out of line with the OP's intent.)
At each point where $gamma$ is regular, define the unit tangent field
$$
T(t) = fracgamma'(t)gamma'(t).
$$
By continuity, the zero set of $gamma'$ is a closed subset of $I$; that is, the domain of $T$ is an open subset of $I$. Without loss of generality, we may assume the domain of $T$ is the complement of a discrete set. (Loosely, if there are intervals on which $gamma' = 0$, "excise them and push their endpoints together").
Claim: The following are equivalent:
(i) $gamma$ has a regular $C^1$ parametrization.
(ii) $T$ has a continuous extension to $I$.
(i) implies (ii): The unit tangent field is independent of monotone reparametrization in the sense that if $Gamma(t) = gamma(tau(t))$ for some differentiable function $tau$ with positive derivative, then the unit tangent field of $Gamma$ at $t$ is the unit tangent field of $gamma$ at $tau(t)$. If $gamma$ has a regular $C^1$ reparametrization $Gamma$, then the unit tangent field of $Gamma$ continuously extends the unit tangent field of $T$.
(ii) implies (i). If $T$ has a continuous extension to $I$ (which we continue to denote $T$), then
$$
Gamma(s) = int_0^s T(t), dt
$$
is a regular parametrization of $gamma$.
For the curve in question, we have
$$
gamma(t) = (t^3, t^6),qquad
gamma'(t) = (3t^2, 6t^5),qquad
T(t) = frac(3t^2, 6t^5)sqrt9t^4 + 36t^10
= frac(1, 2t^3)sqrt1 + 4t^6,
$$
which has a continuous extension at $t = 0$ (given by the same formula).
For a cusp, say, we have
$$
gamma(t) = (t^2, t^3),qquad
gamma'(t) = (2t, 3t^2),qquad
T(t) = frac(2t, 3t^2)sqrt4t^2 + 9t^6
= fract(2, 3t)t,
$$
which does not extend continuously to $0$.
answered Oct 9 '16 at 22:20
Andrew D. Hwang
51.5k445110
add a comment |Â
up vote
7
down vote
up vote
7
down vote
$newcommandRealsmathbfR$Let $I$ be a non-empty open interval of real numbers, and $gamma:I to Reals^n$ a continuously-differentiable path. (This hypothesis is arguably substantial, but appears not out of line with the OP's intent.)
At each point where $gamma$ is regular, define the unit tangent field
$$
T(t) = fracgamma'(t)gamma'(t).
$$
By continuity, the zero set of $gamma'$ is a closed subset of $I$; that is, the domain of $T$ is an open subset of $I$. Without loss of generality, we may assume the domain of $T$ is the complement of a discrete set. (Loosely, if there are intervals on which $gamma' = 0$, "excise them and push their endpoints together").
Claim: The following are equivalent:
(i) $gamma$ has a regular $C^1$ parametrization.
(ii) $T$ has a continuous extension to $I$.
(i) implies (ii): The unit tangent field is independent of monotone reparametrization in the sense that if $Gamma(t) = gamma(tau(t))$ for some differentiable function $tau$ with positive derivative, then the unit tangent field of $Gamma$ at $t$ is the unit tangent field of $gamma$ at $tau(t)$. If $gamma$ has a regular $C^1$ reparametrization $Gamma$, then the unit tangent field of $Gamma$ continuously extends the unit tangent field of $T$.
(ii) implies (i). If $T$ has a continuous extension to $I$ (which we continue to denote $T$), then
$$
Gamma(s) = int_0^s T(t), dt
$$
is a regular parametrization of $gamma$.
For the curve in question, we have
$$
gamma(t) = (t^3, t^6),qquad
gamma'(t) = (3t^2, 6t^5),qquad
T(t) = frac(3t^2, 6t^5)sqrt9t^4 + 36t^10
= frac(1, 2t^3)sqrt1 + 4t^6,
$$
which has a continuous extension at $t = 0$ (given by the same formula).
For a cusp, say, we have
$$
gamma(t) = (t^2, t^3),qquad
gamma'(t) = (2t, 3t^2),qquad
T(t) = frac(2t, 3t^2)sqrt4t^2 + 9t^6
= fract(2, 3t)t,
$$
which does not extend continuously to $0$.
answered Oct 9 '16 at 22:20
Andrew D. Hwang
51.5k445110
$newcommandRealsmathbfR$Let $I$ be a non-empty open interval of real numbers, and $gamma:I to Reals^n$ a continuously-differentiable path. (This hypothesis is arguably substantial, but appears not out of line with the OP's intent.)
At each point where $gamma$ is regular, define the unit tangent field
$$
T(t) = fracgamma'(t)gamma'(t).
$$
By continuity, the zero set of $gamma'$ is a closed subset of $I$; that is, the domain of $T$ is an open subset of $I$. Without loss of generality, we may assume the domain of $T$ is the complement of a discrete set. (Loosely, if there are intervals on which $gamma' = 0$, "excise them and push their endpoints together").
Claim: The following are equivalent:
(i) $gamma$ has a regular $C^1$ parametrization.
(ii) $T$ has a continuous extension to $I$.
(i) implies (ii): The unit tangent field is independent of monotone reparametrization in the sense that if $Gamma(t) = gamma(tau(t))$ for some differentiable function $tau$ with positive derivative, then the unit tangent field of $Gamma$ at $t$ is the unit tangent field of $gamma$ at $tau(t)$. If $gamma$ has a regular $C^1$ reparametrization $Gamma$, then the unit tangent field of $Gamma$ continuously extends the unit tangent field of $T$.
(ii) implies (i). If $T$ has a continuous extension to $I$ (which we continue to denote $T$), then
$$
Gamma(s) = int_0^s T(t), dt
$$
is a regular parametrization of $gamma$.
For the curve in question, we have
$$
gamma(t) = (t^3, t^6),qquad
gamma'(t) = (3t^2, 6t^5),qquad
T(t) = frac(3t^2, 6t^5)sqrt9t^4 + 36t^10
= frac(1, 2t^3)sqrt1 + 4t^6,
$$
which has a continuous extension at $t = 0$ (given by the same formula).
For a cusp, say, we have
$$
gamma(t) = (t^2, t^3),qquad
gamma'(t) = (2t, 3t^2),qquad
T(t) = frac(2t, 3t^2)sqrt4t^2 + 9t^6
= fract(2, 3t)t,
$$
which does not extend continuously to $0$.
answered Oct 9 '16 at 22:20
Andrew D. Hwang
51.5k445110
answered Oct 9 '16 at 22:20
Andrew D. Hwang
51.5k445110
answered Oct 9 '16 at 22:20
Andrew D. Hwang
51.5k445110
answered Oct 9 '16 at 22:20
Andrew D. Hwang
51.5k445110
51.5k445110
add a comment |Â
add a comment |Â
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1
What about natural parametrization?
– Michael Galuza
Jul 20 '15 at 6:03
1
If we perform a reparametrization of any curve with parameter $t$ to parameter $tau = tau(t)$ then $v_tau = v_t left|fracdtdtauright|$. In your case $v_t = 3t^2sqrt1 + 4t^6$ so we can for example cancel the first factor by picking $3t^2 = fracdtaudtto tau = t^3$. If $tau = int v_t dt$ is a valid parametriation then $v_tau = 1$ everywhere.
– Winther
Jul 20 '15 at 6:46
@Winther what do you mean by "if $tau$ is a valid parametrization"? How can I tell whether it is a valid parametrization?
– Erel Segal-Halevi
Jul 20 '15 at 8:26
1
I just added that since I don't know too much about this. I was thinking the integral might have singularities, it might have branches so that it can only describe parts of the curve etc. I don't know if this is a show-stopper. But $tau(t)$ as described above seems to be closely related to what Michael mentioned above namely the so-called natural parametrisation (as mentioned here).
– Winther
Jul 20 '15 at 8:42