Unit-speed Reparametrization of a circle












1












$begingroup$


I am studying some differential Geometry from Shifrin's differential Geometry notes.



From what I understand so far it seems like it is possible to unit-speed reparametrize a curve whenever the curve is regular.



In the notes that I am reading the parametrization of the circle is:
$alpha = (acos{t},asin{t}), implies lVert alpha '(t)rVert = a$



And then it is mentioned that if we reparametrize the curve by



$beta(s) = (acos(s/a),asin(s/a)) implies lVertbeta'(s)rVert=1$



So my question how did the author derive that this exact reparametrization have unit speed?, He could impossibly just have guessed it..



And is there any General way to find a unit speed reparametrization of a parametrized curve?






calculus geometry differential-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    When dealing with a problem involving a simple curve like this, it's probably better to use simple algebra and geometry than calculus. And since the circle has uniform speed, you could just treat it as a line segment of length $2pi a$ instead.
    $endgroup$
    – R. Burton
    Dec 20 '18 at 17:14
















1












$begingroup$


I am studying some differential Geometry from Shifrin's differential Geometry notes.



From what I understand so far it seems like it is possible to unit-speed reparametrize a curve whenever the curve is regular.



In the notes that I am reading the parametrization of the circle is:
$alpha = (acos{t},asin{t}), implies lVert alpha '(t)rVert = a$



And then it is mentioned that if we reparametrize the curve by



$beta(s) = (acos(s/a),asin(s/a)) implies lVertbeta'(s)rVert=1$



So my question how did the author derive that this exact reparametrization have unit speed?, He could impossibly just have guessed it..



And is there any General way to find a unit speed reparametrization of a parametrized curve?






calculus geometry differential-geometry







share|cite|improve this question











$endgroup$












  • $begingroup$
    When dealing with a problem involving a simple curve like this, it's probably better to use simple algebra and geometry than calculus. And since the circle has uniform speed, you could just treat it as a line segment of length $2pi a$ instead.
    $endgroup$
    – R. Burton
    Dec 20 '18 at 17:14














1












1








1





$begingroup$


I am studying some differential Geometry from Shifrin's differential Geometry notes.



From what I understand so far it seems like it is possible to unit-speed reparametrize a curve whenever the curve is regular.



In the notes that I am reading the parametrization of the circle is:
$alpha = (acos{t},asin{t}), implies lVert alpha '(t)rVert = a$



And then it is mentioned that if we reparametrize the curve by



$beta(s) = (acos(s/a),asin(s/a)) implies lVertbeta'(s)rVert=1$



So my question how did the author derive that this exact reparametrization have unit speed?, He could impossibly just have guessed it..



And is there any General way to find a unit speed reparametrization of a parametrized curve?






calculus geometry differential-geometry







share|cite|improve this question











$endgroup$




I am studying some differential Geometry from Shifrin's differential Geometry notes.



From what I understand so far it seems like it is possible to unit-speed reparametrize a curve whenever the curve is regular.



In the notes that I am reading the parametrization of the circle is:
$alpha = (acos{t},asin{t}), implies lVert alpha '(t)rVert = a$



And then it is mentioned that if we reparametrize the curve by



$beta(s) = (acos(s/a),asin(s/a)) implies lVertbeta'(s)rVert=1$



So my question how did the author derive that this exact reparametrization have unit speed?, He could impossibly just have guessed it..



And is there any General way to find a unit speed reparametrization of a parametrized curve?






calculus geometry differential-geometry




calculus geometry differential-geometry






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 20 '18 at 17:16







Jensens

















asked Dec 20 '18 at 16:49









JensensJensens

245




245












  • $begingroup$
    When dealing with a problem involving a simple curve like this, it's probably better to use simple algebra and geometry than calculus. And since the circle has uniform speed, you could just treat it as a line segment of length $2pi a$ instead.
    $endgroup$
    – R. Burton
    Dec 20 '18 at 17:14


















  • $begingroup$
    When dealing with a problem involving a simple curve like this, it's probably better to use simple algebra and geometry than calculus. And since the circle has uniform speed, you could just treat it as a line segment of length $2pi a$ instead.
    $endgroup$
    – R. Burton
    Dec 20 '18 at 17:14
















$begingroup$
When dealing with a problem involving a simple curve like this, it's probably better to use simple algebra and geometry than calculus. And since the circle has uniform speed, you could just treat it as a line segment of length $2pi a$ instead.
$endgroup$
– R. Burton
Dec 20 '18 at 17:14




$begingroup$
When dealing with a problem involving a simple curve like this, it's probably better to use simple algebra and geometry than calculus. And since the circle has uniform speed, you could just treat it as a line segment of length $2pi a$ instead.
$endgroup$
– R. Burton
Dec 20 '18 at 17:14










2 Answers
2






active

oldest

votes


















2












$begingroup$

The answer to your question is at the top of p. 8 of my notes. In the case of the circle as originally parametrized, the arclength, starting at $t=0$, is $s(t)=at$. So $t=s/a$. Thus, $beta(s) = alpha(s/a) = big(acos(s/a),asin(s/a)big)$ is a reparametrization by arclength. You can immediately check that $|beta'(s)|=1$, but the general argument is in the notes there.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now this might be a silly question, but when taking the derivative of a unit-speed reparametrization curve, do we just get 0?
    $endgroup$
    – Jensens
    Dec 20 '18 at 20:40






  • 1




    $begingroup$
    No, you get the unit tangent vector along the curve. That's sort of the whole point, as you'll see in the next section.
    $endgroup$
    – Ted Shifrin
    Dec 20 '18 at 20:51





















0












$begingroup$

The speed of a parametrization $r$ is given by $|r'|$ by definition. Intuitively you can think of it in this way: if you visualize your curve in space and assume $t$ to represent time then $r(t+ Delta t)-r(t)$ is the vector that takes the points on the curve $r(t)$ at time $t$ to the point on the curve $Delta t$ time later, which is $r(t+ Delta t)$. If you divide this by $Delta t$ you obtain a vector $frac{r(t+ Delta t)-r(t)}{Delta t}$ whose norm is just "distance" between two points divided by the time it takes to go from one to anoher. By making $Delta t$ very small and by taking the norm one obtains the so called speed of the parametrization.



You can look for unitary speed parametrized curves at https://en.wikipedia.org/wiki/Differential_geometry_of_curves#Length_and_natural_parametrization






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry my question was stated unclearly... I edited to how did the author know that this exact reparametrization have unit speed
    $endgroup$
    – Jensens
    Dec 20 '18 at 17:17










  • $begingroup$
    It's the definition of derivatives of functions $mathbb{R} rightarrow mathbb{R}^n$ : $beta ' (s) = ( beta_1'(s), beta_2'(s) ) = (-a frac{1}{a} sin(s/a), a frac{1}{a} cos(s/a))$. Now if you compute the norm it's $1$
    $endgroup$
    – Leonardo
    Dec 20 '18 at 17:21













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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2












$begingroup$

The answer to your question is at the top of p. 8 of my notes. In the case of the circle as originally parametrized, the arclength, starting at $t=0$, is $s(t)=at$. So $t=s/a$. Thus, $beta(s) = alpha(s/a) = big(acos(s/a),asin(s/a)big)$ is a reparametrization by arclength. You can immediately check that $|beta'(s)|=1$, but the general argument is in the notes there.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now this might be a silly question, but when taking the derivative of a unit-speed reparametrization curve, do we just get 0?
    $endgroup$
    – Jensens
    Dec 20 '18 at 20:40






  • 1




    $begingroup$
    No, you get the unit tangent vector along the curve. That's sort of the whole point, as you'll see in the next section.
    $endgroup$
    – Ted Shifrin
    Dec 20 '18 at 20:51


















2












$begingroup$

The answer to your question is at the top of p. 8 of my notes. In the case of the circle as originally parametrized, the arclength, starting at $t=0$, is $s(t)=at$. So $t=s/a$. Thus, $beta(s) = alpha(s/a) = big(acos(s/a),asin(s/a)big)$ is a reparametrization by arclength. You can immediately check that $|beta'(s)|=1$, but the general argument is in the notes there.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Now this might be a silly question, but when taking the derivative of a unit-speed reparametrization curve, do we just get 0?
    $endgroup$
    – Jensens
    Dec 20 '18 at 20:40






  • 1




    $begingroup$
    No, you get the unit tangent vector along the curve. That's sort of the whole point, as you'll see in the next section.
    $endgroup$
    – Ted Shifrin
    Dec 20 '18 at 20:51
















2












2








2





$begingroup$

The answer to your question is at the top of p. 8 of my notes. In the case of the circle as originally parametrized, the arclength, starting at $t=0$, is $s(t)=at$. So $t=s/a$. Thus, $beta(s) = alpha(s/a) = big(acos(s/a),asin(s/a)big)$ is a reparametrization by arclength. You can immediately check that $|beta'(s)|=1$, but the general argument is in the notes there.






share|cite|improve this answer









$endgroup$



The answer to your question is at the top of p. 8 of my notes. In the case of the circle as originally parametrized, the arclength, starting at $t=0$, is $s(t)=at$. So $t=s/a$. Thus, $beta(s) = alpha(s/a) = big(acos(s/a),asin(s/a)big)$ is a reparametrization by arclength. You can immediately check that $|beta'(s)|=1$, but the general argument is in the notes there.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 17:21









Ted ShifrinTed Shifrin

64k44591




64k44591












  • $begingroup$
    Now this might be a silly question, but when taking the derivative of a unit-speed reparametrization curve, do we just get 0?
    $endgroup$
    – Jensens
    Dec 20 '18 at 20:40






  • 1




    $begingroup$
    No, you get the unit tangent vector along the curve. That's sort of the whole point, as you'll see in the next section.
    $endgroup$
    – Ted Shifrin
    Dec 20 '18 at 20:51




















  • $begingroup$
    Now this might be a silly question, but when taking the derivative of a unit-speed reparametrization curve, do we just get 0?
    $endgroup$
    – Jensens
    Dec 20 '18 at 20:40






  • 1




    $begingroup$
    No, you get the unit tangent vector along the curve. That's sort of the whole point, as you'll see in the next section.
    $endgroup$
    – Ted Shifrin
    Dec 20 '18 at 20:51


















$begingroup$
Now this might be a silly question, but when taking the derivative of a unit-speed reparametrization curve, do we just get 0?
$endgroup$
– Jensens
Dec 20 '18 at 20:40




$begingroup$
Now this might be a silly question, but when taking the derivative of a unit-speed reparametrization curve, do we just get 0?
$endgroup$
– Jensens
Dec 20 '18 at 20:40




1




1




$begingroup$
No, you get the unit tangent vector along the curve. That's sort of the whole point, as you'll see in the next section.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 20:51






$begingroup$
No, you get the unit tangent vector along the curve. That's sort of the whole point, as you'll see in the next section.
$endgroup$
– Ted Shifrin
Dec 20 '18 at 20:51













0












$begingroup$

The speed of a parametrization $r$ is given by $|r'|$ by definition. Intuitively you can think of it in this way: if you visualize your curve in space and assume $t$ to represent time then $r(t+ Delta t)-r(t)$ is the vector that takes the points on the curve $r(t)$ at time $t$ to the point on the curve $Delta t$ time later, which is $r(t+ Delta t)$. If you divide this by $Delta t$ you obtain a vector $frac{r(t+ Delta t)-r(t)}{Delta t}$ whose norm is just "distance" between two points divided by the time it takes to go from one to anoher. By making $Delta t$ very small and by taking the norm one obtains the so called speed of the parametrization.



You can look for unitary speed parametrized curves at https://en.wikipedia.org/wiki/Differential_geometry_of_curves#Length_and_natural_parametrization






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry my question was stated unclearly... I edited to how did the author know that this exact reparametrization have unit speed
    $endgroup$
    – Jensens
    Dec 20 '18 at 17:17










  • $begingroup$
    It's the definition of derivatives of functions $mathbb{R} rightarrow mathbb{R}^n$ : $beta ' (s) = ( beta_1'(s), beta_2'(s) ) = (-a frac{1}{a} sin(s/a), a frac{1}{a} cos(s/a))$. Now if you compute the norm it's $1$
    $endgroup$
    – Leonardo
    Dec 20 '18 at 17:21


















0












$begingroup$

The speed of a parametrization $r$ is given by $|r'|$ by definition. Intuitively you can think of it in this way: if you visualize your curve in space and assume $t$ to represent time then $r(t+ Delta t)-r(t)$ is the vector that takes the points on the curve $r(t)$ at time $t$ to the point on the curve $Delta t$ time later, which is $r(t+ Delta t)$. If you divide this by $Delta t$ you obtain a vector $frac{r(t+ Delta t)-r(t)}{Delta t}$ whose norm is just "distance" between two points divided by the time it takes to go from one to anoher. By making $Delta t$ very small and by taking the norm one obtains the so called speed of the parametrization.



You can look for unitary speed parametrized curves at https://en.wikipedia.org/wiki/Differential_geometry_of_curves#Length_and_natural_parametrization






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Sorry my question was stated unclearly... I edited to how did the author know that this exact reparametrization have unit speed
    $endgroup$
    – Jensens
    Dec 20 '18 at 17:17










  • $begingroup$
    It's the definition of derivatives of functions $mathbb{R} rightarrow mathbb{R}^n$ : $beta ' (s) = ( beta_1'(s), beta_2'(s) ) = (-a frac{1}{a} sin(s/a), a frac{1}{a} cos(s/a))$. Now if you compute the norm it's $1$
    $endgroup$
    – Leonardo
    Dec 20 '18 at 17:21
















0












0








0





$begingroup$

The speed of a parametrization $r$ is given by $|r'|$ by definition. Intuitively you can think of it in this way: if you visualize your curve in space and assume $t$ to represent time then $r(t+ Delta t)-r(t)$ is the vector that takes the points on the curve $r(t)$ at time $t$ to the point on the curve $Delta t$ time later, which is $r(t+ Delta t)$. If you divide this by $Delta t$ you obtain a vector $frac{r(t+ Delta t)-r(t)}{Delta t}$ whose norm is just "distance" between two points divided by the time it takes to go from one to anoher. By making $Delta t$ very small and by taking the norm one obtains the so called speed of the parametrization.



You can look for unitary speed parametrized curves at https://en.wikipedia.org/wiki/Differential_geometry_of_curves#Length_and_natural_parametrization






share|cite|improve this answer









$endgroup$



The speed of a parametrization $r$ is given by $|r'|$ by definition. Intuitively you can think of it in this way: if you visualize your curve in space and assume $t$ to represent time then $r(t+ Delta t)-r(t)$ is the vector that takes the points on the curve $r(t)$ at time $t$ to the point on the curve $Delta t$ time later, which is $r(t+ Delta t)$. If you divide this by $Delta t$ you obtain a vector $frac{r(t+ Delta t)-r(t)}{Delta t}$ whose norm is just "distance" between two points divided by the time it takes to go from one to anoher. By making $Delta t$ very small and by taking the norm one obtains the so called speed of the parametrization.



You can look for unitary speed parametrized curves at https://en.wikipedia.org/wiki/Differential_geometry_of_curves#Length_and_natural_parametrization







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 20 '18 at 17:09









LeonardoLeonardo

3339




3339












  • $begingroup$
    Sorry my question was stated unclearly... I edited to how did the author know that this exact reparametrization have unit speed
    $endgroup$
    – Jensens
    Dec 20 '18 at 17:17










  • $begingroup$
    It's the definition of derivatives of functions $mathbb{R} rightarrow mathbb{R}^n$ : $beta ' (s) = ( beta_1'(s), beta_2'(s) ) = (-a frac{1}{a} sin(s/a), a frac{1}{a} cos(s/a))$. Now if you compute the norm it's $1$
    $endgroup$
    – Leonardo
    Dec 20 '18 at 17:21




















  • $begingroup$
    Sorry my question was stated unclearly... I edited to how did the author know that this exact reparametrization have unit speed
    $endgroup$
    – Jensens
    Dec 20 '18 at 17:17










  • $begingroup$
    It's the definition of derivatives of functions $mathbb{R} rightarrow mathbb{R}^n$ : $beta ' (s) = ( beta_1'(s), beta_2'(s) ) = (-a frac{1}{a} sin(s/a), a frac{1}{a} cos(s/a))$. Now if you compute the norm it's $1$
    $endgroup$
    – Leonardo
    Dec 20 '18 at 17:21


















$begingroup$
Sorry my question was stated unclearly... I edited to how did the author know that this exact reparametrization have unit speed
$endgroup$
– Jensens
Dec 20 '18 at 17:17




$begingroup$
Sorry my question was stated unclearly... I edited to how did the author know that this exact reparametrization have unit speed
$endgroup$
– Jensens
Dec 20 '18 at 17:17












$begingroup$
It's the definition of derivatives of functions $mathbb{R} rightarrow mathbb{R}^n$ : $beta ' (s) = ( beta_1'(s), beta_2'(s) ) = (-a frac{1}{a} sin(s/a), a frac{1}{a} cos(s/a))$. Now if you compute the norm it's $1$
$endgroup$
– Leonardo
Dec 20 '18 at 17:21






$begingroup$
It's the definition of derivatives of functions $mathbb{R} rightarrow mathbb{R}^n$ : $beta ' (s) = ( beta_1'(s), beta_2'(s) ) = (-a frac{1}{a} sin(s/a), a frac{1}{a} cos(s/a))$. Now if you compute the norm it's $1$
$endgroup$
– Leonardo
Dec 20 '18 at 17:21




















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