Mixing
Arc-length parametrisation
$begingroup$
I can not understand how the concept of arc length works. We define
$a(t)=int_{0}^{t}midgamma^{'}(x)mid dx$
for some curve $gamma: I rightarrowmathbb{R}^2$.
If we then normalise the curve we get $a(t)=t$ which we call paramertise by arc length. I can see how the direction of the tangent is whats most relevant for the curves description, but now we change the magnitude of the tangent.
What does this mean and how does it matter? Intuitively Id say that this has to do with "speed". But how does one make sense of this? Is it the lenght of the intervall $I$ ? because any intervall has the same number of points.
calculus differential-geometry
asked Jan 8 at 7:22
user7534user7534
666
$endgroup$
|
show 3 more comments
$begingroup$
I can not understand how the concept of arc length works. We define
$a(t)=int_{0}^{t}midgamma^{'}(x)mid dx$
for some curve $gamma: I rightarrowmathbb{R}^2$.
If we then normalise the curve we get $a(t)=t$ which we call paramertise by arc length. I can see how the direction of the tangent is whats most relevant for the curves description, but now we change the magnitude of the tangent.
What does this mean and how does it matter? Intuitively Id say that this has to do with "speed". But how does one make sense of this? Is it the lenght of the intervall $I$ ? because any intervall has the same number of points.
calculus differential-geometry
asked Jan 8 at 7:22
user7534user7534
666
$endgroup$
$begingroup$
Intuitively (as you ask for an intuitive grasp of the concept) : knowing that speed, generaly speaking, is the derivative of position, integrating a (norm of a) speed gives back a position. Moreover there is a normalizing aspect with the integration from $0$ to $t$ : thus you are with a "normalized" position... Once again it is with "hand waving"...
$endgroup$
– Jean Marie
Jan 8 at 7:39
1
$begingroup$
@Jean Marie I am sorry, I dont understand that at all. The same thing occurs if we integrate from any $t_{0}$.
$endgroup$
– user7534
Jan 8 at 7:45
$begingroup$
If we begin from $t_0$ the "starting time" of the chronometer will be at $t_0$ instead of $0$, that's all.
$endgroup$
– Jean Marie
Jan 8 at 7:48
$begingroup$
Another way to see it is to draw a curve $(x(t),y(t)) $ with a regular scale for $t = 0, 0.1, 0.2,...$ : you will see that, in general, resulting points are not regularly spaced. Thus you need a re-parameterization $t=varphi(s)$ of the curve in order that the points of $(x(varphi(s)),y(varphi(s))$ are regularly spaced.
$endgroup$
– Jean Marie
Jan 8 at 7:52
2
$begingroup$
This is not the main issue : till a certain point, you can take the unit of length you want (e.g., take yards instead of meters). The main issue is to annihilate accelerations or decelarations when you move along your trajectory (i.e., making acceleration = 0 everymwhere)
$endgroup$
– Jean Marie
Jan 8 at 8:00
|
show 3 more comments
$begingroup$
I can not understand how the concept of arc length works. We define
$a(t)=int_{0}^{t}midgamma^{'}(x)mid dx$
for some curve $gamma: I rightarrowmathbb{R}^2$.
If we then normalise the curve we get $a(t)=t$ which we call paramertise by arc length. I can see how the direction of the tangent is whats most relevant for the curves description, but now we change the magnitude of the tangent.
What does this mean and how does it matter? Intuitively Id say that this has to do with "speed". But how does one make sense of this? Is it the lenght of the intervall $I$ ? because any intervall has the same number of points.
calculus differential-geometry
asked Jan 8 at 7:22
user7534user7534
666
$endgroup$
I can not understand how the concept of arc length works. We define
$a(t)=int_{0}^{t}midgamma^{'}(x)mid dx$
for some curve $gamma: I rightarrowmathbb{R}^2$.
If we then normalise the curve we get $a(t)=t$ which we call paramertise by arc length. I can see how the direction of the tangent is whats most relevant for the curves description, but now we change the magnitude of the tangent.
What does this mean and how does it matter? Intuitively Id say that this has to do with "speed". But how does one make sense of this? Is it the lenght of the intervall $I$ ? because any intervall has the same number of points.
calculus differential-geometry
calculus differential-geometry
asked Jan 8 at 7:22
user7534user7534
666
asked Jan 8 at 7:22
user7534user7534
666
asked Jan 8 at 7:22
user7534user7534
666
asked Jan 8 at 7:22
user7534user7534
666
asked Jan 8 at 7:22
user7534user7534
666
666
$begingroup$
Intuitively (as you ask for an intuitive grasp of the concept) : knowing that speed, generaly speaking, is the derivative of position, integrating a (norm of a) speed gives back a position. Moreover there is a normalizing aspect with the integration from $0$ to $t$ : thus you are with a "normalized" position... Once again it is with "hand waving"...
$endgroup$
– Jean Marie
Jan 8 at 7:39
1
$begingroup$
@Jean Marie I am sorry, I dont understand that at all. The same thing occurs if we integrate from any $t_{0}$.
$endgroup$
– user7534
Jan 8 at 7:45
$begingroup$
If we begin from $t_0$ the "starting time" of the chronometer will be at $t_0$ instead of $0$, that's all.
$endgroup$
– Jean Marie
Jan 8 at 7:48
$begingroup$
Another way to see it is to draw a curve $(x(t),y(t)) $ with a regular scale for $t = 0, 0.1, 0.2,...$ : you will see that, in general, resulting points are not regularly spaced. Thus you need a re-parameterization $t=varphi(s)$ of the curve in order that the points of $(x(varphi(s)),y(varphi(s))$ are regularly spaced.
$endgroup$
– Jean Marie
Jan 8 at 7:52
2
$begingroup$
This is not the main issue : till a certain point, you can take the unit of length you want (e.g., take yards instead of meters). The main issue is to annihilate accelerations or decelarations when you move along your trajectory (i.e., making acceleration = 0 everymwhere)
$endgroup$
– Jean Marie
Jan 8 at 8:00
|
show 3 more comments
$begingroup$
Intuitively (as you ask for an intuitive grasp of the concept) : knowing that speed, generaly speaking, is the derivative of position, integrating a (norm of a) speed gives back a position. Moreover there is a normalizing aspect with the integration from $0$ to $t$ : thus you are with a "normalized" position... Once again it is with "hand waving"...
$endgroup$
– Jean Marie
Jan 8 at 7:39
1
$begingroup$
@Jean Marie I am sorry, I dont understand that at all. The same thing occurs if we integrate from any $t_{0}$.
$endgroup$
– user7534
Jan 8 at 7:45
$begingroup$
If we begin from $t_0$ the "starting time" of the chronometer will be at $t_0$ instead of $0$, that's all.
$endgroup$
– Jean Marie
Jan 8 at 7:48
$begingroup$
Another way to see it is to draw a curve $(x(t),y(t)) $ with a regular scale for $t = 0, 0.1, 0.2,...$ : you will see that, in general, resulting points are not regularly spaced. Thus you need a re-parameterization $t=varphi(s)$ of the curve in order that the points of $(x(varphi(s)),y(varphi(s))$ are regularly spaced.
$endgroup$
– Jean Marie
Jan 8 at 7:52
2
$begingroup$
This is not the main issue : till a certain point, you can take the unit of length you want (e.g., take yards instead of meters). The main issue is to annihilate accelerations or decelarations when you move along your trajectory (i.e., making acceleration = 0 everymwhere)
$endgroup$
– Jean Marie
Jan 8 at 8:00
$begingroup$
Intuitively (as you ask for an intuitive grasp of the concept) : knowing that speed, generaly speaking, is the derivative of position, integrating a (norm of a) speed gives back a position. Moreover there is a normalizing aspect with the integration from $0$ to $t$ : thus you are with a "normalized" position... Once again it is with "hand waving"...
$endgroup$
– Jean Marie
Jan 8 at 7:39
$begingroup$
Intuitively (as you ask for an intuitive grasp of the concept) : knowing that speed, generaly speaking, is the derivative of position, integrating a (norm of a) speed gives back a position. Moreover there is a normalizing aspect with the integration from $0$ to $t$ : thus you are with a "normalized" position... Once again it is with "hand waving"...
$endgroup$
– Jean Marie
Jan 8 at 7:39
1
1
$begingroup$
@Jean Marie I am sorry, I dont understand that at all. The same thing occurs if we integrate from any $t_{0}$.
$endgroup$
– user7534
Jan 8 at 7:45
$begingroup$
@Jean Marie I am sorry, I dont understand that at all. The same thing occurs if we integrate from any $t_{0}$.
$endgroup$
– user7534
Jan 8 at 7:45
$begingroup$
If we begin from $t_0$ the "starting time" of the chronometer will be at $t_0$ instead of $0$, that's all.
$endgroup$
– Jean Marie
Jan 8 at 7:48
$begingroup$
If we begin from $t_0$ the "starting time" of the chronometer will be at $t_0$ instead of $0$, that's all.
$endgroup$
– Jean Marie
Jan 8 at 7:48
$begingroup$
Another way to see it is to draw a curve $(x(t),y(t)) $ with a regular scale for $t = 0, 0.1, 0.2,...$ : you will see that, in general, resulting points are not regularly spaced. Thus you need a re-parameterization $t=varphi(s)$ of the curve in order that the points of $(x(varphi(s)),y(varphi(s))$ are regularly spaced.
$endgroup$
– Jean Marie
Jan 8 at 7:52
$begingroup$
Another way to see it is to draw a curve $(x(t),y(t)) $ with a regular scale for $t = 0, 0.1, 0.2,...$ : you will see that, in general, resulting points are not regularly spaced. Thus you need a re-parameterization $t=varphi(s)$ of the curve in order that the points of $(x(varphi(s)),y(varphi(s))$ are regularly spaced.
$endgroup$
– Jean Marie
Jan 8 at 7:52
2
2
$begingroup$
This is not the main issue : till a certain point, you can take the unit of length you want (e.g., take yards instead of meters). The main issue is to annihilate accelerations or decelarations when you move along your trajectory (i.e., making acceleration = 0 everymwhere)
$endgroup$
– Jean Marie
Jan 8 at 8:00
$begingroup$
This is not the main issue : till a certain point, you can take the unit of length you want (e.g., take yards instead of meters). The main issue is to annihilate accelerations or decelarations when you move along your trajectory (i.e., making acceleration = 0 everymwhere)
$endgroup$
– Jean Marie
Jan 8 at 8:00
|
show 3 more comments
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$begingroup$
Intuitively (as you ask for an intuitive grasp of the concept) : knowing that speed, generaly speaking, is the derivative of position, integrating a (norm of a) speed gives back a position. Moreover there is a normalizing aspect with the integration from $0$ to $t$ : thus you are with a "normalized" position... Once again it is with "hand waving"...
$endgroup$
– Jean Marie
Jan 8 at 7:39
1
$begingroup$
@Jean Marie I am sorry, I dont understand that at all. The same thing occurs if we integrate from any $t_{0}$.
$endgroup$
– user7534
Jan 8 at 7:45
$begingroup$
If we begin from $t_0$ the "starting time" of the chronometer will be at $t_0$ instead of $0$, that's all.
$endgroup$
– Jean Marie
Jan 8 at 7:48
$begingroup$
Another way to see it is to draw a curve $(x(t),y(t)) $ with a regular scale for $t = 0, 0.1, 0.2,...$ : you will see that, in general, resulting points are not regularly spaced. Thus you need a re-parameterization $t=varphi(s)$ of the curve in order that the points of $(x(varphi(s)),y(varphi(s))$ are regularly spaced.
$endgroup$
– Jean Marie
Jan 8 at 7:52
2
$begingroup$
This is not the main issue : till a certain point, you can take the unit of length you want (e.g., take yards instead of meters). The main issue is to annihilate accelerations or decelarations when you move along your trajectory (i.e., making acceleration = 0 everymwhere)
$endgroup$
– Jean Marie
Jan 8 at 8:00