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Unexpected result from ArcLength

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4

$begingroup$

I want to determine the arc lenght of a parametric curve $C: x(t),y(t) = cos(t)^p , sin(t)^p $ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.

I set up the following function of $p$:

L[p_] := ArcLength[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, 
 Method -> "NIntegrate", MaxRecursion -> 20]

For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708.

For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100] results in 1.30603. Not close to 2 (it's not even bigger than Pi/2).

Plotting, results in the following:

Plot[L[p], p, 0, 1]

Any ideas? I'm running 11.0.0.0

numerical-integration

share|improve this question

edited 6 hours ago

Henrik Schumacher

56.6k577157

asked 7 hours ago

IvanIvan

1,622821

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I get a warning from NIntegrate ("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluate L for small p.
  $endgroup$
  – MarcoB
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarcoB I don't get any warnings when evaluating L[1/100]
  $endgroup$
  – Ivan
  6 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

I want to determine the arc lenght of a parametric curve $C: x(t),y(t) = cos(t)^p , sin(t)^p $ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.

I set up the following function of $p$:

L[p_] := ArcLength[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, 
 Method -> "NIntegrate", MaxRecursion -> 20]

For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708.

For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100] results in 1.30603. Not close to 2 (it's not even bigger than Pi/2).

Plotting, results in the following:

Plot[L[p], p, 0, 1]

Any ideas? I'm running 11.0.0.0

numerical-integration

share|improve this question

edited 6 hours ago

Henrik Schumacher

56.6k577157

asked 7 hours ago

IvanIvan

1,622821

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  I get a warning from NIntegrate ("Numerical integration converging too slowly; suspect one of the following: singularity...") when trying to evaluate L for small p.
  $endgroup$
  – MarcoB
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @MarcoB I don't get any warnings when evaluating L[1/100]
  $endgroup$
  – Ivan
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I want to determine the arc lenght of a parametric curve $C: x(t),y(t) = cos(t)^p , sin(t)^p $ with $p$ between $0$ and $1$, and $t$ between $0$ and $pi/2$.

I set up the following function of $p$:

L[p_] := ArcLength[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, 
 Method -> "NIntegrate", MaxRecursion -> 20]

For $p=1$ we have a quarter of a circle of radius 1 and we know the arc length is equal to $pi/2$. The above function gives the correct result: 1.5708.

For $p$ close to zero, the curve approaches a square, and we know the result should be very close to $2$. However, the function doesn't even come close to it. Evaluating L[1/100] results in 1.30603. Not close to 2 (it's not even bigger than Pi/2).

Plotting, results in the following:

Plot[L[p], p, 0, 1]

Any ideas? I'm running 11.0.0.0

numerical-integration

share|improve this question

edited 6 hours ago

Henrik Schumacher

56.6k577157

asked 7 hours ago

IvanIvan

1,622821

$endgroup$

L[p_] := ArcLength[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, 
 Method -> "NIntegrate", MaxRecursion -> 20]

share|improve this question

edited 6 hours ago

Henrik Schumacher

56.6k577157

asked 7 hours ago

IvanIvan

1,622821

share|improve this question

edited 6 hours ago

Henrik Schumacher

56.6k577157

asked 7 hours ago

IvanIvan

1,622821

edited 6 hours ago

Henrik Schumacher

56.6k577157

edited 6 hours ago

Henrik Schumacher

56.6k577157

asked 7 hours ago

IvanIvan

1,622821

asked 7 hours ago

IvanIvan

1,622821

3 Answers
3

active

oldest

votes

5

$begingroup$

Seems to be a precision thing.

L[p_] = Cos[t]^p, Sin[t]^p

ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]

1.99447959240474567...

share|improve this answer

answered 5 hours ago

Bill WattsBill Watts

3,5111620

$endgroup$

add a comment | 

3

$begingroup$

Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]

gives this plot at $p=0.01$:

(An unpreprocessing plot was here.)

Yes, looks to be approaching two sides of a square, but the sides are shrinking!

UPDATE:

p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]

So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...

share|improve this answer

edited 6 hours ago

answered 6 hours ago

mjwmjw

5879

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  >but the sides are shrinking. Not really. Check your plotrange
  $endgroup$
  – Ivan
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
  $endgroup$
  – MarcoB
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
  $endgroup$
  – mjw
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
  $endgroup$
  – Henrik Schumacher
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Ivan, yes, point well taken! Could be an artifact. But why is the ArcLength[] returning wrong results? Seems that in both cases Mathematica is undersampling ...
  $endgroup$
  – mjw
  6 hours ago
  
  

  
  
  
  

 | 
show 9 more comments

3

$begingroup$

I can only provide an alternative to bypass ArcLength.

The points pts of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p very close to 0, but at least you may obtain a qualitatively correct plot (so I hope).

Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.

n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
 Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
 ]
Plot[L[p], p, 0.001, 1]

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

56.6k577157

$endgroup$

add a comment | 

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5

$begingroup$

Seems to be a precision thing.

L[p_] = Cos[t]^p, Sin[t]^p

ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]

1.99447959240474567...

share|improve this answer

answered 5 hours ago

Bill WattsBill Watts

3,5111620

$endgroup$

add a comment | 

5

$begingroup$

Seems to be a precision thing.

L[p_] = Cos[t]^p, Sin[t]^p

ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]

1.99447959240474567...

share|improve this answer

answered 5 hours ago

Bill WattsBill Watts

3,5111620

$endgroup$

add a comment | 

$begingroup$

Seems to be a precision thing.

L[p_] = Cos[t]^p, Sin[t]^p

ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]

1.99447959240474567...

share|improve this answer

answered 5 hours ago

Bill WattsBill Watts

3,5111620

$endgroup$

L[p_] = Cos[t]^p, Sin[t]^p

ArcLength[L[1/100], t, 0, π/2, WorkingPrecision -> 1000]

1.99447959240474567...

share|improve this answer

answered 5 hours ago

Bill WattsBill Watts

3,5111620

answered 5 hours ago

Bill WattsBill Watts

3,5111620

answered 5 hours ago

Bill WattsBill Watts

3,5111620

3

$begingroup$

Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]

gives this plot at $p=0.01$:

(An unpreprocessing plot was here.)

Yes, looks to be approaching two sides of a square, but the sides are shrinking!

UPDATE:

p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]

So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...

share|improve this answer

edited 6 hours ago

answered 6 hours ago

mjwmjw

5879

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  >but the sides are shrinking. Not really. Check your plotrange
  $endgroup$
  – Ivan
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
  $endgroup$
  – MarcoB
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
  $endgroup$
  – mjw
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
  $endgroup$
  – Henrik Schumacher
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Ivan, yes, point well taken! Could be an artifact. But why is the ArcLength[] returning wrong results? Seems that in both cases Mathematica is undersampling ...
  $endgroup$
  – mjw
  6 hours ago
  
  

  
  
  
  

 | 
show 9 more comments

3

$begingroup$

Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]

gives this plot at $p=0.01$:

(An unpreprocessing plot was here.)

Yes, looks to be approaching two sides of a square, but the sides are shrinking!

UPDATE:

p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]

So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...

share|improve this answer

edited 6 hours ago

answered 6 hours ago

mjwmjw

5879

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  >but the sides are shrinking. Not really. Check your plotrange
  $endgroup$
  – Ivan
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
  $endgroup$
  – MarcoB
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  No, not that desperate ... By the way, you guys are tough! Imagine a classroom where a student gives an answer/suggestion that is not correct. Or a collaboration with somebody making a mistake. Seems that "Mathematica Stack Exchange" culture does not like people taking risks. May hurt creativity ... But we will have strictly precise, quality answers!!
  $endgroup$
  – mjw
  6 hours ago
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @mjw "Seems that "Mathematica Stack Exchange" culture does not like people taking risks." To the contrary. I was really surprise that posts in this threads were downvoted so quickly. Downvotes on answers are actually very uncommon on this site.
  $endgroup$
  – Henrik Schumacher
  6 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  @Ivan, yes, point well taken! Could be an artifact. But why is the ArcLength[] returning wrong results? Seems that in both cases Mathematica is undersampling ...
  $endgroup$
  – mjw
  6 hours ago
  
  

  
  
  
  

 | 
show 9 more comments

$begingroup$

Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]

gives this plot at $p=0.01$:

(An unpreprocessing plot was here.)

Yes, looks to be approaching two sides of a square, but the sides are shrinking!

UPDATE:

p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]

So yes, the sides are not shrinking, but Mathematica seems to be missing some of the curve ...

share|improve this answer

edited 6 hours ago

answered 6 hours ago

mjwmjw

5879

$endgroup$

Manipulate[ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2], p, 0.01, 1]

p = 0.01; ParametricPlot[Cos[t]^p, Sin[t]^p, t, 0, Pi/2, Axes -> False, Frame -> True, PlotRange -> 0, 1.1, 0, 1.1]

share|improve this answer

edited 6 hours ago

answered 6 hours ago

mjwmjw

5879

answered 6 hours ago

mjwmjw

5879

answered 6 hours ago

mjwmjw

5879

3

$begingroup$

I can only provide an alternative to bypass ArcLength.

The points pts of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p very close to 0, but at least you may obtain a qualitatively correct plot (so I hope).

Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.

n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
 Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
 ]
Plot[L[p], p, 0.001, 1]

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

56.6k577157

$endgroup$

add a comment | 

3

$begingroup$

I can only provide an alternative to bypass ArcLength.

The points pts of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p very close to 0, but at least you may obtain a qualitatively correct plot (so I hope).

Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.

n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
 Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
 ]
Plot[L[p], p, 0.001, 1]

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

56.6k577157

$endgroup$

add a comment | 

$begingroup$

I can only provide an alternative to bypass ArcLength.

The points pts of a quarter circle are scaled such that they lie on the desired curve; afterwards the length of the polygonal line is computed. You will still get problems for values of p very close to 0, but at least you may obtain a qualitatively correct plot (so I hope).

Certainly this method won't provide you with the best possible accuracy. The relative error between the arclength $ell$ of an arc and the length $s$ of a secant is roughly $|ell/s - 1| leq ell^2 , max(|kappa|)$ in the limit of $ell to 0$. Since the maximal curvature in of curve goes to $infty$ for $p to infty$, the quality of this approximation will reduce significantly for $p to 0$.

n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
 Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
 ]
Plot[L[p], p, 0.001, 1]

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

56.6k577157

$endgroup$

n = 10000;
pts = Transpose[Cos[Subdivide[0., Pi/2, n]], Sin[Subdivide[0., Pi/2, n]]];
L[p_] := With[x = pts/Power[Dot[(Abs[pts]^(1/p)), 1., 1.], p],
 Total[Sqrt[Dot[Differences[x]^2, 1., 1.]]]
 ]
Plot[L[p], p, 0.001, 1]

share|improve this answer

edited 6 hours ago

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

56.6k577157

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

56.6k577157

answered 6 hours ago

Henrik SchumacherHenrik Schumacher

56.6k577157