Unexpected result from ArcLength












3












$begingroup$


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}]



enter image description here



Any ideas? I'm running 11.0.0.0






numerical-integration







share|improve this question











$endgroup$








  • 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
    3 hours ago












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
















3












$begingroup$


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}]



enter image description here



Any ideas? I'm running 11.0.0.0






numerical-integration







share|improve this question











$endgroup$








  • 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
    3 hours ago












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














3












3








3





$begingroup$


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}]



enter image description here



Any ideas? I'm running 11.0.0.0






numerical-integration







share|improve this question











$endgroup$




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}]



enter image description here



Any ideas? I'm running 11.0.0.0






numerical-integration




numerical-integration






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 3 hours ago









Henrik Schumacher

56.6k577157




56.6k577157










asked 3 hours ago









IvanIvan

1,617821




1,617821








  • 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
    3 hours ago












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














  • 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
    3 hours ago












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








1




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
3 hours ago






$begingroup$
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
3 hours ago














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




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










3 Answers
3






active

oldest

votes


















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}}]


enter image description here



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






share|improve this answer











$endgroup$













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












  • $begingroup$
    Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
    $endgroup$
    – MarcoB
    3 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
    3 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
    3 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
    3 hours ago





















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}]


enter image description here






share|improve this answer











$endgroup$





















    3












    $begingroup$

    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









    $endgroup$













      Your Answer





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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      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}}]


      enter image description here



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






      share|improve this answer











      $endgroup$













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












      • $begingroup$
        Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
        $endgroup$
        – MarcoB
        3 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
        3 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
        3 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
        3 hours ago


















      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}}]


      enter image description here



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






      share|improve this answer











      $endgroup$













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












      • $begingroup$
        Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
        $endgroup$
        – MarcoB
        3 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
        3 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
        3 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
        3 hours ago
















      3












      3








      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}}]


      enter image description here



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






      share|improve this answer











      $endgroup$



      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}}]


      enter image description here



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







      share|improve this answer














      share|improve this answer



      share|improve this answer








      edited 3 hours ago

























      answered 3 hours ago









      mjwmjw

      5879




      5879












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












      • $begingroup$
        Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
        $endgroup$
        – MarcoB
        3 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
        3 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
        3 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
        3 hours ago




















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












      • $begingroup$
        Yes, but this does not seem to answer the OP's question. Try including smaller values of $p$.
        $endgroup$
        – MarcoB
        3 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
        3 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
        3 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
        3 hours ago


















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






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














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




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




      2




      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
      3 hours ago






      $begingroup$
      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
      3 hours ago






      1




      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
      3 hours ago




      $begingroup$
      @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
      3 hours ago




      1




      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
      3 hours ago






      $begingroup$
      @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
      3 hours ago













      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}]


      enter image description here






      share|improve this answer











      $endgroup$


















        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}]


        enter image description here






        share|improve this answer











        $endgroup$
















          3












          3








          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}]


          enter image description here






          share|improve this answer











          $endgroup$



          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}]


          enter image description here







          share|improve this answer














          share|improve this answer



          share|improve this answer








          edited 3 hours ago

























          answered 3 hours ago









          Henrik SchumacherHenrik Schumacher

          56.6k577157




          56.6k577157























              3












              $begingroup$

              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









              $endgroup$


















                3












                $begingroup$

                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









                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  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









                  $endgroup$



                  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












                  share|improve this answer



                  share|improve this answer










                  answered 2 hours ago









                  Bill WattsBill Watts

                  3,4911620




                  3,4911620






























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