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

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...

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 $ell to 0$. Here $kappa$ denotes the curvature of the curve. Since the maximal curvature of the 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]

Edit

The ratio behind this is that in contrast to the parameterization

γ[t_, p_] = Cos[t]^p, Sin[t]^p;

the parameterization

η[t_, p_] = Cos[t], Sin[t]/ Power[Cos[t]^(1/p) + Sin[t]^(1/p), p];

has finite speed which is always helpful for determining the arclength by integration:

assume = p > 0, 0 < t < Pi/2;
speedγ[t_, p_] = Simplify[Sqrt[D[γ[t, p], t].D[γ[t, p], t]], assume];
speedη[t_, p_] = Simplify[Sqrt[D[η[t, p], t].D[η[t, p], t]], assume];
Quiet@GraphicsRow[
 Plot[Evaluate[Table[speedγ[t, 2^-k], k, 0, 10]], t, 0, Pi/2,
 PlotLabel -> "Speed of γ",
 PlotRange -> 0, 10
 ],
 Plot[Evaluate[Table[speedη[t, 2^-k], k, 0, 10]], t, 0, Pi/2,
 PlotLabel -> "Speed of η",
 PlotRange -> 0, 10
 ]
 ,
 ImageSize -> Large
 ]

Whit this parameterization, one can also employ NIntegrate to compute the arclength, at least for not too small p.

NIntegrate[speedη[t, 1/1000], t, 0, Pi/2]

2.

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.)

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 ...

ADDITIONAL UPDATE:

f[t_][p_] := Cos[t]^p, Sin[t]^p;
p = 0.01;
k = 10^6;
Show[
 ListLinePlot[Transpose@f[[Pi]/(2 k) Range[0, k]][p], PlotStyle -> Red], 
 ListPlot[Transpose@f[[Pi]/(2 k) Range[0, k]][p], PlotStyle -> PointSize[Large]], 
 AspectRatio -> 1, Frame -> True, Axes -> False
]

If we sample $t$ with one million equally spaced points, there are big jumps to the first and the last point!

Plot[f[10^-q][.01][[2]], q, 0, 100, Frame -> True, Axes -> False, FrameLabel -> "-Log10[t]", "f[[2]]"]

This plot shows that we need $t le 10^-100$ for the $y$-value of the curve to be less than $approx 0.1$ when $p=.01$.