We calculate the tangent vector of parametric curve `{t^2,t}`  and then calculate the `ArcLength`  from `0` to `t0` then we get the point `{0,s[t0]}` in y-axis.

```
c[t_] = {t^2, t};
s[t_] = ArcLength[c[τ], {τ, 0, t}];
t0 = .6;
Show[ParametricPlot[c[t], {t, 0, 1}, Mesh -> {{t0}}, 
  MeshStyle -> {PointSize[Large], Red}, MeshFunctions -> (#3 &), 
  MeshShading -> {Red, Automatic}], 
 Graphics[{Arrow[{c[t], c[t] + .3 Normalize[c'[t]]}] /. 
    t -> t0, {Arrow[{{0, s[t0]}, {0, s[t0] + .3}}], Thick, Red, 
    PointSize[Large], Point[{0, s[t0]}], 
    Line[{{0, 0}, {0, s[t0]}}]}}], PlotRange -> All]
```
[![enter image description here][1]][1]


After that we translate the curve along the direction`{0, s[t0]} - c[t0]` and rotate it around `{0,s[t0]}` so that the tangent vector become the new direction `{0,1}` which toward to the y-axis;


```
c[t_] = {t^2, t};
s[t_] = ArcLength[c[τ], {τ, 0, t}];
r[t_, t0_] := 
  RotationTransform[{c'[t0], {0, 1}}, {0, s[t0]}][
   c[t] + {0, s[t0]} - c[t0]];
Manipulate[
 ParametricPlot[r[t, t0], {t, 0, 1}, AspectRatio -> Automatic, 
  PlotRange -> {{0, 1}, {0, 2}}], {t0, 0, 1}]
```

[![enter image description here][2]][2]


# Edition 1

We generalize the idea from above to deal with two parametric curves. Here we use `NDSolve` to handle the re-parametric equation of curve 
$$\begin{cases}\frac{\mathrm{d}s}{\mathrm{d}t}=|r'(t)|\\s(0)=0\end{cases}$$

 and thanks @Daniel Huber provide `FunctionInterpolation` to increasing the speed.

```
r1[t_] = {1.5 + 1.5 Cos[π - t], Sin[π - t]}; t1 = 
 FunctionInterpolation[
   InverseFunction[
     NDSolve[{s1'[t] == Norm[r1'[t]], s1[0] == 0}, 
       s1, {t, 0, 100}][[1, 1, 2]]][x], {x, 0, 50}] // Quiet;
r2[t_] = {t^2, 3 t};
t2 = FunctionInterpolation[
    InverseFunction[
      NDSolve[{s2'[t] == Norm[r2'[t]], s2[0] == 0}, 
        s2, {t, 0, 100}][[1, 1, 2]]][x], {x, 0, 50}] // Quiet;
Animate[Show[ParametricPlot[{r1[t1[s]], r2[t2[s]]}, {s, 0, 20}], 
  Graphics[Arrow[{r1[t1[s]], r1[t1[s]] + D[r1[t1[s]], s]} /. 
     s -> s0]], 
  Graphics[Arrow[{r2[t2[s]], r2[t2[s]] + D[r2[t2[s]], s]} /. 
     s -> s0]]], {s0, 0, 20}, DefaultDuration -> 10]
```


```
r1[t_] = {1.5 + 1.5 Cos[π - t], Sin[π - t]}; t1 = 
 FunctionInterpolation[
   InverseFunction[
     NDSolve[{s1'[t] == Norm[r1'[t]], s1[0] == 0}, 
       s1, {t, 0, 100}][[1, 1, 2]]][x], {x, 0, 50}] // Quiet;
r2[t_] = {t^2, 3 t};
t2 = FunctionInterpolation[
    InverseFunction[
      NDSolve[{s2'[t] == Norm[r2'[t]], s2[0] == 0}, 
        s2, {t, 0, 100}][[1, 1, 2]]][x], {x, 0, 50}] // Quiet;
trans[ss_, ss0_] := 
  RotationTransform[{D[r1[t1[s]], s], D[r2[t2[s]], s]} /. s -> ss0, 
    r2[t2[ss0]]][r1[t1[ss]] + r2[t2[ss0]] - r1[t1[ss0]]];
curves = ParametricPlot[{r1[t1[s]], r2[t2[s]]}, {s, 0, 20}];
Animate[Show[curves, 
  ParametricPlot[trans[ss, ss0], {ss, 0, 3 π}]], {ss0, 0, 20}, 
 DefaultDuration -> 10]
```

# Edition 2
I eventual found that we need not use `InverseFunction` at all, just only use the `NDSolve` since we can rewrite the equation as 

$$\frac{\mathrm{d}t}{\mathrm{d}s}=\frac{1}{|r'(t)|}$$
that is
$$\frac{\mathrm{d}t}{\mathrm{d}s}|r'(t)|=1$$
So we consider the equation

```
t'[s]*Norm[r'[t[s]]] == 1, t[0]==0
```

```
r1[t_] = {1.5 + 1.5 Cos[π - t], Sin[π - t]};
r2[t_] = {t^2, 3 t};
L = 20;
t1 = NDSolve[{t1'[s]*Norm[r1'[t1[s]]] == 1, t1[0] == 0}, 
    t1, {s, 0, L}][[1, 1, 2]];
t2 = NDSolve[{t2'[s]*Norm[r2'[t2[s]]] == 1, t2[0] == 0}, 
    t2, {s, 0, L}][[1, 1, 2]];
trans[s_, s0_] := 
  RotationTransform[{(r1@*t1)'[s0], (r2@*t2)'[s0]}, r2[t2[s0]]][
   r1[t1[s]] + r2[t2[s0]] - r1[t1[s0]]];
curves = ParametricPlot[{r1[t1[s]], r2[t2[s]]}, {s, 0, L}];
Animate[Show[curves, ParametricPlot[trans[s, s0], {s, 0, L}], 
  PlotRange -> {{-1, 18}, {-2, 15}}], {s0, 0, L}, 
 DefaultDuration -> 10]
```
[![enter image description here][3]][3]


  [1]: https://i.sstatic.net/b2SeJ.png
  [2]: https://i.sstatic.net/d4VUl.gif
  [3]: https://i.sstatic.net/ggoTH.gif