# Quadratic scaling of axes

The circular arc $$y=\sqrt{1-x^2}\tag{1}$$ becomes the linear function $$y'=1-x'\tag{2}$$ after substitutions $$x'=x^2$$ and $$y'=y^2$$. Here I restrict to arguments in the range $$[0,1]$$. How can I scale the axes so that eq.(1) is displayed as a line. This would be similar to scaling by logarithmic axes. GraphicsRow[{
Plot[Sqrt[1-x^2],{x,0,1},AspectRatio->1,AxesLabel->{x, y}],
Plot[1-x,{x,0,1},AspectRatio->1,AxesLabel->{x',y'}]
}]


Use ScalingFunctions

Plot[Sqrt[1 - x^2], {x, 0, 1}, AspectRatio -> 1,
AxesLabel -> {"x", "y"},
ScalingFunctions -> {{Re[#^2] &, Re@Sqrt[#] &}, {Re[#^2] &,
Re@Sqrt[#] &}}, Ticks -> {0, 1}] You can play with the Ticks to obtain values in x' and 'y'.

• e.g. Ticks -> {Round[Table[Sqrt[i], {i, 0, 1, 0.1}], 0.01]} provides evenly spaced (but rounded) tick marks. Nov 14, 2021 at 16:45
• Could you say more in the answer to {Re[#^2] &, Re@Sqrt[#] &} ? Nov 14, 2021 at 18:35
• ScalingFunction requires a function and its inverse, hence #^2 and Sqrt[#]. The Re is leftover from when I was debugging and isn't needed. Nov 14, 2021 at 18:48