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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'}]
}]
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1 Answer 1

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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}]

enter image description here

You can play with the Ticks to obtain values in x' and 'y'.

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

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