Given z=x^3 rotate this curve towars the y axis about the z axis so that z=f(x,y) but with the same curve as z=x^3. Write z=f(x,y)=f(x^3,theta). Eventually z=y^3 when theta =90 degrees. I am guessing that the projection onto the x and y axes are streatched: z=ax^3+by^3 may be a solution? Could we say z=r^3.
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When we rotate a point $(x,y)$ by $\theta$ degrees counterclockwise, our new point $(x',y')$ is defined to be $x'=x\cos\theta-y\sin\theta$ and $y'=y\cos\theta+x\sin\theta$ (see here for an explanation of this).
To apply this method to the curve $z=x^3$, replacing $x$ with $x\cos\theta-y\sin\theta$ will rotate the $x$ and $y$ axes by $\theta$ degrees. However, we want the curve to rotate, not the axes. Replacing $x$ with $x\cos(-\theta)-y\sin(-\theta)$, which can be simplified to $x\cos\theta+y\sin\theta$, will rotate the axes in the opposite direction, effectively rotating the curve in the direction we want. The result is: $$z=(x\cos\theta+y\sin\theta)^3.$$ And yes, when $\theta=90^\circ$, $z=(x\cos90^\circ+y\sin90^\circ)^3$ or $z=y^3$ (because $\cos90^\circ=0$ and $\sin90^\circ=1$).
Here is a graph (created at https://www.geogebra.org/3d) of $z=x^3$ (in green) and $z=(x\cos45^\circ+y\sin45^\circ)^3$ (in brown):
Blitzquark
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What if I were to try rotating a surface? say : $$Re(\frac{1}{(1+z^2)})=\frac{1+x^2-y^2}{(1+x^2-y^2)^2+4x^2y^2}$$ Replacing x gives: $$Re(\frac{1}{(1+z^2)})=\frac{1+(x\cos45^\circ+y\sin45^\circ)^2-y^2}{(1+(x\cos45^\circ+y\sin45^\circ)^2-y^2)^2+4(x\cos45^\circ+y\sin45^\circ)^2y^2}$$ – User3910 Apr 23 '18 at 23:55
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Also replace y with $y\cos\theta-x\sin\theta$, since we are rotating about the z-axis; I don't know too much about surfaces, but I believe this would do the trick. – Blitzquark Apr 24 '18 at 01:26
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See https://en.wikipedia.org/wiki/Rotation_of_axes for more information about rotating points, conic sections, etc. – Blitzquark Apr 24 '18 at 01:28
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