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I have a curved surface $z=x^2-y^2$. Given a point (x,y) on the surface, how do I calculate the (3D) distance from it to the origin (0,0,0)? By distance I am referring to the distance on top of the surface, not the straight line that can easily be calculated.

What I have found is that on the line (x,y), the surface behaves like the parabola $(1-y/x) (\sqrt{x^2+y^2})^2=(1-y/x)(x^2+y^2)$. Therefore, I can find the length using the formula $d(x,y)=\frac{1}{2(1-y/x)}\int_{0}^{\sqrt{x^2+y^2}}{\sqrt{1+t^2}dt}$.

However this result cannot be correct: upon using Desmos to plot the equation $d(x,y)=1$ to see how a circle on the hyperbolic surface look like, the result was not a closed shape: enter image description here

It also appears like the distance close to $x=0$ but far away from $(0,0,0)$ is 1 unit distance from the origin, when in fact it's much further away. Also, in the origin it appears as if the distance is 1 unit, when obviously it should be 0.

Have I done some of my calculations wrong, am I misinterpreting the results, or am I missing something?

Thanks in advance.

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  • $\begingroup$ $z=x^2+y^2$ doesn't describe a hyperbolic paraboloid, that's an elliptic paraboloid (formally, in fact it's a surface of rotation). So check your assumptions, first.. $\endgroup$ – Professor Vector Feb 19 '18 at 19:24
  • $\begingroup$ @ProfessorVector You are right, I mistyped. I meant $x^2-y^2$ $\endgroup$ – hrsidkpi Feb 19 '18 at 19:30
  • $\begingroup$ Then, you should move on: the equation $(1-y/x) (\sqrt{x^2+y^2})^2=(1-y/x)(x^2+y^2)$ doesn't really look less suspicious. $\endgroup$ – Professor Vector Feb 19 '18 at 19:37
  • $\begingroup$ @ProfessorVector that is not an equation, it’s just another form of the expression. $\sqrt{x^2+y^2}$ is the distance from the z axis, which is the variable in the parabola I’m measuring the length of. I thought writing it once raised to 2 and would emphasize it, and then I removed the root. $\endgroup$ – hrsidkpi Feb 19 '18 at 19:44

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