Inverse 3D plot

4 views (last 30 days)
Haider al-kanan
Haider al-kanan on 6 May 2016
Edited: Haider al-kanan on 24 Oct 2020
I want to inverse surface z , z=f(x,y).suc

Accepted Answer

Walter Roberson
Walter Roberson on 6 May 2016
This is not well-defined in the way you stated it. z takes two parameters, but inverse takes one parameter. inverse would have to emit two parameters separately.
It would be more consistent if you had two functions, inversex and inversey, and you required
z(inversex(z),inversey(z)) = z
This just might be possible, I guess. But you would not be able to require that
inversex(z(x,y)) == x and inversey(z(x,y)) == y
You would need to allow the inversex and inversey to map to any (x,y) location that would produce the same z.
For example, if z = x^2 + 0*y, then inversex() could return either the positive or negative square root of x, and inversey() could return anything, because z() of that combination together would be 1, but there is no formula that, given (say) 17, can tell you whether that came from x = -sqrt(17) or x = +sqrt(17).

More Answers (1)

CS Researcher
CS Researcher on 6 May 2016

In 2-D you can visualize the transformation as a matrix multiplication. If I is your matrix (x,y), you apply the transformation using a matrix A (it could be rotation, translation, etc.). The transformed matrix becomes

J = AI

If you apply $A^{-1}$ on J you get

 A^{-1}J = A^{-1}AI = I

Since inverse of an invertible matrix (check it out) multiplied by the matrix gives you identity matrix.

Were you looking for something like this?

  1 Comment
Haider al-kanan
Haider al-kanan on 7 May 2016
Thanks for comment, but I don't think that works, here is my question x=0:0.1:10; y=0:0.1:10; [xx,yy]=meshgrid(x,y); zz=xx.^2-yy.^2; I am not sure if your suggestion works on that, could you please clarify to me how is that works?

Sign in to comment.

Categories

Find more on Mathematics in Help Center and File Exchange

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!