Hi, Say I have the parametrization r = (tanhu, sechu*sinv, sechu*cosv) How do I get the cartesian equation of the surface? Thanks

Looks difficult, but if I had to do that, I would start by expressing all trig function in terms of exponential functions. Note that it may not be possible to get a nice closed form expression, but if it is possible, then you are more likely to see the method with exponentials.

Do you understand the principles of what you have to do? You have three equations x = f(u, v), y = g(u, v), and z = h(u, v). Your objective is to combine them algebraically in such a way as to remove the explicit dependence on the parameters u and v. Thus, suppose you know u = U(x, y, z) and v = V(x, y, z). Then a Cartesian equation of the surface would be x = f(U(x, y, z), V(x, y, z)). There are two more corresponding to the y and z coordinates. All easy to write symbolically, but the majority of real-world functions aren't so algebraically congenial. Another problem with using the Cartesian coordinates is you can easily pick values of the coordinates that are not on the surface. For example, in 2-D, if you have a unit circle, blindly picking x = 7 gives you a point that is not on the circle (of course, it's valid for the mapping, but it's not in the set of points that geometrically describe the circle). Besides steveb's suggestion (which is a good one -- and remember to use Euler's equation for the sin and cos too), also grab a math handbook and start throwing trig and hyperbolic trig identities into the equations -- sometimes you get lucky. Oh, and you might be able to use the Gudermannian function, which relates hyperbolic trig and regular trig functions without complex numbers. You don't explain what you're trying to accomplish, but if it's appropriate, an approximation might be useful. You can use a computer tool to plot what this surface looks like. Then, especially if your interest is only over a small region, an approximation might be usable -- and provide easier algebra.

The cartesian answer is meant to be x^2 + y^2 + z^2 = 1. Simple sphere lol I think you have to eliminate all the exponential terms using this. Anyway, thx for the help and directives...

Well, knowing the answer will certainly help you get some insight. At least you know it's possible to get a simple closed form. Keep us posted on how you are doing on this. I'm sure if you post intermediate steps, we can help you further, if you get stuck.

Actually, knowing the answer let's you treat this as a trig identity, and it turns out that that is a rather simple one to prove. However, starting without the answer would be much harder.

There seems to be a lot of information missing. The above expression seems to already be in cartesian form where the vector r is described by the ordered triplet (x,y,z) where; Specifically; While -1< x < 1, there is no such restriction given for y or z and as it appears to stand, r, over all u,v ε R, can span very non-spherical shapes. Can you state the full problem? That might help.

What he is asking is pretty clear. He has the parametric form of the surface; that is, x, y and z in terms of two parameters, as you pointed out. He really should give the range for u and v, but it's not hard to figure out that any values limit the range of x, y and z to +/- 1. Hence, the range of x, y and z will follow from the unlimited range of u and v, so those doesn't need to be specified. So, what he is looking for is the equation of the surface in terms of x, y and z directly, rather than the parametric form. We already now know the answer is the standard equation for the unit sphere x^2+y^2+z^2=1, so any confusion that might have existed at first, is now gone. This pretty much defines what he wants to do. The solution is almost trivial once you know it's a sphere. Had we not known this, then someonesdad's advice to plot the solution now appears (in hindsight) to be the best advice offered. Once you plot it, you will immediately know (or at least strongly suspect) it is a unit sphere. Once you know that, the functional form x^2+y^2+z^2=1 is an obvious first guess. Once you know that, then the proof of this guess becomes a trig identity as follows. Only the most basic trig identities need to be known to prove the above identity. I won't do the final step, since the OP should do that.

Well, I guess that's what I'm getting at. u and v would normally be accompanied by an r, or some other radial scalar, which in this case can be inferred as being 1. Then knowing that the surface is a unit sphere makes it trivial. Too trivial. I'm just wondering if that is how the problem was stated, or whether he looked the answer up in the back of the book.