Expressing Hyperboloid as Parametric Function & Tangent Plane

[LostInSpace]

Hi!
I am supposed to write the hyperboloid x^2 + y^2 - z^2=1 as a parametric funktion and find an expression for the tangent plane in an arbitary point in terms of the parameters.

I think I have figured out that the parametric funktion is
<br /> \left\lbrace\begin{array}{ccl}<br /> x &amp;=&amp; \sqrt{1+t^2}\cos\varphi \\<br /> y &amp;=&amp; \sqrt{1+t^2}\sin\varphi \\<br /> z &amp;=&amp; t<br /> \end{array}\right.<br />

And if z = f(x,y), the tangent plane for the point (x_0, y_0, f(x_0, y_0)) is given by
<br /> f_t(x,y) = f(x_0, y_0) + \frac{\partial f(x_0, y_0)}{\partial x}(x-x_0) + \frac{\partial f(x_0, y_0)}{\partial y}(y - y_0)<br />

which in this case is evaluates to
<br /> f_t(x,y) = \sqrt{x_0^2 + y_0^2 - 1} + \left(\frac{x_0}{\sqrt{x_0^2+y_0^2-1}}\right)(x-x_0) + \left(\frac{y_0}{\sqrt{x_0^2+y_0^2-1}}\right)(y - y_0)<br />

Hope I'm right so far...
How am I supposed to express f_t(x,y) in terms of the parameters (t, \varphi)?

Thanks in advance!

 

[matt grime]

1. You have two t's meaning different things.

2. You need to put x_0 and y_0 in terms of t_0 and \phi_0, that's all, assuming all else is correct. The x and y in the tangent plane are not the x and y in the original equation, that is the solution sets for those equations are different. DO NOT put the parametrization for x and y from the original equation into the equaition for the plane.

NB, I think they might have expected you to use the chain rule to find the tangent vector.

 

[LostInSpace]

Hi, and thanks for responding!
I did that, and the result is
\xi_{t,\varphi}(x,y)=\frac{1}{t}\left[\sqrt{1+t^2}(x\cos\varphi + y\sin\varphi)\right]

This looks quite good... The problem is that I don't know how to verify it. Is it possible to do a parametric and a regular plot in the same graph in mathematica or maple?

Thanks again!

 

[matt grime]

I'm not sure that is correct. Ok, it looks good, but there is no z in there, and that can't be right - you're saying the tangent plane is always parallel to a plane orthogonal to the x-y plane, or equivalently parallel to a plane containing the z axis.

So, from first principles it is easy to define the tangent plane to the surface

f(x,y,z)=0 at a,b,c

as the set of x,y,z satisfying

f_x(a,b,c)(x-a) + f_y(a,b,c)(y-b) + f_z(a,b,c)(z-c)=0


so f in your case is x^2+y^2-z^2-1

 

[LostInSpace]

Well, essentially what I have done, is that I have defined z to be a funktion of x and y - z = \xi_{t,\varphi}(x,y). I did a plot in Maple with t=2, \varphi = 1, which gave me a "tilted" plane, but since I don't know which point those values correspond to, there is no way I can determine whether or not it is correct. How could I verify it?

Maybe I'm wrong, though... yeah, that's probably it :)

 

[matt grime]

You understand what f(x,y,z)=0 means, in that given some equation you can rearrange it to get it equal to zero on one side. You can by the look of it write down partial derivatives accurately, you know the answer is


f_x(a,b,c)(x-a)+f_y(a,b,c)(y-b)+f_z(a,b,c)(z-c)=0

where the subscripts mean partial derivatives (sorry i didn't mention that), why do you need to verify anything? That IS the answer. If you want to check what you're doing is right, find some examples with answers inthe book you're working from (I am presuming here). And have a little faith in your abilities, that's the best advice. You know the theory, if you can do basic manipulation and partial differentiation you can get the right answer, so trust yourself.



If you really have to verify it, find three points you KNOW to be on the plane and not lying on the same line, plug their coordinates in and see if you get a zero. The first thinf to do would be to put in the point at which you'r finding the tangent plane and see if that's on the plane. Of course if that is, and you've got the correct normal vector you're done.


Have some faith in your skills.

 

[LostInSpace]

Thanks for your encouragement!