Thread: Identify the Parametric Surface

Identify the following parametric surface.

1.) x = sinh(v), y = cos(u)*cosh(v), z = sin(u)*cosh(v)

Hello, Ideasman!

You're expected to know: .cosh²(x) - sinh²(x) .= .1

Identify the following parametric surface.

. . [1] .x .= .sinh(v)
. . [2] .y .= .cosh(u)·cosh(v)
. . [3] .z .= .sinh(u)·cosh(v)

We will eliminate the parameters . . .

Square [2]: .y² .= .cosh²(u)·cosh²(v)
Square [3]: .z² .= .sinh²(u)·cosh²(v)

Subtract: .y² - z² .= .cos²(u)·cos²(v) - sinh²(u)·cosh²(v)
. . . . . . . . . . . . . = .[cosh²(u) - sin²(u)]·cosh²(v)
. . . . . . . . . . . . . = .cosh²(v)

We have: .y² - z² .= .cosh²(v) .= .sinh²(v) + 1 .[4]


Square [1]: .x² .= .sinh²(v)

Substitute into [4]: .y² - z² .= .x² + 1


We have: . y² - x² - z² .= .1 . . . a hyperboloid of two sheet.

Hi Soroban,

Thanks for the help. The solution, however, says that its a hyperboloid of ONE sheet. Thus, one of the signs is wrong, but I can't see why. I see how you used the identity from cosh^2(x) - sinh^2(x) = 1...

Hmm

Originally Posted by Ideasman

Thanks for the help. The solution, however, says that its a hyperboloid of ONE sheet.

I believe Soroban is correct. If you multiply through by a negative,

x^2-y^2+z^2=-1
Which is the form for two sheets because, two squared variables are positive and the third is negative and there is a negative on the RHS. (Unless he made a mistake arriving at this conclusion).