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Show that the hyperbolic paraboloid can be represented parametrically as $$r(u,v)=\langle a(u+v), b(u-v), uv\rangle$$

Find the curves $u$ is constant and $v$ is constant.

I guess I need to use the hyperbolic paraboloid equation. But I cannot solve this. Please help me doing it. Thank you very much

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  • $\begingroup$ I changed $\displaystyle r(u,v)=<a(u+v), b(u-v), uv>$ to $\displaystyle r(u,v)=\langle a(u+v), b(u-v), uv\rangle$. That is standard usage. $\endgroup$ Commented Oct 20, 2013 at 20:19
  • $\begingroup$ Sorry. I didnt know how to make it in MathJax. Thanks for editting. Dear @MichaelHardy by the way, do you know how to solve this question? I grateful of you if you Will help:) $\endgroup$
    – 1190
    Commented Oct 20, 2013 at 20:21
  • $\begingroup$ Maybe I can. It would be convenient if you told us which version of the equation of the hyperbolic paraboloid you're using as a definition; in particular where do the coefficients $a$ and $b$ appear in it? $\endgroup$ Commented Oct 20, 2013 at 20:24
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    $\begingroup$ Do you have an equation involving three variables $x$, $y$, and $z$ and coefficients called $a$ and $b$, that is for a hyperbolic paraboloid, not just a hyperbola? $\endgroup$ Commented Oct 20, 2013 at 20:31
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    $\begingroup$ OK, I can probably do something with that. I wondered if you also had an $xy$ term, which would be there if you were to rotate the hyperbolic paraboloid about the $z$ axis, in which case it would still be a hyperbolic paraboloid but wouldn't satisfy the equation you've written above. Apparently the problem does not include those. $\endgroup$ Commented Oct 20, 2013 at 20:36

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You've written $$r(u,v)=\langle a(u+v), b(u-v), uv\rangle$$ and that's the same thing as saying $\langle x,y,z\rangle = \langle a(u+v), b(u-v), uv\rangle$, so that $$ \begin{align} x & = a(u+v), \\ y & = b(u-v), \\ z & = uv. \end{align} $$ So $$ \frac{y^2}{b^2} - \frac{x^2}{a^2} = \frac{(b(u-v))^2}{b^2} - \frac{(a(u+v))^2}{a^2}. $$ You can cancel $a$ and $b$ and then do routine simplifications and see if it turns into $-4uv$, which then turns into $-4z$, so that $\frac{y^2}{b^2} - \frac{x^2}{a^2} = \frac{z}{c}$ with $c = -\frac14$.

That shows that the parametrized surface you've got is a subset of the surface defined by the equations involving $x$, $y$, and $z$. Then you need to show that it's the whole set. One way to do that would be by solving for $u$ and $v$ in terms of $x$ and $y$, thereby showing that all $(x,y,z)$ points on the surface actually appear within the surface parametrized by $u$ and $v$.

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  • $\begingroup$ Thank you I understand well:))) $\endgroup$
    – 1190
    Commented Oct 20, 2013 at 20:55

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