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Question:

Find a parameterization of the paraboloid $900z = 25x^2 + 36y^2$.


My Work

$25x^2 + 36y^2 = 900z$

$\implies (5x)^2 + (6y)^2 = (30\sqrt{z})^2$

Can we can represent this equation using cylindrical coordinates?

$(x,y,z) = (\rho \cos(\theta), \rho \sin(\theta), \zeta)$ where $\rho = $ radius and $2\pi \ge \theta \ge 0$.


I think I'm on the right track here, but I've spent hours unsuccessfully pondering over this problem and doing research. I want to get it into cylindrical coordinates and then parameterise it in terms of $u$ and $v$, but I'm just completely stuck.

I would greatly appreciate it if people could please take the time to show me the correct reasoning and solution for this problem.

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    $\begingroup$ Your parametrization $(x,y) = (\rho \cos(\theta), \rho \sin(\theta))$ cannot work because it would mean that the level curves are circles, whereas they are ellipses. $\endgroup$ – Jean Marie Apr 9 '17 at 10:30
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Whenever you have an equation of the form $z = f(x,y)$, you can use $x$ and $y$ as parameters. So one possible set of parametric equations is $$ x = u \quad ; \quad y = v \quad ; \quad z = \frac{1}{900}(25u^2 + 36v^2) $$ If you really want to use trig functions, then: $$ x = \tfrac15 r \cos \theta \quad ; \quad y = \tfrac16 r \sin \theta \quad ; \quad z = \tfrac{1}{900}r^2 $$ True polar coordinates are going to be messy. The problem is that each curve of the form $z = \text{constant}$ is an ellipse, and the equation of an ellipse in polar coordinates is a bit complicated.

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