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Given the following vector equation, how do I eliminate the parameters $u,v$ to get an equation of a surface in rectangular coordinates?

$$\vec{r}(u,v)=3u\cos(v)\hat{\imath} + 4u\sin(v)\hat{\jmath} + u\hat{k}$$

I can express this as a set of parametric equations:

$$x(u,v)=3u\cos(v)$$ $$y(u,v)=4u\sin(v)$$ $$z(u,v)=u$$

I'm not sure where to go from here -- how do you combine these three into one equation? Is there a general procedure to follow when eliminating parameters?

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    $\begingroup$ What happens if you compute $x^2/9+y^2/16$? Think about it. $\endgroup$ – mrs Feb 27 '13 at 16:33
  • $\begingroup$ Hm, it works ($x^2/9+y^2/16=z^2$). But how do you know to approach the problem that way? $\endgroup$ – kennysong Feb 27 '13 at 17:45
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Short Answer: $16x^2+9y^2=144z^2$

Derivation:

The little trick here is to exploit the fact that $z(u,v)=u$. Our surface is given by

$$(x(u,v),y(u,v),z(u,v)) = (3u\cos v,4u\sin v,u).$$

We know that $\cos^2v+\sin^2v=1$ for all $v$ and we can use this fact. Notice that:

$$\frac{x}{3z} = \frac{3u\cos v}{3u} = \cos v$$

Similarily, we can show that $y/4z = \sin v$ and then use the identity $\cos^2v = \sin^2v\equiv 1$:

$$\left( \frac{x}{3z} \right)^{\! 2} + \left( \frac{y}{4z} \right)^{\! 2} = 1$$

Expanding out all of the powers gives the following chain of events:

$$\left( \frac{x}{3z} \right)^{\! 2} + \left( \frac{y}{4z} \right)^{\! 2} = 1 \implies \frac{x^2}{9z^2}+\frac{y^2}{16z^2}=1 \implies 16x^2+9y^2=144z^2$$

We can check this equation just to make sure it's correct:

$$16x^2+9y^2=16(3u\cos v)^2+9(4u\sin v)^2 = 144u^2\cos^2v + 144u^2\sin^2v=144u^2 = 144z^2$$

GENERAL WARNING

Let $P$ be the set of points given by a parametrisation, and let $E$ be the set of points given by an equation. We have shown that every point in $P$ is a point in $E$, that is to say $P$ is a subset of $E$. It might be the case that there are points in $E$ that the parametrisation does not cover. Think of $P$ as $(x(t),y(t)) = (t,\sqrt{t})$ and $E$ as $x=y^2$. (The points with $x=y^2$ and $y<0$ are not in $P$ although they are in $E$.) To show that an equation and a parametrisation give the same set of points we must show that all the points of the equation's solution are covered by the parametrisation ($E \subseteq P$) and that all of the points of the parametrisation satisfy the equation ($P \subseteq E$). If $P \subseteq E$ and $E \subseteq P$ then $E=P$, just like with ordinary numbers: if $a \le b$ and $b \le a$ then $a=b$.

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  • The mapping $v\mapsto (\cos v,\sin v)$ gives the standard parametrization of the unit circle on the plane around the origin.
  • Rescaling from the origin, (using orthogonal affinities) $3$ times along the $x$-axis and $4$ times along $y$-axis, will give an ellipse, parametrized as $v\mapsto (3\cos v,4\sin v)$. Its equation is thus $\displaystyle\left(\frac x3\right)^2 + \left(\frac y4\right)^2=1$.
  • Then derive the equation of the plane slices for each (fixed) $z=u\in\Bbb R$.
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