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?

  • 2
    $\begingroup$ What happens if you compute $x^2/9+y^2/16$? Think about it. $\endgroup$
    – Mikasa
    Feb 27, 2013 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, 2013 at 17:45

2 Answers 2


Short Answer: $16x^2+9y^2=144z^2$


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:

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


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 $a=b$.

  • 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) $z=u\in\Bbb R$.

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.