Parametric and implicit representation of a cone

Question

http://mathworld.wolfram.com/Cone.html shows the parametric and implicit representation of a cone, I am wondering what the equation would look like if we also consider the bottom circle face for the cone.

Answer 1

Implicit form

More than one equation

The formula you refer to seems to be the following:

$$\frac{x^2+y^2}{c^2}=(z-z_0)^2$$

This is only a single euation, and as such, it describes the cone extended to infinity. Points below the base will be part of that cone, as will be points above the apex, where it continues symmetrically. To restrict this formulation to the cone from base circle to apex, you have to add the inequality

$$0 \leq z \leq z_0$$

So even for the “simple” cone, you need more than a single equation.

Adding the circle

Now let's add the circle. There is an easy trick to formulate that either of two conditions must be satisfied: rephrase the equations so that the right hand sides becomes zero, then multiply them. A product is zero if and only if any of its factors is zero. In this case you obtain the equation

$$\left(\frac{x^2+y^2}{c^2}-(z-z_0)^2\right)z = 0$$

which describes the fact that a point lies either on the cone or in the $z=0$ plane.

Add to that an inequality which restructs the portion of the plane to the circle:

$$ x^2 + y^2 \leq r^2 $$

and you are done. You could even omit half of the inequality given above, and only require

$$ z \leq z_0 $$

as this will omit the mirror image above the apex, while the first inequality will restrict your set of points to the an infinite cylinder which will exclude the parts of the infinite cone that lie below the $z=0$ plane.

Parametric form

When you want to use a single parameter to describe parts of two algebraic surfaces which join in a non-smooth way, you cannot do this with smooth functions. For this reason, you will have to use a case distinction at some point. One possible way would be to use absolute values like this:

\begin{align*} z &= \frac{u + \lvert u\rvert}{2} \\ x &= \frac{h-\lvert u\rvert}{h}r\cos\theta \\ y &= \frac{h-\lvert u\rvert}{h}r\sin\theta \\ u &\in [-h, h]\quad\theta\in[0, 2\pi) \end{align*}

Now different values of $u$ will correspond to different circles. The first half of the way, for negative $u$, radii increase while $z$ stays at zero to form the disc, while for positive $u$ the radii will decrease while $z$ increases to form the cone.

The formula for $z$ might also have been written as

$$z=\begin{cases}u&\text{if }u>0\\0&\text{else}\end{cases}$$

Which is more explicit but doesn't fit the other formulas as well as the above formulation.