Parametric form of a cubic function

Question

After researching about curves in computer graphics (splines in my case), I have come across something I did not know: Explicit functions like: \$y=x^2+2\$ are not the best way to interpolate between points because you could get 2 or more values for y for the same x. So, parametric functions seem much better for this if we make the commonly used t parameter, the distance from origin, time, etc...

After studying a bit how to convert from explicit to parametric form I have been shocked when I have found in some texts they use these parametric cubic functions to interpolate through points: $$\begin{align} x(t)&=at^3 + bt^2 + ct +d \\ y(t)&=2at^3 + 2bt^2 + 2ct +2d \end{align} $$ but wait... this is the exact same form the explicit function was: $$ y(x)= ax^3 + bx^2 + cx + d $$ From what I have seen for other example functions in some math books the parametric funcions where not the same form that its explicit form counterpart.

Am I missing something?

Examples of the texts I'm refering to could be:

https://www.cs.helsinki.fi/group/goa/mallinnus/curves/curves.html

http://escience.anu.edu.au/lecture/cg/Spline/parametricCubicCurves.en.html

Answer 1

It's trivial to make a parametric equation that has the same form as the explicit function.

Take for example: $$ y = x^2 $$ Here \$y\$ is a quadratic polynomial in \$x\$. We can use \$t = x\$ as our parameter to get: $$ p(t) = ( t, t^2 ) $$ Where \$p_x\$ and \$p_y\$ are quadratic polynomials in \$t\$. (Okay, \$x\$ is linear polynomial, which is only a degenerate quadratic, but the salient point is that it's a polynomial of order at most 2)

This is not guaranteed though. Let's use the alternate parameter \$u = t^{1/3}\$ $$ q(u) = (u^3, u^6) $$ This describes the same parabola, but now it's using third & sixth-order polynomials instead of second-order.

So the choice of parameter can wildly affect the form of the parametrized equation. (Some natural choices of parameter, like arc length, often have no closed-form equation at all!)

We can also attack this in reverse. Take for example this parametric equation: $$ f(t) = (t^2, t^2 + t) $$ Here both components \$f_x\$ and \$f_y\$ are quadratic polynomials in \$t\$, but \$y\$ can't be expressed as a quadratic polynomial in \$x\$. (The closest would be \$y = x ± \sqrt x\$ which isn't even a single-valued function)

So, it's not a rule that the parametric equation must or must not have the same form as the explicit version of the equation (and neither is guaranteed to even exist as a closed form expression). Some parametrizations will have a similar form and some won't, depending on the choice of function and parameter.

Answer 2

The first example you are parametric equations: $$\begin{align} x(t)&=at^3 + bt^2 + ct +d \\ y(t)&=2at^3 + 2bt^2 + 2ct +2d \end{align}$$ In the parametric equations, x & y are found independently of each other; this independence is what makes them parametric. In contrast, your second equation: $$ y(x)= ax^3 + bx^2 + cx + d $$ defines y as a function of x. Since y depends on x, the second equation is not parametric. Your examples have both been expressed as polynomials, but they are not the same.