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1.1.1 Plane curves

A plane curve can be expressed in the parametric form as
$\displaystyle x = x(t),\;\;\;y= y(t)\;,$     (1.1)

where the coordinates of the point $ (x,y)$ of the curve are expressed as functions of a parameter $ t$ within a closed interval $ t_1\leq t \leq t_2$ . The functions $ x(t)$ and $ y(t)$ are assumed to be continuous with a sufficient number of continuous derivatives. The parametric curve is said to be of class $ r$ , if the functions have continuous derivatives up to the order $ r$ , inclusively [206]. In vector notation the parametric curve can be specified by a vector-valued function
$\displaystyle {\bf r} = {\bf r}(t)\;.$     (1.2)

Another method of representing a curve analytically is to impose one condition on a variable point $ (x,y)$ by an equation of the form

$\displaystyle f(x,y)=0\;.$     (1.3)

This is an implicit equation for a plane curve. When $ f(x,y)$ is linear in variables $ x$ and $ y$ , (1.3) represents a straight line. If $ f(x,y)$ is of the second degree in $ x$ and $ y$ (i.e. $ a x^2 + 2b xy + c y^2 + 2d x + 2e y + h =0$ ), (1.3) represents a variety of plane curves called conic sections [79]. The implicit equation for a plane curve can also be expressed as an intersection curve between a parametric surface and a plane. We will discuss this formulation in Chap. 5.

The explicit form can be considered as a special case of parametric and implicit forms. If $ t$ can be expressed as a function of $ x$ or $ y$ , we can easily eliminate $ t$ from (1.1) to generate the explicit form

$\displaystyle y=F(x) \;\;\;or\;\;\; x=G(y)\;.$     (1.4)

This is always possible at least locally when $ \frac{dx}{dt}\neq 0$ or $ \frac{dy}{dt}\neq 0$ [412]. Conversely if we set $ x$ or $ y$ in (1.4) to be equal to the parameter $ t$ we obtain the parametric form (1.1). Also if the implicit equation (1.3) can be solved for one variable in terms of the other, we also obtain (1.4). This is always possible at least locally when $ \frac{\partial f}{\partial y}\neq 0$ or $ \frac{\partial f}{\partial x}\neq 0$ [166].

Figure 1.1: Folium of Descartes
\begin{figure}\centerline{ \psfig{figure=fig/descartes.ps,height=2.5in} }\end{figure}

Example 1.1.1. Figure 1.1 shows the Folium of Descartes, introduced by R. Descartes in 1638, with its asymptotic line [227]. It can be expressed in parametric form

$\displaystyle {\bf r}(t) = \left(\frac{3t}{1+t^3}, \frac{3t^2}{1+t^3}\right)^T, \; \; \; -\infty < t < \infty \;\;(t \neq -1)\;,$     (1.5)

where superscript $ T$ denotes transpose of a vector. For $ t< -1$ the curve is located in the fourth quadrant and approaches the origin as $ t$ goes to $ -\infty$ . For $ -1<t<0$ the curve is located in the second quadrant, and $ t=0$ corresponds to the origin. In the first quadrant it forms a loop moving counter-clockwise as $ t$ increases from 0 to $ +\infty$ . Eliminating $ t$ from (1.5), the Folium of Decartes can be also expressed in an implicit form
$\displaystyle f(x,y)=x^3 + y^3 -3xy =0\;.$     (1.6)

We can easily trace the curve using the parametric equation (1.5) by evaluating $ x(t)$ and $ y(t)$ for a discrete sampling of $ t$ , while such tracing is more difficult when using the implicit equation (1.6). However, determining if a point $ (x_0, y_0)$ lies on the curve is easier when using the implicit rather than the parametric equation of the curve. For example, we can verify that the point $ (\frac{3}{2}, \frac{3}{2})$ lies on the curve by substituting $ x=\frac{3}{2}$ and $ y=\frac{3}{2}$ into implicit form and deducing that $ f(\frac{3}{2}, \frac{3}{2})=0$ . However, it is more complex to deduce this using the parametric form. We first set $ x(t)=\frac{3}{2}$ which yields a cubic equation $ t^3-2t+1=0$ . The roots of the cubic equation are 1, $ \frac{-1 \pm \sqrt{5}}{2}$ . Then we substitute each root into $ y(t)$ to see if it becomes equal to $ \frac{3}{2}$ . An alternate way to do this involves the theory of resultants from algebraic geometry that we will see in Sect. 5.4.2.

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December 2009