Plane curves

Parametric curves: Plane curves

Plane curves

The circle on the right-hand side , with the origin $O=(0,0)$ as centre and with radius $2$ , consists of all points $P(x,y)$ such that $x^2+y^2=4$ .

But the points on the circle can also be described by a parametric curve, for which we specify the points on the curve via the coordinate functions $x=2\cos t$ and $y=2\sin t$ . We speak of coordinate functions because the coordinates $x$ and $y$ are functions of the variable $t$ . As $t$ increases from $0$ to $2\pi$ , the point $P$ moves counter clockwise along the circle.

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In the interactive diagram below the figure you can move the point $P$ by dragging the slider between $0$ and $2\pi$ ; the turquoise arc indicates the path along which the point $P=(2\cos t,2\sin t)$ travels from $0$ to $t$ .

GeoGebra

In general, you can describe a plane curve mathematically by two functions $x=f(t)$ and $y=g(t)$ describe a curve in the plane; we speak of a parameterisation or parameter equations of the curve with parameter $t$ . Almost always one uses neat functions $f$ and $g$ as coordinate functions. The parameter $t$ represents time in many applications.

The parameterisation of a curve is not unique: the following example illustrates a new parameterisation of the circle with origin $O=(0,0)$ as centre and with radius $2$ .

Parameter equations of the circle with the origin $O=(0,0)$ as centre and with radius $2$ is:

$x=2\cos t\quad\text{and}\quad y=2\sin t$ Another parameterisation of the circle is:

$x=2\sin t\quad\text{and}\quad y=2\cos t$ The direction of circulation is now clockwise, and at $t=0$ the point is $(0,2)$ .

Yet another parameterisation of the circle is:

$x=2(1-t^2)\quad\text{and}\quad 2t\sqrt{2-t^2}\qquad\text{for}\quad -\sqrt{2}\le t\le \sqrt{2}$ After all, the following is true for this parameterisation:

$\begin{aligned}x^2+y^2&= 4(1- t^2)^2+ 4t^2(2-t^2)\\[0.25cm] &= 4(1- 2t^2+t^4)+ 4t^2(2-t^2)\\[0.25cm] &=4\end{aligned}$

A fourth parameterisation of the circle, but now without the point $(-2,0)$ , uses rational functions as a coordinate functions:

$x=\frac{2-2t^2}{1+t^2}\quad\text{and}\quad y=\frac{4t}{1+t^2}$ This is not a parameterisation of the circle because the point $(-2,0)$ cannot be described in this way; but otherwise all points on the circle are present! After all, the following is true for this parameterisation:

$\begin{aligned}x^2+y^2&= \frac{(2-2t^2)^2}{(1+t^2)^2}+\frac{(4t)^2}{(1+t^2)^2}\\[0.25cm] &= \frac{4-8t^2+4t^4}{(1+t^2)^2}+\frac{16t^2}{(1+t^2)^2}\\[0.25cm] &= \frac{4+8t^2+4t^4}{(1+t^2)^2}\\[0.25cm] &=\frac{4(1+t^2)^2}{(1+t^2)^2} \\[0.25cm] &= 4\end{aligned}$