Parametric equations of a curve

Let the equation of a curve be

$$F(x,y) = 0.$$ (6.7)

If $x$ is given as a function of a third variable, $t$ say, called a parameter, then by virtue of (6.7) $y$ is also a function of $t$, and the same functional relation (6.7) between $x$ and $y$ may generally be expressed by means of equations in the form

$$\begin{cases}x = f(t), \\ y = g(t) \end{cases}$$ (6.8)

each value of $t$ giving a value of $x$ and a value of $y$. Equations (6.8) are called parametric equations of the curve. If we eliminate $t$ between equations (6.8), it is evident that the relation (6.7) must result.

Example 6.5.1   For example, take equation of circle

$$x^2 + y^2= r^2\ {\rm or}\ y = \sqrt{r^2 - x^2}.$$

We have

$$\begin{cases}x= r\cos\, t \\ y = r\sin\, t \end{cases}$$ (6.9)

as parametric equations of the circle, $t$ being the parameter^6.6.

If we eliminate $t$ between equations (6.9) by squaring and adding the results, we have

$$x^2 + y^2 = r^2(\cos^2t + \sin^2t) = r^2,$$

the rectangular equation of the circle. It is evident that if $t$ varies from 0 to $2\pi$, the point $P=(x,y)$ will describe a complete circumference.

In §6.13 we shall discuss the motion of a point $P$, which motion is defined by equations such as

$$\begin{cases}x = f(t), \\ y = g(t) \end{cases}$$

We call these the parametric equations of the path, the time $t$ being the parameter.

Example 6.5.2   Newtonian physics tells us that

$$\begin{cases} x = v_0 \cos \alpha \cdot t, \\ y = -\frac{1}{2} gt^2 + v_0 \sin \alpha \cdot t \end{cases}$$

are really the parametric equations of the trajectory of a projectile^6.7, the time $t$ being the parameter. The elimination of $t$ gives the rectangular equation of the trajectory

$$y = x \tan \alpha - \frac{gx^2}{2v_0^2 \cos^2 \alpha}.$$

Since from (6.8) $y$ is given as a function of $t$, and $t$ as a function of $x$, we have

$$\begin{array}{ll} \frac{dy}{dx} & = \frac{dy}{dt} \cdot \fr... ...} \cdot \frac{1}{\frac{dx}{dt}}\ \ \ {\rm by\ XXVI} \end{array}$$

that is,

$$\frac{dy}{dx} = \frac{\frac{dy}{dt}}{\frac{dx}{dt}} = \frac{g'(t)}{f'(t)}.$$ (6.10)

Hence, if the parametric equations of a curve are given, we can find equations of tangent and normal, lengths of subtangent and subnormal at a given point on the curve, by first finding the value of $\frac{dy}{dx}$ at that point from (6.10) and then substituting in formulas (6.1), (6.2), (6.3), (6.4) of the last section.

Example 6.5.3   Find equations of tangent and normal, lengths of subtangent and subnormal to the ellipse

$$\begin{cases}x = a \cos \phi, \\ y = b \sin \phi, \end{cases}$$ (6.11)

at the point where $\phi = \frac{\pi}{4}$.

As in Figure 6.6 draw the major and minor auxiliary circles of the ellipse. Through two points B and C on the same radius draw lines parallel to the axes of coordinates. These lines will intersect in a point $P=(x,y)$ on the ellipse, because $x = OA = OB\cos\phi = a\cos\phi$ and $y = AP = OD = OC\sin\phi = b\sin\phi$, or, $\frac{x}{a} = \cos \phi$ and $\frac{y}{b} = \sin \phi$. Now squaring and adding, we get

$$\frac{x^2}{a^2} + \frac{y^2}{b^2} = \cos^2 \phi + \sin^2 \phi = 1,$$

the rectangular equation of the ellipse. $\phi$ is sometimes called the eccentric angle of the ellipse at the point P.

Figure 6.6: Auxiliary circles of an ellipse.

Solution. The parameter being $\phi$, $\frac{dx}{d\phi} = - a \sin \phi$, $\frac{dy}{d\phi} = b \cos \phi$.

Substituting $\phi = \frac{\pi}{4}$ in the given equations (6.11), we get $\left ( \frac{a}{\sqrt{2}}, \frac{b}{\sqrt{2}} \right )$ as the point of contact. Hence $\frac{dy}{dx}\vert _{x=x_1,y=y_1} = - \frac{b}{a}$. Substituting in (6.1),

$$y - \frac{b}{\sqrt{2}} = -\frac{b}{a} \left ( x - \frac{a}{\sqrt{2}} \right ),$$

or, $bx + ay = \sqrt{2} ab$, the equation of tangent. Substituting in (6.2),

$$y - \frac{b}{\sqrt{2}} = \frac{a}{b} \left ( x - \frac{a}{\sqrt{2}} \right ),$$

or, $\sqrt{2}(ax - by)= a^2 - b^2$, the equation of normal. Substituting in (6.3) and (6.4), we find

$$\frac{b}{\sqrt{2}} \left ( -\frac{b}{a} \right ) = -\frac{b^2}{a\sqrt{2}} ,$$

the length of subnormal, and

$$\frac{b}{\sqrt{2}} \left ( -\frac{a}{b} \right ) = -\frac{a}{\sqrt{2}},$$

the length of subtangent.

Example 6.5.4   Given equation of the cycloid in parametric form

$$\begin{cases} x = a(\theta - \sin \theta), \\ y = a(1 - \cos \theta), \end{cases}$$

$\theta$ being the variable parameter; find lengths of subtangent, subnormal, tangent, and normal at the point where $\theta = \frac{\pi}{2}$.

The path described by a point on the circumference of a circle which rolls without sliding on a fixed straight line is called the cycloid. Let the radius of the rolling circle be $a$, P the generating point, and M the point of contact with the fixed line OX, which is called the base. If arc PM equals OM in length, then P will touch at O if the circle is rolled to the left. We have, denoting angle POM by $\theta$,

$$\begin{array}{ll} x &= OM - NM = a\theta - a\sin\theta = a(... ...PN = MC - AC = a - a\cos\theta = a(1 - \cos\theta), \end{array}$$

the parametric equations of the cycloid, the angle $\theta$ through which the rolling circle turns being the parameter. $OD = 2\pi a$ is called the base of one arch of the cycloid, and the point V is called the vertex. Eliminating $\theta$, we get the rectangular equation

$$x = a \arccos \left ( \frac{a - y}{a} \right ) - \sqrt{ 2ay - y^2 }.$$

Figure 6.7: Tangent line of a cycloid.

The Sage commands for creating this plot are as follows:

[fontsize=\small,fontfamily=courier,fontshape=tt,frame=single,label=\sage]

sage: t = var("t")
sage: f1 = lambda t: [t-sin(t),1-cos(t)]
sage: p1 = parametric_plot(f1(t), 0.0, 2*pi, rgbcolor=(1,0,0))
sage: f2 = lambda t: [t+RR(pi)/2-1,t+1]
sage: p2 = parametric_plot(f2(t), -1, 1, rgbcolor=(1,0,0))
sage: f3 = lambda t: [-t+RR(pi)/2,t]
sage: p3 = parametric_plot(f3(t), -1, 1, rgbcolor=(1,0,0))
sage: t1 = text("P", (RR(pi)/2-1+0.1,1-0.1))
sage: t2 = text("T", (-0.4,0.1))
sage: t3 = text("N", (RR(pi)/2,0))
sage: show(p1+p2+p3+t1+t2+t3)

Solution:

$$\frac{dx}{d\theta} = a(1 - \cos \theta),\ \ \ \frac{dy}{d\theta} = a \sin \theta.$$

Substituting in (6.10),

$$\frac{dy}{dx} = \frac{\sin \theta}{1 - \cos \theta},$$

the slope at any point. Since $\theta = \frac{\pi}{2}$, the point of contact is $\left ( \frac{\pi a}{2} - a, a \right )$, and $\frac{dy}{dx}\vert _{x=x_1,y=y_1} = 1$.

Substituting in (6.3), (6.4), (6.5), (6.6) of the last section, we get

length of subtangent = $a$,

length of subnormal = $a$,

length of tangent = $a \sqrt{2}$,

length of normal = $a \sqrt{2}$.