HSC Extension 1 and 2 Mathematics/3-Unit/Preliminary/Parametrics

The parametric form of a curve is an algebraic representation which expresses the co-ordinates of each point on the curve as a function of an introduced parameter, most frequently $$t$$. This contrasts with Cartesian form in that parametric equations do not describe an explicit relation between $$x$$ and $$y$$. This relation must be derived in order to convert from parametric to Cartesian form.

In 3-unit, parametrics focuses on a parametric representation of the quadratic, moving from parametric to Cartesian form and vice-versa, and manipulating geometrical aspects of the quadratic with parametrics. Recognition of other parametric forms is also useful, and more forms are introduced and dealt with in the 4-unit topic, conics.

Reasons for using the parametric form
In specific situations (in the school syllabus, primarily Conic sections), the parametric representation can be useful because:
 * points on the curve are represented by a single number, not two, simplifying algebra;
 * some elegant results are possible; for instance, in the standard paramaterisation of the quadratic, the gradient is equal to the parameter, $$t$$;
 * some curves, which cannot be expressed in functional form (for example, the circle, which is neither a function of $$x$$ nor of $$y$$) can be conveniently expressed in parametric form;
 * intuitively, it allows for an easier way to find points on the graph: you can sub in any value for the parameter and instantly find a point, whereas a relational form is not deterministic in the same way.

Furthermore, the parametric form occurs in certain natural phenomena. For instance, using the equations of motion, a thrown ball's location at any time can be calculated using the laws of projectile motion. This is implicitly paramaterising the ball's path by time; to find the shape of the ball's path, (which we know is a parabola) we must use parametrics, to eliminate time, $$t$$, from the equations.

Converting parametric to Cartesian forms
One of the simplest parametric forms is the line:
 * $$\begin{align}

x & = t \\ y & = t\end{align}$$ By inspection, it is obvious that this describes the line $$y = x$$. However, what is the formalised approach for doing this?

We are looking for some relation between $$x$$ and $$y$$ which has no $$t$$ in it. In other words, we want to eliminate $$t$$ from the equations. In the above example, we did this by equating the first and second equations, eliminating $$t$$.

Another example:

 * $$\begin{align}x & = 3t + 1 \\

y & = 9t^2 - 1\end{align}$$ We solve equation 1 for t, and substitute into equation 2:

\begin{align} x & = 3t + 1 \\ t & = \frac{x - 1}{3} \end{align} $$ Sub into equation 2:

\begin{align} y & = 9t^2 - 1 \\ & = 9\left(\frac{x - 1}{3}\right)^2 - 1 \\ & = x^2 - 2x + 1 - 1 \\ & = x^2 - 2x \end{align} $$ Here we have a Cartesian form, as required.

Three standard parametric forms
These parametric forms occur frequently, and should be recognised by 3- and 4-unit students.

Parabola
The standard paramaterisation of the parabola describes one with focal length $$a\,$$ and with its vertex at the origin. In Cartesian form, this is given as
 * $$x^2 = 4ay\,$$

In parametric form, this is given as
 * $$\begin{align}x & = 2at \\

y & = at^2\end{align}$$

Eliminating $$t\,$$ allows us to verify that this is equivalent to the Cartesian form:

\begin{align} x  & = 2at \\ x^2 & = 4a^2t^2 \\ & = 4a (at^2) \\ & = 4ay \end{align} $$ as required.

Circle
The circle with radius $$r\,$$ and center at the origin can be written in Cartesian form as
 * $$x^2 + y^2 = r^2\,$$

Introducing the parameter, $$\theta\,$$, this is:
 * $$\begin{align}x &= r\cos\theta \\

y &= r\sin\theta\end{align}$$ We convert to Cartesian as follows:

\begin{align} x^2      & = r^2\cos^2\theta \\ y^2      & = r^2\sin^2\theta \\ x^2 + y^2 & = r^2\cos^2\theta + r^2\sin^2\theta \\ & = r^2(\cos^2\theta + \sin^2\theta) \end{align} $$ Recalling the trig identity,
 * $$\sin^2\theta + \cos^2\theta = 1\,$$

we conclude
 * $$x^2 + y^2 = r^2\,$$

as required.

Geometrical interpretation
Unlike most parametrics, the conic sections are paramaterised by $$\theta\,$$ instead of $$t\,$$. For the circle, this implies a geometric representation: $$\theta\,$$ represents the angle the point makes with the $$x$$ axis.

Ellipse
Conceptually, an ellipse is just a 'squished' circle. The parametric form makes this clear:
 * $$\begin{align}x &= a\cos\theta\\

y &= b\sin\theta\end{align}$$ The cos and sin are still there, but they are now multiplied by different constants so that the $$x$$ and $$y$$ components are stretched differently. We can turn this into the parametric form in a similar fashion to the circle:

\begin{align} x^2                              & = a^2\cos^2\theta \\ y^2                              & = b^2\sin^2\theta \\ \frac{x^2}{a^2} + \frac{y^2}{b^2} & = \cos^2\theta + \sin^2\theta \\ & = 1 \end{align} $$ which is the standard form of an ellipse, with $$x$$ and $$y$$ intercepts at $$\pm a\,$$ and $$\pm b\,$$, respectively.

Properties of the parabola
3-Unit students are expected to remember the parametric description of a parabola ($$x = 2at, y = at^2$$). They are also expected to know (and/or be able to quickly derive) the equations of the tangent and the normal to the parabola at the point $$t$$ and at the point $$(x_1,y_1)$$.

Derivation of equations of the tangent and the normal
Differentiating each parametric equation with respect to $$t\,$$,
 * $$\begin{align}

\frac{dy}{dt} & = 2at \\ \frac{dx}{dt} & = 2a \end{align}$$ Then the gradient $$\tfrac{dy}{dx}$$ can be obtained by dividing $$\tfrac{dy}{dt}$$ by $$\tfrac{dx}{dt}$$ (this is the chain rule):
 * $$\begin{align}

\frac{dy}{dx} & = \frac{dy}{dt} \frac{dt}{dx} \\ & = \frac{2at}{1} \frac{1}{2a} \\ & = t \end{align}$$

Note that this result could have been derived without the chain rule, by taking the derivative of the Cartesian form (with respect to $$x\,$$) and solving for $$t\,$$. However, the above derivation is faster and more elegant.

The equation of the tangent at P
The gradient of the tangent at a point $$P(2at,at^2)$$ then is $$t$$. The equation of the tangent at $$P$$ is:
 * Using the point-gradient formula $$y-y_1=m(x-x_1)\,$$, the equation is:
 * $$\begin{align}

y-at^2 & = t(x-2at) \\ & = tx - 2at^2 \\ y+at^2 & = tx \\ y-tx+at^2 & = 0 & \mbox{ or } y = xt - at^2 \end{align}$$

The equation of the normal at P
The gradient of the normal at the point $$P$$ is $$\tfrac{-1}{t}$$ (since two perpendicular lines with gradients $$m_1\,$$ and $$m_2\,$$ must have $$m_1 m_2 = -1\,$$). Similarly to the derivation of the equation of the tangent:
 * $$\begin{align}

y - at^2 & = \tfrac{-1}{t}(x - 2at) \\ & = \tfrac{-x}{t} + \tfrac{2at}{t} \\ yt - at^3 & = -x + 2at \\ yt + x - at(t^2 + 2) & = 0 & \mbox{or } x + ty = 2at + at^3 \end{align}$$

Chords of a parabola


Suppose that $$P(2ap,ap^2)$$ and $$Q(2aq,aq^2)$$ are two distinct points on the parabola $$x^2=4ay$$. We can derive the equation for the line $$PQ$$ by finding the gradient and using the point-gradient formula:
 * $$\begin{align}

m &= \frac{ap^2-aq^2}{2ap-2aq} \\ &= \frac{a(p-q)(p+q)}{2a(p-q)} \\ &= \tfrac{1}{2}(p+q) \end{align}$$ so the chord is
 * $$\begin{align}

y - ap^2 &= \tfrac{1}{2}(p+q)(x-2ap) \\ &= \tfrac{1}{2}(p+q)x - ap^2 - apq \\ y &= \tfrac{1}{2}(p+q)x - apq \end{align}$$

Intersection of tangents
The points $$P(2ap,ap^2)$$ and $$Q(2aq,aq^2)$$ lie on the parabola $$x^2=4ay$$. The intersection of the tangents at $$P$$ and $$Q$$ can be found in terms of $$p$$ and $$q$$:
 * The tangents are:
 * $$\begin{align}

y & = px - ap^2 \\ y & = qx - aq^2 \end{align}$$
 * Subtracting these,
 * $$\begin{align}

y - y & = px - ap^2 - (qx - aq^2) \\ 0 & = px - ap^2 - qx + aq^2 \\ 0 & = px - qx + aq^2 - ap^2 \\ x(p-q) & = a(p^2 - q^2) \\ x(p-q) & = a(p+q)(p-q) \\ x & = a(p+q) \end{align}$$
 * Substituting into the original tangent formula,
 * $$\begin{align}

y & = p(a(p+q)) - ap^2 \\ & = ap(p+q) - ap^2 \\ & = ap^2 + apq - ap^2 \\ & = apq \end{align}$$ So the point of intersection is described by $$T(a(p+q),apq)\,$$.