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

Jump to: navigation, search

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

Diagram of a circle, indicating the geometrical meaning of the parameter

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

Diagram of a parabola, showing the normal and tangent at a point 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

Diagram of a parabola, a chord PQ

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 $T(a(p+q),apq)\,$.