Parametric equations

Parametric equations describe both coordinates using a third variable called a parameter. Instead of writing $y$ directly in terms of $x$, we write

$$x=f(t),\qquad y=g(t).$$

Each permitted value of $t$ generates one point $(x,y)$. As $t$ changes, that point traces a curve. The parameter may represent time, angle or another useful quantity, but it does not need to have a physical meaning.

Prerequisites

You should be able to:

• rearrange equations and substitute expressions
• solve simultaneous equations
• use identities such as $\sin^2 t+\cos^2 t=1$
• interpret domains, ranges and inequalities
• recognise equations of straight lines, circles and parabolas

Review functions, domains and ranges or straight lines if these foundations are uncertain.

A parameter generates coordinates

Consider

$$x=2t+1,\qquad y=t^2-3.$$

Substituting several values gives:

$t$	$x$	$y$	Point
$-2$	$-3$	$1$	$(-3,1)$
$-1$	$-1$	$-2$	$(-1,-2)$
$0$	$1$	$-3$	$(1,-3)$
$1$	$3$	$-2$	$(3,-2)$
$2$	$5$	$1$	$(5,1)$

The coordinates must come from the same parameter value. Pairing the $x$ coordinate from $t=1$ with the $y$ coordinate from $t=2$ does not produce a point specified by these equations.

Check 1

For $x=3t-2$ and $y=4-t^2$, find the point corresponding to $t=-1$.

Answer

$$x=3(-1)-2=-5,\qquad y=4-(-1)^2=3.$$

The point is $(-5,3)$.

Eliminating the parameter

To find a Cartesian equation, use one parametric equation to express the parameter in terms of a coordinate, then substitute into the other equation.

Example 1: a parabola

Given

$$x=2t+1,\qquad y=t^2-3,$$

rearrange the first equation:

$$t=\frac{x-1}{2}.$$

Substitute into $y=t^2-3$:

$$y=\left(\frac{x-1}{2}\right)^2-3 =\frac{(x-1)^2}{4}-3.$$

This Cartesian equation reveals a parabola with vertex $(1,-3)$. If $t$ can be any real number, then $x$ can also be any real number.

Example 2: eliminate using an identity

For

$$x=4\cos t,\qquad y=4\sin t,$$

we have

$$\frac{x}{4}=\cos t,\qquad \frac{y}{4}=\sin t.$$

Squaring and adding removes $t$:

$$\frac{x^2}{16}+\frac{y^2}{16} =\cos^2t+\sin^2t =1,$$

so

$$x^2+y^2=16.$$

This is a circle of radius $4$. See circles for its Cartesian geometry.

Example 3: do not introduce false points

Let

$$x=t^2,qquad y=t^3.$$

Since $x^3=t^6$ and $y^2=t^6$, elimination gives

$$y^2=x^3.$$

But the parametrisation also requires $x=t^2\geq0$. Here that restriction is already forced by $y^2=x^3$, because the left side cannot be negative. In other problems, squaring can create Cartesian points that the original parameter never reaches, so restrictions must always be checked separately.

Parameter restrictions determine the traced curve

The Cartesian equation alone may not describe exactly the same set of points as the parametric equations.

Suppose

$$x=t+1,\qquad y=t^2,qquad 0\leq t\leq3.$$

Eliminating $t$ gives

$$y=(x-1)^2.$$

However, $0\leq t\leq3$ implies

$$1\leq x\leq4.$$

Only the section of the parabola between $(1,0)$ and $(4,9)$ is traced. Writing only $y=(x-1)^2$ loses essential information.

For trigonometric parametrisations, an interval may trace a whole curve once, part of it, or the same curve repeatedly. For $x=a\cos t$, $y=a\sin t$:

• $0\leq t\leq2\pi$ traces the circle once
• $0\leq t\leq\pi$ traces the upper semicircle
• $0\leq t\leq4\pi$ traces the circle twice

Check 2

Eliminate $t$ from

$$x=2t-1,\qquad y=t^2+2,qquad -1\leq t\leq2,$$

and state the corresponding restriction on $x$.

Answer

Since $t=\frac{x+1}{2}$,

$$y=\left(\frac{x+1}{2}\right)^2+2.$$

The endpoint values of $t$ give

$$-3\leq x\leq3.$$

Orientation and direction of travel

Parametric equations contain directional information. Orientation describes the direction in which the curve is traced as the parameter increases.

For

$$x=3\cos t,\qquad y=3\sin t,$$

the point begins at $(3,0)$ when $t=0$, reaches $(0,3)$ when $t=\frac{\pi}{2}$, and then reaches $(-3,0)$ when $t=\pi$. The circle is traced anticlockwise.

Changing only the sign of $y$ gives

$$x=3\cos t,\qquad y=-3\sin t.$$

The Cartesian equation remains $x^2+y^2=9$, but the point moves clockwise. Equal Cartesian equations do not guarantee equal parametrisations.

When drawing a parametric curve:

1. calculate key points, including parameter endpoints
2. find the Cartesian shape if elimination is manageable
3. apply all parameter restrictions
4. add arrows showing increasing parameter values

Constructing parametric forms

A Cartesian curve usually has many valid parametrisations.

Example 4: a straight line

The line through $(2,-1)$ with direction vector $\begin{pmatrix}3\\4\end{pmatrix}$ can be written as

$$x=2+3t,\qquad y=-1+4t,$$

where $t\in\mathbb R$. Eliminating $t$ gives

$$\frac{x-2}{3}=\frac{y+1}{4},$$

or $4x-3y=11$.

For the line segment from $(2,-1)$ to $(5,3)$, the same formulas work with $0\leq t\leq1$.

Example 5: a parabola

For $y=(x-2)^2+1$, the simplest choice is

$$x=t+2,\qquad y=t^2+1.$$

Choosing $x=t$ and $y=(t-2)^2+1$ is equally valid. A good parameter makes the intended calculation or interpretation simple.

Example 6: a circle

For $(x-a)^2+(y-b)^2=r^2$, a standard parametrisation is

$$x=a+r\cos t,\qquad y=b+r\sin t.$$

This form builds the centre and radius directly into the coordinates. The interval chosen for $t$ determines how much of the circle is traced.

More realistic choices of parameter, assumptions and domains are developed in parametric modelling.

Intersections involving parametric curves

To intersect a parametric curve with a Cartesian curve, substitute the expressions for $x$ and $y$ into the Cartesian equation and solve for the parameter. Then calculate the coordinates.

Example 7: intersection with a line

The curve

$$x=t^2+1,\qquad y=2t$$

meets the line $y=x-1$. Substitute:

$$2t=(t^2+1)-1=t^2.$$

Therefore

$$t(t-2)=0,$$

so $t=0$ or $t=2$. The intersection points are

$$(1,0)\quad\text{and}\quad(5,4).$$

Always return to coordinates. A parameter value is not itself an intersection point.

Two parametrically defined curves

If one curve uses parameter $t$ and another uses parameter $s$, do not assume $t=s$. At an intersection their coordinates agree, so solve

$$x_1(t)=x_2(s),\qquad y_1(t)=y_2(s)$$

as simultaneous equations for the two independent parameters.

Check 3

Find where

$$x=t+2,\qquad y=t^2$$

meets $y=2x-3$.

Answer

Substitution gives

$$t^2=2(t+2)-3=2t+1.$$

Hence

$$t^2-2t-1=0,$$

so $t=1\pm\sqrt2$. The points are

$$\left(3+\sqrt2,3+2\sqrt2\right)$$

and

$$\left(3-\sqrt2,3-2\sqrt2\right).$$

The exact values should be retained unless decimals are requested.

Common misconceptions

The parameter is not automatically $x$

In $x=f(t)$ and $y=g(t)$, $t$ independently determines both coordinates. Treating it as another name for $x$ destroys that relationship.

Eliminating the parameter can lose restrictions

State the Cartesian equation and the coordinate restrictions implied by the parameter interval. Check whether algebraic operations such as squaring introduced extra points.

The same curve can have different orientations

Cartesian equations specify a set of points. Parametric equations can additionally specify where a point starts, which direction it moves, and how repeatedly the curve is traced.

Different curves need different parameters

At an intersection, two curves occupy the same coordinate. Their parameter values do not need to be equal.

A table is evidence, not a proof of the whole shape

A few generated points help reveal a curve, but elimination or another exact argument is needed to establish its equation and restrictions.

Final check

For

$$x=1+2\cos t,\qquad y=-1+2\sin t,qquad 0\leq t\leq\pi,$$

answer the following:

1. What is the Cartesian equation?
2. Which part of the curve is traced?
3. What are the initial and final points?
4. What is the orientation?

Answers

Using $\cos^2t+\sin^2t=1$ gives

$$(x-1)^2+(y+1)^2=4.$$

Because $0\leq t\leq\pi$, $\sin t\geq0$, so $y\geq-1$. The upper semicircle is traced. It starts at $(3,-1)$, passes through $(1,1)$ and finishes at $(-1,-1)$, moving anticlockwise.

Once the geometry is secure, continue to parametric differentiation to calculate gradients, tangents, normals and second derivatives without first eliminating the parameter.