Parametric modelling

A parametric model describes two or more related quantities using a third quantity called a parameter. It is especially useful when position changes with time, when a curve is easier to trace than to write as $y=f(x)$, or when one Cartesian equation would hide direction and restrictions.

For example,

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

describes a moving point. At $t=0$ it is at $(1,-4)$, and each increase of one in $t$ moves it $3$ units right and $2$ units up.

This lesson focuses on building and evaluating models. Review parametric equations for the underlying notation and algebra.

Learning goals

By the end of this lesson, you should be able to:

• choose a meaningful parameter;
• construct parametric equations from geometry or motion;
• interpret the parameter, its units and its domain;
• identify direction and starting position;
• eliminate the parameter when useful;
• distinguish a full Cartesian curve from the part traced by a model;
• test a model against data, assumptions and physical constraints.

What makes a parametric model?

A two-dimensional parametric model has the form

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

where $D$ is the permitted domain of the parameter.

The model contains three different kinds of information:

1. Path: the set of points $(x,y)$ that can be reached.
2. Direction: the order in which those points are traced as $t$ increases.
3. Timing: if $t$ represents time, how quickly the point moves along the path.

A Cartesian equation such as $y=2x-6$ gives the path, but it usually does not preserve direction or timing.

First interpretation

Consider

$$x=1+4t,\qquad y=5-2t,\qquad 0\leq t\leq3.$$

At the endpoints,

$$t=0:(x,y)=(1,5),$$ $$t=3:(x,y)=(13,-1).$$

As $t$ increases, $x$ increases and $y$ decreases. The model traces the line segment from $(1,5)$ towards $(13,-1)$.

Eliminating $t$ gives

$$t=\frac{x-1}{4}$$

and hence

$$y=5-2\left(\frac{x-1}{4}\right) =\frac{11}{2}-\frac{x}{2}.$$

However, the model does not trace the whole line. Since $0\leq t\leq3$,

$$1\leq x\leq13.$$

The parameter domain is part of the answer, not an optional note.

Choosing a parameter

A useful parameter should simplify the relationships and, where possible, have a clear interpretation.

Common choices include:

• time $t$ for motion;
• an angle $\theta$ for circles and periodic motion;
• distance travelled $s$ along a straight path;
• a dimension such as radius $r$ in a changing geometric shape;
• a dimensionless proportion $u$ with $0\leq u\leq1$ between two endpoints.

The parameter need not be $t$. Its symbol should match its meaning.

Example: a point dividing a line segment

Let $A=(2,-1)$ and $B=(8,11)$. A point moving from $A$ to $B$ can be modelled using a proportion $u$:

$$\begin{aligned} x&=2+u(8-2)=2+6u,\\ y&=-1+u(11-(-1))=-1+12u, \end{aligned} \qquad 0\leq u\leq1.$$

At $u=0$, the point is at $A$. At $u=1$, it is at $B$. At $u=1/2$, it is halfway between them:

$$(x,y)=(5,5).$$

This construction generalises. For points $A=(x_1,y_1)$ and $B=(x_2,y_2)$,

$$x=x_1+u(x_2-x_1),\qquad y=y_1+u(y_2-y_1),\qquad 0\leq u\leq1.$$

Constructing models from motion

For motion at constant velocity, position equals initial position plus velocity multiplied by time.

If an object starts at $(x_0,y_0)$ and has constant velocity $(v_x,v_y)$, then

$$\boxed{x=x_0+v_xt,\qquad y=y_0+v_yt.}$$

Example: a rescue boat

A boat starts at the point $(4,7)$ km relative to a harbour and travels with velocity

$$\begin{pmatrix}6\\-2\end{pmatrix} \text{ km h}^{-1}.$$

After $t$ hours its position is

$$x=4+6t,\qquad y=7-2t,$$

with $t\geq0$ while the velocity remains valid.

After $1.5$ hours,

$$x=4+6(1.5)=13,$$ $$y=7-2(1.5)=4.$$

The boat is at $(13,4)$ km.

To eliminate $t$,

$$t=\frac{x-4}{6},$$

so

$$y=7-2\left(\frac{x-4}{6}\right) =\frac{25-x}{3}.$$

This Cartesian equation identifies the route, but only the parametric model says that the boat starts at $(4,7)$ and moves in the direction $(6,-2)$ at the stated rate.

Units are evidence

In

$x=x_0+v_xt,$

the units must be consistent:

$$\text{km}=\text{km}+(\text{km h}^{-1})(\text{h}).$$

If the units do not balance, the model is wrong or incompletely converted.

Curved paths

Circular motion

A circle with centre $(a,b)$ and radius $r$ has the parametrisation

$$\boxed{x=a+r\cos\theta,\qquad y=b+r\sin\theta.}$$

The identity $\cos^2\theta+\sin^2\theta=1$ confirms that

$$(x-a)^2+(y-b)^2=r^2.$$

For $0\leq\theta\leq2\pi$, the point starts at $(a+r,b)$ and traces the circle anticlockwise once. Replacing $\sin\theta$ by $-\sin\theta$ reverses the direction.

Example: part of a Ferris wheel

A passenger moves on a wheel of radius $10$ m whose centre is $(0,12)$. Suppose the passenger starts at the lowest point and moves anticlockwise. One suitable model is

$$x=10\sin\theta,\qquad y=12-10\cos\theta.$$

At $\theta=0$,

$$(x,y)=(0,2),$$

the lowest point. At $\theta=\pi/2$,

$$(x,y)=(10,12).$$

If only the first quarter-turn is being modelled, the correct domain is

$$0\leq\theta\leq\frac{\pi}{2}.$$

The full circle equation alone would incorrectly include positions outside the stated interval of motion.

Elliptical motion

An ellipse centred at $(a,b)$ with horizontal semi-axis $p$ and vertical semi-axis $q$ can be modelled by

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

Eliminating $t$ gives

$$\frac{(x-a)^2}{p^2}+\frac{(y-b)^2}{q^2}=1.$$

For example,

$$x=3+5\cos t,\qquad y=-2+2\sin t$$

has centre $(3,-2)$, horizontal semi-axis $5$ and vertical semi-axis $2$.

Time, angle and speed

The same path can be traced with different timing.

Compare

$$\mathcal A:quad x=\cos t,\qquad y=\sin t$$

with

$$\mathcal B:quad x=\cos(3t),\qquad y=\sin(3t).$$

Both trace the unit circle anticlockwise. Model $\mathcal B$ completes a revolution three times as quickly because its angular displacement is $3t$.

Now compare

$$\mathcal C:quad x=\cos t,\qquad y=-\sin t.$$

It traces the same circle clockwise. Eliminating the parameter from all three models gives $x^2+y^2=1$, so Cartesian conversion loses timing and direction.

If the parameter is time, the velocity components are

$$\frac{dx}{dt}\quad\text{and}\quad\frac{dy}{dt}.$$

The speed is

$$\sqrt{\left(\frac{dx}{dt}\right)^2+ \left(\frac{dy}{dt}\right)^2}.$$

These derivatives are developed fully in parametric differentiation.

Converting between parametric and Cartesian forms

Eliminating the parameter can reveal the familiar shape, make intersections easier to find, or allow comparison with another curve. It must be done without losing restrictions.

Example: a parabola with restricted motion

Let

$$x=2t+1,\qquad y=t^2-4,qquad -1\leq t\leq3.$$

From the first equation,

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

Substitution gives

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

Since $-1\leq t\leq3$ and $x=2t+1$ increases with $t$,

$$-1\leq x\leq7.$$

The Cartesian model is therefore

$$y=\frac{(x-1)^2}{4}-4,qquad -1\leq x\leq7,$$

traced from left to right.

When conversion is not one-to-one

Consider

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

Then

$$y^2=t^6=x^3,$$

so the Cartesian equation is

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

Writing $t=\sqrt{x}$ would silently assume $t\geq0$ and lose the lower branch where $t<0$ and $y<0$. Squaring can also introduce points unless restrictions are checked. Always test the converted equation against the original parametrisation.

Constructing a model from conditions

Example: a quadratic flight path

A simplified model places a ball at

$$x=12t,$$ $$y=1+18t-5t^2,$$

where $t$ is measured in seconds and distances in metres.

Interpret the constants:

• $x(0)=0$ and $y(0)=1$, so the initial position is $(0,1)$;
• horizontal position increases at $12$ m s$^{-1}$;
• the vertical model initially rises because the coefficient of $t$ is positive;
• the negative $t^2$ term eventually makes the height decrease.

The physical domain ends when the ball reaches the ground, not when the algebraic formula ceases to exist. Set $y=0$:

$$1+18t-5t^2=0.$$

Thus

$$t=\frac{18\pm\sqrt{18^2+20}}{10} =\frac{9\pm2\sqrt{21.5}}{5}.$$

Only the positive root is physically meaningful, approximately

$t=3.65\text{ s}.$

The model domain is therefore approximately

$0\leq t\leq3.65.$

Eliminating $t$ using $t=x/12$ gives

$$y=1+\frac32x-\frac{5}{144}x^2,$$

with the corresponding horizontal restriction

$$0\leq x\lesssim43.8.$$

The quadratic continues below ground level, but that part of the graph is not part of the flight model.

Assumptions and limitations

A mathematical formula is not automatically a good model. State what has been simplified and decide where the simplification is reasonable.

For the flight model above, possible assumptions include:

• constant gravitational acceleration;
• negligible air resistance and wind;
• motion in a vertical plane;
• no spin or lift;
• flat ground at $y=0$;
• a point-like ball;
• accurate initial position and velocity.

These assumptions limit the model. Air resistance may make horizontal speed decrease, wind may move the ball sideways, and uneven ground changes the landing condition.

A structured model check

Ask five questions.

1. Are the variables defined?

State what $x$, $y$ and the parameter represent, including units.

2. Is the domain realistic?

A model of a journey may apply only between departure and arrival. A model of height must not be interpreted after impact.

3. Do special values make sense?

Check the start, end and one or more intermediate values. For periodic models, check quarter-turns or half-turns.

4. Are dimensions and scales consistent?

Do not add metres to metres per second. Check whether computed sizes are plausible.

5. What is omitted?

Identify effects that the formula does not represent and explain how they might affect predictions.

Comparing two models

Suppose two cyclists have positions

$$A:quad x=2+5t,\qquad y=1+2t$$

and

$$B:quad x=14-t,\qquad y=7,$$

where $t$ is in hours.

They meet only if both coordinates agree at the same time. From the $y$ coordinates,

$$1+2t=7,$$

so $t=3$. At this time,

$$x_A=2+5(3)=17,$$

but

$$x_B=14-3=11.$$

They do not meet. Their Cartesian paths might intersect at different times, but simultaneous position requires one common parameter value satisfying every coordinate equation.

Common misconceptions

• Omitting the domain: the equations and the parameter interval together define the model.
• Treating the parameter as an axis: $t$ controls both coordinates but is not normally plotted as a third axis in a plane curve.
• Assuming $t$ always means time: it may represent angle, distance or a dimensionless proportion.
• Finding a Cartesian curve and claiming equivalence: conversion can lose direction, timing, repeated tracing and restrictions.
• Choosing both square-root signs carelessly: recovering $t$ from $t^2=x$ requires attention to the original domain.
• Confusing path intersection with collision: moving objects collide only if they occupy the same point at the same time.
• Extending a physical model indefinitely: a formula may remain algebraically valid after its assumptions have failed.
• Ignoring units: inconsistent units expose an invalid construction.
• Describing limitations vaguely: say which assumption fails and how that could change the prediction.

Check your understanding

1. Interpret a linear model

A particle has position

$$x=7-3t,\qquad y=2+4t,qquad 0\leq t\leq5.$$

Find its starting point, endpoint and direction of travel.

Answer

At $t=0$, the particle is at $(7,2)$. At $t=5$, it is at

$$(7-15,2+20)=(-8,22).$$

As $t$ increases, $x$ decreases and $y$ increases, so it travels left and upwards from $(7,2)$ to $(-8,22)$.

2. Keep the restriction

Eliminate $t$ from

$$x=3t-2,\qquad y=t^2+1,qquad 0\leq t\leq4.$$

Answer

Since

$$t=\frac{x+2}{3},$$

we obtain

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

The domain $0\leq t\leq4$ gives

$$-2\leq x\leq10.$$

The curve is traced from left to right.

3. Build a segment model

Construct a parametric model for the line segment from $P=(-3,4)$ to $Q=(5,-2)$.

Answer

Using $0\leq u\leq1$,

$$x=-3+u(5-(-3))=-3+8u,$$ $$y=4+u(-2-4)=4-6u.$$

4. Identify a circular model

Describe

$$x=4+3\cos t,\qquad y=1-3\sin t,qquad 0\leq t\leq\pi.$$

Answer

The path lies on the circle

$$(x-4)^2+(y-1)^2=9.$$

It starts at $(7,1)$ when $t=0$, passes through $(4,-2)$ when $t=\pi/2$, and ends at $(1,1)$ when $t=\pi$. It traces the lower semicircle clockwise.

5. Same path, different model

Explain one difference between

$$x=2\cos t,\quad y=2\sin t$$

and

$$x=2\cos(4t),\quad y=2\sin(4t).$$

Answer

Both trace the circle $x^2+y^2=4$ anticlockwise. If $t$ is time, the second model travels around the circle four times as quickly. Over the same parameter interval, it may also trace the circle more times.

6. Test a physical domain

A height is modelled by

$$h=2+14t-4.9t^2.$$

Why is $t\geq0$ alone an inadequate model domain?

Answer

The formula eventually gives negative heights. If the object lands when $h=0$, the domain should end at the positive solution of

$$2+14t-4.9t^2=0.$$

The exact endpoint depends on the physical event being modelled, but values after impact should not be included without a new model.

7. Distinguish intersection from collision

Two particles move according to

$$A:quad x=t,\quad y=t^2$$

and

$$B:quad x=2-s,\quad y=1,$$

where both parameters measure time from the same starting instant. What must be true for a collision?

Answer

The particles must have the same coordinates at the same time, so set $s=t$ and solve

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

The first equation gives $t=1$, which also satisfies $t^2=1$. Therefore they collide at $(1,1)$ at time $1$, provided $t=1$ lies in both model domains.

Merely finding an intersection between the two Cartesian paths would not be sufficient.

What to learn next

Use parametric equations to strengthen conversion and curve interpretation. Then study parametric differentiation to find gradients, tangents, stationary points, velocity and speed directly from a parametric model.