Parametric equations in modellingAQA A-Level Maths: Revision notes

Section 1

Why use a parameter

In many real situations the position $(x,y)$ of an object depends on time. A parametric model writes $x$ and $y$ separately as functions of time $t$ , which tells you where the object is and when. A Cartesian equation gives only the shape of the path. Before using a model, note the units, the allowed range of $t$ and any assumptions, such as ignoring air resistance.

Key termsparametric modelpath

Exam tip

Write down the meaning of $x$ , $y$ and $t$ with units before you start calculating.

Section 2

Projectile motion

A ball kicked from the ground may be modelled by $x=12t$ , $y=16t-5t^2$ .

Time of flight: solve $y=0$ : $t(16-5t)=0$ , so $t=3.2$ s.
Range: substitute into $x$ : $12\times3.2=38.4$ m.
Greatest height: halfway through the flight, $t=1.6$ , so $y=16(1.6)-5(1.6)^2=12.8$ m. The vertical motion is quadratic, so use symmetry about the highest point, or completing the square.

Key termstime of flightrange

Common mistake

Using $t=0$ as the landing time. $y=0$ has two roots; the launch is $t=0$ and the landing is the other.

Section 3

Circular motion

A point on a wheel of radius $r$ with centre $(0,h)$ can be modelled by $x=r\sin\omega t$ , $y=h-r\cos\omega t$ , where $\omega$ is the angle turned per second in radians.

Period: one revolution takes $t=\frac{2\pi}{\omega}$ .
Greatest and least height: $\cos$ takes values between $-1$ and $1$ , so $y$ lies between $h-r$ and $h+r$ . For $r=10$ , $h=12$ , $\omega=0.2$ : period $10\pi$ s, heights from $2$ m to $22$ m.

Key termsperiodangular speed

Common mistake

Multiplying instead of dividing: the period is $2\pi\div\omega$ , not $2\pi\times\omega$ .

Section 4

Eliminating the parameter and conic paths

Eliminate $t$ to find the path. For $x=80\cos\theta$ , $y=50\sin\theta$ with $\theta=0.1t$ , use $\cos^2\theta+\sin^2\theta=1$ : $\frac{x^2}{6400}+\frac{y^2}{2500}=1.$ This is an oval (an ellipse). To find when the runner has $x=40$ : $80\cos\theta=40$ , $\theta=\frac\pi3$ , so $t=\frac{10\pi}{3}\approx10.5$ s.

Key termsellipse

Exam tip

Always convert back to the context: give the answer in seconds or metres, not just a value of $\theta$ .

Section 5

Paths crossing versus objects colliding

Two objects collide only if they are at the same point at the same time. Eliminate the parameter to find where the paths cross, then compare the times each object reaches that point. Aircraft $A$ : $x=5t$ , $y=20-2t$ ; aircraft $B$ : $x=10+4t$ , $y=3t-7$ . The paths cross at $(30,8)$ , but $A$ gets there at $t=6$ and $B$ at $t=5$ . They do not collide, but at $t=5$ they are $\sqrt{29}\approx5.4$ m apart.

Key termscollision

Common mistake

Treating a crossing of paths as a collision without checking the times.

Section 6

Evaluating a model

Models make assumptions, and good answers say how those assumptions affect results.

Air resistance and spin are ignored in a projectile model.
Objects are treated as points, so a real object may be larger than the gap predicted.
The model is only valid for the stated range of $t$ . Comment on whether the answer is reasonable, such as a pod height between $2$ m and $22$ m.

Key termsassumption

Exam tip

For a 'comment' or 'evaluate' question, link the limitation to its effect on the result.

That's the notes covered.

Carry on to the next subtopic.

Exam questions on Parametric equations in modelling

A ball is kicked from level ground. At time $t$ seconds after the kick, its horizontal distance from the starting point is $x=12t$ metres and its height is $y=16t-5t^2$ metres. The model applies until the ball lands.
Find the greatest height reached by the ball.2 marks
A Ferris wheel has its centre $12$ m above the ground. The position of a passenger pod is modelled by $x=10\sin(0.2t)$ , $y=12-10\cos(0.2t)$ , where $x$ is the horizontal distance in metres from the vertical line through the centre, $y$ is the height above the ground in metres and $t$ is the time in seconds after the pod passes its lowest point. The angle $0.2t$ is in radians.
State the greatest height of the pod above the ground, and justify your answer.2 marks
A runner follows an oval track. Relative to the centre of the track, the runner's position at time $t$ seconds is $x=80\cos\theta$ , $y=50\sin\theta$ , where $\theta=0.1t$ radians and $x$ and $y$ are in metres.
Show that the runner's path has the Cartesian equation $\dfrac{x^2}{6400}+\dfrac{y^2}{2500}=1$ .3 marks

See the full worksheet

Written by the Exaim team, led by Shaun Daswani (Head of Upper Secondary, Improve ME Institute; MSc Financial Mathematics, Imperial College London; BSc, UCL) and Jason Daswani (operational lead, Improve ME Institute; LSE).