C4PureStretch

Modelling with Parametric Equations

Parametric equations model motion where the parameter is usually time. They let you describe an object’s horizontal and vertical position separately, then answer questions about range, height and timing.

30 min Coordinate Geometry

The big picture

The parameter domain is part of a motion model. Algebraic equations can predict positions before launch or after impact; reject those times if the model describes only the free flight. For a target height there may be two valid times, one on ascent and one on descent. Use the horizontal position or the direction of vertical velocity to distinguish them.

Edexcel A level Maths: 8.5 Modelling with Parametric EquationsCurated by Kepler, not produced by us. Watch it, then work through the notes below.

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What you'll be able to do

Interpret t as time in a model
Find position at a given time
Find when an object reaches a height or position
Find greatest height and horizontal range

Time as the parameter

With $x$ and $y$ both functions of time $t$ , substitute a time to get the position, or set $x$ or $y$ to a value and solve for $t$ to find when an event happens.

Example 1A ball has

x = 20t

y = 10t - 5t^2

. When does it land (

y = 0

10t - 5t^2 = 0 \Rightarrow 5t(2 - t) = 0

t = 0

(start) or

t = 2

AnswerIt lands at

t = 2

s, at

x = 40

Tip — Set y = 0 for landing/ground level; the t = 0 root is the launch.

Greatest height

The greatest height occurs when the vertical component is at its maximum — found by symmetry of the y(t) parabola or by differentiating $y$ with respect to $t$ and setting it to zero.

Keep the physically valid time interval

The parameter domain is part of a motion model. Algebraic equations can predict positions before launch or after impact; reject those times if the model describes only the free flight.

For a target height there may be two valid times, one on ascent and one on descent. Use the horizontal position or the direction of vertical velocity to distinguish them.

Example 1A ball has

x=12t

y=15t-5t^2

, for its flight above level ground. When is it 10 m high?

15t-5t^2=10

gives

t^2-3t+2=0

2The roots are t = 1 and t = 2 s. Ground impact is at t = 3 s, so both are in the flight interval.

3The horizontal positions are x = 12 and x = 24 m.

dy/dt=15-10t

, positive at 1 and negative at 2, identifying ascent and descent.

AnswerAt 1 s on ascent (12 m horizontally) and 2 s on descent (24 m horizontally).

Think like an examiner

Write the defining equations and show the substitution before evaluating.
Check the domain, units and interpretation of the final result.

Common misconceptions

Taking t = 0 as the landing time.

t = 0 is the launch; the other root is the landing.

Using x to find greatest height.

Greatest height comes from the y-equation.

Key results to remember

y = 0 \Rightarrow \text{landing time}

Ground level.

\tfrac{dy}{dt} = 0 \Rightarrow \text{greatest height}

Vertical maximum.

Key takeaways

The parameter is usually time.
Set y = 0 for landing; the non-zero root is the time of flight.
Greatest height from the maximum of y(t).

Test yourself

Answer a few questions, then review anything you missed.

Next step

Keep practising

Start with the quiz above to check the key ideas. You can also choose another topic from the practice library.

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