Parametric Modelling — Edexcel A-level Mathematics (Pure)

SPEC 3.4 • COORDINATE GEOMETRY

Parametric Equations in Modelling

Using parametric equations to model real-world motion in a variety of contexts

Modelling with Parametrics

Parametric equations are powerful tools for modelling real-world motion because the parameter t usually represents time.

Key Principle

At time t, the position is (x, y) where x = f(t) and y = g(t). This tells us where an object is at any moment.

Common Modelling Contexts

Projectiles - balls, stones, ski jumpers, dolphins

Vehicles - planes taking off, boats crossing rivers

Circular motion - Ferris wheels, figure skaters

Other curves - rollercoasters, hill climbs, bowl designs

Projectile Motion

For a projectile launched with initial speed v at angle θ to the horizontal:

Standard Projectile Equations

x = (v cos θ) t

y = (v sin θ) t - \frac{1}{2} g t^{2}

Or with initial height h:

y = h + (v sin θ) t - \frac{1}{2} g t^{2}

Key Techniques

Maximum height: Set dy/dt = 0 and solve for t

Time of flight: Set y = 0 (ground level) and solve for t

Range: Substitute time of flight into x equation

01Worked Example

A plane's position at time t seconds after take-off is modelled by:

x = (v cos θ) t

m,

y = (v sin θ) t

m, where v is speed and θ is angle of elevation.

When the plane has travelled 600 m horizontally, it has climbed 120 m. The plane's speed is 50 m s⁻¹.

(a) Find the angle of elevation θ. (b) Find the parametric equations. (c) Find the height after 10 seconds.

(a) Angle of elevation:

tan θ = 120/600 = 1/5

θ = tan⁻¹(1/5) = 11.3° (1 d.p.)

(b) Parametric equations:

x = 50 cos(11.3°)t = 49.0t m (3 s.f.)

y = 50 sin(11.3°)t = 9.80t m (3 s.f.)

(c) Height after 10 seconds:

y = 9.80 × 10 = 98 m

Showing Motion is a Straight Line

Some motion appears linear when there's no acceleration (constant velocity in both directions).

Method

To show motion is a straight line, eliminate t to get a Cartesian equation of form y = mx + c.

Example: Plane (from Example 1)

x = 49t, so t = x/49

y = 9.8t = 9.8 × (x/49) = x/5

This is a straight line through the origin with gradient 1/5.

Domain Restrictions

The domain t > 0 for a climbing plane is not realistic - the plane would climb forever at constant angle. In reality, it must level off!

02Worked Example

A stone is thrown from a 25 m high cliff at 5 m s⁻¹ at 45°. Position after t seconds:

x = \frac{5 2}{2} t

m,

y = - 4.9 t^{2} + \frac{5 2}{2} t + 25

m, for 0 ≤ t ≤ k

(a) Find k (when stone hits ground). (b) Find horizontal distance travelled.

(a) Stone hits ground when y = 0:

-4.9t² + (5√2/2)t + 25 = 0

Using quadratic formula:

t = (-3.536 ± √(12.5 + 490))/(-9.8)

t = -1.92 or t = 2.648...

Since t ≥ 0, k = 2.65 (2 d.p.)

(b) Horizontal distance when t = 2.648:

x = (5√2/2) × 2.648 = 3.536 × 2.648

x = 9.36 m (2 d.p.)

Circular Motion

For motion in a circle of radius r with centre at (a, b):

Circular Motion Equations

x = a + r cos (ω t) y = b + r sin (ω t)

Or with different forms: x = r sin t, y = b − r cos t (for wheel starting at bottom)

Period

The period of a cos(bt) or a sin(bt) is:

T = \frac{2 π}{ω}

Verifying a Circle

Show (x − a)² + (y − b)² = r²

by eliminating t using sin²t + cos²t = 1

03Worked Example

A car on a Ferris wheel at time t minutes has position:

x = 12 sin t

,

y = 12 - 12 cos t

where x is horizontal distance and y is height above ground in metres.

(a) Show motion is a circle with radius 12 m. (b) Find maximum height. (c) Find time for one revolution.

(a) Show it's a circle:

From the equations: sin t = x/12, cos t = (12 − y)/12

Using sin²t + cos²t = 1:

(x/12)² + ((12 − y)/12)² = 1

x² + (y − 12)² = 144 - a circle, centre (0, 12), radius 12 m ✓

(b) Maximum height:

Maximum when cos t = −1, so y = 12 − 12(−1) = 24 m

(c) Time for one revolution:

Period of sin t and cos t is 2π

One revolution takes 2π minutes ≈ 6.28 minutes

Complex Motion Paths

Some motion involves different frequencies in x and y, creating complex patterns like figure-of-eights.

Figure-of-Eight Motion

When x and y have different periods, the object traces complex paths. To complete one full pattern, both x and y must return to their starting values.

Finding When Path Returns

Find the period of x: T₁ = 2π/ω₁

Find the period of y: T₂ = 2π/ω₂

Full pattern completes at LCM of the two periods

Self-Intersection

The path intersects itself when the object is at the same (x, y) position at two different times.

04Worked Example

A figure skater's motion is modelled by:

x = 8 cos 20 t

,

y = 12 sin (10 t - \frac{π}{3})

, t ≥ 0 (t in minutes, x, y in metres)

(a) Find coordinates at t = 0. (b) Find where the skater crosses the y-axis.

(a) At t = 0:

x = 8 cos 0 = 8

y = 12 sin(−π/3) = 12 × (−√3/2) = −6√3

Position: (8, −6√3) ≈ (8, −10.4)

(b) Crosses y-axis when x = 0:

8 cos 20t = 0

cos 20t = 0

20t = π/2, 3π/2, 5π/2, 7π/2, ...

t = π/40, 3π/40, 5π/40, 7π/40, ...

Substituting into y: y-coordinates are ±3.11 and ±11.59

Crosses at: (0, −3.11), (0, 11.59), (0, 3.11), (0, −11.59)

Other Modelling Contexts

Parametric equations can model many real-world situations beyond standard projectiles and circles:

ContextTypical Form

Boat crossing riverx = at, y = bt (linear, affected by current)

Rollercoasterx = a + ln t, y = b sin(ct) (logarithmic + periodic)

Hill climb profilex = a tan t, y = b sin 2t

Bowl cross-sectionx = t − a sin t, y = 1 − b cos t (modified cycloid)

Always Check

Consider whether the domain given is realistic for the physical situation being modelled.

05Worked Example

A dolphin leaping from water has position:

x = 2 t

m,

y = - 4.9 t^{2} + 10 t

m

where x is horizontal distance and y is height above sea level.

(a) Find time to complete jump. (b) Find horizontal distance. (c) Show path is quadratic. (d) Find maximum height.

(a) Jump complete when y = 0:

−4.9t² + 10t = 0

t(−4.9t + 10) = 0

t = 0 or t = 10/4.9 = 2.04 seconds

(b) Horizontal distance:

x = 2 × 2.04 = 4.08 m

(c) Cartesian equation:

t = x/2, substitute: y = −4.9(x/2)² + 10(x/2)

y = −1.225x² + 5x (quadratic) ✓

(d) Maximum height:

dy/dt = −9.8t + 10 = 0, so t = 10/9.8 ≈ 1.02 s

y = −4.9(1.02)² + 10(1.02) = 5.10 m

06Worked Example

A rollercoaster path is modelled by:

x = 5 + ln t

,

y = 5 sin 2 t

, for 0 < t ≤ π/2

One unit represents 5 m in real life.

(a) Find coordinates when t = π/6. (b) Find maximum height of rollercoaster.

(a) When t = π/6:

x = 5 + ln(π/6) = 5 + ln(0.5236) = 5 − 0.647 = 4.35

y = 5 sin(2 × π/6) = 5 sin(π/3) = 5 × (√3/2) = 4.33

Position: (4.35, 4.33) units

(b) Maximum height:

Maximum of y = 5 sin 2t is when sin 2t = 1

This occurs when 2t = π/2, i.e., t = π/4

Maximum y = 5 × 1 = 5 units

In real life: 5 × 5 m = 25 m

Finding Gradients of Paths

In modelling, the gradient at a point tells us the direction of motion or steepness of a path.

Gradient Formula

\frac{d y}{d x} = \frac{d y / d t}{d x / d t}

Interpreting Gradients

dy/dx > 0: Path is ascending (going uphill)

dy/dx < 0: Path is descending (going downhill)

dy/dx = 0: Horizontal tangent (peak or trough)

dx/dt = 0: Vertical tangent (turning back)

Average Gradient

Average gradient between two points = (change in y)/(change in x) = Δy/Δx

Practice

Question 1:

x = 103 t

m,

y = - 4.9 t^{2} + 10 t

m

Find the horizontal distance travelled when it hits the ground.

Find when ball lands (y = 0):

−4.9t² + 10t = 0

t(−4.9t + 10) = 0

t = 0 or t = 10/4.9 = 2.04 seconds

Horizontal distance:

x = 10√3 × 2.04 = 17.3 × 2.04

x = 35.3 m

Practice

Question 2:

x = 0.9 t

m,

y = - 3.2 t

m

where x is distance east and y is distance north. The river is 75 m wide (north-south).

(a) Find time to cross. (b) Show motion is a straight line and find its gradient.

(a) Cross river when |y| = 75:

|−3.2t| = 75

3.2t = 75

t = 23.4 seconds

(b) Cartesian equation:

From x = 0.9t: t = x/0.9

y = −3.2 × (x/0.9) = −3.56x

This is a straight line with gradient −3.56

(Negative because boat drifts south while moving east)

Practice

Question 3:

x = 50 tan t

m,

y = 20 sin 2 t

m, for 0 < t ≤ π/2

Find the value of t at the highest point of the climb.

Maximum height when dy/dt = 0:

y = 20 sin 2t

dy/dt = 40 cos 2t = 0

cos 2t = 0

2t = π/2 (first positive solution in domain)

t = π/4

Check: At t = π/4:

y = 20 sin(π/2) = 20 × 1 = 20 m (maximum)

Summary

Key Modelling Skills

TaskMethod

Find position at time tSubstitute t into both equations

When does object land/reach ground?Set y = 0 and solve for t

Find maximum heightSet dy/dt = 0, solve for t, then find y

Show motion is a line/curveEliminate t to get Cartesian equation

Show motion is a circleUse sin²t + cos²t = 1

Exam Reminders

Always check if the domain is realistic for the context

Include units in your final answers

Interpret your answers in the context of the problem