What is 3.4: Parametric modelling?

In a parametric model, the parameter often represents time and the pair (x(t),y(t)) gives the object's position at that time.

How many marks are 3.4 questions worth?

3.4 practice drills on this page are worth 2–7 marks each; marks in a real exam vary by question.

Which past exam papers has 3.4 appeared in?

We have not yet indexed a verified past-paper appearance for 3.4 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

3.4

Use parametric equations in modelling in a variety of contexts.

Draft — not yet indexed

Parametric modelling

Worked answers and methods for 3.4 on Edexcel A-level Maths 9MA0.

Explanation

In a parametric model, the parameter often represents time and the pair $(x(t),y(t))$ gives the object's position at that time.
Translate a contextual event into equations: meeting a boundary fixes one coordinate, while a collision requires both coordinates of two objects to agree at the same time.
Eliminating the parameter can reveal the path, but the parameter range still controls the physically modelled section of that path.
Check units, permitted times and whether a calculated event occurs within the model's domain.
A common error is to find where two paths cross without checking that the objects arrive there simultaneously.

Worked example

An arch is modelled by $x=6t$ , $y=12t-3t^2$ for $0\le t\le4$ , with $x$ and $y$ in metres. Find a Cartesian equation for the arch and its horizontal span.

1.Since $x=6t$ , $t=x/6$ .
2.Substitute into $y$ : $y=12(x/6)-3(x/6)^2=2x-x^2/12$ .
3.The parameter interval gives $0\le x\le24$ .
4.The arch meets ground level at its two endpoints, so its horizontal span is $24-0=24$ m.

Answer: $y=2x-\dfrac{x^2}{12}$ for $0\le x\le24$ ; Span $24$ m

Common mistakes

Don't solve for a geometric intersection but assign the two objects different parameter values at the same time.
Don't treat the parameter as a physical coordinate rather than eliminating it and interpreting the stated interval.

Exam tip

In a parametric model, connect the parameter endpoints to the physical endpoints before reporting dimensions.

Worked practice

Tier 1 · Easy

A particle's position after

t

seconds is modelled by

x=12t

y=20t-5t^2

, where distances are in metres. Find its position when

t=1.5

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$(18,18.75)$ metres	2
Notes
Substitute $t=1.5$ : $x=12(1.5)=18$ and $y=20(1.5)-5(1.5)^2=30-11.25=18.75$ .

(2 marks)

Tier 2 · Standard

A robot moves with position

x=2t+1

y=t^2

for

0\leq t\leq4

, where distances are in metres and

t

is in seconds. Find the time at which it crosses the line

y=3x-5

and give the exact coordinates of the crossing point.

(4)

(Total for Question 2 is 4 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	$t=3-\sqrt7$ seconds Crossing point $\left(7-2\sqrt7,\,16-6\sqrt7\right)$	4
Notes
At a crossing, $t^2=3(2t+1)-5$ , so $t^2-6t+2=0$ and $t=3\pm\sqrt7$ . Only $3-\sqrt7$ lies in the modelled interval $0\leq t\leq4$ . Substitution gives $x=2(3-\sqrt7)+1=7-2\sqrt7$ and $y=(3-\sqrt7)^2=16-6\sqrt7$ .

(4 marks)

Tier 3 · Hard

Two particles move for

t\ge0

seconds. Particle

P

has position

(3t+2,t^2+2)

and particle

Q

has position

(12-2t,t+4)

. Determine whether they collide and, if they do, find the time and position of the collision.

(4)

(Total for Question 3 is 4 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	They collide at $t=2$ seconds at $(8,6)$ .	4
Notes
Equal $x$ -coordinates require $3t+2=12-2t$ , so $5t=10$ and $t=2$ . Check the other coordinate at this same time: for $P$ , $y=2^2+2=6$ ; for $Q$ , $y=2+4=6$ . Both positions are therefore $(8,6)$ at $t=2$ , so a collision occurs. The linear $x$ equation has only one solution, so there is no other collision.

(4 marks)

Tier 1 · Easy

A survey drone is modelled as having position

(x,y)=(2+6t,15-3t)

at time

t

seconds, where distances are in metres and

0\leq t\leq5

. Find the time and position at which it reaches the line

y=6

(3)

(Total for Question 4 is 3 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$t=3$ seconds $(20,6)$ metres	3
Notes
Set $15-3t=6$ , giving $t=3$ , which lies in the modelled interval. Then $x=2+6(3)=20$ , so the position is $(20,6)$ metres.

(3 marks)

Tier 2 · Standard

An inspection vehicle moves with position

(x,y)=(4t,t^2-6t+10)

for

0\leq t\leq5

. Coordinates are measured in metres and

t

denotes seconds. Find both crossings of the marker line

y=2

, including their times and positions, and the elapsed time between them.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$t=2$ and $t=4$ seconds $(8,2)$ and $(16,2)$ $2$ seconds	4
Notes
At the marker line, $t^2-6t+10=2$ , so $t^2-6t+8=0=(t-2)(t-4)$ . Both values lie in the modelled interval. Since $x=4t$ , the positions are $(8,2)$ and $(16,2)$ , separated by $4-2=2$ seconds.

(4 marks)

Tier 3 · Hard

For

0\leq t\leq7

, a drone's position is modelled by

x=2t+1

and

y=t^2-6t+13

, with distances in kilometres and

t

in hours. A signal corridor is the set of points on or below the line

y=x

. Find when the drone enters and leaves the corridor, the two boundary positions, and the total time spent in the corridor.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Enters at $t=2$ hours and leaves at $t=6$ hours $(5,5)$ and $(13,13)$ $4$ hours	5
Notes
Inside the corridor, $t^2-6t+13\leq2t+1$ , so $t^2-8t+12\leq0$ . Since $(t-2)(t-6)\leq0$ , this holds for $2\leq t\leq6$ , within the modelled interval. The boundary positions are $(5,5)$ at $t=2$ and $(13,13)$ at $t=6$ . The time inside is $6-2=4$ hours.

(5 marks)

Tier 2 · Standard

A survey vessel has position

(x,y)=(3+4t,1+3t)

at time

t

hours, where distances are in kilometres and

0\leq t\leq5

. A beacon is at

(14,10)

. Find the exact time at which the vessel is closest to the beacon and its minimum distance from the beacon.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$t=\dfrac{71}{25}$ hours Minimum distance $=\dfrac35$ km	5
Notes
The squared distance from the beacon is $D^2=(4t-11)^2+(3t-9)^2=25t^2-142t+202$ . Completing the square gives $D^2=25(t-71/25)^2+9/25$ . Since $71/25$ lies in the modelled interval, the minimum occurs then and is $D=\sqrt{9/25}=3/5$ km.

(5 marks)

Tier 3 · Hard

Warehouse robot

P

follows the route

x=2s+1

y=s+4

, where

s

is the number of minutes after

P

starts. Robot

Q

follows the route

x=11-u

y=2u-1

, where

u

is the number of minutes after

Q

starts. Find the intersection of the two routes and the corresponding values of

s

and

u

. Hence determine how much earlier or later

Q

must start for the robots to meet there.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	$s=3$ , $u=4$ Intersection $(7,7)$ $Q$ must start $1$ minute before $P$	5
Notes
At the route intersection, $2s+1=11-u$ and $s+4=2u-1$ . These equations are $2s+u=10$ and $s-2u=-5$ , giving $u=4$ and $s=3$ . Substitution gives the common position $(7,7)$ . Robot $Q$ needs $4$ minutes to reach it while $P$ needs $3$ minutes, so $Q$ must start $1$ minute before $P$ for them to arrive together.

(5 marks)

Tier 3 · Hard

A tracking beacon moves with position

(x,y)=(8\cos t,6\sin t)

at time

t

hours, where

0\leq t\leq2\pi

. The beacon is inside a monitored region when

x\geq0

and

y\geq3

. Find the complete interval of times for which the beacon is inside the region. Give the exact entry and exit positions and the fraction of the modelled time spent inside the region.

(6)

(Total for Question 9 is 6 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	$\dfrac{\pi}{6}\leq t\leq\dfrac{\pi}{2}$ Entry $\left(4\sqrt3,3\right)$ ; exit $(0,6)$ Fraction $=\dfrac16$	6
Notes
The condition $y\geq3$ gives $6\sin t\geq3$ , so $\sin t\geq1/2$ and therefore $\pi/6\leq t\leq5\pi/6$ on the stated interval. The condition $x\geq0$ gives $\cos t\geq0$ , so $0\leq t\leq\pi/2$ or $3\pi/2\leq t\leq2\pi$ . Intersecting these sets gives $\pi/6\leq t\leq\pi/2$ . Substitution gives entry position $(4\sqrt3,3)$ and exit position $(0,6)$ . The duration is $\pi/3$ hours out of $2\pi$ hours, so the fraction is $1/6$ .

(6 marks)

Q10

Tier 3 · Hard

10.

During a six-hour survey, a craft follows

(x,y)=(-6+2t,-8+3t)

, with

t

measured in hours and coordinates in kilometres (

0\leq t\leq6

). It can communicate with a relay at the origin while it is at most

5

km from the relay. Find the exact times at which communication starts and ends, and hence the total communication time. Eliminate

t

to find a Cartesian equation of the craft's route, stating the restriction on

x

(7)

(Total for Question 10 is 7 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Communication starts at $t=\dfrac{36-\sqrt{321}}{13}$ and ends at $t=\dfrac{36+\sqrt{321}}{13}$ Total time $=\dfrac{2\sqrt{321}}{13}$ hours Route: $3x-2y+2=0$ , where $-6\leq x\leq6$	7
Notes
The squared distance from the relay is $D^2=(-6+2t)^2+(-8+3t)^2=13t^2-72t+100$ . Communication is possible when $D^2\leq25$ , so $13t^2-72t+75\leq0$ . The boundary roots are $t=(36\pm\sqrt{321})/13$ , both in the modelled interval, and the inequality holds between them. Their difference is $2\sqrt{321}/13$ hours. From $x=-6+2t$ , $t=(x+6)/2$ . Substitution into $y=-8+3t$ gives $y=3x/2+1$ , or $3x-2y+2=0$ . As $0\leq t\leq6$ , the corresponding restriction is $-6\leq x\leq6$ .

(7 marks)

Verified exam appearances

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Other points in 3 Coordinate geometry in the (x, y) plane

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