A-Level · Mathematics · AQA · Mark scheme decoded
AQA A-Level Mathematics: Parametric Equations in Modelling — mark scheme explained
The short answer
Parametric equations are a powerful tool in coordinate geometry, allowing us to describe curves and paths that might be difficult or impossible to express using a single Cartesian equation.
The question
Convert the parametric equations x = t 3 and y = t 2 to Cartesian form.
[Paraphrased for study — not reproduced from any exam paper.]
Mark scheme, decoded
How the examiner actually awards the marks on this topic.
Gradora's own decode of the marking approach — not the exam board's published mark scheme.
How marks are awarded
For questions involving conversion between parametric and Cartesian forms, ensure you show all steps clearly. For finding tangents and normals, include the calculation of dy/dx and the use of point-slope form. In modelling problems, clearly state your assumptions and the equations used.
What the command words demand
- Convert
- Change parametric equations to Cartesian form or vice versa.
- Find
- Determine specific values or equations, such as the slope of a tangent or normal.
- Derive
- Show the steps to obtain a formula or equation from given information.
- Model
- Use parametric equations to describe real-world scenarios, such as projectile motion.
Model answer
A full-mark response to the question above, worked through step by step.
Timing: Allocate about 5-7 minutes for each question involving parametric equations to ensure you have enough time to show all necessary steps and calculations.
- Solve the second equation for t: y = t 2 → t = ±√y1 mark
- Substitute t = √y into the first equation: x = (√y) 3 = y 3/21 mark
- Square both sides to obtain a Cartesian relation valid for both roots: x 2 = y 32 marks
Final answer: x 2 = y 3
Work through every step correctly and you earn all 4 marks.
Another worked example
Find the equation of the tangent to the parametric curve x = t 2 , y = t + 1 at the point where t = 2.
- Calculate dx/dt and dy/dt: dx/dt = 2t, dy/dt = 11 mark
- Find dy/dx: dy/dx = (dy/dt) / (dx/dt) = 1 / (2t)1 mark
- Substitute t = 2 into the parametric equations to find the coordinates of the point: x = 2 2 = 4, y = 2 + 1 = 31 mark
- Calculate the slope of the tangent at t = 2: dy/dx = 1 / (2 × 2) = 1/41 mark
- Use the point-slope form to find the equation of the tangent: y - 3 = (1/4)(x - 4)1 mark
- Simplify: y = (1/4)x + 21 mark
Final answer: y = (1/4)x + 2
Work through every step correctly and you earn all 6 marks.
Common mistakes
Forgetting to eliminate the parameter t when converting parametric equations to Cartesian form.
Why it happens: Students sometimes stop at finding dx/dt and dy/dt without completing the conversion process.
Fix: Always solve one of the parametric equations for t and substitute it into the other equation to eliminate t.
Incorrectly calculating the derivative dy/dx for parametric equations.
Why it happens: Students may forget to use the chain rule (dy/dt) / (dx/dt) or make algebraic errors in simplification.
Fix: Double-check your calculations and ensure you are using the correct formula: dy/dx = (dy/dt) / (dx/dt).
Using the wrong point when finding the equation of a tangent or normal.
Why it happens: Students may substitute the parameter t into the parametric equations incorrectly, leading to incorrect coordinates.
Fix: Always verify the coordinates by substituting the given value of t into both x and y parametric equations.
Forgetting to find the slope of the normal as the negative reciprocal of the tangent's slope.
Why it happens: Students may confuse the process for finding tangents and normals, leading to incorrect slopes.
Fix: Remember that the slope of the normal is always the negative reciprocal of the slope of the tangent: m normal = -1 / (dy/dx).
Incorrectly applying trigonometric values in projectile motion problems.
Why it happens: Students may use incorrect values for sin(θ) and cos(θ), leading to errors in the parametric equations.
Fix: Always double-check your trigonometric values, especially for common angles like 30°, 45°, and 60°.
Failing to consider both positive and negative roots when eliminating the parameter t.
Why it happens: Students may overlook the possibility of multiple solutions when solving for t, leading to incomplete or incorrect Cartesian equations.
Fix: When solving for t, always consider both positive and negative roots if applicable.
Using the wrong formula for the maximum height in projectile motion problems.
Why it happens: Students may use incorrect formulas or forget to set the vertical velocity to zero when finding the maximum height.
Fix: Always remember that the maximum height is found by setting the vertical velocity (dy/dt) to zero and solving for t, then substituting back into the y equation.
Where the marks go
The question types you’ll meet on this topic and the marks each one carries — so you know what to expect and where to focus.
| Question type | What you’re asked to do | Marks |
|---|---|---|
| Parametric To Cartesian | Eliminate the parameter t to express the relationship between x and y directly. | 4 |
| Parametric Tangent Equation | Find the tangent line to a parametric curve at a given parameter value. | 6 |
| Projectile Motion Modelling | Form parametric equations for a projectile's motion and calculate its maximum height reached. | 6 |
| Parametric Normal Equation | Find the equation of the normal to a parametric curve at a given parameter value. | 6 |
| Total across these question types | 22 | |
Question types and mark tariffs are Gradora’s guidance based on how this topic is typically examined — not the board’s official paper structure.