Vector and cartesian equations of curves (WACE Year 12 Mathematics Specialist Unit 3)
“Write vector and parametric equations of curves and convert between vector, parametric and cartesian forms”
A vector equation traces a curve as the parameter varies. Read off the parametric equations , , then eliminate to obtain a cartesian relation between and that names the curve.
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What this dot point is asking
SCSA wants you to move fluently between the vector form of a curve, its parametric component equations, and its cartesian equation, and to recognise the standard curves these describe.
From vector to parametric form
A curve in the plane can be written as a position vector that depends on a parameter:
As ranges over an interval, the tip of sweeps out the curve. The two component functions and are the parametric equations of the same curve.
Three-dimensional curves
The same idea extends to space, where a vector function traces a curve in three dimensions. A single Cartesian equation can no longer capture such a curve; instead it is described by two relations between , and , or most naturally by keeping the parametric form. A helix, for instance, has , , : the first two components trace a circle while the third climbs steadily, so the path spirals up a cylinder of radius . Parametric form is the only practical way to describe curves like this.
Eliminating the parameter
To get the cartesian equation, eliminate the parameter. Solve one parametric equation for and substitute into the other, or use an identity (such as ) that links the two components directly.
The resulting relation between and , free of , is the cartesian equation. Always note any restriction the parameter places on the range of or , since the parametric curve may be only part of the full cartesian graph.
Recognising standard curves
Some parametrisations recur. A linear traces a straight line. The pair , traces a circle of radius , since squaring and adding gives . The pair , traces an ellipse.
A translated circle uses , , which gives , a circle of radius centred at . A parabola can be parametrised as , , which eliminates to . Knowing these standard forms lets you name a curve at a glance, which is often half the marks in a description question.
Direction of tracing and the parameter range
The vector form carries more information than the cartesian equation: it also records the direction and the part of the curve that is traced. As increases, the tip of moves along the curve in a definite sense; for , the motion is anticlockwise starting from . A restricted range such as traces only the upper half of the circle, so the cartesian equation must be accompanied by the inequality to describe the same set of points. Always translate the parameter restriction into a restriction on or .
Why both forms are useful
The cartesian equation is best for recognising the shape and for checking whether a point lies on the curve. The vector or parametric form is best for describing motion along the curve, for differentiating to get velocity (the next dot point), and for handling curves that are not functions, such as a full circle, where a single rule cannot capture both halves. Fluency in moving between the two is a core SCSA expectation.
Step 1: Write down the parametric equations
Reading off the components of the position vector gives the two parametric equations. This separates the single vector equation into two scalar equations that can be manipulated algebraically.
Step 2: Isolate the trigonometric ratios
To eliminate the parameter , we need each trigonometric ratio on its own. Dividing both equations by 3 achieves this.
Step 3: Apply the Pythagorean identity to eliminate
Since holds for all , substituting the expressions from Step 2 removes entirely and leaves a relation between and only.
Step 4: Interpret the result
Recognising the standard form identifies the shape. Noting the direction of increasing in the parametric form confirms the sense of traversal.
The curve is a circle of radius centred at the origin, traced anticlockwise as increases.
Final answer: (a circle of radius centred at the origin).
- Forgetting the parameter restriction
- A limited range of may trace only an arc; state the resulting range of and .
- Wrong identity for elimination
- For trigonometric parametrisations, use ; substituting clumsily often fails.
- Confusing a curve with a line
- A linear vector equation gives a line; a trigonometric one gives a circle or ellipse. Check the form first.
Exam-style questions
Questions in the style of SCSA exam questions on this dot point, each with a worked answer. They are written by ExamExplained unless tagged "Past paper"; the year shows the paper a question is modelled on.
WACE 2022-style6 marksCalculator-assumed. A curve is given by for . (a) Find its cartesian equation. (b) Describe the curve fully.Show worked answer →
A translated-circle parametrisation.
(a) Parametric equations: and . Isolate the trig terms: and . Using : , so .
(b) A circle of radius centred at , traced once anticlockwise as runs from to .
Markers reward isolating the trig ratios, the Pythagorean elimination, the centre and radius, and the direction of tracing.
WACE 2024-style5 marksCalculator-free. The line through in the direction has vector equation . Find its cartesian equation.Show worked answer →
Eliminating the parameter from a line.
Parametric equations: and . Solve the second for : . Substitute into the first: .
Rearrange: , or equivalently . Markers reward writing the component equations, solving for , the substitution, and a tidy cartesian form. The gradient matches the direction .