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Vector and cartesian equations of curves (WACE Year 12 Mathematics Specialist Unit 3)

Syllabus dot point

“Write vector and parametric equations of curves and convert between vector, parametric and cartesian forms”

7 min read

Quick answer

A vector equation r(t)=x(t) i+y(t) j\mathbf{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j} traces a curve as the parameter tt varies. Read off the parametric equations x=x(t)x = x(t), y=y(t)y = y(t), then eliminate tt to obtain a cartesian relation between xx and yy that names the curve.

Jump to a section
  1. What this dot point is asking
  2. From vector to parametric form
  3. Three-dimensional curves
  4. Eliminating the parameter
  5. Recognising standard curves
  6. Direction of tracing and the parameter range
  7. Why both forms are useful
  8. Exam-style questions

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:

r(t)=x(t) i+y(t) j.\mathbf{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j}.

As tt ranges over an interval, the tip of r(t)\mathbf{r}(t) sweeps out the curve. The two component functions x=x(t)x = x(t) and y=y(t)y = y(t) are the parametric equations of the same curve.

Three-dimensional curves

The same idea extends to space, where a vector function r(t)=x(t) i+y(t) j+z(t) k\mathbf{r}(t) = x(t)\,\mathbf{i} + y(t)\,\mathbf{j} + z(t)\,\mathbf{k} 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 xx, yy and zz, or most naturally by keeping the parametric form. A helix, for instance, has x=acos⁡tx = a\cos t, y=asin⁡ty = a\sin t, z=btz = bt: the first two components trace a circle while the third climbs steadily, so the path spirals up a cylinder of radius aa. Parametric form is the only practical way to describe curves like this.

Eliminating the parameter

Converting to cartesian form

To get the cartesian equation, eliminate the parameter. Solve one parametric equation for tt and substitute into the other, or use an identity (such as cos⁡2t+sin⁡2t=1\cos^2 t + \sin^2 t = 1) that links the two components directly.

The resulting relation between xx and yy, free of tt, is the cartesian equation. Always note any restriction the parameter places on the range of xx or yy, since the parametric curve may be only part of the full cartesian graph.

Recognising standard curves

Some parametrisations recur. A linear r(t)=a+tb\mathbf{r}(t) = \mathbf{a} + t\mathbf{b} traces a straight line. The pair x=acos⁡tx = a\cos t, y=asin⁡ty = a\sin t traces a circle of radius aa, since squaring and adding gives x2+y2=a2x^2 + y^2 = a^2. The pair x=acos⁡tx = a\cos t, y=bsin⁡ty = b\sin t traces an ellipse.

A translated circle uses x=h+acos⁡tx = h + a\cos t, y=k+asin⁡ty = k + a\sin t, which gives (x−h)2+(y−k)2=a2(x - h)^2 + (y - k)^2 = a^2, a circle of radius aa centred at (h,k)(h, k). A parabola can be parametrised as x=at2x = at^2, y=2aty = 2at, which eliminates to y2=4axy^2 = 4ax. 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 tt increases, the tip of r(t)\mathbf{r}(t) moves along the curve in a definite sense; for x=acos⁡tx = a\cos t, y=asin⁡ty = a\sin t the motion is anticlockwise starting from (a,0)(a, 0). A restricted range such as 0≤t≤π0 \le t \le \pi traces only the upper half of the circle, so the cartesian equation must be accompanied by the inequality y≥0y \ge 0 to describe the same set of points. Always translate the parameter restriction into a restriction on xx or yy.

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 y=f(x)y = f(x) rule cannot capture both halves. Fluency in moving between the two is a core SCSA expectation.

Convert r(t)=3cos⁡t i+3sin⁡t j\mathbf{r}(t) = 3\cos t\,\mathbf{i} + 3\sin t\,\mathbf{j} to cartesian form

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.

x=3cos⁡t,y=3sin⁡t.x = 3\cos t, \qquad y = 3\sin t.

Step 2: Isolate the trigonometric ratios

To eliminate the parameter tt, we need each trigonometric ratio on its own. Dividing both equations by 3 achieves this.

cos⁡t=x3,sin⁡t=y3.\cos t = \frac{x}{3}, \qquad \sin t = \frac{y}{3}.

Step 3: Apply the Pythagorean identity to eliminate tt

Since cos⁡2t+sin⁡2t=1\cos^2 t + \sin^2 t = 1 holds for all tt, substituting the expressions from Step 2 removes tt entirely and leaves a relation between xx and yy only.

(x3)2+(y3)2=1  ⟹  x2+y2=9.\left(\frac{x}{3}\right)^2 + \left(\frac{y}{3}\right)^2 = 1 \;\Longrightarrow\; x^2 + y^2 = 9.

Step 4: Interpret the result

Recognising the standard form x2+y2=r2x^2 + y^2 = r^2 identifies the shape. Noting the direction of increasing tt in the parametric form confirms the sense of traversal.

The curve is a circle of radius 33 centred at the origin, traced anticlockwise as tt increases.

Final answer: x2+y2=9x^2 + y^2 = 9 (a circle of radius 33 centred at the origin).

Common traps
Forgetting the parameter restriction
A limited range of tt may trace only an arc; state the resulting range of xx and yy.
Wrong identity for elimination
For trigonometric parametrisations, use cos⁡2+sin⁡2=1\cos^2 + \sin^2 = 1; 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 marks
Calculator-assumed. A curve is given by r(t)=(2+3cos⁡t) i+(−1+3sin⁡t) j\mathbf{r}(t) = (2 + 3\cos t)\,\mathbf{i} + (-1 + 3\sin t)\,\mathbf{j} for 0≤t<2π0 \le t < 2\pi. (a) Find its cartesian equation. (b) Describe the curve fully.
Show worked answer →

A translated-circle parametrisation.

(a) Parametric equations: x=2+3cos⁡tx = 2 + 3\cos t and y=−1+3sin⁡ty = -1 + 3\sin t. Isolate the trig terms: cos⁡t=x−23\cos t = \dfrac{x - 2}{3} and sin⁡t=y+13\sin t = \dfrac{y + 1}{3}. Using cos⁡2t+sin⁡2t=1\cos^2 t + \sin^2 t = 1: (x−23)2+(y+13)2=1\left(\dfrac{x - 2}{3}\right)^2 + \left(\dfrac{y + 1}{3}\right)^2 = 1, so (x−2)2+(y+1)2=9(x - 2)^2 + (y + 1)^2 = 9.

(b) A circle of radius 33 centred at (2,−1)(2, -1), traced once anticlockwise as tt runs from 00 to 2π2\pi.

Markers reward isolating the trig ratios, the Pythagorean elimination, the centre and radius, and the direction of tracing.

WACE 2024-style5 marks
Calculator-free. The line through A(1,4)A(1, 4) in the direction (2,−1)(2, -1) has vector equation r(t)=(1,4)+t(2,−1)\mathbf{r}(t) = (1, 4) + t(2, -1). Find its cartesian equation.
Show worked answer →

Eliminating the parameter from a line.

Parametric equations: x=1+2tx = 1 + 2t and y=4−ty = 4 - t. Solve the second for tt: t=4−yt = 4 - y. Substitute into the first: x=1+2(4−y)=1+8−2y=9−2yx = 1 + 2(4 - y) = 1 + 8 - 2y = 9 - 2y.

Rearrange: x+2y=9x + 2y = 9, or equivalently y=9−x2y = \dfrac{9 - x}{2}. Markers reward writing the component equations, solving for tt, the substitution, and a tidy cartesian form. The gradient −12-\dfrac{1}{2} matches the direction (2,−1)(2, -1).

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