A Cartesian equation tells you where a curve lives. But parametric equations also tell you when a point is there, and which way it's travelling. That extra information, direction of tracing, initial and terminal points, transforms a static shape into a dynamic story. In this lesson you'll build the skill of reading that story from a table of values, and telling it back through a well-labelled sketch.
As $t$ increases from $0$, how does the point $(\cos t, \sin t)$ move around the unit circle? Without working it out algebraically make a prediction: clockwise or anticlockwise? Where does the point start? Where does it go first?
Sketching parametric curves comes down to two core skills: build a table of values to find key points, and use the Cartesian equation to confirm the shape. Add direction arrows and label initial/terminal points and you're done.
Every parametric sketch lives on two roads: substitute parameter values to generate key coordinates and plot them, then identify the Cartesian shape to confirm what you're drawing and add direction arrows.
To sketch a curve given parametrically, follow these five steps:
Five-step sketching method: (1) find Cartesian form, (2) key points via table, (3) direction, (4) restrictions, (5) label initial/terminal points; Always choose t values that make the trig functions equal to 0, 1, e.g. multiples of /2
Pause, copy the five-step parametric sketching method into your book: (1) find Cartesian form; (2) table of key points at $t = 0, \pi/2, \pi, 3\pi/2, 2\pi$; (3) mark direction of tracing; (4) note domain restrictions; (5) label initial and terminal points.
Quick check: For $x = t - 1$, $y = 2t + 3$ with $0 \le t \le 3$, what is the initial point (at $t = 0$)?
We just saw the five-step sketching strategy: find Cartesian form, build a key-point table, restrict the domain, and label endpoints. That raises a question: the Cartesian form shows the path but not which way the point moves, how do you add this directional information to the sketch? This card answers it → evaluate $(x,y)$ at a few increasing $t$-values and mark arrows on the curve showing the direction of tracing.
The direction of tracing describes how the point $(x, y)$ moves as $t$ increases. This information is not visible in the Cartesian equation, it is unique to the parametric representation.
To determine direction:
Key insight: A parametric curve can double back on itself, the same $x$ value can be visited multiple times (with different $y$ or $t$ values). This is impossible to represent in a standard Cartesian function $y = f(x)$.
The unit circle traced anticlockwise from $(1, 0)$ as $t$ increases from $0$ to $2\pi$.
Direction of tracing is determined by how (x, y) changes as t increases; x = t, y = t: starts at (1,0), moves anticlockwise (since is initially increasing)
Pause, copy the direction rule into your book: determine tracing direction by computing $(x,y)$ at two nearby $t$-values; for $x = \cos t$, $y = \sin t$, the point starts at $(1,0)$ and moves anticlockwise as $t$ increases.
Did you get this? True or false: the parametric equations $x = \cos t$, $y = \sin t$ trace the unit circle in an anticlockwise direction as $t$ increases.
Worked examples · 3 in a row, reveal as you go
Sketch the curve $x = t^2$, $y = t^3$ for $-2 \le t \le 2$, indicating direction.
| $t$ | $x = t^2$ | $y = t^3$ |
|---|---|---|
| $-2$ | $4$ | $-8$ |
| $-1$ | $1$ | $-1$ |
| $0$ | $0$ | $0$ |
| $1$ | $1$ | $1$ |
| $2$ | $4$ | $8$ |
Sketch the curve $x = 2\cos t$, $y = 3\sin t$ for $0 \le t \le 2\pi$, indicating direction.
For $x = t - 1$, $y = 2t + 3$ with $0 \le t \le 3$, find the initial and terminal points, the Cartesian equation, and sketch.
Fill the gap: For $x = 3\cos t$, $y = 3\sin t$ ($0 \le t \le \frac{\pi}{2}$), the initial point is and the terminal point is .
Misconceptions to fix · the 3 traps that cost marks
Did you get this? True or false: for $x = \cos t$, $y = -\sin t$ ($0 \le t \le 2\pi$), the curve is traced in a clockwise direction.
Activities · practice with the ideas
For $x = t - 1$, $y = 2t + 3$ with $0 \le t \le 3$, find the initial and terminal points and state the direction of tracing.
Describe the curve $x = 3\cos t$, $y = 3\sin t$ for $0 \le t \le \pi/2$, including its shape, key points, and direction.
For $x = t^2 - 1$, $y = t$ with $t \ge 0$, find the Cartesian equation and describe the portion traced.
Explain why the parametric curve $x = t^2$, $y = t^3$ for $-2 \le t \le 2$ doubles back on itself, but $y = x^{3/2}$ does not.
What change would you make to $x = 2\cos t$, $y = 3\sin t$ to trace the ellipse in a clockwise direction?
Odd one out: Which of these parametric curves does NOT trace an entire closed loop (i.e. does not return to its starting point)?
Earlier you were asked: as $t$ increases from $0$, how does $(\cos t, \sin t)$ move around the unit circle?
At $t = 0$: point is $(1, 0)$, the rightmost point. As $t$ increases, $\cos t$ decreases (point moves left) and $\sin t$ increases (point moves up). So the initial motion is anticlockwise. The point completes the full circle at $t = 2\pi$, returning to $(1, 0)$. Knowing the direction of tracing is the key advantage of parametric form over the Cartesian equation $x^2 + y^2 = 1$, which shows the shape but not the motion.
Pick your answer, then rate your confidence. That tells the system what to drill next. Each retry pulls a fresh mix from the bank.
Q1. For $x = t - 1$, $y = 2t + 3$ with $0 \le t \le 3$, find the initial and terminal points and sketch the curve indicating its direction. (2 marks)
Q2. Describe the curve $x = 3\cos t$, $y = 3\sin t$ for $0 \le t \le \pi/2$, including its shape, the key points at each endpoint, and the direction of tracing. (2 marks)
Q3. For $x = t^2 - 1$, $y = t$ with $t \ge 0$, find the Cartesian equation, state any restrictions, and sketch the curve. (2 marks)
Activity 1: 1. Initial $(-1,3)$, terminal $(2,9)$, direction left-to-right upwards · 2. Quarter-circle of radius 3, centre origin, from $(3,0)$ to $(0,3)$, anticlockwise · 3. $x = y^2 - 1$ ($y \ge 0$), right branch of a parabola, starting at $(-1, 0)$ when $t=0$ · 4. For $t < 0$, $x = t^2 > 0$ so the curve visits positive $x$ values with negative $y$ (below $x$-axis), then revisits the same $x$ values with positive $y$ after $t=0$. The Cartesian equation $y = x^{3/2}$ only captures $y \ge 0$. · 5. Change $y = 3\sin t$ to $y = -3\sin t$.
Q1 (2 marks): Initial point $(-1, 3)$ at $t=0$; terminal point $(2, 9)$ at $t=3$ [1]. Cartesian: $t = x+1$, so $y = 2(x+1)+3 = 2x+5$; line segment from $(-1,3)$ to $(2,9)$, traced left-to-right [1].
Q2 (2 marks): The curve is the first-quadrant arc of a circle $x^2 + y^2 = 9$, radius 3 [1]. Initial point $(3, 0)$ at $t=0$; terminal point $(0, 3)$ at $t=\pi/2$; traced anticlockwise [1].
Q3 (2 marks): $t = y$ (since $y = t$), so $x = y^2 - 1$ [1]. Restriction $t \ge 0 \Rightarrow y \ge 0$: this is the upper half of the parabola $x = y^2 - 1$, starting at $(-1, 0)$ and moving right and upwards as $t$ increases [1].
A full module quiz covering every lesson in this module, not just this one. Set aside a decent block of time and treat it like a real assessment.
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