Year 12 Maths Extension 1 (2027) Calculus

Derivatives of parametric functions

Q: How do you differentiate parametric equations?

Differentiate x and y with respect to the parameter, then divide dy by dt by dx by dt.

Q: How do you find the gradient at a point?

Find dy by dx in terms of the parameter, then substitute the parameter value at that point.

Q: When is a parametric tangent horizontal or vertical?

Horizontal when dy by dt is zero and dx by dt is not; vertical when dx by dt is zero and dy by dt is not.

Q: Why divide the two derivatives?

By the chain rule dy by dx equals dy by dt times dt by dx, and dt by dx is the reciprocal of dx by dt.

Q: Is parametric differentiation in Extension 1?

Yes, it is part of the Year 12 Calculus topic, outcome ME1-12-04.

20 practice questions 4 video lessons Theory + worked examples

Learn to find derivatives of parametric functions for NSW Year 12 Mathematics Extension 1 (NESA outcome ME1-12-04). When a curve is defined by separate equations for $x$ and $y$ in terms of a parameter, the chain rule links their rates of change.

You will learn to differentiate parametric equations by combining the rates for $y$ and $x$, then find gradients and tangents to parametric curves — a core calculus skill for curves and motion in the Extension 1 course.

Practice 20 questions

Practice questions

Every question with a fully worked solution.

Start practising

Watch 4 video(s)

Test video only Watch
Test Video 2 Watch
Derivatives of Parametric Functions Watch
Parametric Equations : Differentiation : ExamSolutions Watch

Theory

A parametric curve gives $x$ and $y$ separately as functions of a parameter. This NSW Year 12 Mathematics Extension 1 topic (NESA outcome ME1-12-04) differentiates such curves with the chain rule: $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ .

A parametric curve gives the two coordinates separately as functions of a parameter (usually $t$ or $θ$ ): $x = x (t)$ and $y = y (t)$ . As the parameter runs, the point $(x (t), y (t))$ traces the curve.

To find the gradient, differentiate each coordinate with respect to the parameter and divide: this is the chain rule $\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x}$ , where $\frac{d t}{d x}$ is the reciprocal of $\frac{d x}{d t}$ .

The result is usually in terms of the parameter, so to find the gradient at a point you substitute the parameter value — not $x$ .

NESA link. Part of the Year 12 Calculus topic, outcome ME1-12-04 ("selects and applies differentiation and integration techniques to solve problems") with MAO-WM-01. The syllabus asks students to find the derivative of a parametrically defined function using the chain rule and solve related problems.

x = a \cos θ, y = b \sin θ

: horizontal tangents at top and bottom, vertical at the sides.

x = t^{2} - 4, y = t^{3} - 3 t

: horizontal tangents at

t = \pm 1

, a vertical tangent at

t = 0

\frac{d y}{d x} = \frac{\frac{d y}{d t}}{\frac{d x}{d t}}, \frac{d x}{d t} \neq 0.

dy/dx = (dy/dt) / (dx/dt)

The tangent at a point uses $y - y_{1} = m (x - x_{1})$ with $m = \frac{d y}{d x}$ ; the normal uses $- \frac{1}{m}$ . Horizontal and vertical tangents depend on which derivative is zero:

Tangent	Condition
Horizontal	$\frac{d y}{d t} = 0, \frac{d x}{d t} \neq 0$
Vertical	$\frac{d x}{d t} = 0, \frac{d y}{d t} \neq 0$

How to differentiate a parametric curve

Differentiate each coordinate with respect to the parameter to get $\frac{d x}{d t}$ and $\frac{d y}{d t}$ .
Divide to get $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ , leaving the answer in the parameter.
For a point, substitute the parameter value (not $x$ ) to get the numerical gradient.
For a tangent, also find the point $(x, y)$ at that parameter, then use $y - y_{1} = m (x - x_{1})$ .

Example 1 — Gradient in the parameter

A curve is given by

x = 4 t, y = 2 t^{2} - t

. Find

\frac{d y}{d x}

Solution

$\frac{d x}{d t}$	$=$	$4$
$\frac{d y}{d t}$	$=$	$4 t - 1$
$\frac{d y}{d x}$	$=$	$\frac{4 t - 1}{4}$

dy/dx = (4t - 1)/4

So $\frac{d y}{d x} = \frac{4 t - 1}{4}$ .

Example 2 — Gradient at a point

A curve is given by

x = 5 \cos θ, y = 3 \sin θ

. Find the gradient at

θ = \frac{π}{6}

Solution

$\frac{d x}{d θ}$	$=$	$- 5 \sin θ$
$\frac{d y}{d θ}$	$=$	$3 \cos θ$
$\frac{d y}{d x}$	$=$	$- \frac{3}{5} \cot θ = - \frac{3 \sqrt{3}}{5}$

gradient = -3/5 cot(pi/6) = -3 sqrt(3)/5

The gradient is $- \frac{3 \sqrt{3}}{5}$ .

Example 3 — A tangent line

A curve is given by

x = 2 t, y = t^{3}

. Find the tangent at

t = 2

Solution

$\frac{d y}{d x}$	$=$	$\frac{3 t^{2}}{2}$
$t = 2$	$\Rightarrow$	$m = 6, (x, y) = (4, 8)$
$y - 8$	$=$	$6 (x - 4)$

tangent: y = 6x - 16

Tangent: $y = 6 x - 16$ .

Example 4 — Horizontal and vertical tangents

For

x = t^{2} - 4, y = t^{3} - 3 t

, find where the tangent is horizontal and where it is vertical.

Solution

$\frac{d x}{d t} = 2 t$ , $\frac{d y}{d t} = 3 t^{2} - 3$ .

Horizontal: $3 t^{2} - 3 = 0$	$\Rightarrow$	$t = \pm 1$
Vertical: $2 t = 0$	$\Rightarrow$	$t = 0$

horizontal at t = plus/minus 1; vertical at t = 0

Horizontal at $t = \pm 1$ ; vertical at $t = 0$ .

Common pitfalls

Multiplying instead of dividing.

\frac{d y}{d x}

\frac{d y}{d t}

divided by

\frac{d x}{d t}

— never multiply them.

Substituting $x$ . The answer is in the parameter, so substitute the

t

- or

θ

-value for a gradient, not

x

Vertical tangents. At a vertical tangent

\frac{d x}{d t} = 0

, so

\frac{d y}{d x}

is undefined — not zero.

Forgetting the point. A tangent line needs both the gradient and the actual point

(x, y)

at that parameter.

Frequently asked questions

How do you differentiate parametric equations?

Differentiate $x$ and $y$ with respect to the parameter, then divide: $\frac{d y}{d x} = \frac{d y / d t}{d x / d t}$ .

How do you find the gradient at a point on a parametric curve?

Find $\frac{d y}{d x}$ in terms of the parameter, then substitute the parameter value at that point.

When is a parametric tangent horizontal or vertical?

Horizontal when $\frac{d y}{d t} = 0$ (and $\frac{d x}{d t} \neq 0$ ); vertical when $\frac{d x}{d t} = 0$ (and $\frac{d y}{d t} \neq 0$ ).

Why divide the two derivatives?

By the chain rule $\frac{d y}{d x} = \frac{d y}{d t} \cdot \frac{d t}{d x}$ , and $\frac{d t}{d x}$ is the reciprocal of $\frac{d x}{d t}$ .

Is parametric differentiation in Extension 1?

Yes — it is part of the Year 12 Calculus topic, outcome ME1-12-04.

← Previous subtopic

This is the first subtopic

Next subtopic →

Derivatives of inverse functions

Derivatives of parametric functions

📖 Theory

How to differentiate a parametric curve

Common pitfalls

Frequently asked questions

Theory